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Chapter 8

Logarithms

Class - 9 Concise Mathematics Selina

Exercise 8(A)

Question 1(a)

The value of log2 8\text{log}_{\sqrt{2}} \space 8 is :

  1. 6

  2. 4

  3. 3

  4. 8

Answer

Let value of log2 8\text{log}_{\sqrt{2}} \space 8 be x.

∴log2 8=x⇒(2)x=8⇒(2)x=(2)6⇒x=6.\therefore \text{log}_{\sqrt{2}} \space 8 = x \\[1em] \Rightarrow (\sqrt{2})^x = 8 \\[1em] \Rightarrow (\sqrt{2})^x = (\sqrt{2})^6 \\[1em] \Rightarrow x = 6.

Hence, Option 1 is the correct option.

Question 1(b)

If log4 x = 2.5, the value of x is :

  1. 12.5

  2. 32

  3. 10

  4. 20

Answer

Given,

⇒ log4 x = 2.5

⇒ x = 42.5

⇒ x = 42 + 0.5

⇒ x = 42.40.5

⇒ x = 16.(4)1216.(4)^{\dfrac{1}{2}}

⇒ x = 16.(22)1216.(2^2)^{\dfrac{1}{2}}

⇒ x = 16 × 2 = 32.

Hence, Option 2 is the correct option.

Question 1(c)

If log3 x=4\text{log}_{\sqrt{3}} \space x = 4, the value of x is :

  1. 12

  2. 6

  3. 9

  4. 24

Answer

Given,

⇒log3 x=4⇒x=(3)4⇒x=9.\Rightarrow \text{log}_{\sqrt{3}} \space x = 4 \\[1em] \Rightarrow x = (\sqrt{3})^4 \\[1em] \Rightarrow x = 9.

Hence, Option 3 is the correct option.

Question 1(d)

If logx 64 = 1.5, the value of x is :

  1. 48

  2. 32

  3. 64

  4. 16

Answer

Given,

⇒ logx 64 = 1.5

⇒ 64 = x1.5

⇒ 64 = (x)0.5 + 0.5 + 0.5

⇒ 64 = x0.5.x0.5.x0.5

⇒ 64 = x×x×x\sqrt{x} \times \sqrt{x} \times \sqrt{x}

⇒ 43=(x)34^3 = (\sqrt{x})^3

⇒ x=4\sqrt{x} = 4

Squaring both sides, we get :

⇒ (x)2=42(\sqrt{x})^2 = 4^2

⇒ x = 16.

Hence, Option 4 is the correct option.

Question 1(e)

If log2 (x2 - 4) = 5, the value of x is :

  1. ±6

  2. 6

  3. -6

  4. ±12

Answer

Given,

⇒ log2 (x2 - 4) = 5

⇒ x2 - 4 = 25

⇒ x2 - 4 = 32

⇒ x2 = 32 + 4

⇒ x2 = 36

⇒ x = 36\sqrt{36}

⇒ x = ±6\pm 6.

Hence, Option 1 is the correct option.

Question 1(f)

If log10 x = a, the value of 10a - 1 in terms of x is :

  1. 10x

  2. x10\dfrac{x}{10}

  3. 10x\dfrac{10}{x}

  4. 110x\dfrac{1}{10x}

Answer

Given,

⇒ log10 x = a

⇒ x = 10a .......(1)

We need to find the value of:

⇒ 10a - 1

⇒ 10a.10-1

Substituting value of 10a from equation (1), in above equation, we get :

⇒ x.10-1

⇒ x10\dfrac{x}{10}.

Hence, Option 2 is the correct option.

Question 2

Express each of the following in logarithmic form :

(i) 53 = 125

(ii) 3-2 = 19\dfrac{1}{9}

(iii) 10-3 = 0.001

(iv) (81)34=27(81)^{\dfrac{3}{4}} = 27

Answer

(i) Given,

⇒ 53 = 125

⇒ log5125 = 3.

Hence, required logarithmic form is log5 125 = 3.

(ii) Given,

⇒3−2=19⇒log3 19=−2.\Rightarrow 3^{-2} = \dfrac{1}{9} \\[1em] \Rightarrow \text{log}_{3} \space {\dfrac{1}{9}} = -2.

Hence, required logarithmic form is log3 19=−2.\text{log}_{3} \space {\dfrac{1}{9}} = -2.

(iii) Given,

⇒ 10-3 = 0.001

⇒ log10 0.001 = -3.

Hence, required logarithmic form is log10 0.001 = -3.

(iv) Given,

⇒(81)34=27⇒log81 27=34.\Rightarrow (81)^{\dfrac{3}{4}} = 27 \\[1em] \Rightarrow \text{log}_{81} \space 27 = \dfrac{3}{4}.

Hence, required logarithmic form is log81 27=34.\text{log}_{81} \space 27 = \dfrac{3}{4}.

Question 3

Express each of the following in exponential form :

(i) log8 0.125 = -1

(ii) log10 0.01 = -2

(iii) loga A = x

(iv) log10 1 = 0

Answer

(i) Given,

⇒ log8 0.125 = -1

⇒ 8-1 = 0.125

Hence, required exponential form is 8-1 = 0.125

(ii) Given,

⇒ log10 0.01 = -2

⇒ 10-2 = 0.01

Hence, required exponential form is 10-2 = 0.01

(iii) Given,

⇒ loga A = x

⇒ ax = A.

Hence, required exponential form is ax = A.

(iv) Given,

⇒ log10 1 = 0

⇒ 100 = 1.

Hence, required exponential form is 100 = 1.

Question 4

Solve for x : log10 x = -2.

Answer

Given,

⇒ log10 x = -2

⇒ x = 10-2 = 1100\dfrac{1}{100} = 0.01

Hence, x = 0.01

Question 5(i)

Find the logarithm of 100 to the base 10.

Answer

Let,

⇒ log10 100 = x

⇒ 100 = 10x

⇒ 102 = 10x

⇒ x = 2.

Hence, required value = 2.

Question 5(ii)

Find the logarithm of 0.1 to the base 10.

Answer

Let,

⇒ log10 0.1 = x

⇒ (10)x = 0.1

⇒ 10x = 110\dfrac{1}{10}

⇒ 10x = (10-1)

⇒ x = -1.

Hence, required value = -1.

Question 5(iii)

Find the logarithm of 0.001 to the base 10.

Answer

Let,

⇒ log10 (0.001) = x

⇒ (10)x = 0.001

⇒ 10x=1100010^x = \dfrac{1}{1000}

⇒ 10x = 1103\dfrac{1}{10^3}

⇒ 10x = 10-3

⇒ x = -3.

Hence, required value = -3.

Question 5(iv)

Find the logarithm of 32 to the base 4.

