Convert the following to logarithmic form:
(i) 52 = 25
(ii) a5 = 64
(iii) 7x = 100
(iv) 90 = 1
(v) 61 = 6
(vi) 3-2 = 1 9 \dfrac{1}{9} 9 1
(vii) 10-2 = 0.01
(viii) ( 81 ) 3 4 (81)^{\dfrac{3}{4}} ( 81 ) 4 3
Answer
(i) 52 = 25
⇒ log5 25 = 2.
(ii) a5 = 64
⇒ loga 64 = 5.
(iii) 7x = 100
⇒ log7 100 = x.
(iv) 90 = 1
⇒ log9 1 = 0.
(v) 61 = 6
⇒ log6 6 = 1.
(vi) 3-2 = 1 9 \dfrac{1}{9} 9 1
⇒ log3 1 9 \dfrac{1}{9} 9 1
(vii) 10-2 = 0.01
⇒ log10 0.01 = -2.
(viii) ( 81 ) 3 4 (81)^{\dfrac{3}{4}} ( 81 ) 4 3
⇒ log81 27 = 3 4 \dfrac{3}{4} 4 3
Convert the following into exponential form:
(i) log2 32 = 5
(ii) log3 81 = 4
(iii) log3 1 3 \dfrac{1}{3} 3 1
(iv) log8 4 = 2 3 \dfrac{2}{3} 3 2
(v) log8 32 = 5 3 \dfrac{5}{3} 3 5
(vi) log10 (0.001) = -3
(vii) log2 0.25 = -2
(viii) loga ( 1 a ) \Big(\dfrac{1}{a}\Big) ( a 1 )
Answer
(i) log2 32 = 5
⇒ 25 = 32
(ii) log3 81 = 4
⇒ 34 = 81
(iii) log3 1 3 \dfrac{1}{3} 3 1
⇒ 3-1 = 1 3 \dfrac{1}{3} 3 1
(iv) log8 4 = 2 3 \dfrac{2}{3} 3 2
⇒ ( 8 ) 2 3 (8)^{\dfrac{2}{3}} ( 8 ) 3 2
(v) log8 32 = 5 3 \dfrac{5}{3} 3 5
⇒ ( 8 ) 5 3 = 32 (8)^{\dfrac{5}{3}} = 32 ( 8 ) 3 5 = 32
(vi) log10 (0.001) = -3
⇒ 10-3 = 0.001
(vii) log2 0.25 = -2
⇒ (2)-2 = 0.25
(viii) loga ( 1 a ) \Big(\dfrac{1}{a}\Big) ( a 1 )
⇒ a-1 = 1 a \dfrac{1}{a} a 1
By converting to exponential form, find the values of :
(i) log2 16
(ii) log5 125
(iii) log4 8
(iv) log9 27
(v) log10 (0.01)
(vi) log7 1 7 \dfrac{1}{7} 7 1
(vii) log0.5 256
(viii) log2 0.25
Answer
(i) log2 16 = x
⇒ 2x = 16
⇒ 2x = 24
∴ x = 4.
Hence, log2 16 = 4.
(ii) log5 125 = x
⇒ 5x = 125
⇒ 5x = 53
∴ x = 3.
Hence, log5 125 = 3.
(iii) log4 8 = x
⇒ 4x = 8
⇒ (22 )x = 23
⇒ 22x = 23
⇒ 2x = 3
⇒ x = 3 2 . \dfrac{3}{2}. 2 3 .
Hence, log4 8 = 3 2 \dfrac{3}{2} 2 3
(iv) log9 27 = x
⇒ 9x = 27
⇒ (32 )x = 33
⇒ 32x = 33
⇒ 2x = 3
⇒ x = 3 2 . \dfrac{3}{2}. 2 3 .
Hence, log9 27 = 3 2 \dfrac{3}{2} 2 3
(v) log10 (0.01) = x
⇒ 10x = 0.01
⇒ 10x = 10-2
∴ x = -2.
Hence, log10 (0.01) = -2.
(vi) log7 1 7 \dfrac{1}{7} 7 1
⇒ 7x = 1 7 \dfrac{1}{7} 7 1
⇒ 7x = 7-1
∴ x = -1.
Hence, log7 1 7 \dfrac{1}{7} 7 1
(vii) log0.5 256 = x
⇒ (0.5)x = 256
⇒ ( 5 10 ) x \Big(\dfrac{5}{10}\Big)^x ( 10 5 ) x 8
⇒ ( 1 2 ) x \Big(\dfrac{1}{2}\Big)^x ( 2 1 ) x 8
⇒ (2)-x = (2)8
∴ -x = 8 ⇒ x = -8.
Hence, log0.5 256 = -8.
(viii) log2 0.25 = x
⇒ 2x = 0.25
⇒ 2x = 25 100 \dfrac{25}{100} 100 25
⇒ 2x = 1 4 = 1 2 2 \dfrac{1}{4} = \dfrac{1}{2^2} 4 1 = 2 2 1
⇒ 2x = 2-2
⇒ x = -2.
Hence, log2 0.25 = -2.
Solve the following equation for x:
log3 x = 2
Answer
Given,
⇒ log3 x = 2
⇒ x = 32 = 9.
Hence, x = 9.
Solve the following equation for x:
logx 25 = 2
Answer
Given,
⇒ logx 25 = 2
⇒ 25 = x2
⇒ 52 = x2
⇒ x = 5.
Hence, x = 5.
Solve the following equation for x:
log10 x = -2
Answer
Given,
⇒ log10 x = -2
⇒ x = 10-2
⇒ x = 1 10 2 = 1 100 \dfrac{1}{10^2} = \dfrac{1}{100} 1 0 2 1 = 100 1
Hence, x = 0.01.
Solve the following equation for x:
log4 x = 1 2 \dfrac{1}{2} 2 1
Answer
Given,
⇒ log4 x = 1 2 \dfrac{1}{2} 2 1
⇒ x = 4 1 2 = 4 4^{\dfrac{1}{2}} = \sqrt{4} 4 2 1 = 4
Hence, x = 2.
Solve the following equation for x:
logx 11 = 1
Answer
Given,
⇒ logx 11 = 1
⇒ x1 = 11
⇒ x = 11.
Hence, x = 11.
Solve the following equation for x:
logx 1 4 \dfrac{1}{4} 4 1
Answer
Given,
⇒ logx 1 4 \dfrac{1}{4} 4 1
⇒ x − 1 = 1 4 ⇒ 1 x = 1 4 ⇒ x = 4. \Rightarrow x^{-1} = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{4} \\[1em] \Rightarrow x = 4. ⇒ x − 1 = 4 1 ⇒ x 1 = 4 1 ⇒ x = 4.
Hence, x = 4.
Solve the following equation for x:
log81 x = 3 2 \dfrac{3}{2} 2 3
Answer
Given,
⇒ log81 x = 3 2 \dfrac{3}{2} 2 3
⇒ x = ( 81 ) 3 2 ⇒ x = ( 9 ) 2 3 2 ⇒ x = 9 3 = 729. \Rightarrow x = (81)^{\dfrac{3}{2}} \\[1em] \Rightarrow x = (9)^2{\dfrac{3}{2}} \\[1em] \Rightarrow x = 9^3 = 729. ⇒ x = ( 81 ) 2 3 ⇒ x = ( 9 ) 2 2 3 ⇒ x = 9 3 = 729.
Hence, x = 729.
Solve the following equation for x:
log9 x = 2.5
Answer
Given,
⇒ log9 x = 2.5
⇒ x = 92.5
⇒ x = ( 3 2 ) 5 2 = 3 2 × 5 2 (3^2)^{\dfrac{5}{2}} = 3^{2 \times \dfrac{5}{2}} ( 3 2 ) 2 5 = 3 2 × 2 5
⇒ x = 35 = 243.
Hence, x = 243.
Solve the following equation for x:
log4 x = -1.5
Answer
Given,
⇒ log4 x = -1.5
⇒ x = 4-1.5
⇒ x = 4 − 3 2 = ( 2 2 ) − 3 2 4^{-\dfrac{3}{2}} = (2^2)^{-\dfrac{3}{2}} 4 − 2 3 = ( 2 2 ) − 2 3
⇒ x = 2-3 = 1 2 3 = 1 8 \dfrac{1}{2^3} = \dfrac{1}{8} 2 3 1 = 8 1
Hence, x = 1 8 \dfrac{1}{8} 8 1
Solve the following equation for x:
log√5 x = 2
Answer
Given,
⇒ log√5 x = 2
x = ( 5 ) 2 (\sqrt{5})^2 ( 5 ) 2
Hence, x = 5.
Solve the following equation for x:
logx 0.001 = -3
Answer
Given,
⇒ logx 0.001 = -3
⇒ x-3 = 0.001
⇒ x-3 = 1 1000 = 1 10 3 \dfrac{1}{1000} = \dfrac{1}{10^3} 1000 1 = 1 0 3 1
⇒ x-3 = 10-3
⇒ x = 10.
Hence, x = 10.
Solve the following equation for x:
log√3 (x + 1) = 2
Answer
Given,
⇒ log√3 (x + 1) = 2
⇒ (x + 1) = ( 3 ) 2 (\sqrt{3})^2 ( 3 ) 2
⇒ (x + 1) = 3
⇒ x = 2.
Hence, x = 2.
Solve the following equation for x:
log4 (2x + 3) = 3 2 \dfrac{3}{2} 2 3
Answer
Given,
⇒ log4 (2x + 3) = 3 2 \dfrac{3}{2} 2 3
⇒ 2x + 3 = 4 3 2 4^{\dfrac{3}{2}} 4 2 3
⇒ 2x + 3 = ( 2 2 ) 3 2 (2^2)^{\dfrac{3}{2}} ( 2 2 ) 2 3
⇒ 2x + 3 = 23
⇒ 2x + 3 = 8
⇒ 2x = 5
⇒ x = 5 2 \dfrac{5}{2} 2 5
Hence, x = 5 2 \dfrac{5}{2} 2 5
Solve the following equation for x:
log 2 3 \text{log}_{\sqrt[3]{2}} log 3 2
Answer
Given,
⇒ log 2 3 x = 3 ⇒ x = ( 2 3 ) 3 ⇒ x = ( 2 1 3 ) 3 ⇒ x = 2. \Rightarrow \text{log}_{\sqrt[3]{2}}x = 3 \\[1em] \Rightarrow x = (\sqrt[3]{2})^3 \\[1em] \Rightarrow x = (2^{\dfrac{1}{3}})^3 \\[1em] \Rightarrow x = 2. ⇒ log 3 2 x = 3 ⇒ x = ( 3 2 ) 3 ⇒ x = ( 2 3 1 ) 3 ⇒ x = 2.
Hence, x = 2.
Solve the following equation for x:
log2 (x2 - 1) = 3
Answer
Given,
⇒ log 2 ( x 2 − 1 ) \text{log}_2(x^2 - 1) log 2 ( x 2 − 1 )
⇒ x2 - 1 = 23
⇒ x2 - 1 = 8
⇒ x2 = 9
⇒ x = 9 = ± 3 \sqrt{9} = \pm3 9 = ± 3
Hence, x = ±3.
Solve the following equation for x:
log x = -1
Answer
Given,
⇒ log x = -1
⇒ x = 10-1
⇒ x = 1 10 . \dfrac{1}{10}. 10 1 .
Hence, x = 1 10 \dfrac{1}{10} 10 1
Solve the following equation for x:
log(2x - 3) = 1
Answer
Given,
⇒ log(2x - 3) = 1
⇒ 2x - 3 = 101
⇒ 2x - 3 = 10
⇒ 2x = 13
⇒ x = 13 2 = 6 1 2 . \dfrac{13}{2} = 6\dfrac{1}{2}. 2 13 = 6 2 1 .
Hence, x = 6 1 2 . 6\dfrac{1}{2}. 6 2 1 .
Solve the following equation for x:
log x = -2, 0, 1 3 \dfrac{1}{3} 3 1
Answer
Given,
⇒ log x = -2
⇒ x = 10-2 = 1 10 2 = 1 100 \dfrac{1}{10^2} = \dfrac{1}{100} 1 0 2 1 = 100 1
⇒ log x = 0
⇒ x = 100 = 1.
⇒ log x = 1 3 \dfrac{1}{3} 3 1
⇒ x = 10 1 3 = 10 3 . 10^{\dfrac{1}{3}} = \sqrt[3]{10}. 1 0 3 1 = 3 10 .
Hence, x = 1 100 , 1 , 10 3 . \dfrac{1}{100}, 1, \sqrt[3]{10}. 100 1 , 1 , 3 10 .
Given log10 a = b, express 102b - 3 in terms of a.
Answer
Given,
⇒ log10 a = b
∴ a = 10b .
Simplifying 102b - 3 we get,
⇒ 102b - 3 = 102b .10-3
= 10 2 b 10 3 = ( 10 b ) 2 10 3 \dfrac{10^{2b}}{10^3} = \dfrac{(10^b)^2}{10^3} 1 0 3 1 0 2 b = 1 0 3 ( 1 0 b ) 2
= a 2 10 3 = a 2 1000 \dfrac{a^2}{10^3} = \dfrac{a^2}{1000} 1 0 3 a 2 = 1000 a 2
Hence, 102b - 3 = a 2 1000 . \dfrac{a^2}{1000}. 1000 a 2 .
Given log10 x = a, log10 y = b and log10 z = c,
(i) write down 102a - 3 in terms of x.
(ii) write down 103b - 1 in terms of y.
(iii) if log10 P = 2a + b 2 \dfrac{b}{2} 2 b
Answer
(i) Given,
⇒ log10 x = a
∴ x = 10a .
Simplifying 102a - 3 we get,
⇒ 102a - 3 = 102a .10-3
= 10 2 a 10 3 = ( 10 a ) 2 10 3 \dfrac{10^{2a}}{10^3} = \dfrac{(10^a)^2}{10^3} 1 0 3 1 0 2 a = 1 0 3 ( 1 0 a ) 2
= x 2 10 3 = x 2 1000 \dfrac{x^2}{10^3} = \dfrac{x^2}{1000} 1 0 3 x 2 = 1000 x 2
Hence, 102a - 3 = x 2 1000 . \dfrac{x^2}{1000}. 1000 x 2 .
(ii) Given,
⇒ log10 y = b
∴ y = 10b .
Simplifying 103b - 1 we get,
⇒ 103b - 1 = 103b .10-1
= 10 3 b 10 1 = ( 10 b ) 3 10 \dfrac{10^{3b}}{10^1} = \dfrac{(10^b)^3}{10} 1 0 1 1 0 3 b = 10 ( 1 0 b ) 3
= y 3 10 \dfrac{y^3}{10} 10 y 3
Hence, 103b - 1 = y 3 10 . \dfrac{y^3}{10}. 10 y 3 .
(iii) Given,
log10 z = c
⇒ z = 10c .
log10 P = 2a + b 2 \dfrac{b}{2} 2 b
⇒ P = 10 2 a + b 2 − 3 c = 10 2 a .10 b 2 .10 − 3 c = ( 10 a ) 2 . ( 10 b ) 1 2 . ( 10 c ) − 3 = ( x ) 2 . ( y ) 1 2 . ( z ) − 3 = x 2 y z 3 . \Rightarrow P = 10^{2a + \frac{b}{2} - 3c} \\[1em] = 10^{2a}.10^{\frac{b}{2}}.10^{-3c} \\[1em] = (10^a)^2.(10^b)^{\frac{1}{2}}.(10^c)^{-3} \\[1em] = (x)^2.(y)^{\frac{1}{2}}.(z)^{-3} \\[1em] = \dfrac{x^2\sqrt{y}}{z^3}. ⇒ P = 1 0 2 a + 2 b − 3 c = 1 0 2 a .1 0 2 b .1 0 − 3 c = ( 1 0 a ) 2 . ( 1 0 b ) 2 1 . ( 1 0 c ) − 3 = ( x ) 2 . ( y ) 2 1 . ( z ) − 3 = z 3 x 2 y .
Hence, P = x 2 y z 3 . \dfrac{x^2\sqrt{y}}{z^3}. z 3 x 2 y .
If log10 x = a and log10 y = b, find the value of xy.
Answer
Given,
log10 x = a
⇒ x = 10a .
log10 y = b
⇒ y = 10b
xy = 10a .10b = 10a + b .
Hence, xy = 10a + b .
Given log10 a = m and log10 b = n, express a 3 b 2 \dfrac{a^3}{b^2} b 2 a 3
Answer
Given,
log10 a = m
⇒ a = 10m
log10 b = n
⇒ b = 10n
⇒ a 3 b 2 = ( 10 m ) 3 ( 10 n ) 2 = 10 3 m 10 2 n = 10 3 m − 2 n . \Rightarrow \dfrac{a^3}{b^2} = \dfrac{(10^m)^3}{(10^n)^2} \\[1em] = \dfrac{10^{3m}}{10^{2n}} = 10^{3m - 2n}. ⇒ b 2 a 3 = ( 1 0 n ) 2 ( 1 0 m ) 3 = 1 0 2 n 1 0 3 m = 1 0 3 m − 2 n .
Hence, a 3 b 2 \dfrac{a^3}{b^2} b 2 a 3 3m - 2n .
Given log10 x = 2a and log10 y = b 2 \dfrac{b}{2} 2 b
(i) write 10a in terms of x.
(ii) write 102b + 1 in terms of y.
(iii) if log10 P = 3a - 3b, express P in terms of x and y.
Answer
(i) Given,
log10 x = 2a
⇒ x = 102a
⇒ x = (10a )2
⇒ 10a = x \sqrt{x} x
Hence, 10a = x \sqrt{x} x
(ii) Given,
log10 y = b 2 \dfrac{b}{2} 2 b
⇒ y = 10 b 2 ⇒ y = ( 10 b ) 1 2 \Rightarrow y = 10^{\dfrac{b}{2}} \\[1em] \Rightarrow y = (10^b)^{\dfrac{1}{2}} \\[1em] ⇒ y = 1 0 2 b ⇒ y = ( 1 0 b ) 2 1
Squaring both sides we get,
⇒ y2 = 10b
Simplifying 102b + 1 we get,
102b + 1 = 102b .10
= (10b )2 .10
= (y2 )2 .10
= 10y4 .