Answer

Let,

⇒ log4 32 = x

⇒ 32 = 4x

⇒ (2)5 = (22)x

⇒ (2)5 = (2)2x

⇒ 2x = 5

⇒ x = 52\dfrac{5}{2}.

Hence, required value = 52\dfrac{5}{2}.

Question 5(v)

Find the logarithm of 0.125 to the base 2.

Answer

Let,

⇒ log2 (0.125) = x

⇒ 0.125 = 2x

⇒ 1251000=2x\dfrac{125}{1000} = 2^x

⇒ 18=2x\dfrac{1}{8} = 2^x

⇒ 123=2x\dfrac{1}{2^3} = 2^x

⇒ 2-3 = 2x

⇒ x = -3.

Hence, required value = -3.

Question 5(vi)

Find the logarithm of 116\dfrac{1}{16} to the base 4.

Answer

Let,

⇒log(4) 116=x⇒116=4x⇒142=4x⇒4x=4−2⇒x=−2.\Rightarrow \text{log}_{(4)} \space \dfrac{1}{16} = x \\[1em] \Rightarrow \dfrac{1}{16} = 4^x\\[1em] \Rightarrow \dfrac{1}{4^2} = 4^x \\[1em] \Rightarrow 4^x = 4^{-2} \\[1em] \Rightarrow x = -2.

Hence, required value = -2.

Question 5(vii)

Find the logarithm of 27 to the base 9.

Answer

Let,

⇒ log9 27 = x

⇒ 27 = 9x

⇒ 33 = (32)x

⇒ 33 = 32x

⇒ 2x = 3

⇒ x = 32\dfrac{3}{2}.

Hence, required value = 32\dfrac{3}{2}.

Question 5(viii)

Find the logarithm of 181\dfrac{1}{81} to the base 27.

Answer

Let,

⇒log27 (181)=x⇒181=27x⇒(134)=(33)x⇒3−4=33x⇒3x=−4⇒x=−43.\Rightarrow \text{log}_{27} \space \Big(\dfrac{1}{81}\Big) = x \\[1em] \Rightarrow \dfrac{1}{81} = 27^x \\[1em] \Rightarrow \Big(\dfrac{1}{3^4}\Big) = (3^3)^x \\[1em] \Rightarrow 3^{-4} = 3^{3x} \\[1em] \Rightarrow 3x = -4 \\[1em] \Rightarrow x = -\dfrac{4}{3}.

Hence, required value = −43-\dfrac{4}{3}.

Question 6(i)

State, true or false :

If log10 x = a, then 10x = a

Answer

Given,

⇒ log10 x = a

⇒ 10a = x

Hence, the statement "If log10 x = a, then 10x = a" is false.

Question 6(ii)

State, true or false :

If xy = z, then y = logz x

Answer

Given,

⇒ xy = z

⇒ logx z = y.

Hence, the statement "If xy = z, then y = logz x" is false.

Question 6(iii)

State, true or false :

log2 8 = 3 and log8 2 = 13\dfrac{1}{3}.

Answer

Given,

⇒ log2 8 = 3

⇒ 23 = 8

⇒ 8 = 8, which is true.

Given,

⇒ log8 2 = 13\dfrac{1}{3}

⇒ 8138^{\dfrac{1}{3}} = 2

⇒ (23)13(2^3)^{\dfrac{1}{3}} = 2

⇒ 2 = 2, which is true.

Hence, the statement "log2 8 = 3 and log8 2 = 13\dfrac{1}{3}" is True.

Question 7(i)

Find x, if log3 x = 0

Answer

Given,

⇒ log3 x = 0

⇒ x = 30

⇒ x = 1.

Hence, x = 1.

Question 7(ii)

Find x, if logx 2 = -1

Answer

Given,

⇒ logx 2 = -1

⇒ (x)-1 = 2

⇒ 1x=2\dfrac{1}{x} = 2

⇒ x = 12\dfrac{1}{2}.

Hence, x = 12\dfrac{1}{2}.

Question 7(iii)

Find x, if log9 243 = x

Answer

Given,

⇒ log9 243 = x

⇒ 243 = 9x

⇒ 35 = (32)x

⇒ 35 = 32x

⇒ 2x = 5

⇒ x = 52=212\dfrac{5}{2} = 2\dfrac{1}{2}.

Hence, x = 2122\dfrac{1}{2}.

Question 7(iv)

Find x, if log5 (x - 7) = 1

Answer

Given,

⇒ log5 (x - 7) = 1

⇒ x - 7 = 51

⇒ x - 7 = 5

⇒ x = 5 + 7 = 12.

Hence, x = 12.

Question 7(v)

Find x, if log4 32 = x - 4

Answer

Given,

⇒ log4 32 = x - 4

⇒ 32 = 4x - 4

⇒ 25 = (22)x - 4

⇒ 25 = 22(x - 4)

⇒ 25 = 22x - 8

⇒ 5 = 2x - 8

⇒ 2x = 8 + 5

⇒ 2x = 13

⇒ x = 132=612\dfrac{13}{2} = 6\dfrac{1}{2}.

Hence, x = 6126\dfrac{1}{2}.

Question 7(vi)

Find x, if log7 (2x2 - 1) = 2

Answer

Given,

⇒ log7 (2x2 - 1) = 2

⇒ 2x2 - 1 = 72

⇒ 2x2 - 1 = 49

⇒ 2x2 = 49 + 1

⇒ 2x2 = 50

⇒ x2 = 502\dfrac{50}{2}

⇒ x2 = 25

⇒ x = 25=±5\sqrt{25} = \pm 5.

Hence, x = ±5\pm 5.

Question 8(i)

Evaluate log10 0.01

Answer

Let,

⇒log10 (0.01)=x⇒0.01=10x⇒1100=10x⇒1102=10x⇒10−2=10x⇒x=−2.\Rightarrow \text{log}_{10} \space (0.01) = x \\[1em] \Rightarrow 0.01 = 10^x \\[1em] \Rightarrow \dfrac{1}{100} = 10^x \\[1em] \Rightarrow \dfrac{1}{10^2} = 10^x \\[1em] \Rightarrow 10^{-2} = 10^x \\[1em] \Rightarrow x = -2.

Hence, log10 0.01 = -2.

Question 8(ii)

Evaluate log2 (1 ÷ 8)

Answer

Let,

⇒log2 (1÷8)=x⇒1÷8=2x⇒18=2x⇒123=2x⇒2−3=2x⇒x=−3.\Rightarrow \text{log}_{2} \space (1 ÷ 8) = x \\[1em] \Rightarrow 1 ÷ 8 = 2^x \\[1em] \Rightarrow \dfrac{1}{8} = 2^x \\[1em] \Rightarrow \dfrac{1}{2^3} = 2^x \\[1em] \Rightarrow 2^{-3} = 2^x \\[1em] \Rightarrow x = -3.

Hence, log2 (1 ÷ 8) = -3.

Question 8(iii)

Evaluate log5 1

Answer

Let,

⇒ log5 1 = x

⇒ 1 = 5x

⇒ 50 = 5x

⇒ x = 0.

Hence, log5 1 = 0.

Question 8(iv)

Evaluate log5 125

Answer

Let,

⇒ log5 125 = x

⇒ 125 = 5x

⇒ 53 = 5x

⇒ x = 3.

Hence, log5 125 = 3.

Question 8(v)

Evaluate log16 8

Answer

Let,

⇒ log16 8 = x

⇒ 8 = 16x

⇒ 23 = (24)x

⇒ 23 = 24x

⇒ 4x = 3

⇒ x = 34\dfrac{3}{4}.

Hence, log16 8 = 34\dfrac{3}{4}.

Question 8(vi)

Evaluate log0.5 16

Answer

Let,

⇒log0.5 16=x⇒16=(0.5)x⇒16=(510)x⇒24=(12)x⇒24=(2−1)x⇒24=2−x⇒−x=4⇒x=−4.\Rightarrow \text{log}_{0.5} \space 16 = x \\[1em] \Rightarrow 16 = (0.5)^x \\[1em] \Rightarrow 16 = \Big(\dfrac{5}{10}\Big)^x \\[1em] \Rightarrow 2^4 = \Big(\dfrac{1}{2}\Big)^x \\[1em] \Rightarrow 2^4 = (2^{-1})^x \\[1em] \Rightarrow 2^4 = 2^{-x} \\[1em] \Rightarrow -x = 4 \\[1em] \Rightarrow x = -4.

Hence, log0.5 16 = -4.

Question 9

If loga m = n, express an - 1 in terms of a and m.

Answer

Given,

⇒ loga m = n

⇒ m = an

We need to find the value of:

an - 1

⇒ an.a-1

⇒ m.a-1

⇒ ma\dfrac{m}{a}.

Hence, an - 1 = ma\dfrac{m}{a}.

Question 10

Given log2 x = m and log5 y = n.

(i) Express 2m - 3 in terms of x.

(ii) Express 53n + 2 in terms of y.

Answer

Given,

⇒ log2 x = m and log5 y = n

⇒ x = 2m ......(1)

and,

⇒ y = 5n .......(2)

(i) Given,

⇒ 2m - 3

⇒ 2m.2-3

⇒ 2m23\dfrac{2^m}{2^3}

Substituting value of x from equation (1) in above equation, we get :

⇒ x8\dfrac{x}{8}.

Hence, 2m - 3 = x8\dfrac{x}{8}.

(ii) Given,

⇒ 53n + 2

⇒ (5n)3.52

Substituting value of 5n from equation (2) in above equation, we get :

⇒ 25y3

Hence, 53n + 2 = 25y3.

Exercise 8(B)

Question 1(a)

The value of 3 + log5 5-2

  1. 1

  2. 5

  3. 3 - 15\dfrac{1}{5}

  4. 3 + 15\dfrac{1}{5}

Answer

Given,

⇒ 3 + log5 5-2

⇒ 3 + (-2log5 5)

⇒ 3 + (-2 × 1)

⇒ 3 - 2

⇒ 1.

Hence, Option 1 is the correct option.

Question 1(b)

The value of log5 75 - log5 3 is :

  1. 72

  2. 2

  3. 25

  4. 5

Answer

Given,

⇒log5 75−log5 3⇒log5 753⇒log5 25⇒log5 52⇒2×log5 5⇒2×1⇒2.\Rightarrow \text{log}_{5} \space 75 - \text{log}_{5} \space 3 \\[1em] \Rightarrow \text{log}_{5} \space {\dfrac{75}{3}} \\[1em] \Rightarrow \text{log}_{5} \space 25 \\[1em] \Rightarrow \text{log}_{5} \space 5^2 \\[1em] \Rightarrow 2 \times \text{log}_{5} \space 5 \\[1em] \Rightarrow 2 \times 1 \\[1em] \Rightarrow 2.

Hence, Option 2 is the correct option.

Question 1(c)

The value of log5 125 ÷ log5 5\sqrt{5} is :

  1. 5

  2. 120

  3. 6

  4. 60

Answer

Given,

⇒log5 125÷log5 5⇒log5 53÷log5 (5)12⇒3×log5 5÷12×log5 5⇒3×1÷12×1⇒3÷12⇒3×2⇒6.\Rightarrow \text{log}_{5} \space 125 ÷ \text{log}_{5} \space \sqrt{5} \\[1em] \Rightarrow \text{log}_{5} \space 5^3 ÷ \text{log}_{5} \space (5)^{\dfrac{1}{2}} \\[1em] \Rightarrow 3 \times \text{log}_{5} \space 5 ÷ \dfrac{1}{2} \times \text{log}_{5} \space 5 \\[1em] \Rightarrow 3 \times 1 ÷ \dfrac{1}{2} \times 1 \\[1em] \Rightarrow 3 ÷ \dfrac{1}{2} \\[1em] \Rightarrow 3 \times 2 \\[1em] \Rightarrow 6.

Hence, Option 3 is the correct option.

Question 1(d)

The value of (x)4logx a(\sqrt{x})^{\text{4log}_{x} \space a} is :

  1. a

  2. ax

  3. 12ax\dfrac{1}{2}ax

  4. a2

Answer

Given,

⇒(x)4logx a⇒(x12)4logx a⇒(x)12×4logx a⇒(x)2logx a⇒(x)logx a2⇒a2.\Rightarrow (\sqrt{x})^{\text{4log}_{x} \space a} \\[1em] \Rightarrow (x^{\dfrac{1}{2}})^{\text{4log}_{x} \space a} \\[1em] \Rightarrow (x)^{\dfrac{1}{2} \times \text{4log}_{x} \space a} \\[1em] \Rightarrow (x)^{\text{2log}_{x} \space a} \\[1em] \Rightarrow (x)^{\text{log}_{x} \space a^2} \\[1em] \Rightarrow a^2.

Hence, Option 4 is the correct option.