Hence, 102b + 1 = 10y4 .
(iii) Given,
log10 P = 3a - 2b
⇒ P = 103a - 2b
⇒ P = 103a .10-2b
= (10a )3 .(10b )-2
= ( x ) 3 × ( y 2 ) − 2 = ( x 3 ) 1 2 × 1 ( y 2 ) 2 = x 3 2 y 4 = (\sqrt{x})^3 \times (y^2)^{-2} \\[1em] = (x^3)^{\dfrac{1}{2}} \times \dfrac{1}{(y^2)^2} \\[1em] = \dfrac{x^{\dfrac{3}{2}}}{y^4} = ( x ) 3 × ( y 2 ) − 2 = ( x 3 ) 2 1 × ( y 2 ) 2 1 = y 4 x 2 3
Hence, P = x 3 2 y 4 \dfrac{x^{\dfrac{3}{2}}}{y^4} y 4 x 2 3
If log2 y = x and log3 z = x, find 72x in terms of y and z.
Answer
Given,
log2 y = x and log3 z = x
⇒ y = 2x and z = 3x .
(72)x = (23 .32 )x
= (2)3x .(3)2x
= (2x )3 .(3x )2
= (y)3 .(z)2
Hence, (72)x = y3 .z2 .
If log2 x = a and log5 y = a, write 1002a - 1 in terms of x and y.
Answer
Given,
log2 x = a and log5 y = a
⇒ x = 2a and y = 5a
1002a - 1 = 1002a .100-1
= (100a )2 .100-1
= [(22 .52 )a ]2 .100-1
= [(22a ).(52a )]2 .100-1
= [(2a )2 .(5a )2 ]2 .(100)-1
= [x2 .y2 ]2 .100-1
= x 4 y 4 100 \dfrac{x^4y^4}{100} 100 x 4 y 4
Hence, 1002a - 1 = x 4 y 4 100 \dfrac{x^4y^4}{100} 100 x 4 y 4
Simplify the following :
log a3 - log a2
Answer
Given,
⇒ log a3 - log a2
⇒ 3log a - 2log a
⇒ log a.
Hence, log a3 - log a2 = log a
Simplify the following :
log a3 ÷ log a2
Answer
Given,
log a3 ÷ log a2
⇒ log a 3 log a 2 ⇒ 3 log a 2 log a ⇒ 3 2 . \Rightarrow \dfrac{\text{log a}^3}{\text{log a}^2} \\[1em] \Rightarrow \dfrac{3\text{log a}}{2\text{log a}} \\[1em] \Rightarrow \dfrac{3}{2}. ⇒ log a 2 log a 3 ⇒ 2 log a 3 log a ⇒ 2 3 .
Hence, log a3 ÷ log a2 = 3 2 . \dfrac{3}{2}. 2 3 .
Simplify the following :
log 4 log 2 \dfrac{\text{log} 4}{\text{log} 2} log 2 log 4
Answer
Given,
⇒ log 4 log 2 ⇒ log 2 2 log 2 ⇒ 2 log 2 log 2 ⇒ 2. \Rightarrow \dfrac{\text{log 4}}{\text{log 2}} \\[1em] \Rightarrow \dfrac{\text{log 2}^2}{\text{log 2}} \\[1em] \Rightarrow \dfrac{\text{2 log 2}}{\text{log 2}} \\[1em] \Rightarrow 2. ⇒ log 2 log 4 ⇒ log 2 log 2 2 ⇒ log 2 2 log 2 ⇒ 2.
Hence, log 4 log 2 \dfrac{\text{log} 4}{\text{log} 2} log 2 log 4
Simplify the following :
log 8 log 9 log 27 \dfrac{\text{log 8 log 9}}{\text{log 27}} log 27 log 8 log 9
Answer
Given,
⇒ log 8 log 9 log 27 ⇒ log 2 3 log 3 2 log 3 3 ⇒ 3log 2. 2log 3 3 log 3 ⇒ 6log 2.log 3 3log 3 ⇒ 2 log 2 = log 2 2 = log 4 . \Rightarrow \dfrac{\text{log 8 log 9}}{\text{log 27}} \\[1em] \Rightarrow \dfrac{\text{log 2}^3 \text{log 3}^2}{\text{log 3}^3} \\[1em] \Rightarrow \dfrac{\text{3log 2. 2log 3}}{3\text{log} 3} \\[1em] \Rightarrow \dfrac{\text{6log 2.log 3}}{\text{3log 3}} \\[1em] \Rightarrow 2\text{log 2} = \text{log 2}^2 \\[1em] = \text{log 4}. ⇒ log 27 log 8 log 9 ⇒ log 3 3 log 2 3 log 3 2 ⇒ 3 log 3 3log 2. 2log 3 ⇒ 3log 3 6log 2.log 3 ⇒ 2 log 2 = log 2 2 = log 4 .
Hence, log 8 log 9 log 27 \dfrac{\text{log 8 log 9}}{\text{log 27}} log 27 log 8 log 9
Simplify the following :
log 27 log 3 \dfrac{\text{log 27}}{\text{log } \sqrt{3}} log 3 log 27
Answer
Given,
⇒ log 27 log 3 ⇒ log 3 3 log 3 1 2 ⇒ 3log 3 1 2 log 3 ⇒ 3 1 2 ⇒ 6. \Rightarrow \dfrac{\text{log 27}}{\text{log }\sqrt{3}} \\[1em] \Rightarrow \dfrac{\text{log 3}^3}{\text{log 3}^{\dfrac{1}{2}}} \\[1em] \Rightarrow \dfrac{\text{3log 3}}{\dfrac{1}{2}\text{log 3}} \\[1em] \Rightarrow \dfrac{3}{\dfrac{1}{2}} \\[1em] \Rightarrow 6. ⇒ log 3 log 27 ⇒ log 3 2 1 log 3 3 ⇒ 2 1 log 3 3log 3 ⇒ 2 1 3 ⇒ 6.
Hence, log 27 log 3 \dfrac{\text{log 27}}{\text{log }\sqrt{3}} log 3 log 27
Simplify the following :
log 9 - log 3 log 27 \dfrac{\text{log 9 - log 3}}{\text{ log 27}} log 27 log 9 - log 3
Answer
Given,
⇒ log 9 - log 3 log 27 ⇒ log 3 2 − log 3 log 3 3 ⇒ 2log 3 - log 3 3log 3 ⇒ log 3 3log 3 ⇒ 1 3 . \Rightarrow \dfrac{\text{log 9 - log 3}}{\text{log 27}} \\[1em] \Rightarrow \dfrac{\text{log 3}^2 - \text{log 3}}{\text{log 3}^3} \\[1em] \Rightarrow \dfrac{\text{2log 3 - log 3}}{\text{3log 3}} \\[1em] \Rightarrow \dfrac{\text{log 3}}{\text{3log 3}} \\[1em] \Rightarrow \dfrac{1}{3}. ⇒ log 27 log 9 - log 3 ⇒ log 3 3 log 3 2 − log 3 ⇒ 3log 3 2log 3 - log 3 ⇒ 3log 3 log 3 ⇒ 3 1 .
Hence, log 9 - log 3 log 27 = 1 3 . \dfrac{\text{log 9 - log 3}}{\text{log 27}} = \dfrac{1}{3}. log 27 log 9 - log 3 = 3 1 .
Evaluate the following:
log( 10 ÷ 10 3 ) (10 ÷ \sqrt[3]{10}) ( 10 ÷ 3 10 )
Answer
Given,
⇒ log ( 10 ÷ 10 3 ) ⇒ log ( 10 10 3 ) ⇒ log ( 10 10 1 3 ) ⇒ log ( 10 ) 1 − 1 3 ⇒ log ( 10 ) 2 3 ⇒ 2 3 log 10 ⇒ 2 3 . \Rightarrow \text{log}(10 ÷ \sqrt[3]{10}) \\[1em] \Rightarrow \text{log}\Big(\dfrac{10}{\sqrt[3]{10}}\Big) \\[1em] \Rightarrow \text{log}\Big(\dfrac{10}{10^{\dfrac{1}{3}}}\Big) \\[1em] \Rightarrow \text{log}(10)^{1 - \dfrac{1}{3}} \\[1em] \Rightarrow \text{log}(10)^{\dfrac{2}{3}} \\[1em] \Rightarrow \dfrac{2}{3}\text{log 10} \\[1em] \Rightarrow \dfrac{2}{3}. ⇒ log ( 10 ÷ 3 10 ) ⇒ log ( 3 10 10 ) ⇒ log ( 1 0 3 1 10 ) ⇒ log ( 10 ) 1 − 3 1 ⇒ log ( 10 ) 3 2 ⇒ 3 2 log 10 ⇒ 3 2 .
Hence, log ( 10 ÷ 10 3 ) = 2 3 \text{log}(10 ÷ \sqrt[3]{10}) = \dfrac{2}{3} log ( 10 ÷ 3 10 ) = 3 2
Evaluate the following:
2 + 1 2 \dfrac{1}{2} 2 1 -3
Answer
Given,
⇒ 2 + 1 2 log ( 10 ) − 3 ⇒ 2 + − 3 2 log 10 ⇒ 2 − 3 2 × 1 ⇒ 2 − 3 2 = 1 2 . \Rightarrow 2 + \dfrac{1}{2}\text{log} (10)^{-3} \\[1em] \Rightarrow 2 + \dfrac{-3}{2}\text{log}10 \\[1em] \Rightarrow 2 - \dfrac{3}{2} \times 1 \\[1em] \Rightarrow 2 - \dfrac{3}{2} = \dfrac{1}{2}. ⇒ 2 + 2 1 log ( 10 ) − 3 ⇒ 2 + 2 − 3 log 10 ⇒ 2 − 2 3 × 1 ⇒ 2 − 2 3 = 2 1 .
Hence, 2 + 1 2 log ( 10 ) − 3 = 1 2 2 + \dfrac{1}{2}\text{log} (10)^{-3} = \dfrac{1}{2} 2 + 2 1 log ( 10 ) − 3 = 2 1
Evaluate the following:
2log 5 + log 8 - 1 2 \dfrac{1}{2} 2 1
Answer
Given,
⇒ 2log 5 + log 8 − 1 2 log 4 ⇒ log 5 2 + log 8 − 1 2 × log 2 2 ⇒ log 25 + log 8 − 1 2 × 2 log 2 ⇒ log 25 + log 8 - log 2 ⇒ log 25 × 8 2 ⇒ log 100 ⇒ log 10 2 = 2 log 10 ⇒ 2 × 1 = 2. \Rightarrow \text{2log 5 + log 8} - \dfrac{1}{2}\text{log 4} \\[1em] \Rightarrow \text{log 5}^2 + \text{log 8} - \dfrac{1}{2} \times \text{log 2}^2 \\[1em] \Rightarrow \text{log 25 + log 8} - \dfrac{1}{2} \times 2\text{log 2} \\[1em] \Rightarrow \text{log 25 + log 8 - log 2} \\[1em] \Rightarrow \text{log} \dfrac{25 \times 8}{2} \\[1em] \Rightarrow \text{log 100} \\[1em] \Rightarrow \text{log} 10^2 = 2\text{log 10} \Rightarrow 2 \times 1 = 2. ⇒ 2log 5 + log 8 − 2 1 log 4 ⇒ log 5 2 + log 8 − 2 1 × log 2 2 ⇒ log 25 + log 8 − 2 1 × 2 log 2 ⇒ log 25 + log 8 - log 2 ⇒ log 2 25 × 8 ⇒ log 100 ⇒ log 1 0 2 = 2 log 10 ⇒ 2 × 1 = 2.
Hence, 2log 5 + log 8 − 1 2 log 4 = 2. \text{2log 5 + log 8} - \dfrac{1}{2}\text{log 4} = 2. 2log 5 + log 8 − 2 1 log 4 = 2.
Evaluate the following:
2log 103 + 3log 10-2 - 1 3 log 5 − 3 + 1 2 log 4 \dfrac{1}{3}\text{log 5}^{-3} + \dfrac{1}{2}\text{log 4} 3 1 log 5 − 3 + 2 1 log 4
Answer
Given,
⇒ 2 log 10 3 + 3 log 10 − 2 − 1 3 log 5 − 3 + 1 2 log 4 ⇒ 2 × 3 log 10 + 3 × − 2 log 10 − 1 3 × ( − 3 ) log 5 + 1 2 log 2 2 ⇒ 2 × 3 × 1 + 3 × ( − 2 ) × 1 − ( − 1 ) log 5 + 1 2 × 2 × log 2 ⇒ 6 − 6 + log 5 + log 2 ⇒ log 5 × 2 ⇒ log 10 1. \Rightarrow 2\text{log} 10^3 + 3\text{log} 10^{-2} - \dfrac{1}{3}\text{log 5}^{-3} + \dfrac{1}{2}\text{log 4} \\[1em] \Rightarrow 2 \times 3\text{log 10} + 3 \times -2\text{log 10} - \dfrac{1}{3} \times (-3) \text{log 5} + \dfrac{1}{2}\text{log 2}^2 \\[1em] \Rightarrow 2 \times 3 \times 1 + 3 \times (-2) \times 1 - (-1)\text{log 5} + \dfrac{1}{2} \times 2 \times \text{log 2} \\[1em] \Rightarrow 6 - 6 + \text{log 5} + \text{log 2} \\[1em] \Rightarrow \text{log 5 × 2} \\[1em] \Rightarrow \text{log 10} \\[1em] 1. ⇒ 2 log 1 0 3 + 3 log 1 0 − 2 − 3 1 log 5 − 3 + 2 1 log 4 ⇒ 2 × 3 log 10 + 3 × − 2 log 10 − 3 1 × ( − 3 ) log 5 + 2 1 log 2 2 ⇒ 2 × 3 × 1 + 3 × ( − 2 ) × 1 − ( − 1 ) log 5 + 2 1 × 2 × log 2 ⇒ 6 − 6 + log 5 + log 2 ⇒ log 5 × 2 ⇒ log 10 1.
Hence, 2 log 10 3 + 3 log 10 − 2 − 1 3 log 5 − 3 + 1 2 log 4 2\text{log} 10^3 + 3\text{log} 10^{-2} - \dfrac{1}{3}\text{log 5}^{-3} + \dfrac{1}{2}\text{log 4} 2 log 1 0 3 + 3 log 1 0 − 2 − 3 1 log 5 − 3 + 2 1 log 4 .
Evaluate the following:
2log 2 + log 5 - 1 2 log 36 − log 1 30 \dfrac{1}{2}\text{log 36} - \text{log}\dfrac{1}{30} 2 1 log 36 − log 30 1
Answer
Given,
⇒ 2log 2 + log 5 − 1 2 log 36 - log 1 30 ⇒ 2log 2 + log 5 − 1 2 log 6 2 − (log 1 - log 30) ⇒ 2log 2 + log 5 − 1 2 × 2 × log 6 - log 1 + log 30 ⇒ 2 log 2 + log 5 − log 6 − 0 + log ( 5 × 6 ) ⇒ log 2 2 + log 5 - log 6 + log 5 + log 6 ⇒ log 4 + log 5 + log 5 ⇒ log ( 4 × 5 × 5 ) ⇒ log 100 = 2. \Rightarrow \text{2log 2 + log 5} - \dfrac{1}{2}\text{log 36 - log}\dfrac{1}{30} \\[1em] \Rightarrow \text{2log 2 + log 5} - \dfrac{1}{2} \text{log 6}^2 - \text{(log 1 - log 30)} \\[1em] \Rightarrow \text{2log 2 + log 5} - \dfrac{1}{2} \times 2 \times \text{log 6 - log 1 + log 30} \\[1em] \Rightarrow 2\text{log } 2 + \text{log } 5 - \text{log } 6 - 0 + \text{log } (5 \times 6) \\[1em] \Rightarrow \text{log } 2^2 + \text{log 5 - log 6 + log 5 + log 6} \\[1em] \Rightarrow \text{log 4 + log 5 + log 5} \\[1em] \Rightarrow \text{log } (4 \times 5 \times 5) \\[1em] \Rightarrow \text{log 100} = 2. ⇒ 2log 2 + log 5 − 2 1 log 36 - log 30 1 ⇒ 2log 2 + log 5 − 2 1 log 6 2 − (log 1 - log 30) ⇒ 2log 2 + log 5 − 2 1 × 2 × log 6 - log 1 + log 30 ⇒ 2 log 2 + log 5 − log 6 − 0 + log ( 5 × 6 ) ⇒ log 2 2 + log 5 - log 6 + log 5 + log 6 ⇒ log 4 + log 5 + log 5 ⇒ log ( 4 × 5 × 5 ) ⇒ log 100 = 2.
Hence, 2log 2 + log 5 − 1 2 log 36 - log 1 30 \text{2log 2 + log 5} - \dfrac{1}{2}\text{log 36 - log}\dfrac{1}{30} 2log 2 + log 5 − 2 1 log 36 - log 30 1
Evaluate the following:
2log 5 + log 3 + 3log 2 - 1 2 \dfrac{1}{2} 2 1
Answer
Given,
⇒ 2log 5 + log 3 + 3log 2 - 1 2 \dfrac{1}{2} 2 1
⇒ log 52 + log 3 + log 23 - 1 2 \dfrac{1}{2} 2 1 2 - log 102
⇒ log 25 + log 3 + log 8 - 1 2 × 2 \dfrac{1}{2} \times 2 2 1 × 2
⇒ log 25 + log 3 + log 8 - log 6 - log 100
⇒ log 25 × 3 × 8 6 × 100 \dfrac{25 \times 3 \times 8}{6 \times 100} 6 × 100 25 × 3 × 8
⇒ log 600 600 \dfrac{600}{600} 600 600
⇒ log 1
⇒ 0.
Hence, 2log 5 + log 3 + 3log 2 - 1 2 \dfrac{1}{2} 2 1
Evaluate the following:
log 2 + 16log 16 15 + 12 log 25 24 + 7 log 81 80 \dfrac{16}{15} + 12\text{log } \dfrac{25}{24} + 7\text{log }\dfrac{81}{80} 15 16 + 12 log 24 25 + 7 log 80 81
Answer
Given,
⇒ log 2 + 16log 16 15 + 12 log 25 24 + 7 log 81 80 ⇒ log 2 + 16(log 16 - log 15) + 12(log 25 - log 24) + 7(log 81 - log 80) ⇒ log 2 + 16log 16 - 16log 15 + 12log 25 - 12log 24 + 7log 81 - 7log 80 ⇒ log 2 + 16log 2 4 − 16log 3.5 + 12log 5 2 − 12log 2 3 .3 + 7log 3 4 − 7log 2 4 .5 ⇒ log 2 + 16.4log 2 - 16(log 3 + log 5) + 12.2log 5 - 12(log 2 3 + log 3) + 7.4log 3 − 7(log 2 4 + log 5) ⇒ log 2 + 64log 2 - 16log 3 - 16log 5 + 24log 5 - 12.3log 2 - 12log 3 + 28log 3 - 28log 2 - 7log 5 ⇒ 65log 2 - 36log 2 - 28log 2 - 16log 3 - 12log 3 + 28log 3 - 16log 5 + 24log 5 - 7log 5 ⇒ log 2 + log 5 ⇒ log 2.5 ⇒ log 10 = 1. \Rightarrow \text{log 2 + 16log }\dfrac{16}{15} + 12\text{log } \dfrac{25}{24} + 7\text{log }\dfrac{81}{80} \\[1em] \Rightarrow \text{log 2 + 16(log 16 - log 15) + 12(log 25 - log 24) + 7(log 81 - log 80)} \\[1em] \Rightarrow \text{log 2 + 16log 16 - 16log 15 + 12log 25 - 12log 24 + 7log 81 - 7log 80} \\[1em] \Rightarrow \text{log 2 + 16log 2}^4 - \text{16log 3.5 + 12log 5}^2 - \text{12log 2}^3.3 + \text{7log 3}^4 - \text{7log 2}^4.5 \\[1em] \Rightarrow \text{log 2 + 16.4log 2 - 16(log 3 + log 5) + 12.2log 5 - 12(log 2}^3 + \text{log 3) + 7.4log 3} - \text{7(log 2}^4 + \text{log 5)} \\[1em] \Rightarrow \text{log 2 + 64log 2 - 16log 3 - 16log 5 + 24log 5 - 12.3log 2 - 12log 3 + 28log 3 - 28log 2 - 7log 5} \\[1em] \Rightarrow \text{65log 2 - 36log 2 - 28log 2 - 16log 3 - 12log 3 + 28log 3 - 16log 5 + 24log 5 - 7log 5} \\[1em] \Rightarrow \text{log 2 + log 5} \\[1em] \Rightarrow \text{log 2.5} \\[1em] \Rightarrow \text{log 10} = 1. ⇒ log 2 + 16log 15 16 + 12 log 24 25 + 7 log 80 81 ⇒ log 2 + 16(log 16 - log 15) + 12(log 25 - log 24) + 7(log 81 - log 80) ⇒ log 2 + 16log 16 - 16log 15 + 12log 25 - 12log 24 + 7log 81 - 7log 80 ⇒ log 2 + 16log 2 4 − 16log 3.5 + 12log 5 2 − 12log 2 3 .3 + 7log 3 4 − 7log 2 4 .5 ⇒ log 2 + 16.4log 2 - 16(log 3 + log 5) + 12.2log 5 - 12(log 2 3 + log 3) + 7.4log 3 − 7(log 2 4 + log 5) ⇒ log 2 + 64log 2 - 16log 3 - 16log 5 + 24log 5 - 12.3log 2 - 12log 3 + 28log 3 - 28log 2 - 7log 5 ⇒ 65log 2 - 36log 2 - 28log 2 - 16log 3 - 12log 3 + 28log 3 - 16log 5 + 24log 5 - 7log 5 ⇒ log 2 + log 5 ⇒ log 2.5 ⇒ log 10 = 1.
Hence, log 2 + 16log 16 15 + 12 log 25 24 + 7 log 81 80 = 1 \text{log 2 + 16log }\dfrac{16}{15} + 12\text{log } \dfrac{25}{24} + 7\text{log }\dfrac{81}{80} = 1 log 2 + 16log 15 16 + 12 log 24 25 + 7 log 80 81 = 1
Evaluate the following:
2log10 5 + log10 8 - 1 2 \dfrac{1}{2} 2 1 10 4
Answer
Given,
⇒ 2log10 5 + log10 8 - 1 2 \dfrac{1}{2} 2 1 10 4
⇒ log10 52 + log10 8 - log 10 4 1 2 \text{log}_{10}4^{\dfrac{1}{2}} log 10 4 2 1
⇒ log10 25 + log10 8 - log10 2
⇒ log10 25 × 8 2 \dfrac{25 \times 8}{2} 2 25 × 8
⇒ log10 100
⇒ log10 102
⇒ 2log10 10
⇒ 2.
Hence, 2log10 5 + log10 8 - 1 2 \dfrac{1}{2} 2 1 10 4 = 2.
Express the following as a single logarithm:
2log 3 - 1 2 \dfrac{1}{2} 2 1
Answer
Given,
⇒ 2log 3 − 1 2 log 2 4 + log 2 2 .3 ⇒ 2log 3 − 1 2 × 4 × log 2 + log 2 2 + log 3 ⇒ 2log 3 − 2 log 2 + 2 log 2 + log 3 ⇒ 2log 3 + log 3 ⇒ 3log 3 ⇒ log 3 3 = log 27 . \Rightarrow \text{2log 3} - \dfrac{1}{2}\text{log 2}^4 + \text{log 2}^2.3 \\[1em] \Rightarrow \text{2log 3} - \dfrac{1}{2} \times 4 \times \text{log 2} + \text{log 2}^2 + \text{log 3} \\[1em] \Rightarrow \text{2log 3} - 2\text{log 2} + 2\text{log 2} + \text{log 3} \\[1em] \Rightarrow \text{2log 3 + log 3} \\[1em] \Rightarrow \text{3log 3} \\[1em] \Rightarrow \text{log 3}^3 = \text{log 27}. ⇒ 2log 3 − 2 1 log 2 4 + log 2 2 .3 ⇒ 2log 3 − 2 1 × 4 × log 2 + log 2 2 + log 3 ⇒ 2log 3 − 2 log 2 + 2 log 2 + log 3 ⇒ 2log 3 + log 3 ⇒ 3log 3 ⇒ log 3 3 = log 27 .
Hence, 2log 3 - 1 2 \dfrac{1}{2} 2 1
Express the following as a single logarithm:
2log10 5 - log10 2 + 3log10 4 + 1
Answer
Given,
⇒ 2log10 5 - log10 2 + 3log10 4 + 1
⇒ log10 52 - log10 2 + log10 43 + log10 10
⇒ log10 25 + log10 64 + log10 10 - log10 2
⇒ log10 25 × 64 × 10 2 \dfrac{25 \times 64 \times 10}{2} 2 25 × 64 × 10
⇒ log10 16000 2 = log 10 8000 \dfrac{16000}{2} = \text{log}_{10}8000 2 16000 = log 10 8000
Hence, 2log10 5 - log10 2 + 3log10 4 + 1 = log10 8000 .
Express the following as a single logarithm:
1 2 \dfrac{1}{2} 2 1
Answer
Given,
⇒ 1 2 log 36 + 2log 8 - log 1.5 ⇒ 1 2 log 6 2 + log 8 2 − log 15 10 ⇒ 1 2 × 2 × log 6 + log 64 - (log 15 - log 10) ⇒ log 6 + log 64 - log 15 + log 10 ⇒ log 6 × 64 × 10 15 ⇒ log 256 . \Rightarrow \dfrac{1}{2}\text{log 36 + 2log 8 - log 1.5} \\[1em] \Rightarrow \dfrac{1}{2}\text{log 6}^2 + \text{log 8}^2 - \text{log} \dfrac{15}{10} \\[1em] \Rightarrow \dfrac{1}{2} \times 2 \times \text{log 6 + log 64 - (log 15 - log 10)} \\[1em] \Rightarrow \text{log 6 + log 64 - log 15 + log 10} \\[1em] \Rightarrow \text{log} \dfrac{6 \times 64 \times 10}{15} \\[1em] \Rightarrow \text{log 256}. ⇒ 2 1 log 36 + 2log 8 - log 1.5 ⇒ 2 1 log 6 2 + log 8 2 − log 10 15 ⇒ 2 1 × 2 × log 6 + log 64 - (log 15 - log 10) ⇒ log 6 + log 64 - log 15 + log 10 ⇒ log 15 6 × 64 × 10 ⇒ log 256 .
Hence, 1 2 log 36 + 2log 8 - log 1.5 \dfrac{1}{2}\text{log 36 + 2log 8 - log 1.5} 2 1 log 36 + 2log 8 - log 1.5
Express the following as a single logarithm:
1 2 \dfrac{1}{2} 2 1
Answer
Given,
⇒ 1 2 log 25 - 2log 3 + 1 ⇒ 1 2 log 5 2 − 2log 3 + log 10 ⇒ 1 2 × 2 log 5 − log 3 2 + log 10 ⇒ log 5 - log 9 + log 10 ⇒ log 5 × 10 9 ⇒ log 50 9 . \Rightarrow \dfrac{1}{2}\text{log 25 - 2log 3 + 1} \\[1em] \Rightarrow \dfrac{1}{2}\text{log 5}^2 - \text{2log 3 + log 10} \\[1em] \Rightarrow \dfrac{1}{2} \times 2\text{log 5} - \text{log 3}^2 + \text{log 10} \\[1em] \Rightarrow \text{log 5 - log 9 + log 10} \\[1em] \Rightarrow \text{log }\dfrac{5 \times 10}{9} \\[1em] \Rightarrow \text{log }\dfrac{50}{9}. ⇒ 2 1 log 25 - 2log 3 + 1 ⇒ 2 1 log 5 2 − 2log 3 + log 10 ⇒ 2 1 × 2 log 5 − log 3 2 + log 10 ⇒ log 5 - log 9 + log 10 ⇒ log 9 5 × 10 ⇒ log 9 50 .
Hence, 1 2 log 25 - 2log 3 + 1 = log 50 9 . \dfrac{1}{2}\text{log 25 - 2log 3 + 1} = \text{log }\dfrac{50}{9}. 2 1 log 25 - 2log 3 + 1 = log 9 50 .
Express the following as a single logarithm:
1 2 \dfrac{1}{2} 2 1
Answer
Given,
⇒ 1 2 log 9 + 2log 3 - log 6 + log 2 - 2 ⇒ 1 2 log 3 2 + log 3 2 − log 6 + log 2 - 2log 10 ⇒ 1 2 × 2 × log 3 + log 9 - log 6 + log 2 - log 10 2 ⇒ log 3 + log 9 + log 2 - log 6 - log 100 ⇒ log 3 × 9 × 2 6 × 100 ⇒ log 9 100 . \Rightarrow \dfrac{1}{2}\text{log 9 + 2log 3 - log 6 + log 2 - 2} \\[1em] \Rightarrow \dfrac{1}{2}\text{log 3}^2 + \text{log 3}^2 - \text{log 6 + log 2 - 2log 10} \\[1em] \Rightarrow \dfrac{1}{2} \times 2 \times \text{log 3 + log 9 - log 6 + log 2 - log 10}^2 \\[1em] \Rightarrow \text{log 3 + log 9 + log 2 - log 6 - log 100} \\[1em] \Rightarrow \text{log }\dfrac{3 \times 9 \times 2}{6 \times 100} \\[1em] \Rightarrow \text{log } \dfrac{9}{100}. ⇒ 2 1 log 9 + 2log 3 - log 6 + log 2 - 2 ⇒ 2 1 log 3 2 + log 3 2 − log 6 + log 2 - 2log 10 ⇒ 2 1 × 2 × log 3 + log 9 - log 6 + log 2 - log 10 2 ⇒ log 3 + log 9 + log 2 - log 6 - log 100 ⇒ log 6 × 100 3 × 9 × 2 ⇒ log 100 9 .
Hence, 1 2 log 9 + 2log 3 - log 6 + log 2 - 2 = log 9 100 \dfrac{1}{2}\text{log 9 + 2log 3 - log 6 + log 2 - 2} = \text{log }\dfrac{9}{100} 2 1 log 9 + 2log 3 - log 6 + log 2 - 2 = log 100 9
Prove the following:
log10 4 ÷ log10 2 = log3 9
Answer
Given,
log10 4 ÷ log10 2 = log3 9
Simplifying L.H.S. we get,
⇒ log10 4 ÷ log10 2 = log10 22 ÷ log10 2
= 2 log 10 2 log 10 2 \dfrac{2\text{log}_{10}2}{\text{log}_{10}2} log 10 2 2 log 10 2
= 2.
Simplifying R.H.S. we get,
⇒ log3 9 = log3 32
= 2log3 3
= 2.
Since, L.H.S. = R.H.S.
Hence, proved that log10 4 ÷ log10 2 = log3 9.
Prove the following:
log10 25+ log10 4 = log5 25
Answer
Given,
log10 25+ log10 4 = log5 25
Simplifying L.H.S. we get,
⇒ log10 25+ log10 4 = log10 (25 × 4)
= log10 100
= log10 102
= 2log10 10 = 2.
Simplifying R.H.S. we get,
⇒ log5 25 = log5 52
= 2log5 5
= 2(1) = 2.
Since, L.H.S. = R.H.S.,
Hence, proved that log10 25+ log10 4 = log5 25.
If x = (100)a , y = (10000)b and z = (10)c , express log 10 y x 2 z 3 \dfrac{10\sqrt{y}}{x^2z^3} x 2 z 3 10 y
Answer
Given,
x = (100)a = (102 )a = 102a ,
y = (10000)b = (104 )b = 104b ,
z = (10)c .
⇒ log 10 y x 2 z 3 ⇒ log 10 y − log x 2 . z 3 ⇒ log 10 + log y − (log x 2 + log z 3 ) ⇒ 1 + log y 1 2 − 2log x - 3log z ⇒ 1 + 1 2 log 10 4 b − 2 log 10 2 a − 3 log 10 c ⇒ 1 + 1 2 × 4 b × log 10 - 4a.log 10 - 3c.log 10 ⇒ 1 + 2 b ( 1 ) − 4 a ( 1 ) − 3 c ( 1 ) ⇒ 1 + 2 b − 4 a − 3 c . \Rightarrow \text{log }\dfrac{10\sqrt{y}}{x^2z^3} \\[1em] \Rightarrow \text{log 10}\sqrt{y} - \text{log x}^2.z^3 \\[1em] \Rightarrow \text{log 10 + log }\sqrt{y} - \text{(log x}^2 + \text{log z}^3) \\[1em] \Rightarrow 1 + \text{log y}^{\dfrac{1}{2}} - \text{2log x - 3log z} \\[1em] \Rightarrow 1 + \dfrac{1}{2}\text{log 10}^{4b} - \text{2 log 10}^{2a} - \text{3 log 10}^c \\[1em] \Rightarrow 1 + \dfrac{1}{2} \times 4b \times \text{log 10 - 4a.log 10 - 3c.log 10} \\[1em] \Rightarrow 1 + 2b(1) - 4a(1) - 3c(1) \\[1em] \Rightarrow 1 + 2b - 4a - 3c. ⇒ log x 2 z 3 10 y ⇒ log 10 y − log x 2 . z 3 ⇒ log 10 + log y − (log x 2 + log z 3 ) ⇒ 1 + log y 2 1 − 2log x - 3log z ⇒ 1 + 2 1 log 10 4 b − 2 log 10 2 a − 3 log 10 c ⇒ 1 + 2 1 × 4 b × log 10 - 4a.log 10 - 3c.log 10 ⇒ 1 + 2 b ( 1 ) − 4 a ( 1 ) − 3 c ( 1 ) ⇒ 1 + 2 b − 4 a − 3 c .
Hence, log 10 y x 2 z 3 \dfrac{10\sqrt{y}}{x^2z^3} x 2 z 3 10 y
If a = log10 x, find the following in terms of a:
(i) x
(ii) log10 x 2 5 \sqrt[5]{x^2} 5 x 2
(iii) log10 3x
Answer
(i) Given,
a = log10 x
⇒ x = 10a .
(ii) Given,
⇒ log 10 ( x 2 ) 1 5 ⇒ 1 5 × 2 log 10 x ⇒ 1 5 × 2 log 10 10 a ⇒ 2 a 5 log 10 10 ⇒ 2 a 5 × 1 ⇒ 2 a 5 . \Rightarrow \text{log}_{10}(x^2)^{\dfrac{1}{5}} \\[1em] \Rightarrow \dfrac{1}{5} \times 2\text{log}_{10}x \\[1em] \Rightarrow \dfrac{1}{5} \times 2\text{log}_{10}10^a \\[1em] \Rightarrow \dfrac{2a}{5}\text{log}_{10}10 \\[1em] \Rightarrow \dfrac{2a}{5} \times 1 \\[1em] \Rightarrow \dfrac{2a}{5}. ⇒ log 10 ( x 2 ) 5 1 ⇒ 5 1 × 2 log 10 x ⇒ 5 1 × 2 log 10 1 0 a ⇒ 5 2 a log 10 10 ⇒ 5 2 a × 1 ⇒ 5 2 a .
Hence, log10 x 2 5 = 2 a 5 \sqrt[5]{x^2} = \dfrac{2a}{5} 5 x 2 = 5 2 a
(iii) Given,
a = log10 x
Now,
⇒ log10 3x
⇒ log10 (3 × x)
⇒ log10 3 + log10 x
⇒ log10 3 + a.
Hence, log10 3x = log10 3 + a.