Question 1(e)

If log 125log 15\dfrac{\text{log }125}{\text{log } \dfrac{1}{5}} = log x, the value of x is :

  1. 0.001

  2. 0.01

  3. 25

  4. 5

Answer

Given,

⇒log 125log 15=log x⇒log 53log 5−1=log x⇒3log 5−1log 5=log x⇒−3=log x⇒x=10−3⇒x=1103⇒x=11000⇒x=0.001\Rightarrow \dfrac{\text{log }125}{\text{log } \dfrac{1}{5}} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{log }5^3}{\text{log } 5^{-1}} = \text{log x} \\[1em] \Rightarrow \dfrac{3\text{log }5}{-1\text{log } 5} = \text{log x} \\[1em] \Rightarrow -3 = \text{log x} \\[1em] \Rightarrow x = 10^{-3} \\[1em] \Rightarrow x = \dfrac{1}{10^3} \\[1em] \Rightarrow x = \dfrac{1}{1000} \\[1em] \Rightarrow x = 0.001

Hence, Option 1 is the correct option.

Question 1(f)

If log (x - 5) + log (x + 5) = 2 log 12, the positive value of x is :

  1. 5

  2. 13

  3. 4.8

  4. 12

Answer

Given,

⇒ log (x - 5) + log (x + 5) = 2 log 12

⇒ log (x - 5)(x + 5) = log 122

⇒ log (x2 + 5x - 5x - 25) = log 144

⇒ log (x2 - 25) = log 144

⇒ x2 - 25 = 144

⇒ x2 = 144 + 25

⇒ x2 = 169

⇒ x = 169\sqrt{169}

⇒ x = ±13\pm 13.

So, positive value of x = 13.

Hence, Option 2 is the correct option.

Question 2(i)

Express in terms of log 2 and log 3 :

log 36

Answer

Simplifying the expression,

⇒ log 36

⇒ log (22 × 32)

⇒ log 22 + log 32

⇒ 2 log 2 + 2 log 3.

Hence, log 36 = 2 log 2 + 2 log 3.

Question 2(ii)

Express in terms of log 2 and log 3 :

log 144

Answer

Simplifying the expression,

⇒ log 144

⇒ log (24 × 32)

⇒ log 24 + log 32

⇒ 4 log 2 + 2 log 3.

Hence, log 144 = 4 log 2 + 2 log 3.

Question 2(iii)

Express in terms of log 2 and log 3 :

log 4.5

Answer

Simplifying the expression,

⇒ log 4.5

⇒ log 4510\dfrac{45}{10}

⇒ log 92\dfrac{9}{2}

⇒ log 9 - log 2

⇒ log 32 - log 2

⇒ 2 log 3 - log 2.

Hence, log 4.5 = 2 log 3 - log 2.

Question 2(iv)

Express in terms of log 2 and log 3 :

log 2651−log 91119\text{log } \dfrac{26}{51} - \text{log } \dfrac{91}{119}

Answer

Simplifying the expression,

⇒log 2651−log 91119⇒log (2651÷91119)⇒log (2651×11991)⇒log 30944641⇒log 23⇒log 2−log 3.\Rightarrow \text{log } \dfrac{26}{51} - \text{log } \dfrac{91}{119} \\[1em] \Rightarrow \text{log } \Big(\dfrac{26}{51} ÷ \dfrac{91}{119}\Big) \\[1em] \Rightarrow \text{log } \Big(\dfrac{26}{51} \times \dfrac{119}{91}\Big) \\[1em] \Rightarrow \text{log } \dfrac{3094}{4641} \\[1em] \Rightarrow \text{log } \dfrac{2}{3} \\[1em] \Rightarrow \text{log } 2 - \text{log 3}.

Hence, log 2651−log 91119=log 2 - log 3\text{log } \dfrac{26}{51} - \text{log } \dfrac{91}{119} = \text{log 2 - log 3}.

Question 2(v)

Express in terms of log 2 and log 3 :

log 7516−2 log 59+log 32243\text{log } \dfrac{75}{16} - \text{2 log } \dfrac{5}{9} + \text{log } \dfrac{32}{243}

Answer

Simplifying the expression,

⇒log 7516−2 log 59+log 32243\Rightarrow \text{log } \dfrac{75}{16} - \text{2 log } \dfrac{5}{9} + \text{log } \dfrac{32}{243} \\[1em]

⇒ (log 75 - log 16) - 2 (log 5 - log 9) + (log 32 - log 243)

⇒ log 75 - log 16 - 2 log 5 + 2 log 9 + log 32 - log 243

⇒ log 75 - log 16 - log 52 + log 92 + log 32 - log 243

⇒ log 75 - log 16 - log 25 + log 81 + log 32 - log 243

⇒ log 75 + log 81 + log 32 - log 16 - log 25 - log 243

⇒ log 75 + log 81 + log 32 - (log 16 + log 25 + log 243)

⇒ log (75 × 81 × 32) - log (16 × 25 × 243)

⇒ log 75×81×3216×25×243\dfrac{75 \times 81 \times 32}{16 \times 25 \times 243}

⇒ log 2.

Hence, log 7516−2 log 59+log 32243\text{log } \dfrac{75}{16} - \text{2 log } \dfrac{5}{9} + \text{log } \dfrac{32}{243} = log 2.

Question 3(i)

Express the following in a form free from logarithm :

2 log x - log y = 1

Answer

Given,

⇒ 2 log x - log y = 1

⇒ log x2 - log y = 1

⇒ log x2y\dfrac{x^2}{y} = 1

⇒ x2y\dfrac{x^2}{y} = 101

⇒ x2 = 10y.

Hence, required equation is x2 = 10y.

Question 3(ii)

Express the following in a form free from logarithm :

2 log x + 3 log y = log a

Answer

Given,

⇒ 2 log x + 3 log y = log a

⇒ log x2 + log y3 = log a

⇒ log (x2.y3) = log a

⇒ x2.y3 = a

Hence, required equation is x2y3 = a.

Question 3(iii)

Express the following in a form free from logarithm :

a log x - b log y = 2 log 3

Answer

Given,

⇒ a log x - b log y = 2 log 3

⇒ log xa - log yb = log 32

⇒ log (xayb)\Big(\dfrac{x^a}{y^b}\Big) = log 32

⇒ log (xayb)\Big(\dfrac{x^a}{y^b}\Big) = log 9

⇒ (xayb)=9\Big(\dfrac{x^a}{y^b}\Big) = 9

⇒ xa = 9yb.

Hence, required equation is xa = 9yb.

Question 4(i)

Evaluate :

log 5 + log 8 - 2 log 2

Answer

Evaluating the expression,

⇒ log 5 + log 8 - 2 log 2

⇒ log 5 + log 8 - log 22

⇒ log 5×822\dfrac{5 \times 8}{2^2}

⇒ log 404\dfrac{40}{4}

⇒ log 10

⇒ 1.

Hence, log 5 + log 8 - 2 log 2 = 1.

Question 4(ii)

Evaluate :

log10 8 + log10 25 + 2 log10 3 - log10 18

Answer

Evaluating the expression,

⇒ log10 8 + log10 25 + 2 log10 3 - log10 18

⇒ log10 8 + log10 25 + log10 32 - log10 18

⇒ log10 8×25×3218\dfrac{8 \times 25 \times 3^2}{18}

⇒ log10 180018\dfrac{1800}{18}

⇒ log10 100

⇒ log10 102

⇒ 2 × log10 10

⇒ 2 × 1

⇒ 2.

Hence, log10 8 + log10 25 + 2 log10 3 - log10 18 = 2.

Question 4(iii)

Evaluate :

log 4+13 log 125−15 log 32\text{log 4} + \dfrac{1}{3}\text{ log 125} - \dfrac{1}{5}\text{ log 32}

Answer

Evaluating the expression,

⇒log 4+13 log 125−15 log 32⇒log 4+13 log 53−15 log 25⇒log 4 + log (5)(3×13)−log (2)(5×15)⇒log 4 + log 5 - log 2⇒log 4×52⇒log 202⇒log 10⇒1.\Rightarrow \text{log 4} + \dfrac{1}{3}\text{ log 125} - \dfrac{1}{5}\text{ log 32} \\[1em] \Rightarrow \text{log 4} + \dfrac{1}{3}\text{ log 5}^3 - \dfrac{1}{5}\text{ log 2}^5 \\[1em] \Rightarrow \text{log 4 + log (5)}^{\Big(3 \times \dfrac{1}{3}\Big)} - \text{log (2)}^{\Big(5 \times \dfrac{1}{5}\Big)} \\[1em] \Rightarrow \text{log 4 + log 5 - log 2} \\[1em] \Rightarrow \text{log } \dfrac{4 \times 5}{2} \\[1em] \Rightarrow \text{log } \dfrac{20}{2} \\[1em] \Rightarrow \text{log } 10 \\[1em] \Rightarrow 1.

Hence, log 4+13 log 125−15 log 32\text{log 4} + \dfrac{1}{3}\text{ log 125} - \dfrac{1}{5}\text{ log 32} = 1.

Question 5

Prove that :

2 log 1518−log 25162+log 49=log 2\text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} = \text{log 2}

Answer

To prove:

2 log 1518−log 25162+log 49=log 2\text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} = \text{log 2}

Solving L.H.S. of the equation, we get :

⇒2 log 1518−log 25162+log 49⇒log (1518)2+log 49−log 25162⇒log 225324+log 49−log 25162⇒log 225324×4925162⇒log 900291625162⇒log 900×16225×2916⇒log 14580072900⇒log 2.\Rightarrow \text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} \\[1em] \Rightarrow \text{log } \Big(\dfrac{15}{18}\Big)^2 + \text{log } \dfrac{4}{9} - \text{log } \dfrac{25}{162} \\[1em] \Rightarrow \text{log } \dfrac{225}{324} + \text{log } \dfrac{4}{9} - \text{log } \dfrac{25}{162} \\[1em] \Rightarrow \text{log } \dfrac{\dfrac{225}{324} \times \dfrac{4}{9}}{\dfrac{25}{162}} \\[1em] \Rightarrow \text{log } \dfrac{\dfrac{900}{2916}}{\dfrac{25}{162}} \\[1em] \Rightarrow \text{log } \dfrac{900 \times 162}{25 \times 2916} \\[1em] \Rightarrow \text{log } \dfrac{145800}{72900} \\[1em] \Rightarrow \text{log } 2.

Since, L.H.S. = R.H.S.

Hence, proved that 2 log 1518−log 25162+log 49=log 2\text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} = \text{log 2}.

Question 6

Find x, if :

x - log 48 + 3 log 2 = 13\dfrac{1}{3} log 125 - log 3

Answer

Given,

⇒ x - log 48 + 3 log 2 = 13\dfrac{1}{3} log 125 - log 3

⇒ x - log 48 + log 23 = log (125)13(125)^{\dfrac{1}{3}} - log 3

⇒ x - log 48 + log 8 = log (53)13(5^3)^{\dfrac{1}{3}} - log 3

⇒ x - log 48 + log 8 = log 5 - log 3

⇒ x = log 5 + log 48 - log 3 - log 8

⇒ x = (log 5 + log 48) - (log 3 + log 8)

⇒ x = log (5 × 48) - log (3 × 8)

⇒ x = log 240 - log 24

⇒ x = log 24024\dfrac{240}{24}

⇒ x = log 10

⇒ x = 1.

Hence, x = 1.

Question 7

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.

Question 8(i)

Solve for x :

log10 (x - 10) = 1

Answer

Given,

⇒ log10 (x - 10) = 1

⇒ x - 10 = 101

⇒ x - 10 = 10

⇒ x = 10 + 10 = 20.

Hence, x = 20.

Question 8(ii)

Solve for x :

log (x2 - 21) = 2

Answer

Given,

⇒ log (x2 - 21) = 2

⇒ x2 - 21 = 102

⇒ x2 - 21 = 100

⇒ x2 = 100 + 21

⇒ x2 = 121

⇒ x = 121\sqrt{121}

⇒ x = ±11\pm 11.

Hence, x = ±11\pm 11.

Question 8(iii)

Solve for x :

log (x - 2) + log (x + 2) = log 5

Answer

Given,

⇒ log (x - 2) + log (x + 2) = log 5

⇒ log (x - 2)(x + 2) = log 5

⇒ log (x2 + 2x - 2x - 4) = log 5

⇒ log (x2 - 4) = log 5

⇒ x2 - 4 = 5

⇒ x2 = 5 + 4

⇒ x2 = 9

⇒ x = 9=±3\sqrt{9} = \pm 3.

Since, x cannot be negative as that will make (x - 2) and (x + 2) negative.

Hence, x = 3.

Question 8(iv)

Solve for x :

log (x + 5) + log (x - 5) = 4 log 2 + 2 log 3

Answer

Given,

⇒ log (x + 5) + log (x - 5) = 4 log 2 + 2 log 3

⇒ log (x + 5)(x - 5) = log 24 + log 32

⇒ log (x2 - 5x + 5x - 25) = log (24 × 32)

⇒ log (x2 - 25) = log (16 × 9)

⇒ log (x2 - 25) = log 144

⇒ x2 - 25 = 144

⇒ x2 = 144 + 25

⇒ x2 = 169

⇒ x = 169=±13\sqrt{169} = \pm 13.

Since, x cannot be negative as that will make (x + 5) and (x - 5) negative.

Hence, x = 13.