If a = log2 3 \dfrac{2}{3} 3 2 3 5 \dfrac{3}{5} 5 3 5 2 \sqrt{\dfrac{5}{2}} 2 5
(i) a + b + c
(ii) 5a + b + c
Answer
(i) Given,
⇒ a + b + c = log 2 3 + log 3 5 + 2 log 5 2 = log 2 - log 3 + log 3 - log 5 + 2 × log ( 5 2 ) 1 2 = log 2 − log 5 + 2 × 1 2 × (log 5 - log 2) = log 2 - log 5 + log 5 - log 2 = 0. \Rightarrow a + b + c = \text{log }\dfrac{2}{3} + \text{log }\dfrac{3}{5} + 2\text{log }\sqrt{\dfrac{5}{2}} \\[1em] = \text{log 2 - log 3 + log 3 - log 5} + 2 \times \text{log}\Big(\dfrac{5}{2}\Big)^{\dfrac{1}{2}} \\[1em] = \text{log } 2 - \text{log } 5 + 2 \times \dfrac{1}{2} \times \text{(log 5 - log 2)} \\[1em] = \text{log 2 - log 5 + log 5 - log 2} \\[1em] = 0. ⇒ a + b + c = log 3 2 + log 5 3 + 2 log 2 5 = log 2 - log 3 + log 3 - log 5 + 2 × log ( 2 5 ) 2 1 = log 2 − log 5 + 2 × 2 1 × (log 5 - log 2) = log 2 - log 5 + log 5 - log 2 = 0.
Hence, a + b + c = 0.
(ii) Given,
⇒ 5a + b + c = 50 = 1.
Hence, 5a + b + c = 1.
If x = log 3 5 , y = log 5 4 and z = 2log 3 2 \dfrac{3}{5}, \text{ y = log }\dfrac{5}{4}\text{ and z = 2log }\dfrac{\sqrt{3}}{2} 5 3 , y = log 4 5 and z = 2log 2 3
(i) x + y - z
(ii) 3x + y - z
Answer
Given,
⇒ x + y − z = log 3 5 + log 5 4 − 2 log 3 2 = log 3 - log 5 + log 5 - log 4 - 2(log 3 − log 2) = log 3 - log 4 - 2log 3 1 2 + 2log 2 = log 3 - log 2 2 − 2 × 1 2 log 3 + 2log 2 = log 3 - 2log 2 - log 3 + 2log 2 = 0. \Rightarrow x + y - z = \text{log }\dfrac{3}{5} + \text{log }\dfrac{5}{4} - 2\text{log }\dfrac{\sqrt{3}}{2} \\[1em] = \text{log 3 - log 5 + log 5 - log 4 - 2(log }\sqrt{3} - \text{log 2)} \\[1em] = \text{log 3 - log 4 - 2log 3}^{\dfrac{1}{2}} + \text{2log 2} \\[1em] = \text{log 3 - log 2}^2 - 2 \times \dfrac{1}{2}\text{log 3 + 2log 2} \\[1em] = \text{log 3 - 2log 2 - log 3 + 2log 2} \\[1em] = 0. ⇒ x + y − z = log 5 3 + log 4 5 − 2 log 2 3 = log 3 - log 5 + log 5 - log 4 - 2(log 3 − log 2) = log 3 - log 4 - 2log 3 2 1 + 2log 2 = log 3 - log 2 2 − 2 × 2 1 log 3 + 2log 2 = log 3 - 2log 2 - log 3 + 2log 2 = 0.
Hence, x + y - z = 0.
(ii) Given,
3x + y - z = 30 = 1.
Hence, 3x + y - z = 1.
If x = log10 12, y = log4 2 × log10 9 and z = log10 0.4, find the values of
(i) x - y - z
(ii) 7x - y - z
Answer
(i) Given,
x - y - z = log10 12 - log4 2 × log10 9 - log10 0.4
= log 10 3.4 − log 4 ( 4 ) 1 2 × log 10 3 2 − log 10 4 10 = log 10 3 + log 10 4 − 1 2 log 4 4 × 2 log 10 3 − ( log 10 4 − log 10 10 ) = log 10 3 + log 10 4 − 1 × log 10 3 − log 10 4 + log 10 10 = log 10 3 − log 10 3 + 1 = 1. = \text{log}_{10} \space 3.4 - \text{log}_{4} \space (4)^{\dfrac{1}{2}} \times \text{log}_{10} \space 3^2 - \text{log}_{10} \space \dfrac{4}{10} \\[1em] = \text{log}_{10} \space 3 + \text{log}_{10} \space 4 - \dfrac{1}{2}\text{log}_{4}4 \space \times 2\text{log}_{10} \space 3 - (\text{log}_{10} \space 4 - \text{log}_{10} \space 10) \\[1em] = \text{log}_{10} \space 3 + \text{log}_{10} \space 4 - 1 \times \text{log}_{10} \space 3 - \text{log}_{10} \space 4 + \text{log}_{10} \space 10 \\[1em] = \text{log}_{10} \space 3 - \text{log}_{10} \space 3 + 1 \\[1em] = 1. = log 10 3.4 − log 4 ( 4 ) 2 1 × log 10 3 2 − log 10 10 4 = log 10 3 + log 10 4 − 2 1 log 4 4 × 2 log 10 3 − ( log 10 4 − log 10 10 ) = log 10 3 + log 10 4 − 1 × log 10 3 − log 10 4 + log 10 10 = log 10 3 − log 10 3 + 1 = 1.
Hence, x - y - z = 1.
(ii) Given,
7x - y - z = 71 = 7.
Hence, 7x - y - z = 1.
If log V + log 3 = log π + log 4 + 3log r, find V in terms of other quantities.
Answer
Given,
log V + log 3 = log π + log 4 + 3log r
⇒ log V = log π + log 4 + 3log r - log 3
⇒ log V = log π + log 4 + log r3 - log 3
⇒ log V = log π 4 r 3 3 \dfrac{π4r^3}{3} 3 π 4 r 3
⇒ V = 4 3 π r 3 \dfrac{4}{3}πr^3 3 4 π r 3
Given 3(log 5 - log 3) - (log 5 - 2log 6) = 2 - log n, find n.
Answer
Given,
3(log 5 - log 3) - (log 5 - 2log 6) = 2 - log n
⇒ 3log 5 - 3log 3 - log 5 + 2log 6 = 2 - log n
⇒ 2log 5 - log 33 + log 62 = 2 - log n
⇒ log 52 - log 27 + log 36 = 2 - log n
⇒ log 25 - log 27 + log 36 = 2log 10 - log n
⇒ log25 × 36 27 \dfrac{25 \times 36}{27} 27 25 × 36 2 - log n
⇒ log100 3 \dfrac{100}{3} 3 100
⇒ log 100 - log 3 = log 100 - log n
⇒ log n = log 3
⇒ n = 3.
Hence, n = 3.
Given that log10 y + 2log10 x = 2, express y in terms of x.
Answer
Given,
log10 y + 2log10 x = 2
⇒ log10 y + log10 x2 = 2
⇒ log10 yx2 = 2log10 10
⇒ log10 yx2 = log10 102
⇒ log10 yx2 = log10 100
⇒ yx2 = 100
⇒ y = 100 x 2 . \dfrac{100}{x^2}. x 2 100 .
Hence, y = 100 x 2 \dfrac{100}{x^2} x 2 100
Express log10 2 + 1 in the form of log10 x.
Answer
Given,
⇒ log10 2 + 1
⇒ log10 2 + log10 10
⇒ log10 2 × 10
⇒ log10 20
Hence, log10 2 + 1 = log10 20.
If a2 = log10 x, b3 = log10 y and a 2 2 − b 3 3 \dfrac{a^2}{2} - \dfrac{b^3}{3} 2 a 2 − 3 b 3 10 z, express z in terms of x and y.
Answer
Given,
a2 = log10 x
b3 = log10 y
Substituting above values in a 2 2 − b 3 3 = log 10 z \dfrac{a^2}{2} - \dfrac{b^3}{3} = \text{log}_{10}z 2 a 2 − 3 b 3 = log 10 z
⇒ log 10 x 2 − log 10 y 3 = log 10 z ⇒ 3 log 10 x − 2 log 10 y 6 = log 10 z ⇒ log 10 x 3 − log 10 y 2 6 = log 10 z ⇒ 1 6 log 10 x 3 y 2 = log 10 z ⇒ log 10 ( x 3 y 2 ) 1 6 = log 10 z ⇒ log 10 ( x 1 2 y 1 3 ) = log 10 z ⇒ log 10 x y 3 = log 10 z ⇒ z = x y 3 . \Rightarrow \dfrac{\text{log}_{10}x}{2} - \dfrac{\text{log}_{10}y}{3} = \text{log}_{10}z \\[1em] \Rightarrow \dfrac{3\text{log}_{10}x - 2\text{log}_{10}y}{6} = \text{log}_{10}z \\[1em] \Rightarrow \dfrac{\text{log}_{10}x^3 - \text{log}_{10}y^2}{6} = \text{log}_{10}z \\[1em] \Rightarrow \dfrac{1}{6}\text{log}_{10}\dfrac{x^3}{y^2} = \text{log}_{10}z \\[1em] \Rightarrow \text{log}_{10}\Big(\dfrac{x^3}{y^2}\Big)^{\dfrac{1}{6}} = \text{log}_{10}z \\[1em] \Rightarrow \text{log}_{10}\Big(\dfrac{x^\dfrac{1}{2}}{y^\dfrac{1}{3}}\Big) = \text{log}_{10}z \\[1em] \Rightarrow \text{log}_{10}\dfrac{\sqrt{x}}{\sqrt[3]y} = \text{log}_{10}z \\[1em] \Rightarrow z = \dfrac{\sqrt{x}}{\sqrt[3]y}. ⇒ 2 log 10 x − 3 log 10 y = log 10 z ⇒ 6 3 log 10 x − 2 log 10 y = log 10 z ⇒ 6 log 10 x 3 − log 10 y 2 = log 10 z ⇒ 6 1 log 10 y 2 x 3 = log 10 z ⇒ log 10 ( y 2 x 3 ) 6 1 = log 10 z ⇒ log 10 ( y 3 1 x 2 1 ) = log 10 z ⇒ log 10 3 y x = log 10 z ⇒ z = 3 y x .
Hence, z = x y 3 . z = \dfrac{\sqrt{x}}{\sqrt[3]y}. z = 3 y x .
Given that log m = x + y and log n = x - y, express the value of log m2 n in terms of x and y.
Answer
Given,
log m = x + y .......(i)
log n = x - y ........(ii)
Multiplying (i) by 2 we get,
2log m = log m2 = 2x + 2y ......(iii)
Adding (ii) and (iii) we get,
⇒ log m2 + log n = 2x + 2y + (x - y)
⇒ log m2 n = 3x + y.
Hence, log m2 n = 3x + y.
Given that log x = m + n and log y = m - n, express the value of log( 10 x y 2 ) \Big(\dfrac{10x}{y^2}\Big) ( y 2 10 x )
Answer
Given,
⇒ log ( 10 x y 2 ) = log 10 x − log y 2 \Rightarrow \text{log }\Big(\dfrac{10x}{y^2}\Big) = \text{log }10x - \text{log } y^2 ⇒ log ( y 2 10 x ) = log 10 x − log y 2
= log 10 + log x - 2log y
= 1 + m + n - 2(m - n)
= 1 + m + n - 2m + 2n
= 1 - m + 3n.
Hence, log( 10 x y 2 ) \Big(\dfrac{10x}{y^2}\Big) ( y 2 10 x )
If log x 2 = log y 3 \dfrac{\text{log x}}{2} = \dfrac{\text{log y}}{3} 2 log x = 3 log y y 4 x 6 \dfrac{y^4}{x^6} x 6 y 4
Answer
Given,
log x 2 = log y 3 \dfrac{\text{log x}}{2} = \dfrac{\text{log y}}{3} 2 log x = 3 log y
⇒ 3log x = 2log y
⇒ log x3 = log y2
⇒ x3 = y2
Squaring both sides we get,
⇒ x6 = y4
⇒ x 6 y 4 = 1 \dfrac{x^6}{y^4} = 1 y 4 x 6 = 1
Hence, x 6 y 4 = 1 \dfrac{x^6}{y^4} = 1 y 4 x 6 = 1
Solve for x:
log x + log 5 = 2log 3
Answer
Given,
⇒ log x + log 5 = 2log 3
⇒ log x = log 32 - log 5
⇒ log x = log 9 - log 5
⇒ log x = log 9 5 \dfrac{9}{5} 5 9
⇒ x = 9 5 \dfrac{9}{5} 5 9
Hence, x = 9 5 x = \dfrac{9}{5} x = 5 9
Solve for x:
log3 x - log3 2 = 1
Answer
Given,
⇒ log3 x - log3 2 = 1
⇒ log 3 x 2 = log 3 3 \text{log}_3 \dfrac{x}{2} = \text{log}_3 3 log 3 2 x = log 3 3
⇒ x 2 = 3 \dfrac{x}{2} = 3 2 x = 3
⇒ x = 6.
Hence, x = 6.
Solve for x:
x = log 125 log 25 \dfrac{\text{log 125}}{\text{log 25}} log 25 log 125
Answer
Given,
⇒ x = log 125 log 25 ⇒ x = log 5 3 log 5 2 ⇒ x = 3log 5 2log 5 ⇒ x = 3 2 . \Rightarrow x = \dfrac{\text{log 125}}{\text{log 25}} \\[1em] \Rightarrow x = \dfrac{\text{log 5}^3}{\text{log 5}^2} \\[1em] \Rightarrow x = \dfrac{\text{3log 5}}{\text{2log 5}} \\[1em] \Rightarrow x = \dfrac{3}{2}. ⇒ x = log 25 log 125 ⇒ x = log 5 2 log 5 3 ⇒ x = 2log 5 3log 5 ⇒ x = 2 3 .
Hence, x = 3 2 \dfrac{3}{2} 2 3
Solve for x:
log 8 log 2 × log 3 log 3 = 2 log x \dfrac{\text{log 8}}{\text{log 2}} \times \dfrac{\text{log 3}}{\text{log}\sqrt{3}} = 2\text{log} x log 2 log 8 × log 3 log 3 = 2 log x
Answer
Given,
⇒ log 8 log 2 × log 3 l o g 3 = 2 l o g x ⇒ log 2 3 log 2 × log 3 log 3 1 2 = 2 l o g x ⇒ 3log 2 log 2 × log 3 1 2 log 3 = 2 log x ⇒ 3 × 2 = 2 log x ⇒ log x = 6 2 ⇒ log x = 3 ⇒ x = 10 3 = 1000. \Rightarrow \dfrac{\text{log 8}}{\text{log 2}} \times \dfrac{\text{log 3}}{log\sqrt{3}} = 2log x \\[1em] \Rightarrow \dfrac{\text{log 2}^3}{\text{log 2}} \times \dfrac{\text{log 3}}{\text{log} 3^{\dfrac{1}{2}}} = 2log x \\[1em] \Rightarrow \dfrac{\text{3log 2}}{\text{log 2}} \times \dfrac{\text{log 3}}{\dfrac{1}{2}\text{log 3}} = \text{2 log x} \\[1em] \Rightarrow 3 \times 2 = 2\text{log } x \\[1em] \Rightarrow \text{log } x = \dfrac{6}{2} \\[1em] \Rightarrow \text{log } x = 3 \\[1em] \Rightarrow x = 10^3 = 1000. ⇒ log 2 log 8 × l o g 3 log 3 = 2 l o gx ⇒ log 2 log 2 3 × log 3 2 1 log 3 = 2 l o gx ⇒ log 2 3log 2 × 2 1 log 3 log 3 = 2 log x ⇒ 3 × 2 = 2 log x ⇒ log x = 2 6 ⇒ log x = 3 ⇒ x = 1 0 3 = 1000.
Hence, x = 1000.
Given 2log10 x + 1 = log10 250, find
(i) x
(ii) log10 2x
Answer
(i) Given,
2log10 x + 1 = log10 250
⇒ log10 x2 + log10 10 = log10 250
⇒ log10 x2 = log10 250 - log10 10
⇒ log10 x2 = log 10 250 10 \text{log}_{10}\dfrac{250}{10} log 10 10 250
⇒ log10 x2 = log10 25
⇒ x2 = 25
⇒ x = 5.
Hence, x = 5.
(ii) Given,
log10 2x
⇒ log10 2(5) = log10 10 = 1.
Hence, log10 2x = 1.
If log x log 5 = log y 2 log 2 = log 9 log 1 3 \dfrac{\text{log x}}{\text{log 5}} = \dfrac{\text{log y}^2}{\text{log 2}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}} log 5 log x = log 2 log y 2 = log 3 1 log 9
Answer
Given,
log x log 5 = log y 2 log 2 = log 9 log 1 3 \dfrac{\text{log x}}{\text{log 5}} = \dfrac{\text{log y}^2}{\text{log 2}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}} log 5 log x = log 2 log y 2 = log 3 1 log 9
Considering,
⇒ log x log 5 = log 9 log 1 3 ⇒ log x = log 9 × log 5 log 1 3 ⇒ log x = log 9 × log 5 log 1 - log 3 ⇒ log x = log 3 2 × log 5 0 - log 3 ⇒ log x = 2log 3 × log 5 -log 3 ⇒ log x = -2log 5 ⇒ log x = log 5 − 2 ⇒ x = 5 − 2 = 1 25 . \Rightarrow \dfrac{\text{log x}}{\text{log 5}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}} \\[1em] \Rightarrow \text{log x} = \dfrac{\text{log 9 × log 5}}{\text{log}\dfrac{1}{3}} \\[1em] \Rightarrow \text{log x} = \dfrac{\text{log 9 × log 5}}{\text{log 1 - log 3}} \\[1em] \Rightarrow \text{log x} = \dfrac{\text{log 3}^2 × \text{log } 5}{\text{0 - log 3}} \\[1em] \Rightarrow \text{log x} = \dfrac{\text{2log 3 × log 5}}{\text{-log 3}} \\[1em] \Rightarrow \text{log x} = \text{-2log 5} \\[1em] \Rightarrow \text{log x} = \text{log 5}^{-2} \\[1em] \Rightarrow x = 5^{-2} = \dfrac{1}{25}. ⇒ log 5 log x = log 3 1 log 9 ⇒ log x = log 3 1 log 9 × log 5 ⇒ log x = log 1 - log 3 log 9 × log 5 ⇒ log x = 0 - log 3 log 3 2 × log 5 ⇒ log x = -log 3 2log 3 × log 5 ⇒ log x = -2log 5 ⇒ log x = log 5 − 2 ⇒ x = 5 − 2 = 25 1 .