Question 9(i)

Solve for x :

log 81log 27\dfrac{\text{log 81}}{\text{log 27}} = x

Answer

Given,

⇒log 81log 27=x⇒log 34log 33=x⇒4 log 33 log 3=x⇒x=43=113.\Rightarrow \dfrac{\text{log 81}}{\text{log 27}} = x \\[1em] \Rightarrow \dfrac{\text{log 3}^4}{\text{log 3}^3} = x \\[1em] \Rightarrow \dfrac{\text{4 log 3}}{\text{3 log 3}} = x \\[1em] \Rightarrow x = \dfrac{4}{3} = 1\dfrac{1}{3}.

Hence, x = 1131\dfrac{1}{3}.

Question 9(ii)

Solve for x :

log 128log 32\dfrac{\text{log 128}}{\text{log 32}} = x

Answer

Given,

⇒log 128log 32=x⇒log 27log 25=x⇒7 log 25 log 2=x⇒x=75=1.4\Rightarrow \dfrac{\text{log 128}}{\text{log 32}} = x \\[1em] \Rightarrow \dfrac{\text{log 2}^7}{\text{log 2}^5} = x \\[1em] \Rightarrow \dfrac{\text{7 log 2}}{\text{5 log 2}} = x \\[1em] \Rightarrow x = \dfrac{7}{5} = 1.4

Hence, x = 1.4

Question 9(iii)

Solve for x :

log 64log 8\dfrac{\text{log 64}}{\text{log 8}} = log x

Answer

Given,

⇒log 64log 8=log x⇒log 26log 23=log x⇒6 log 23 log 2=log x⇒log x=2⇒x=102=100.\Rightarrow \dfrac{\text{log 64}}{\text{log 8}} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{log 2}^6}{\text{log 2}^3} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{6 log 2}}{\text{3 log 2}} = \text{log x} \\[1em] \Rightarrow \text{log x} = 2 \\[1em] \Rightarrow x = 10^2 = 100.

Hence, x = 100.

Question 9(iv)

Solve for x :

log 225log 15\dfrac{\text{log 225}}{\text{log 15}} = log x

Answer

Given,

⇒log 225log 15=log x⇒log 152log 15=log x⇒2 log 15log 15=log x⇒log x=2⇒x=102=100.\Rightarrow \dfrac{\text{log 225}}{\text{log 15}} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{log 15}^2}{\text{log 15}} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{2 log 15}}{\text{log 15}} = \text{log x} \\[1em] \Rightarrow \text{log x} = 2 \\[1em] \Rightarrow x = 10^2 = 100.

Hence, x = 100.

Question 10

Given log x = m + n and log y = m - n, express the value of log 10xy2\dfrac{10x}{y^2} in terms of m and n.

Answer

Given,

⇒ log x = m + n

⇒ x = 10m + n ........(1)

Given,

⇒ log y = m - n

⇒ y = 10m - n ........(2)

Substituting value of x and y from equation (1) and equation (2) in log 10xy2\dfrac{10x}{y^2}, we get :

⇒log10xy2=log 10×10m+n(10m−n)2=log 10m+n+1102(m−n)=log 10m+n+1−2(m−n)=log 10m+n+1−2m+2n=log 103n−m+1=(3n−m+1) log10=(3n−m+1)×1=3n−m+1=1−m+3n\Rightarrow \text{log} \dfrac{10x}{y^2} = \text{log } \dfrac{10 \times 10^{m + n}}{(10^{m - n})^2} \\[1em] = \text{log } \dfrac{10^{m + n + 1}}{10^{2(m - n)}} \\[1em] = \text{log } 10^{m + n + 1 - 2(m - n)} \\[1em] = \text{log } 10^{m + n + 1 - 2m + 2n} \\[1em] = \text{log } 10^{3n - m + 1} \\[1em] = (3n - m + 1) \text{ log} 10 \\[1em] = (3n - m + 1) \times 1 \\[1em] = 3n - m + 1 \\[1em] = 1 - m + 3n

Hence, log 10xy2\dfrac{10x}{y^2} = 1 - m + 3n.

Question 11(i)

State, true or false :

log 1 × log 1000 = 0

Answer

Given,

log 1 × log 1000 = 0

Solving L.H.S. of the above equation, we get :

⇒ log 1 × log 1000

⇒ 0 × log 1000

⇒ 0.

Since, L.H.S. = R.H.S.

Hence, the statement "log 1 × log 1000 = 0" is true.

Question 11(ii)

State, true or false :

log xlog y\dfrac{\text{log x}}{\text{log y}} = log x - log y

Answer

Given,

log xlog y\dfrac{\text{log x}}{\text{log y}} = log x - log y

Solving R.H.S. of the above equation, we get :

⇒ log x - log y

⇒ log xy\dfrac{x}{y}.

Since, L.H.S. ≠ R.H.S.

Hence, the statement log xlog y=log x−log y\dfrac{\text{log x}}{\text{log y}} = \text{log x} - \text{log y} is false.

Question 11(iii)

State, true or false :

If log 25log 5\dfrac{\text{log 25}}{\text{log 5}} = log x, then x = 2

Answer

Given,

log 25log 5\dfrac{\text{log 25}}{\text{log 5}} = log x

Solving the equation, we get :

⇒log 25log 5=log x⇒log 52log 5=log x⇒2 log 5log 5=log x⇒log x=2⇒x=102=100.\Rightarrow \dfrac{\text{log 25}}{\text{log 5}} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{log 5}^2}{\text{log 5}} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{2 log 5}}{\text{log 5}} = \text{log x} \\[1em] \Rightarrow \text{log x} = 2 \\[1em] \Rightarrow x = 10^2 = 100.

Since, x is not equal to 2.

Hence, the statement log 25log 5\dfrac{\text{log 25}}{\text{log 5}} = log x, then x = 2 is false.

Question 11(iv)

State, true or false :

log x × log y = log x + log y

Answer

Given,

log x × log y = log x + log y

Solving the R.H.S. of the above equation, we get :

⇒ log x + log y

⇒ log xy

Since, L.H.S. ≠ R.H.S.

Hence, the statement log x × log y = log x + log y is false.

Question 12(i)

If log10 2 = a and log10 3 = b; express log 12 in terms of 'a' and 'b'.

Answer

Given,

log102 = a and log103 = b

Simplifying the expression :

⇒ log 12

⇒ log (22 × 3)

⇒ log 22 + log 3

⇒ 2 log 2 + log 3

⇒ 2a + b.

Hence, log 12 = 2a + b.

Question 12(ii)

If log10 2 = a and log10 3 = b; express log 2.25 in terms of 'a' and 'b'.

Answer

Given,

log102 = a and log103 = b

Simplifying the expression :

⇒ log 2.25

⇒ log 225100\dfrac{225}{100}

⇒ log 94\dfrac{9}{4}

⇒ log 9 - log 4

⇒ log 32 - log 22

⇒ 2 log 3 - 2 log 2

⇒ 2b - 2a.

Hence, log 2.25 = 2b - 2a.

Question 12(iii)

If log10 2 = a and log10 3 = b; express log 2142\dfrac{1}{4} in terms of 'a' and 'b'.

Answer

Given,

log10 2 = a and log10 3 = b

Simplifying the expression :

⇒log 214⇒log 94⇒log 9 - log 4⇒log 32−log 22⇒2 log 3 - 2 log 2⇒2b - 2a.\Rightarrow \text{log } 2\dfrac{1}{4} \\[1em] \Rightarrow \text{log } \dfrac{9}{4} \\[1em] \Rightarrow \text{log 9 - log 4} \\[1em] \Rightarrow \text{log 3}^2 - \text{log 2}^2 \\[1em] \Rightarrow \text{2 log 3 - 2 log 2} \\[1em] \Rightarrow \text{2b - 2a}.

Hence, log 214\text{log } 2\dfrac{1}{4} = 2b - 2a.

Question 12(iv)

If log10 2 = a and log10 3 = b; express log 5.4 in terms of 'a' and 'b'.

Answer

Given,

log10 2 = a and log10 3 = b

Simplifying the expression :

⇒log 5.4⇒log 5410⇒log 54 - log 10⇒log (2×33)−1⇒log 2+log 33−1⇒log 2 + 3 log 3−1⇒a+3b−1.\Rightarrow \text{log } 5.4 \\[1em] \Rightarrow \text{log } \dfrac{54}{10} \\[1em] \Rightarrow \text{log 54 - log 10} \\[1em] \Rightarrow \text{log }(2 \times 3^3) - 1 \\[1em] \Rightarrow \text{log 2} + \text{log }3^3 - 1 \\[1em] \Rightarrow \text{log 2 + 3 log 3} - 1 \\[1em] \Rightarrow a + 3b - 1.

Hence, log 5.4 = a + 3b - 1.

Question 12(v)

If log10 2 = a and log10 3 = b; express log 60 in terms of 'a' and 'b'.

Answer

Given,

log10 2 = a and log10 3 = b

Simplifying the expression :

⇒ log 60

⇒ log (2 × 3 × 10)

⇒ log 2 + log 3 + log 10

⇒ a + b + 1.

Hence, log 60 = a + b + 1.

Question 12(vi)

If log10 2 = a and log10 3 = b; express log 3183\dfrac{1}{8} in terms of 'a' and 'b'.

Answer

Given,

log10 2 = a and log10 3 = b

Simplifying the expression :

⇒log 318⇒log 258⇒log 25 - log 8⇒log 52−log 23⇒2 log 5 - 3 log 2⇒2 log 102−3 log 2⇒2 (log 10 - log 2)−3 log 2⇒2 log 10 - 2 log 2 - 3 log 2⇒2×1−5 log 2⇒2−5a.\Rightarrow \text{log } 3\dfrac{1}{8} \\[1em] \Rightarrow \text{log } \dfrac{25}{8} \\[1em] \Rightarrow \text{log 25 - log 8} \\[1em] \Rightarrow \text{log 5}^2 - \text{log 2}^3 \\[1em] \Rightarrow \text{2 log 5 - 3 log 2} \\[1em] \Rightarrow \text{2 log } \dfrac{10}{2} - \text{3 log 2} \\[1em] \Rightarrow \text{2 (log 10 - log 2)} - \text{3 log 2} \\[1em] \Rightarrow \text{2 log 10 - 2 log 2 - 3 log 2} \\[1em] \Rightarrow 2 \times 1 - \text{5 log 2} \\[1em] \Rightarrow 2 - 5a.

Hence, log 318\text{log } 3\dfrac{1}{8} = 2 - 5a.

Question 13(i)

If log 2 = 0.3010 and log 3 = 0.4771; find the value of log 12.

Answer

Simplifying the expression,

⇒ log 12

⇒ log (22 × 3)

⇒ log 22 + log 3

⇒ 2 log 2 + log 3

⇒ 2 × 0.3010 + 0.4771

⇒ 0.6020 + 0.4771

⇒ 1.0791

Hence, log 12 = 1.0791

Question 13(ii)

If log 2 = 0.3010 and log 3 = 0.4771; find the value of log 1.2.

Answer

Simplifying the expression,

⇒ log 1.2

⇒ log 1210\dfrac{12}{10}

⇒ log 12 - log 10

⇒ log (22 × 3) - log 10

⇒ log 22 + log 3 - log 10

⇒ 2 log 2 + log 3 - 1

⇒ 2 × 0.3010 + 0.4771 - 1

⇒ 0.6020 + 0.4771 - 1

⇒ 1.0791 - 1

⇒ 0.0791

Hence, log 1.2 = 0.0791

Question 13(iii)

If log 2 = 0.3010 and log 3 = 0.4771; find the value of log 3.6.

Answer

Simplifying the expression,

⇒ log 3.6

⇒ log 3610\dfrac{36}{10}

⇒ log 36 - log 10

⇒ log (12 × 3) - log 10

⇒ log 12 + log 3 - log 10

⇒ log (22 × 3) + log 3 - log 10

⇒ log 22 + log 3 + log 3 - log 10

⇒ 2 log 2 + 2 log 3 - 1

⇒ 2 × 0.3010 + 2 × 0.4771 - 1

⇒ 0.6020 + 0.9542 - 1

⇒ 1.5562 - 1

⇒ 0.5562

Hence, log 3.6 = 0.5562

Question 13(iv)

If log 2 = 0.3010 and log 3 = 0.4771; find the value of log 15.

Answer

Simplifying the expression,

⇒ log 15

⇒ log (3 × 5)

⇒ log 3 + log 5

⇒ log 3 + log 102\dfrac{10}{2}

⇒ log 3 + log 10 - log 2

⇒ 0.4771 + 1 - 0.3010

⇒ 1.1761

Hence, log 15 = 1.1761

Question 13(v)

If log 2 = 0.3010 and log 3 = 0.4771; find the value of log 25.

Answer

Simplifying the expression,

⇒ log 25

⇒ log 52

⇒ 2 log 5

⇒ 2 log 102\dfrac{10}{2}

⇒ 2(log 10 - log 2)

⇒ 2(1 - 0.3010)

⇒ 2 × 0.699

⇒ 1.398

Hence, log 25 = 1.398

Question 13(vi)

If log 2 = 0.3010 and log 3 = 0.4771; find the value of 23\dfrac{2}{3} log 8.