Considering,
⇒ log y 2 log 2 = log 9 log 1 3 ⇒ log y 2 = log 9 × log 2 log 1 - log 3 ⇒ log y 2 = log 3 2 × log 2 0 - log 3 ⇒ log y 2 = 2log 3 × log 2 -log 3 ⇒ log y 2 = -2log 2 ⇒ log y 2 = log 2 − 2 ⇒ y 2 = 2 − 2 ⇒ y = 1 2 2 = 1 2 . \Rightarrow \dfrac{\text{log y}^2}{\text{log 2}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}} \\[1em] \Rightarrow \text{log y}^2 = \dfrac{\text{log 9} \times \text{log 2}}{\text{log 1 - log 3}} \\[1em] \Rightarrow \text{log y}^2 = \dfrac{\text{log 3}^2 \times \text{log 2}}{\text{0 - log 3}} \\[1em] \Rightarrow \text{log y}^2 = \dfrac{\text{2log 3} \times \text{log 2}}{\text{-log 3}} \\[1em] \Rightarrow \text{log y}^2 = \text{-2log 2} \\[1em] \Rightarrow \text{log y}^2 = \text{log 2}^{-2} \\[1em] \Rightarrow y^2 = 2^{-2} \\[1em] \Rightarrow y = \sqrt{\dfrac{1}{2^2}} = \dfrac{1}{2}. ⇒ log 2 log y 2 = log 3 1 log 9 ⇒ log y 2 = log 1 - log 3 log 9 × log 2 ⇒ log y 2 = 0 - log 3 log 3 2 × log 2 ⇒ log y 2 = -log 3 2log 3 × log 2 ⇒ log y 2 = -2log 2 ⇒ log y 2 = log 2 − 2 ⇒ y 2 = 2 − 2 ⇒ y = 2 2 1 = 2 1 .
Hence, x = 1 25 and y = 1 2 . x = \dfrac{1}{25} \text{and y} = \dfrac{1}{2}. x = 25 1 and y = 2 1 .
Prove the following:
3log 4 = 4log 3
Answer
Given,
3log 4 = 4log 3
Taking log on both sides we get,
⇒ log 3log 4 = log 4log 3
⇒ log 4.log 3 = log 3.log 4
Since, L.H.S. = R.H.S.,
Hence, proved that 3log 4 = 4log 3 .
Prove the following:
27log 2 = 8log 3
Answer
Given,
27log 2 = 8log 3
Taking log on both sides we get,
⇒ log 27log 2 = log 8log 3
⇒ log 2.log 27 = log 3.log 8
⇒ log 2.log 33 = log 3.log 23
⇒ 3.log 2.log 3 = 3log 3.log 2
Since, L.H.S. = R.H.S. hence proved,
Hence, proved that 27log 2 = 8log 3 .
Solve the following equation:
log(2x + 3) = log 7
Answer
Given,
⇒ log(2x + 3) = log 7
⇒ 2x + 3 = 7
⇒ 2x = 4
⇒ x = 2.
Hence, x = 2.
Solve the following equation:
log(x + 1) + log(x - 1) = log 24
Answer
Given,
⇒ log(x + 1) + log(x - 1) = log 24
⇒ log((x + 1)(x - 1)) = log 24
⇒ log(x2 - 1) = log 24
⇒ x2 - 1 = 24
⇒ x2 = 25
⇒ x = 5.
Hence, x = 5.
Solve the following equation:
log(10x + 5) - log(x - 4) = 2
Answer
Given,
log(10x + 5) - log(x - 4) = 2
⇒ log ( 10 x + 5 ) − log ( x − 4 ) = 2 log 10 ⇒ log 10 x + 5 x − 4 = log 10 2 ⇒ 10 x + 5 x − 4 = 100 ⇒ 10 x + 5 = 100 ( x − 4 ) ⇒ 10 x + 5 = 100 x − 400 ⇒ 100 x − 10 x = 405 ⇒ 90 x = 405 ⇒ x = 405 90 ⇒ x = 4.5 \Rightarrow \text{log}(10x + 5) - \text{log}(x - 4) = 2\text{log } 10 \\[1em] \Rightarrow \text{log}\dfrac{10x + 5}{x - 4} = \text{log } 10^2 \\[1em] \Rightarrow \dfrac{10x + 5}{x - 4} = 100 \\[1em] \Rightarrow 10x + 5 = 100(x - 4) \\[1em] \Rightarrow 10x + 5 = 100x - 400 \\[1em] \Rightarrow 100x - 10x = 405 \\[1em] \Rightarrow 90x = 405 \\[1em] \Rightarrow x = \dfrac{405}{90} \\[1em] \Rightarrow x = 4.5 ⇒ log ( 10 x + 5 ) − log ( x − 4 ) = 2 log 10 ⇒ log x − 4 10 x + 5 = log 1 0 2 ⇒ x − 4 10 x + 5 = 100 ⇒ 10 x + 5 = 100 ( x − 4 ) ⇒ 10 x + 5 = 100 x − 400 ⇒ 100 x − 10 x = 405 ⇒ 90 x = 405 ⇒ x = 90 405 ⇒ x = 4.5
Hence, x = 4.5
Solve the following equation:
log10 5 + log10 (5x + 1) = log10 (x + 5) + 1
Answer
Given,
⇒ log10 5 + log10 (5x + 1) = log10 (x + 5) + 1
⇒ log10 5(5x + 1) = log10 (x + 5) + log10 10
⇒ log10 (25x + 5) = log10 10(x + 5)
⇒ log10 (25x + 5) = log10 (10x + 50)
⇒ 25x + 5 = 10x + 50
⇒ 25x - 10x = 50 - 5
⇒ 15x = 45
⇒ x = 3.
Hence, x = 3.
Solve the following equation:
log(4y - 3) = log(2y + 1) - log 3
Answer
Given,
⇒ log(4y - 3) = log(2y + 1) - log 3
⇒ log ( 4 y − 3 ) = log 2 y + 1 3 \text{log}(4y - 3) = \text{log}\dfrac{2y + 1}{3} log ( 4 y − 3 ) = log 3 2 y + 1
⇒ 2 y + 1 3 = 4 y − 3 \dfrac{2y + 1}{3} = 4y - 3 3 2 y + 1 = 4 y − 3
⇒ 2y + 1 = 3(4y - 3)
⇒ 2y + 1 = 12y - 9
⇒ 12y - 2y = 1 + 9
⇒ 10y = 10
⇒ y = 1.
Hence, y = 1.
Solve the following equation:
log10 (x + 2) + log10 (x - 2) = log10 3 + 3log10 4
Answer
Given,
⇒ log10 (x + 2) + log10 (x - 2) = log10 3 + 3log10 4
⇒ log10 (x + 2)(x - 2) = log10 3 + log10 43
⇒ log10 (x2 - 4) = log10 3 + log10 64
⇒ log10 (x2 - 4) = log10 (3 × 64)
⇒ log10 (x2 - 4) = log10 192
⇒ x2 - 4 = 192
⇒ x2 = 196
⇒ x = 196 \sqrt{196} 196
Hence, x = 14.
Solve the following equation:
log(3x + 2) + log(3x - 2) = 5log 2
Answer
Given,
⇒ log(3x + 2) + log(3x - 2) = 5log 2
⇒ log(3x + 2)(3x - 2) = log 25
⇒ log(9x2 - 4) = log 32
⇒ 9x2 - 4 = 32
⇒ 9x2 = 36
⇒ x2 = 4
⇒ x = 2.
Hence, x = 2.
Solve for x: log3 (x + 1) - 1 = 3 + log3 (x - 1)
Answer
Given,
log3 (x + 1) - 1 = 3 + log3 (x - 1)
Since log3 3 = 1, the above equation can be written as
⇒ log3 (x + 1) - log3 3 = 3log3 3 + log3 (x - 1)
⇒ log3 (x + 1) - log3 3 = log3 33 + log3 (x - 1)
⇒ log 3 ( x + 1 ) 3 = log 3 [ 3 3 × ( x − 1 ) ] \text{log}_3\dfrac{(x + 1)}{3} = \text{log}_3\space[3^3 \times (x - 1)] log 3 3 ( x + 1 ) = log 3 [ 3 3 × ( x − 1 )]
⇒ log 3 ( x + 1 ) 3 = log 3 [ 27 ( x − 1 ) ] \text{log}_3\dfrac{(x + 1)}{3} = \text{log}_3\space[27(x - 1)] log 3 3 ( x + 1 ) = log 3 [ 27 ( x − 1 )]
⇒ x + 1 3 = 27 ( x − 1 ) \dfrac{x + 1}{3} = 27(x - 1) 3 x + 1 = 27 ( x − 1 )
⇒ x + 1 = 81(x - 1)
⇒ x + 1 = 81x - 81
⇒ 81x - x = 1 + 81
⇒ 80x = 82
⇒ x = 82 80 = 41 40 = 1 1 40 . \dfrac{82}{80} = \dfrac{41}{40} = 1\dfrac{1}{40}. 80 82 = 40 41 = 1 40 1 .
Hence, x = 1 1 40 1\dfrac{1}{40} 1 40 1
Solve for x: 5log x + 3log x = 3log x + 1 - 5log x - 1 .
Answer
Given,
5log x + 3log x = 3log x + 1 - 5log x - 1
⇒ 5log x + 3log x = 3log x .31 - 5log x .5-1
⇒ 5log x + 5log x .5-1 = 3log x .31 - 3log x
⇒ 5log x (1 + 5-1 ) = 3log x (3 - 1)
⇒ 5log x ( 1 + 1 5 ) \Big(1 + \dfrac{1}{5}\Big) ( 1 + 5 1 ) log x
⇒ 5log x .6 5 \dfrac{6}{5} 5 6 log x
⇒ 5 log x 3 log x = 5 × 2 6 ⇒ ( 5 3 ) log x = 5 3 ⇒ log x = 1 ⇒ log x = log 10 ⇒ x = 10. \Rightarrow \dfrac{5^{\text{log } x}}{3^{\text{log } x}} = \dfrac{5 \times 2}{6} \\[1em] \Rightarrow \Big(\dfrac{5}{3}\Big)^{\text{log } x} = \dfrac{5}{3} \\[1em] \Rightarrow \text{log } x = 1 \\[1em] \Rightarrow \text{log } x = \text{log } 10 \\[1em] \Rightarrow x = 10. ⇒ 3 log x 5 log x = 6 5 × 2 ⇒ ( 3 5 ) log x = 3 5 ⇒ log x = 1 ⇒ log x = log 10 ⇒ x = 10.
Hence, x = 10.
If logx − y 2 = 1 2 \dfrac{x - y}{2} = \dfrac{1}{2} 2 x − y = 2 1 2 + y2 = 6xy.
Answer
Given,
⇒ log x − y 2 = 1 2 ( log x + log y ) ⇒ 2 log x − y 2 = log xy ⇒ ( x − y 2 ) 2 = x y ⇒ ( x − y ) 2 = 4 x y ⇒ x 2 + y 2 − 2 x y = 4 x y ⇒ x 2 + y 2 = 6 x y . \Rightarrow \text{log}\dfrac{x - y}{2} = \dfrac{1}{2}(\text{log x + log y}) \\[1em] \Rightarrow 2\text{log}\dfrac{x - y}{2} = \text{log xy} \\[1em] \Rightarrow \Big(\dfrac{x - y}{2}\Big)^2 = xy \\[1em] \Rightarrow (x - y)^2 = 4xy \\[1em] \Rightarrow x^2 + y^2 - 2xy = 4xy \\[1em] \Rightarrow x^2 + y^2 = 6xy. ⇒ log 2 x − y = 2 1 ( log x + log y ) ⇒ 2 log 2 x − y = log xy ⇒ ( 2 x − y ) 2 = x y ⇒ ( x − y ) 2 = 4 x y ⇒ x 2 + y 2 − 2 x y = 4 x y ⇒ x 2 + y 2 = 6 x y .
Hence, proved that x2 + y2 = 6xy.
If x2 + y2 = 23xy, prove that logx + y 5 = 1 2 \dfrac{x + y}{5} = \dfrac{1}{2} 5 x + y = 2 1
Answer
Given,
x2 + y2 = 23xy
Above equation can be written as,
⇒ x 2 + y 2 = 25 x y − 2 x y ⇒ x 2 + y 2 + 2 x y = 25 x y ⇒ x 2 + y 2 + 2 x y 25 = x y ⇒ ( x + y 5 ) 2 = x y \Rightarrow x^2 + y^2 = 25xy - 2xy \\[1em] \Rightarrow x^2 + y^2 + 2xy = 25xy \\[1em] \Rightarrow \dfrac{x^2 + y^2 + 2xy}{25} = xy \\[1em] \Rightarrow \Big(\dfrac{x + y}{5}\Big)^2 = xy ⇒ x 2 + y 2 = 25 x y − 2 x y ⇒ x 2 + y 2 + 2 x y = 25 x y ⇒ 25 x 2 + y 2 + 2 x y = x y ⇒ ( 5 x + y ) 2 = x y
Taking log on both sides we get,
⇒ log ( x + y 5 ) 2 = log x y ⇒ 2 log x + y 5 = log x + log y ⇒ log x + y 5 = 1 2 ( log x + log y ) \Rightarrow \text{log }\Big(\dfrac{x + y}{5}\Big)^2 = \text{log }xy \\[1em] \Rightarrow 2\text{log }\dfrac{x + y}{5} = \text{log x + log y} \\[1em] \Rightarrow \text{log }\dfrac{x + y}{5} = \dfrac{1}{2}(\text{log x + log y}) \\[1em] ⇒ log ( 5 x + y ) 2 = log x y ⇒ 2 log 5 x + y = log x + log y ⇒ log 5 x + y = 2 1 ( log x + log y )
Hence, proved that log x + y 5 = 1 2 ( log x + log y ) \text{log}\dfrac{x + y}{5} = \dfrac{1}{2}(\text{log x + log y}) log 5 x + y = 2 1 ( log x + log y )
If p = log10 20 and q = log10 25, find the value of x if
2log10 (x + 1) = 2p - q
Answer
Given,
2log10 (x + 1) = 2p - q
⇒ 2log10 (x + 1) = 2log10 20 - log10 25
⇒ log10 (x + 1)2 = log10 202 - log10 25
⇒ log10 (x + 1)2 = log10 400 - log10 25
⇒ log10 (x + 1)2 = log 10 400 25 \text{log}_{10}\space\dfrac{400}{25} log 10 25 400
⇒ log10 (x + 1)2 = log10 16
⇒ (x + 1)2 = 16
⇒ x2 + 1 + 2x = 16
x2 + 2x - 15 = 0
x2 + 5x - 3x - 15 = 0
x(x + 5) - 3(x + 5) = 0
(x - 3)(x + 5) = 0
x = 3 or -5.
But x ≠ -5 as then (x + 1) will be negative.
Hence, x = 3.
Show that:
1 log 2 42 + 1 log 3 42 + 1 log 7 42 = 1 \dfrac{1}{\text{log}_2\space42} + \dfrac{1}{\text{log}_3\space42} + \dfrac{1}{\text{log}_7\space42} = 1 log 2 42 1 + log 3 42 1 + log 7 42 1 = 1
Answer
Given,
1 log 2 42 + 1 log 3 42 + 1 log 7 42 = 1 \dfrac{1}{\text{log}_2\space42} + \dfrac{1}{\text{log}_3\space42} + \dfrac{1}{\text{log}_7\space42} = 1 log 2 42 1 + log 3 42 1 + log 7 42 1 = 1
Simplifying L.H.S. we get,
1 log 2 42 + 1 log 3 42 + 1 log 7 42 \dfrac{1}{\text{log}_2\space42} + \dfrac{1}{\text{log}_3\space42} + \dfrac{1}{\text{log}_7\space42} log 2 42 1 + log 3 42 1 + log 7 42 1 42 2 + log42 3 + log42 7
= log42 (2 × 3 × 7)
= log42 42
= 1.
Since, L.H.S. = R.H.S.,
Hence, proved that 1 log 2 42 + 1 log 3 42 + 1 log 7 42 = 1 \dfrac{1}{\text{log}_2\space42} + \dfrac{1}{\text{log}_3\space42} + \dfrac{1}{\text{log}_7\space42} = 1 log 2 42 1 + log 3 42 1 + log 7 42 1 = 1
Show that:
1 log 8 36 + 1 log 9 36 + 1 log 18 36 = 2 \dfrac{1}{\text{log}_8\space36} + \dfrac{1}{\text{log}_9\space36} + \dfrac{1}{\text{log}_{18}\space36} = 2 log 8 36 1 + log 9 36 1 + log 18 36 1 = 2
Answer
Given,
1 log 8 36 + 1 log 9 36 + 1 log 18 36 = 2 \dfrac{1}{\text{log}_8\space36} + \dfrac{1}{\text{log}_9\space36} + \dfrac{1}{\text{log}_{18}\space36} = 2 log 8 36 1 + log 9 36 1 + log 18 36 1 = 2
Simplifying L.H.S. we get,
1 log 8 36 + 1 log 9 36 + 1 log 18 36 \dfrac{1}{\text{log}_8\space36} + \dfrac{1}{\text{log}_9\space36} + \dfrac{1}{\text{log}_{18}\space36} log 8 36 1 + log 9 36 1 + log 18 36 1 36 8 + log36 9 + log36 18
= log36 (8 × 9 × 18)
= log36 (36)2
= 2log36 36
= 2.
Since, L.H.S. = R.H.S.,
Hence, proved that 1 log 8 36 + 1 log 9 36 + 1 log 18 36 = 2 \dfrac{1}{\text{log}_8\space36} + \dfrac{1}{\text{log}_9\space36} + \dfrac{1}{\text{log}_{18}\space36} = 2 log 8 36 1 + log 9 36 1 + log 18 36 1 = 2
Prove the following identities:
1 log a abc + 1 log b abc + 1 log c abc = 1 \dfrac{1}{\text{log}_a\text{ abc}} + \dfrac{1}{\text{log}_b\text{ abc}} + \dfrac{1}{\text{log}_c\text{ abc}} = 1 log a abc 1 + log b abc 1 + log c abc 1 = 1
Answer
Given,
1 log a abc + 1 log b abc + 1 log c abc = 1 \dfrac{1}{\text{log}_a\text{ abc}} + \dfrac{1}{\text{log}_b\text{ abc}} + \dfrac{1}{\text{log}_c\text{ abc}} = 1 log a abc 1 + log b abc 1 + log c abc 1 = 1
Simplifying L.H.S. we get,
⇒ 1 log a abc + 1 log b abc + 1 log _ c abc ⇒ log abc a + log abc b + log abc c ⇒ log abc ( a × b × c ) ⇒ log abc abc ⇒ 1. \Rightarrow \dfrac{1}{\text{log}_a\text{ abc}} + \dfrac{1}{\text{log}_b\text{ abc}} + \dfrac{1}{\text{log}\_c\text{ abc}} \\[1em] \Rightarrow \text{log}_\text{abc}\text{ a} + \text{log}_\text{abc}\text{ b} + \text{log}_\text{abc}\text{ c} \\[1em] \Rightarrow \text{log}_\text{abc}\space(\text{a }\times \text{b } \times \text{c}) \\[1em] \Rightarrow \text{log}_\text{abc}\text{ abc} \\[1em] \Rightarrow 1. ⇒ log a abc 1 + log b abc 1 + log _ c abc 1 ⇒ log abc a + log abc b + log abc c ⇒ log abc ( a × b × c ) ⇒ log abc abc ⇒ 1.
Since, L.H.S. = R.H.S.,
Hence, proved that 1 log a abc + 1 log b abc + 1 log c abc = 1 \dfrac{1}{\text{log}_a\text{ abc}} + \dfrac{1}{\text{log}_b\text{ abc}} + \dfrac{1}{\text{log}_c\text{ abc}} = 1 log a abc 1 + log b abc 1 + log c abc 1 = 1
Prove the following identities:
logb a . logc b . logd c = logd a
Answer
Given,
logb a . logc b . logd c = logd a
Simplifying L.H.S. we get,
⇒ log b a . log c b . log d c ⇒ log a log b × log b log c × log c log d ⇒ log a log d ⇒ log _ d a . \Rightarrow \text{log}_\text{b}\text{ a}\space.\space\text{log}_\text{c}\text{ b}\space.\space\text{log}_\text{d}\text{ c} \\[1em] \Rightarrow \dfrac{\text{log a}}{\text{log b}} \times \dfrac{\text{log b}}{\text{log c}} \times \dfrac{\text{log c}}{\text{log d}} \\[1em] \Rightarrow \dfrac{\text{log a}}{\text{log d}} \\[1em] \Rightarrow \text{log}\_\text{d}\text{ a}. ⇒ log b a . log c b . log d c ⇒ log b log a × log c log b × log d log c ⇒ log d log a ⇒ log _ d a .
Hence, proved that logb a . logc b . logd c = logd a.
Given that loga x = 1 α \dfrac{1}{α} α 1 b x = 1 β \dfrac{1}{β} β 1 c x = 1 γ \dfrac{1}{γ} γ 1 abc x.
Answer
Given,
⇒ 1 α = log a x = log x log a ⇒ log a = α log x ⇒ 1 β = log b x = log x log b ⇒ log b = β log x ⇒ 1 γ = log c x = log x log c ⇒ log c = γ log x \Rightarrow \dfrac{1}{α} = \text{log}_a\space x = \dfrac{\text{log x}}{\text{log a}} \\[1em] \Rightarrow \text{log a} = α\text{log x} \\[1em] \\[1em] \Rightarrow \dfrac{1}{β} = \text{log}_b\space x = \dfrac{\text{log x}}{\text{log b}} \\[1em] \Rightarrow \text{log b} = β\text{log x} \\[1em] \\[1em] \Rightarrow \dfrac{1}{γ} = \text{log}_c\space x = \dfrac{\text{log x}}{\text{log c}} \\[1em] \Rightarrow \text{log c} = γ\text{log x} \\[1em] \\[1em] ⇒ α 1 = log a x = log a log x ⇒ log a = α log x ⇒ β 1 = log b x = log b log x ⇒ log b = β log x ⇒ γ 1 = log c x = log c log x ⇒ log c = γ log x
Solving logabc x we get,
⇒ log abc x = log x log abc = log x log a + log b + log c = log x αlog x + βlog x + γlog x = log x log x(α + β + γ) = 1 ( α + β + γ ) . \Rightarrow \text{log}_\text{abc}\space x = \dfrac{\text{log x}}{\text{log abc}} \\[1em] = \dfrac{\text{log x}}{\text{log a + log b + log c}} \\[1em] = \dfrac{\text{log x}}{\text{αlog x + βlog x + γlog x}} \\[1em] = \dfrac{\text{log x}}{\text{log x(α + β + γ)}} \\[1em] = \dfrac{1}{(\text{α} + \text{β} + \text{γ})}. ⇒ log abc x = log abc log x = log a + log b + log c log x = αlog x + βlog x + γlog x log x = log x(α + β + γ) log x = ( α + β + γ ) 1 .
Hence, logabc x = 1 ( α + β + γ ) \dfrac{1}{(\text{α} + \text{β} + \text{γ})} ( α + β + γ ) 1
Solve for x:
log3 x + log9 x + log81 x = 7 4 \dfrac{7}{4} 4 7
Answer
Given,
⇒ log 3 x + log 9 x + log 81 x = 7 4 ⇒ 1 log x 3 + 1 log x 9 + 1 log x 81 = 7 4 ⇒ 1 log x 3 + 1 log x 3 2 + 1 log x 3 4 = 7 4 ⇒ 1 log x 3 + 1 2 log x 3 + 1 4 log x 3 = 7 4 ⇒ 1 log x 3 [ 1 + 1 2 + 1 4 ] = 7 4 ⇒ 1 log x 3 × 7 4 = 7 4 ⇒ 1 log x 3 = 7 4 × 4 7 ⇒ 1 log x 3 = 1 ⇒ log 3 x = log 3 3 ⇒ x = 3. \Rightarrow \text{log}_3\space x + \text{log}_9\space x + \text{log}_{81} \space x = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3} + \dfrac{1}{\text{log}_x\space 9} + \dfrac{1}{\text{log}_x\space 81} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3} + \dfrac{1}{\text{log}_x\space 3^2} + \dfrac{1}{\text{log}_x\space 3^4} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3} + \dfrac{1}{2\text{log}_x\space 3} + \dfrac{1}{4\text{log}_x\space 3} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3}\Big[1 + \dfrac{1}{2} + \dfrac{1}{4}\Big] = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3} \times \dfrac{7}{4} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3} = \dfrac{7}{4} \times \dfrac{4}{7} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3} = 1 \\[1em] \Rightarrow \text{log}_3\space x = \text{log}_3\space 3 \\[1em] \Rightarrow x = 3. ⇒ log 3 x + log 9 x + log 81 x = 4 7 ⇒ log x 3 1 + log x 9 1 + log x 81 1 = 4 7 ⇒ log x 3 1 + log x 3 2 1 + log x 3 4 1 = 4 7 ⇒ log x 3 1 + 2 log x 3 1 + 4 log x 3 1 = 4 7 ⇒ log x 3 1 [ 1 + 2 1 + 4 1 ] = 4 7 ⇒ log x 3 1 × 4 7 = 4 7 ⇒ log x 3 1 = 4 7 × 7 4 ⇒ log x 3 1 = 1 ⇒ log 3 x = log 3 3 ⇒ x = 3.
Hence, x = 3.
Solve for x:
log2 x + log8 x + log32 x = 23 15 \dfrac{23}{15} 15 23
Answer
Given,
⇒ log 2 x + log 8 x + log 32 x = 23 15 ⇒ 1 log x 2 + 1 log x 8 + 1 log x 32 = 23 15 ⇒ 1 log x 2 + 1 log x 2 3 + 1 log x 2 5 = 23 15 ⇒ 1 log x 2 + 1 3 log x 2 + 1 5 log x 2 = 23 15 ⇒ 1 log x 2 [ 1 + 1 3 + 1 5 ] = 23 15 ⇒ log 2 x [ 15 + 5 + 3 15 ] = 23 15 ⇒ log 2 x × 23 15 = 23 15 ⇒ log 2 x = 23 15 × 15 23 ⇒ log 2 x = 1 ⇒ log 2 x = log 2 2 ⇒ x = 2. \Rightarrow \text{log}_2 \space x + \text{log}_8 \space x + \text{log}_{32} \space x = \dfrac{23}{15} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x \space 2} + \dfrac{1}{\text{log}_x \space 8} + \dfrac{1}{\text{log}_x \space 32} = \dfrac{23}{15} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x \space 2} + \dfrac{1}{\text{log}_x \space 2^3} + \dfrac{1}{\text{log}_x \space 2^5} = \dfrac{23}{15} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x \space 2} + \dfrac{1}{3\text{log}_x \space 2} + \dfrac{1}{5\text{log}_x \space 2} = \dfrac{23}{15} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x \space 2}\Big[1 + \dfrac{1}{3} + \dfrac{1}{5}\Big] = \dfrac{23}{15} \\[1em] \Rightarrow \text{log}_2 \space x\Big[\dfrac{15 + 5 + 3}{15}\Big] = \dfrac{23}{15} \\[1em] \Rightarrow \text{log}_2 \space x \times \dfrac{23}{15} = \dfrac{23}{15} \\[1em] \Rightarrow \text{log}_2 \space x = \dfrac{23}{15} \times \dfrac{15}{23} \\[1em] \Rightarrow \text{log}_2 \space x = 1 \\[1em] \Rightarrow \text{log}_2 \space x = \text{log}_2 \space 2 \\[1em] \Rightarrow x = 2. ⇒ log 2 x + log 8 x + log 32 x = 15 23 ⇒ log x 2 1 + log x 8 1 + log x 32 1 = 15 23 ⇒ log x 2 1 + log x 2 3 1 + log x 2 5 1 = 15 23 ⇒ log x 2 1 + 3 log x 2 1 + 5 log x 2 1 = 15 23 ⇒ log x 2 1 [ 1 + 3 1 + 5 1 ] = 15 23 ⇒ log 2 x [ 15 15 + 5 + 3 ] = 15 23 ⇒ log 2 x × 15 23 = 15 23 ⇒ log 2 x = 15 23 × 23 15 ⇒ log 2 x = 1 ⇒ log 2 x = log 2 2 ⇒ x = 2.
Hence, x = 2.
Multiple Choice Questions
If log 3 27 = x \text{log}_{\sqrt{3}} \space 27 = x log 3 27 = x
3
4
6
9
Answer
Given,
⇒ x = log 3 27 = log 10 27 log 10 3 = log 10 3 3 log 10 3 1 2 = 3 log 10 3 1 2 log 10 3 = 3 1 2 = 3 × 2 = 6. \Rightarrow x = \text{log}_{\sqrt{3}} \space 27 \\[1em] = \dfrac{\text{log}_{10} \space 27}{\text{log}_{10} \space \sqrt{3}} \\[1em] = \dfrac{\text{log}_{10} \space 3^3}{\text{log}_{10} \space 3^{\dfrac{1}{2}}} \\[1em] = \dfrac{3\text{log}_{10} \space 3}{\dfrac{1}{2}\text{log}_{10} \space 3} \\[1em] = \dfrac{3}{\dfrac{1}{2}} \\[1em] = 3 \times 2 = 6. ⇒ x = log 3 27 = log 10 3 log 10 27 = log 10 3 2 1 log 10 3 3 = 2 1 log 10 3 3 log 10 3 = 2 1 3 = 3 × 2 = 6.
Hence, Option 3 is the correct option.
If log5 (0.04) = x, then the value of x is
2
4
-4
-2
Answer
Given,
log5 (0.04) = x
⇒ 5x = 0.04
⇒ 5x = 4 100 = 1 25 \dfrac{4}{100} = \dfrac{1}{25} 100 4 = 25 1
⇒ 5x = 1 5 2 \dfrac{1}{5^2} 5 2 1
⇒ 5x = 5-2
⇒ x = -2.
Hence, Option 4 is the correct option.
If log0.5 64 = x, then the value of x is
-4
-6
4
6
Answer
Given,
log0.5 64 = x
⇒ ( 0.5 ) x = 64 ⇒ ( 5 10 ) x = 64 ⇒ ( 1 2 ) x = ( 2 ) 6 ⇒ ( 2 − 1 ) x = ( 2 ) 6 ⇒ 2 − x = ( 2 ) 6 ⇒ − x = 6 ⇒ x = − 6. \Rightarrow (0.5)^x = 64 \\[1em] \Rightarrow \Big(\dfrac{5}{10}\Big)^x = 64 \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^x = (2)^6 \\[1em] \Rightarrow (2^{-1})^x = (2)^6 \\[1em] \Rightarrow 2^{-x} = (2)^6 \\[1em] \Rightarrow -x = 6 \\[1em] \Rightarrow x = -6. ⇒ ( 0.5 ) x = 64 ⇒ ( 10 5 ) x = 64 ⇒ ( 2 1 ) x = ( 2 ) 6 ⇒ ( 2 − 1 ) x = ( 2 ) 6 ⇒ 2 − x = ( 2 ) 6 ⇒ − x = 6 ⇒ x = − 6.
Hence, Option 2 is the correct option.
If log 5 3 x \text{log}_{\sqrt[3]{5}} \space x log 3 5 x
1 5 \dfrac{1}{5} 5 1
− 1 5 -\dfrac{1}{5} − 5 1
-1
5
Answer
Given,
⇒ log 5 3 x = − 3 ⇒ x = ( 5 3 ) − 3 ⇒ x = [ ( 5 ) 1 3 ] − 3 ⇒ x = ( 5 ) 1 3 × ( − 3 ) ⇒ x = ( 5 ) − 1 = 1 5 . \Rightarrow \text{log}_{\sqrt[3]{5}} \space x = -3 \\[1em] \Rightarrow x = (\sqrt[3]{5})^{-3} \\[1em] \Rightarrow x = [(5)^{\dfrac{1}{3}}]^{-3} \\[1em] \Rightarrow x = (5)^{\dfrac{1}{3} \times (-3)} \\[1em] \Rightarrow x = (5)^{-1} = \dfrac{1}{5}. ⇒ log 3 5 x = − 3 ⇒ x = ( 3 5 ) − 3 ⇒ x = [( 5 ) 3 1 ] − 3 ⇒ x = ( 5 ) 3 1 × ( − 3 ) ⇒ x = ( 5 ) − 1 = 5 1 .
Hence, Option 1 is the correct option.
If log (3x + 1) = 2, then the value of x is
1 3 \dfrac{1}{3} 3 1
99
33
19 3 \dfrac{19}{3} 3 19
Answer
Given,
log (3x + 1) = 2
⇒ 102 = 3x + 1
⇒ 100 = 3x + 1
⇒ 3x = 100 - 1
⇒ 3x = 99
⇒ x = 33.
Hence, Option 3 is the correct option.
The value of 2 + log10 (0.01) is
4
3
1
0
Answer
Given,
2 + log10 (0.01)
⇒ 2 + log10 (10-2 )
⇒ 2 + (-2)log10 10
⇒ 2 + (-2)1
⇒ 2 + (-2)
⇒ 0
Hence, Option 4 is the correct option.
The value of log 8 − log 2 log 32 \dfrac{\text{log} \space 8 - \text{log} \space 2}{\text{log} \space 32} log 32 log 8 − log 2
2 5 \dfrac{2}{5} 5 2
1 4 \dfrac{1}{4} 4 1
− 2 5 -\dfrac{2}{5} − 5 2
1 3 \dfrac{1}{3} 3 1
Answer
Given,
⇒ log 8 − log 2 log 32 ⇒ log 2 3 − log 2 log 2 5 ⇒ 3 log 2 − log 2 5 log 2 ⇒ 2 log 2 5 log 2 ⇒ 2 5 . \Rightarrow \dfrac{\text{log} \space 8 - \text{log} \space 2}{\text{log} \space 32} \\[1em] \Rightarrow \dfrac{\text{log} \space 2^3 - \text{log} \space 2}{\text{log} \space 2^5} \\[1em] \Rightarrow \dfrac{3\text{log} \space 2 - \text{log} \space 2}{5\text{log} \space 2} \\[1em] \Rightarrow \dfrac{2 \text{log} \space 2}{5\text{log} \space 2} \\[1em] \Rightarrow \dfrac{2}{5}. ⇒ log 32 log 8 − log 2 ⇒ log 2 5 log 2 3 − log 2 ⇒ 5 log 2 3 log 2 − log 2 ⇒ 5 log 2 2 log 2 ⇒ 5 2 .
Hence, Option 1 is the correct option.
Consider the following two statements :
Statement 1: log5 150 = log5 25 + log5 125
Statement 2: loga (b + c) = loga b + loga c
Which of the following is valid?
Both the statements are true.
Both the statements are false.
Statement 1 is true, and Statement 2 is false.
Statement 1 is false, and Statement 2 is true.
Answer
According to statement 1; log5 150 = log5 25 + log5 125
Solving R.H.S.,
⇒ log5 25 + log5 125
⇒ log5 (25 x 125)
⇒ log5 3125.
As, log5 150 ≠ log5 3125
∴ Statement 1 is false.
According to statement 2; loga (b + c) = loga b + loga c
Solving R.H.S.
⇒ loga b + loga c
⇒ loga (b x c)
⇒ loga bc.
As, loga (b + c) ≠ loga bc
∴ Statement 2 is false.
∴ Both statements are false.
Hence, option 2 is the correct option.
Assertion Reason Type Questions
Assertion (A): log2 16 = 4.
Reason (R): loga (bc) = loga b + loga c
Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer
loga (bc) = loga b + loga c
This is a fundamental property of logarithms, known as the product rule. It states that the logarithm of a product is the sum of the logarithms of the individual factors.
∴ Reason (R) is true.
Given, log2 16
⇒ log2 (4 x 4)
⇒ log2 4 + log2 4
⇒ log2 22 + log2 22
⇒ 2log2 2 + 2log2 2
⇒ 2 + 2
⇒ 4.
∴ Assertion (A) is true.
∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason (or explanation) for Assertion (A).
Hence, option 3 is the correct option.
Assertion (A): log3 ( 1 9 ) \Big(\dfrac{1}{9}\Big) ( 9 1 )
Reason (R): loga ( 1 b ) \Big(\dfrac{1}{b}\Big) ( b 1 ) a b.
Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer
Given,
⇒ loga ( 1 b ) \Big(\dfrac{1}{b}\Big) ( b 1 )
⇒ loga b-1
⇒ (-1) x loga b
⇒ -loga b
∴ Reason (R) is true.
Given, log3 ( 1 9 ) \Big(\dfrac{1}{9}\Big) ( 9 1 )
⇒ log3 ( 1 3 2 ) \Big(\dfrac{1}{3^2}\Big) ( 3 2 1 )
⇒ log3 3-2
⇒ -2log3 3
⇒ -2 x 1
⇒ -2.
∴ Assertion (A) is true.
∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason (or explanation) for Assertion (A).
Hence, option 3 is the correct option.
Assertion (A): l o g 2 2 5 log_{\sqrt{2}} 2^5 l o g 2 2 5
Reason (R): log am bn = n m \dfrac{n}{m} m n a b.
Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer
Given,
log am bn
By changing the base, we get :
⇒ log b n log a m ⇒ n log b m log a ⇒ n m log a b \Rightarrow \dfrac{\text{ log } b^n}{\text{ log } a^m}\\[1em] \Rightarrow \dfrac{n\text{ log } b}{m\text{ log } a}\\[1em] \Rightarrow \dfrac{n}{m} \text{ log } _a b ⇒ log a m log b n ⇒ m log a n log b ⇒ m n log a b
∴ Reason (R) is true.
Given,
⇒ log 2 2 5 ⇒ log 2 5 log 2 ⇒ log 2 5 log 2 1 2 ⇒ 5 × log 2 1 2 × log 2 ⇒ 10 1 × 1 ⇒ 10. \Rightarrow \text{log }_{\sqrt{2} } 2^5\\[1em] \Rightarrow \dfrac{\text{log } 2^5}{\text{log }\sqrt{2}} \\[1em] \Rightarrow \dfrac{\text{log } 2^5}{\text{log }2^{\dfrac{1}{2}}} \\[1em] \Rightarrow \dfrac{5 \times \text{ log } 2}{\dfrac{1}{2} \times \text{ log } 2} \\[1em] \Rightarrow \dfrac{10}{1} \times 1\\[1em] \Rightarrow 10. ⇒ log 2 2 5 ⇒ log 2 log 2 5 ⇒ log 2 2 1 log 2 5 ⇒ 2 1 × log 2 5 × log 2 ⇒ 1 10 × 1 ⇒ 10.
∴ Assertion (A) is true.
∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason (or explanation) for Assertion (A).
Hence, option 3 is the correct option.
Assertion (A): If log x = log 8 log 0.25 \dfrac{\text{ log } 8}{\text{ log } 0.25} log 0.25 log 8
Reason (R): If log a b = log a c, then b = c.
Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer
According to reason,
If log a b = log a c, then b = c.
If the logarithms of two numbers are equal, and they share the same base, then the numbers themselves must be equal.
∴ Reason (R) is true.
According to assertion,
⇒ log x = log 8 log 0.25 ⇒ log x = log 8 log ( 25 100 ) ⇒ log x = log 2 3 log ( 1 4 ) ⇒ log x = log 2 3 log ( 1 2 2 ) ⇒ log x = log 2 3 log 2 − 2 ⇒ log x = 3 log 2 − 2 log 2 ⇒ log x = − 3 2 ⇒ x = ( 10 ) − 3 2 . \Rightarrow \text{log x} = \dfrac{\text{ log } 8}{\text{ log } 0.25}\\[1em] \Rightarrow \text{log x} = \dfrac{\text{ log } 8}{\text{ log } \Big(\dfrac{25}{100}\Big)}\\[1em] \Rightarrow \text{log x} = \dfrac{\text{ log } 2^3}{\text{ log } \Big(\dfrac{1}{4}\Big)}\\[1em] \Rightarrow \text{log x} = \dfrac{\text{ log } 2^3}{\text{ log } \Big(\dfrac{1}{2^2}\Big)}\\[1em] \Rightarrow \text{log x} = \dfrac{\text{ log } 2^3}{\text{ log } 2^{-2}}\\[1em] \Rightarrow \text{log x} = \dfrac{3\text{ log } 2}{-2\text{ log } 2}\\[1em] \Rightarrow \text{log x} = \dfrac{-3}{2} \\[1em] \Rightarrow x = (10)^{-\dfrac{3}{2}}. ⇒ log x = log 0.25 log 8 ⇒ log x = log ( 100 25 ) log 8 ⇒ log x = log ( 4 1 ) log 2 3 ⇒ log x = log ( 2 2 1 ) log 2 3 ⇒ log x = log 2 − 2 log 2 3 ⇒ log x = − 2 log 2 3 log 2 ⇒ log x = 2 − 3 ⇒ x = ( 10 ) − 2 3 .
∴ Assertion (A) is false.
∴ Assertion (A) is false, Reason (R) is true.
Hence, option 2 is the correct option.
Expand log a x 7 y 8 ÷ z 4 3 \text{log}_a \space {\sqrt[3]{x^7y^8 ÷ \sqrt[4]{z}}} log a 3 x 7 y 8 ÷ 4 z
Answer
Given,
⇒ log a x 7 y 8 ÷ z 4 3 ⇒ log a ( x 7 y 8 ÷ z 4 ) 1 3 ⇒ 1 3 log a ( x 7 y 8 ÷ z 4 ) ⇒ 1 3 [ log a x 7 y 8 − log a z 4 ] ⇒ 1 3 [ log a x 7 + log a y 8 − log a z 1 4 ] ⇒ 1 3 [ 7 log a x + 8 log a y − 1 4 log a z ] ⇒ 7 3 log a x + 8 3 log a y − 1 12 log a z . \Rightarrow \text{log}_a \space {\sqrt[3]{x^7y^8 ÷ \sqrt[4]{z}}} \\[1em] \Rightarrow \text{log}_a \space {(x^7y^8 ÷ \sqrt[4]{z})^{\dfrac{1}{3}}} \\[1em] \Rightarrow \dfrac{1}{3}\text{log}_a \space {(x^7y^8 ÷ \sqrt[4]{z})} \\[1em] \Rightarrow \dfrac{1}{3}[\text{log}_a \space x^7y^8 - \text{log}_a \space \sqrt[4]{z}] \\[1em] \Rightarrow \dfrac{1}{3}[\text{log}_a \space x^7 + \text{log}_a \space y^8 - \text{log}_a \space z^{\dfrac{1}{4}}] \\[1em] \Rightarrow \dfrac{1}{3}[7\text{log}_a \space x + 8\text{log}_a \space y - \dfrac{1}{4}\text{log}_a \space z] \\[1em] \Rightarrow \dfrac{7}{3}\text{log}_a \space x + \dfrac{8}{3}\text{log}_a \space y - \dfrac{1}{12}\text{log}_a \space z. ⇒ log a 3 x 7 y 8 ÷ 4 z ⇒ log a ( x 7 y 8 ÷ 4 z ) 3 1 ⇒ 3 1 log a ( x 7 y 8 ÷ 4 z ) ⇒ 3 1 [ log a x 7 y 8 − log a 4 z ] ⇒ 3 1 [ log a x 7 + log a y 8 − log a z 4 1 ] ⇒ 3 1 [ 7 log a x + 8 log a y − 4 1 log a z ] ⇒ 3 7 log a x + 3 8 log a y − 12 1 log a z .
Hence, log a x 7 y 8 ÷ z 4 3 = 7 3 log a x + 8 3 log a y − 1 12 log a z . \text{log}_a \space {\sqrt[3]{x^7y^8 ÷ \sqrt[4]{z}}} = \dfrac{7}{3}\text{log}_a \space x + \dfrac{8}{3}\text{log}_a \space y - \dfrac{1}{12}\text{log}_a \space z. log a 3 x 7 y 8 ÷ 4 z = 3 7 log a x + 3 8 log a y − 12 1 log a z .
Find the value of log 3 3 3 − log 5 ( 0.04 ) . \text{log}_{\sqrt{3}} \space 3\sqrt{3} - \text{log}_5 \space (0.04). log 3 3 3 − log 5 ( 0.04 ) .
Answer
Given,
⇒ log 3 3 3 − log 5 ( 0.04 ) ⇒ log 3 ( 3 ) 3 − log 5 ( 4 100 ) ⇒ 3 log 3 3 − log 5 ( 1 25 ) ⇒ 3 ( 1 ) − log 5 ( 5 ) − 2 ⇒ 3 − ( − 2 ) log 5 5 ⇒ 3 + 2 ( 1 ) ⇒ 5. \Rightarrow \text{log}_{\sqrt{3}} \space 3\sqrt{3} - \text{log}_5 \space (0.04) \\[1em] \Rightarrow \text{log}_{\sqrt{3}} \space (\sqrt{3})^3 - \text{log}_5 \space \Big(\dfrac{4}{100}\Big) \\[1em] \Rightarrow 3\text{log}_{\sqrt{3}}\sqrt{3} - \text{log}_5\Big(\dfrac{1}{25}\Big) \\[1em] \Rightarrow 3(1) - \text{log}_5(5)^{-2} \\[1em] \Rightarrow 3 - (-2)\text{log}_55 \\[1em] \Rightarrow 3 + 2(1) \\[1em] \Rightarrow 5. ⇒ log 3 3 3 − log 5 ( 0.04 ) ⇒ log 3 ( 3 ) 3 − log 5 ( 100 4 ) ⇒ 3 log 3 3 − log 5 ( 25 1 ) ⇒ 3 ( 1 ) − log 5 ( 5 ) − 2 ⇒ 3 − ( − 2 ) log 5 5 ⇒ 3 + 2 ( 1 ) ⇒ 5.
Hence, log 3 3 3 − log 5 ( 0.04 ) \text{log}_{\sqrt{3}} \space 3\sqrt{3} - \text{log}_5 \space (0.04) log 3 3 3 − log 5 ( 0.04 )
Prove the following:
( log x ) 2 − ( log y ) 2 = log x y . log x y (\text{log} \space x)^2 - (\text{log} \space y)^2 = \text{log} \space \dfrac{x}{y}.\text{log} \space xy ( log x ) 2 − ( log y ) 2 = log y x . log x y
Answer
Given,
( log x ) 2 − ( log y ) 2 = log x y . log x y (\text{log} \space x)^2 - (\text{log} \space y)^2 = \text{log} \space \dfrac{x}{y}.\text{log} \space xy ( log x ) 2 − ( log y ) 2 = log y x . log x y
Simplifying L.H.S. of above equation we get,
⇒ ( log x ) 2 − ( log y ) 2 ⇒ ( log x − log y ) ( log x + log y ) ⇒ log x y . log x y \Rightarrow (\text{log} \space x)^2 - (\text{log} \space y)^2 \\[1em] \Rightarrow (\text{log} \space x - \text{log} \space y)(\text{log} \space x + \text{log} \space y) \\[1em] \Rightarrow \text{log} \space\dfrac{x}{y}.\text{log} \space xy ⇒ ( log x ) 2 − ( log y ) 2 ⇒ ( log x − log y ) ( log x + log y ) ⇒ log y x . log x y
Since, L.H.S. = R.H.S.,
Hence, proved that ( log x ) 2 − ( log y ) 2 = log x y . log x y (\text{log} \space x)^2 - (\text{log} \space y)^2 = \text{log} \space \dfrac{x}{y}.\text{log} \space xy ( log x ) 2 − ( log y ) 2 = log y x . log x y
Prove the following:
2 log 11 13 + log 130 77 − log 55 91 = log 2. 2\text{log} \space \dfrac{11}{13} + \text{log} \space \dfrac{130}{77} - \text{log} \space \dfrac{55}{91} = \text{log} \space 2. 2 log 13 11 + log 77 130 − log 91 55 = log 2.
Answer
Given,
2 log 11 13 + log 130 77 − log 55 91 = log 2. 2\text{log} \space \dfrac{11}{13} + \text{log} \space \dfrac{130}{77} - \text{log} \space \dfrac{55}{91} = \text{log} \space 2. 2 log 13 11 + log 77 130 − log 91 55 = log 2.
Simplifying L.H.S. of the above equation we get,
⇒ 2 log 11 13 + log 130 77 − log 55 91 ⇒ 2 ( log 11 − log 13 ) + ( log 130 − log 77 ) − ( log 55 − log 91 ) ⇒ 2 ( log 11 − log 13 ) + ( log 13.10 − log 11.7 ) − ( log 11.5 − log 13.7 ) ⇒ 2 ( log 11 − log 13 ) + ( log 13 + log 10 − ( log 11 + log 7 ) − ( log 11 + log 5 − ( log 13 + log 7 ) ) ) ⇒ 2 log 11 − 2 log 13 + log 13 + log 10 − log 11 − log 7 − log 11 − log 5 + log 13 + log 7 ⇒ 2 log 11 − 2 log 11 + 2 log 13 − 2 log 13 + log 10 − log 5 ⇒ log 10 − log 5 ⇒ log 2.5 − log 5 ⇒ log 2 + log 5 − log 5 ⇒ log 2. \Rightarrow 2\text{log} \space \dfrac{11}{13} + \text{log} \space \dfrac{130}{77} - \text{log} \space \dfrac{55}{91} \\[1em] \Rightarrow 2(\text{log} \space 11 - \text{log} \space 13) + (\text{log} \space 130 - \text{log} \space 77) - (\text{log} \space 55 - \text{log} \space 91) \\[1em] \Rightarrow 2(\text{log} \space 11 - \text{log} \space 13) + (\text{log} \space 13.10 - \text{log} \space 11.7) - (\text{log} \space 11.5 - \text{log} \space 13.7) \\[1em] \Rightarrow 2(\text{log} \space 11 - \text{log} \space 13) + (\text{log} \space 13 + \text{log} \space 10 - (\text{log} \space 11 + \text{log} \space 7) - (\text{log} \space 11 + \text{log} \space 5 - (\text{log} \space 13 + \text{log} \space 7))) \\[1em] \Rightarrow 2\text{log} \space 11 - 2\text{log} \space 13 + \text{log} \space 13 + \text{log} \space 10 - \text{log} \space 11 - \cancel{\text{log} \space 7} - \text{log} \space 11 - \text{log} \space 5 + \text{log} \space 13 + \cancel{\text{log} \space 7} \\[1em] \Rightarrow \cancel{2\text{log} \space 11} - \cancel{2\text{log} \space 11} + \cancel{2\text{log} \space 13} - \cancel{2\text{log} \space 13} + \text{log} \space 10 - \text{log} \space 5 \\[1em] \Rightarrow \text{log} \space 10 - \text{log} \space 5 \\[1em] \Rightarrow \text{log} \space 2.5 - \text{log} \space 5 \\[1em] \Rightarrow \text{log} \space 2 + \cancel{\text{log} \space 5} - \cancel{\text{log} \space 5} \\[1em] \Rightarrow \text{log} \space 2. ⇒ 2 log 13 11 + log 77 130 − log 91 55 ⇒ 2 ( log 11 − log 13 ) + ( log 130 − log 77 ) − ( log 55 − log 91 ) ⇒ 2 ( log 11 − log 13 ) + ( log 13.10 − log 11.7 ) − ( log 11.5 − log 13.7 ) ⇒ 2 ( log 11 − log 13 ) + ( log 13 + log 10 − ( log 11 + log 7 ) − ( log 11 + log 5 − ( log 13 + log 7 ))) ⇒ 2 log 11 − 2 log 13 + log 13 + log 10 − log 11 − log 7 − log 11 − log 5 + log 13 + log 7 ⇒ 2 log 11 − 2 log 11 + 2 log 13 − 2 log 13 + log 10 − log 5 ⇒ log 10 − log 5 ⇒ log 2.5 − log 5 ⇒ log 2 + log 5 − log 5 ⇒ log 2.
Since, L.H.S. = R.H.S.,
Hence, proved that 2 log 11 13 + log 130 77 − log 55 91 = log 2. 2\text{log} \space \dfrac{11}{13} + \text{log} \space \dfrac{130}{77} - \text{log} \space \dfrac{55}{91} = \text{log} \space 2. 2 log 13 11 + log 77 130 − log 91 55 = log 2.
If log (m + n) = log m + log n, show that n = m m − 1 . \dfrac{m}{m - 1}. m − 1 m .
Answer
Given,
log (m + n) = log m + log n
⇒ log (m + n) = log mn
⇒ m + n = mn
⇒ m = mn - n
⇒ m = n(m - 1)
⇒ n = m m − 1 \dfrac{m}{m - 1} m − 1 m
Hence, proved that n = m m − 1 \dfrac{m}{m - 1} m − 1 m
If log x + y 2 = 1 2 ( log x + log y ) \text{log} \space \dfrac{x + y}{2} = \dfrac{1}{2}(\text{log} \space x + \text{log} \space y) log 2 x + y = 2 1 ( log x + log y )
Answer
Given,
⇒ log x + y 2 = 1 2 ( log x + log y ) ⇒ log x + y 2 = 1 2 ( log x y ) ⇒ log x + y 2 = log ( x y ) 1 2 ⇒ x + y 2 = ( x y ) 1 2 ⇒ x + y = 2 ( x y ) 1 2 \Rightarrow \text{log} \space \dfrac{x + y}{2} = \dfrac{1}{2}(\text{log} \space x + \text{log} \space y) \\[1em] \Rightarrow \text{log} \space \dfrac{x + y}{2} = \dfrac{1}{2}(\text{log} \space xy) \\[1em] \Rightarrow \text{log} \space \dfrac{x + y}{2} = \text{log} \space (xy)^{\dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{x + y}{2} = (xy)^{\dfrac{1}{2}} \\[1em] \Rightarrow x + y = 2(xy)^{\dfrac{1}{2}} \\[1em] ⇒ log 2 x + y = 2 1 ( log x + log y ) ⇒ log 2 x + y = 2 1 ( log x y ) ⇒ log 2 x + y = log ( x y ) 2 1 ⇒ 2 x + y = ( x y ) 2 1 ⇒ x + y = 2 ( x y ) 2 1
Squaring both sides we get,
⇒ ( x + y ) 2 = 4 ( x y ) ⇒ ( x 2 + y 2 + 2 x y ) = 4 x y ⇒ x 2 + y 2 + 2 x y − 4 x y = 0 ⇒ x 2 + y 2 − 2 x y = 0 ⇒ ( x − y ) 2 = 0 ⇒ x − y = 0 ⇒ x = y . \Rightarrow (x + y)^2 = 4(xy) \\[1em] \Rightarrow (x^2 + y^2 + 2xy) = 4xy \\[1em] \Rightarrow x^2 + y^2 + 2xy - 4xy = 0 \\[1em] \Rightarrow x^2 + y^2 - 2xy = 0 \\[1em] \Rightarrow (x - y)^2 = 0 \\[1em] \Rightarrow x - y = 0 \\[1em] \Rightarrow x = y. ⇒ ( x + y ) 2 = 4 ( x y ) ⇒ ( x 2 + y 2 + 2 x y ) = 4 x y ⇒ x 2 + y 2 + 2 x y − 4 x y = 0 ⇒ x 2 + y 2 − 2 x y = 0 ⇒ ( x − y ) 2 = 0 ⇒ x − y = 0 ⇒ x = y .
Hence, proved that x = y.
If a, b are positive real numbers, a > b and a2 + b2 = 27ab, prove that
log ( a − b 5 ) = 1 2 ( log a + log b ) \text{log} \space\Big(\dfrac{a - b}{5}\Big) = \dfrac{1}{2}(\text{log} \space a + \text{log} \space b) log ( 5 a − b ) = 2 1 ( log a + log b )
Answer
Given,
a2 + b2 = 27ab
⇒ a2 + b2 = 2ab + 25ab
⇒ a2 + b2 - 2ab = 25ab
⇒ a b = a 2 + b 2 − 2 a b 25 = ( a − b 5 ) 2 \Rightarrow ab = \dfrac{a^2 + b^2 - 2ab}{25} = \Big(\dfrac{a - b}{5}\Big)^2 ⇒ ab = 25 a 2 + b 2 − 2 ab = ( 5 a − b ) 2
Taking log on both sides:
⇒ log a b = log ( a − b 5 ) 2 ⇒ log a + log b = 2 log ( a − b 5 ) ⇒ log ( a − b 5 ) = 1 2 ( log a + log b ) . \Rightarrow \text{log} \space ab = \text{log} \space \Big(\dfrac{a - b}{5}\Big)^2 \\[1em] \Rightarrow \text{log} \space a + \text{log} \space b = 2 \text{log} \space \Big(\dfrac{a - b}{5}\Big) \\[1em] \Rightarrow \text{log} \space \Big(\dfrac{a - b}{5}\Big) = \dfrac{1}{2}(\text{log} \space a + \text{log} \space b). ⇒ log ab = log ( 5 a − b ) 2 ⇒ log a + log b = 2 log ( 5 a − b ) ⇒ log ( 5 a − b ) = 2 1 ( log a + log b ) .
Hence, proved that log ( a − b 5 ) = 1 2 ( log a + log b ) . \text{log} \space \Big(\dfrac{a - b}{5}\Big) = \dfrac{1}{2}(\text{log} \space a + \text{log} \space b). log ( 5 a − b ) = 2 1 ( log a + log b ) .
Solve the following equation for x:
logx 1 49 \dfrac{1}{49} 49 1
Answer
Given,
⇒ log x 1 49 = − 2 ⇒ 1 7 2 = x − 2 ⇒ 7 − 2 = x − 2 ⇒ x = 7. \Rightarrow \text{log}_x \space \dfrac{1}{49} = -2 \\[1em] \Rightarrow \dfrac{1}{7^2} = x^{-2} \\[1em] \Rightarrow 7^{-2} = x^{-2} \\[1em] \Rightarrow x = 7. ⇒ log x 49 1 = − 2 ⇒ 7 2 1 = x − 2 ⇒ 7 − 2 = x − 2 ⇒ x = 7.
Hence, x = 7.
Solve the following equation for x:
log x 1 4 2 = − 5 \text{log}_x \space \dfrac{1}{4\sqrt{2}} = -5 log x 4 2 1 = − 5
Answer
Given,
⇒ log x 1 4 2 = − 5 ⇒ log x 1 2 2 .2 1 2 = − 5 ⇒ − 1 5 log x 1 2 2 + 1 2 = 1 ⇒ − 1 5 log x 1 2 5 2 = 1 ⇒ − 1 5 log x 2 − 5 2 = 1 ⇒ − 1 5 × − 5 2 log x 2 = 1 ⇒ 1 2 log x 2 = 1 ⇒ log x 2 = 2 ⇒ x 2 = 2 ⇒ x = 2 . \Rightarrow \text{log}_x \space \dfrac{1}{4\sqrt{2}} = -5 \\[1em] \Rightarrow \text{log}_x \space \dfrac{1}{2^2.2^{\dfrac{1}{2}}} = -5 \\[1em] \Rightarrow -\dfrac{1}{5}\text{log}_x \space \dfrac{1}{2^{2 + \frac{1}{2}}} = 1 \\[1em] \Rightarrow -\dfrac{1}{5}\text{log}_x \space \dfrac{1}{2^{\dfrac{5}{2}}} = 1 \\[1em] \Rightarrow -\dfrac{1}{5}\text{log}_x \space 2^{-\dfrac{5}{2}} = 1 \\[1em] \Rightarrow -\dfrac{1}{5} \times -\dfrac{5}{2} \text{log}_x \space 2 = 1 \\[1em] \Rightarrow \dfrac{1}{2}\text{log}_x \space 2 = 1 \\[1em] \Rightarrow \text{log}_x \space 2 = 2 \\[1em] \Rightarrow x^2 = 2 \\[1em] \Rightarrow x = \sqrt{2}. ⇒ log x 4 2 1 = − 5 ⇒ log x 2 2 . 2 2 1 1 = − 5 ⇒ − 5 1 log x 2 2 + 2 1 1 = 1 ⇒ − 5 1 log x 2 2 5 1 = 1 ⇒ − 5 1 log x 2 − 2 5 = 1 ⇒ − 5 1 × − 2 5 log x 2 = 1 ⇒ 2 1 log x 2 = 1 ⇒ log x 2 = 2 ⇒ x 2 = 2 ⇒ x = 2 .
Hence, x = 2 \sqrt{2} 2
Solve the following equation for x:
log x 1 243 = 10 \text{log}_x \space \dfrac{1}{243} = 10 log x 243 1 = 10
Answer
Given,
⇒ log x 1 243 = 10 ⇒ log x 1 − log x 243 = 10 ⇒ 0 − log x ( 3 ) 5 = 10 ⇒ − 5 log x 3 = 10 ⇒ − log x 3 = 10 5 ⇒ log x 3 = − 2 ⇒ x − 2 = 3 ⇒ 1 x 2 = 3 ⇒ 1 x = 3 ⇒ x = 1 3 . \Rightarrow \text{log}_x \space \dfrac{1}{243} = 10 \\[1em] \Rightarrow \text{log}_x \space 1 - \text{log}_x \space 243 = 10 \\[1em] \Rightarrow 0 - \text{log}_x \space (3)^5 = 10 \\[1em] \Rightarrow -5\text{log}_x \space 3 = 10 \\[1em] \Rightarrow -\text{log}_x \space 3 = \dfrac{10}{5} \\[1em] \Rightarrow \text{log}_x \space 3 = -2 \\[1em] \Rightarrow x^{-2} = 3 \\[1em] \Rightarrow \dfrac{1}{x^2} = 3 \\[1em] \Rightarrow \dfrac{1}{x} = \sqrt{3} \\[1em] \Rightarrow x = \dfrac{1}{\sqrt{3}}. ⇒ log x 243 1 = 10 ⇒ log x 1 − log x 243 = 10 ⇒ 0 − log x ( 3 ) 5 = 10 ⇒ − 5 log x 3 = 10 ⇒ − log x 3 = 5 10 ⇒ log x 3 = − 2 ⇒ x − 2 = 3 ⇒ x 2 1 = 3 ⇒ x 1 = 3 ⇒ x = 3 1 .
Hence, x = 1 3 \dfrac{1}{\sqrt{3}} 3 1
Solve the following equation for x:
log4 32 = x - 4
Answer
Given,
⇒ log4 32 = x - 4
⇒ 4x - 4 = 32
⇒ (22 )x - 4 = 25
⇒ 22x - 8 = 25
⇒ 2x - 8 = 5
⇒ 2x = 13
⇒ x = 13 2 = 6 1 2 \dfrac{13}{2} = 6\dfrac{1}{2} 2 13 = 6 2 1
Hence, x = 6 1 2 6\dfrac{1}{2} 6 2 1
Solve the following equation for x:
log7 (2x2 - 1) = 2
Answer
Given,
⇒ log7 (2x2 - 1) = 2
⇒ 2x2 - 1 = 72
⇒ 2x2 - 1 = 49
⇒ 2x2 = 50
⇒ x2 = 25
⇒ x = 5, -5.
Hence, x = 5, -5.
Solve the following equation for x:
log (x2 - 21) = 2
Answer
Given,
⇒ log (x2 - 21) = 2
⇒ x2 - 21 = 102
⇒ x2 = 100 + 21
⇒ x2 = 121
⇒ x = 11, -11.
Hence, x = 11, -11.
Solve the following equation for x:
log6 (x - 2)(x + 3) = 1
Answer
Given,
⇒ log6 (x - 2)(x + 3) = 1
⇒ (x - 2)(x + 3) = 61
⇒ (x2 + 3x - 2x - 6) = 6
⇒ x2 + x - 6 = 6
⇒ x2 + x - 12 = 0
⇒ x2 + 4x - 3x - 12 = 0
⇒ x(x + 4) - 3(x + 4) = 0
⇒ (x - 3)(x + 4) = 0
⇒ x - 3 = 0 or x + 4 = 0
⇒ x = 3 or x = -4.
Hence, x = 3, -4.
Solve the following equation for x:
log6 (x - 2) + log6 (x + 3) = 1
Answer
Given,
⇒ log6 (x - 2) + log6 (x + 3) = 1
⇒ log6 (x - 2)(x + 3) = 1
⇒ (x - 2)(x + 3) = 61
⇒ (x2 + 3x - 2x - 6) = 6
⇒ x2 + x - 6 = 6
⇒ x2 + x - 12 = 0
⇒ x2 + 4x - 3x - 12 = 0
⇒ x(x + 4) - 3(x + 4) = 0
⇒ (x - 3)(x + 4) = 0
⇒ x - 3 = 0 or x + 4 = 0
⇒ x = 3 or x = -4.
In this case x ≠ -4 as (x + 3) and (x - 2) will be negative and log of only positive numbers are defined.
Hence, x = 3.
Solve the following equation for x:
log (x + 1) + log (x - 1) = log 11 + 2log 3
Answer
Given,
⇒ log (x + 1) + log (x - 1) = log 11 + 2log 3
⇒ log (x + 1)(x - 1) = log 11 + log 32
⇒ log (x2 - 1) = log (11 x 9)
⇒ log (x2 - 1) = log 99
⇒ (x2 - 1) = 99
⇒ x2 = 99 + 1
⇒ x2 = 100
⇒ x = 10, -10
In this case x ≠ -10 as (x + 1) and (x - 1) will be negative and log of only positive numbers are defined.
Hence, x = 10.
Solve for x and y:
log x 3 = log y 2 \dfrac{\text{log} \space x}{3} = \dfrac{\text{log} \space y}{2} 3 log x = 2 log y log ( x y ) = 5 \text{log} \space (xy) = 5 log ( x y ) = 5
Answer
Given,
log x 3 = log y 2 \dfrac{\text{log} \space x}{3} = \dfrac{\text{log} \space y}{2} 3 log x = 2 log y
⇒ 2 log x = 3 log y
⇒ 2 log x - 3 log y = 0 .......(i)
Given,
log (xy) = 5
⇒ log x + log y = 5 ......(ii)
Multiplying (ii) by 2 we get,
⇒ 2 log x + 2 log y = 10 ......(iii)
Subtracting (i) from (iii) we get,
⇒ 2 log x + 2 log y - (2 log x - 3 log y) = 10 - 0
⇒ 2 log x - 2 log x + 2 log y + 3 log y = 10
⇒ 5 log y = 10
⇒ log y = 2
⇒ y = 102 = 100.
Substituting value of y in (ii) we get,
⇒ log x + log 100 = 5
⇒ log x + 2 = 5
⇒ log x = 3
⇒ x = 103 = 1000.
Hence, x = 1000 and y = 100.
If a = 1 + logx yz, b = 1 + logy zx and c = 1 + logz xy, then show that
ab + bc + ca = abc.
Answer
a = 1 + logx yz = logx x + logx yz = logx xyz.
∴ 1 a = log x y z x \therefore \dfrac{1}{a} = \text{log}_{xyz} \space x ∴ a 1 = log x yz x
b = 1 + logy zx = logy y + logy zx = logy xyz.
∴ 1 b = log x y z y \therefore \dfrac{1}{b} = \text{log}_{xyz} \space y ∴ b 1 = log x yz y
c = 1 + logz xy = logz z + logz xy = logz xyz.
∴ 1 c = log x y z z \therefore \dfrac{1}{c} = \text{log}_{xyz} \space z ∴ c 1 = log x yz z
⇒ 1 a + 1 b + 1 c = log x y z x + log x y z y + log x y z z ⇒ b c + a c + a b a b c = log x log xyz + log y log xyz + log z log xyz ⇒ b c + a c + a b a b c = log x + log y + log z log xyz ⇒ b c + a c + a b a b c = log xyz log xyz ⇒ b c + a c + a b a b c = 1 ⇒ a b + b c + c a = a b c . \Rightarrow \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \text{log}_{xyz} \space x + \text{log}_{xyz} \space y + \text{log}_{xyz} \space z \\[1em] \Rightarrow \dfrac{bc + ac + ab}{abc} = \dfrac{\text{log x}}{\text{log xyz}} + \dfrac{\text{log y}}{\text{log xyz}} + \dfrac{\text{log z}}{\text{log xyz}} \\[1em] \Rightarrow \dfrac{bc + ac + ab}{abc} = \dfrac{\text{log x + log y + log z}}{\text{log xyz}} \\[1em] \Rightarrow \dfrac{bc + ac + ab}{abc} = \dfrac{\text{log xyz}}{\text{log xyz}} \\[1em] \Rightarrow \dfrac{bc + ac + ab}{abc} = 1 \\[1em] \Rightarrow ab + bc + ca = abc. ⇒ a 1 + b 1 + c 1 = log x yz x + log x yz y + log x yz z ⇒ ab c b c + a c + ab = log xyz log x + log xyz log y + log xyz log z ⇒ ab c b c + a c + ab = log xyz log x + log y + log z ⇒ ab c b c + a c + ab = log xyz log xyz ⇒ ab c b c + a c + ab = 1 ⇒ ab + b c + c a = ab c .
Hence, proved that ab + bc + ca = abc.
If 1 log a x + 1 log b x = 2 log c x \dfrac{1}{\text{ log }_a x} + \dfrac{1}{\text{ log }_b x} = \dfrac{2}{\text{ log }_c x} log a x 1 + log b x 1 = log c x 2 2 = ab.
Answer
Given,
⇒ 1 log a x + 1 log b x = 2 log c x ⇒ 1 log x log a + 1 log x log b = 2 log x log c ⇒ log a log x + log b log x = 2 log c log x ⇒ log a + log b log x = 2 log c log x ⇒ log ( a × b ) log x = 2 log c log x ⇒ log a b log x = 2 log c log x ⇒ log a b = 2 log c ⇒ log a b = log c 2 ⇒ a b = c 2 . \Rightarrow \dfrac{1}{\text{ log }_a x} + \dfrac{1}{\text{ log }_b x} = \dfrac{2}{\text{ log }_c x}\\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{log }x}{\text{log }a}} + \dfrac{1}{\dfrac{\text{log } x}{\text{log } b}} = \dfrac{2}{\dfrac{\text{log }x}{\text{log }c}} \\[1em] \Rightarrow \dfrac{\text{ log } a}{\text{ log } x} + \dfrac{\text{ log } b}{\text{ log } x} = \dfrac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \dfrac{\text{ log } a + \text{ log } b}{\text{ log } x} = \dfrac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \dfrac{\text{ log } (a \times b)}{\text{ log } x} = \dfrac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \dfrac{\text{ log } ab}{\text{ log } x} = \dfrac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \text{ log } ab = 2\text{ log } c\\[1em] \Rightarrow \text{ log } ab = \text{ log } c^2\\[1em] \Rightarrow ab = c^2. ⇒ log a x 1 + log b x 1 = log c x 2 ⇒ log a log x 1 + log b log x 1 = log c log x 2 ⇒ log x log a + log x log b = log x 2 log c ⇒ log x log a + log b = log x 2 log c ⇒ log x log ( a × b ) = log x 2 log c ⇒ log x log ab = log x 2 log c ⇒ log ab = 2 log c ⇒ log ab = log c 2 ⇒ ab = c 2 .
Hence, proved that c2 = ab.