Answer

Simplifying the expression,

⇒23 log 8⇒23 log 23⇒23×3 log 2⇒2log 2⇒2×0.3010⇒0.6020\Rightarrow \dfrac{2}{3} \text{ log 8} \\[1em] \Rightarrow \dfrac{2}{3} \text{ log 2}^3 \\[1em] \Rightarrow \dfrac{2}{3} \times 3 \text{ log 2} \\[1em] \Rightarrow 2 \text{log 2} \\[1em] \Rightarrow 2 \times 0.3010 \\[1em] \Rightarrow 0.6020

Hence, 23\dfrac{2}{3} log 8 = 0.6020

Question 14

Given 2 log10 x + 1 = log10 250, find :

(i) x

(ii) log10 2x

Answer

(i) Given,

⇒ 2 log10 x + 1 = log10 250

⇒ log10 x2 + log10 10 = log10 250

⇒ log10 (x2 × 10) = log10 250

⇒ log10 (10x2) = log10 250

⇒ 10x2 = 250

⇒ x2 = 25010\dfrac{250}{10}

⇒ x2 = 25

⇒ x = 25\sqrt{25} = 5.

Hence, x = 5.

(ii) Substituting value of x in log10 2x, we get :

⇒ log10 2x

⇒ log10 2(5)

⇒ log10 10

⇒ 1.

Hence, log10 2x = 1.

Question 15

Given 3 log x + 12\dfrac{1}{2} log y = 2, express y in terms of x.

Answer

Given,

⇒3 log x+12log y=2⇒log x3+log y12=2⇒log x3y=2⇒x3y=102\Rightarrow \text{3 log x} + \dfrac{1}{2} \text{log y} = 2 \\[1em] \Rightarrow \text{log } x^3 + \text{log } y^{\dfrac{1}{2}} = 2 \\[1em] \Rightarrow \text{log } x^3\sqrt{y} = 2 \\[1em] \Rightarrow x^3\sqrt{y} = 10^2

Squaring both sides, we get :

⇒(x3y)2=(102)2⇒x6y=104⇒y=104x6=10000x−6.\Rightarrow (x^3\sqrt{y})^2 = (10^2)^2 \\[1em] \Rightarrow x^6y = 10^4 \\[1em] \Rightarrow y = \dfrac{10^4}{x^6} = 10000x^{-6}.

Hence, y = 10000x-6.

Exercise 8(C)

Question 1(a)

If log2 (log3 x) = 4, the value of x is :

  1. 316

  2. 163

  3. 3 × 16

  4. 16 ÷ 3

Answer

Given,

⇒ log2 (log3 x) = 4

⇒ log3 x = 24

⇒ log3 x = 16

⇒ x = 316.

Hence, Option 1 is the correct option.

Question 1(b)

If log (5x - 4) - log (x + 1) = log 4, the value of x is :

  1. 6

  2. 8

  3. 4

  4. 12

Answer

Given,

⇒ log (5x - 4) - log (x + 1) = log 4

⇒ log 5x−4x+1\dfrac{5x - 4}{x + 1} = log 4

⇒ 5x−4x+1=4\dfrac{5x - 4}{x + 1} = 4

⇒ 5x - 4 = 4(x + 1)

⇒ 5x - 4 = 4x + 4

⇒ 5x - 4x = 4 + 4

⇒ x = 8.

Hence, Option 2 is the correct option.

Question 1(c)

If log x - log (2x - 1) = 1, the value of x is :

  1. 1910\dfrac{19}{10}

  2. 19 × 10

  3. 1019\dfrac{10}{19}

  4. 119×10\dfrac{1}{19 \times 10}

Answer

Given,

⇒log x−log (2x−1)=1⇒log x2x−1=log 10⇒x2x−1=10⇒x=10(2x−1)⇒x=20x−10⇒20x−x=10⇒19x=10⇒x=1019.\Rightarrow \text{log } x - \text{log } (2x - 1) = 1 \\[1em] \Rightarrow \text{log } \dfrac{x}{2x - 1} = \text{log } 10 \\[1em] \Rightarrow \dfrac{x}{2x - 1} = 10 \\[1em] \Rightarrow x = 10(2x - 1) \\[1em] \Rightarrow x = 20x - 10 \\[1em] \Rightarrow 20x - x = 10 \\[1em] \Rightarrow 19x = 10 \\[1em] \Rightarrow x = \dfrac{10}{19}.

Hence, Option 3 is the correct option.

Question 1(d)

If x = log 35, y = log 54 and z = 2 log 32\dfrac{3}{5}, \text{ y = log } \dfrac{5}{4}\text{ and z = 2 log } \dfrac{\sqrt{3}}{2}, the value of x + y - z is :

  1. 1

  2. -1

  3. 2

  4. 0

Answer

Solving,

⇒x+y−z=log 35+log 54− 2 log 32=log 35+log 54−log (32)2=log 35+log 54−log 34=log 35×5434=log 3434=log 1=0.\Rightarrow x + y - z = \text{log } \dfrac{3}{5} + \text{log } \dfrac{5}{4} - \text{ 2 log } \dfrac{\sqrt{3}}{2} \\[1em] = \text{log } \dfrac{3}{5} + \text{log } \dfrac{5}{4} - \text{log } \Big(\dfrac{\sqrt{3}}{2}\Big)^2 \\[1em] = \text{log } \dfrac{3}{5} + \text{log } \dfrac{5}{4} - \text{log } \dfrac{3}{4} \\[1em] = \text{log } \dfrac{\dfrac{3}{5} \times \dfrac{5}{4}}{\dfrac{3}{4}} \\[1em] = \text{log } \dfrac{\dfrac{3}{4}}{\dfrac{3}{4}} \\[1em] = \text{log } 1 \\[1em] = 0.

Hence, Option 4 is the correct option.

Question 1(e)

If log v + log 3 = log π + log 4 + 3 log r, the value of v in terms of r and other constants is :

  1. 43πr3\dfrac{4}{3}πr^3

  2. 4πr2

  3. πr3

  4. 43πr2\dfrac{4}{3}πr^2

Answer

Given,

⇒ log v + log 3 = log π + log 4 + 3 log r

⇒ log 3v = log π + log 4 + log r3

⇒ log 3v = log 4πr3

⇒ 3v = 4πr3

⇒ v = 43πr3\dfrac{4}{3}πr^3

Hence, Option 1 is the correct option.

Question 1(f)

If log y + 2 log x = 2, the value of y in terms of x is :

  1. x2 ÷ 100

  2. 100 ÷ x2

  3. x ÷ 10

  4. 10 ÷ x

Answer

Given,

⇒ log y + 2 log x = 2

⇒ log y + log x2 = 2

⇒ log yx2 = 2

⇒ yx2 = 102

⇒ yx2 = 100

⇒ y = 100 ÷ x2.

Hence, Option 2 is the correct option.

Question 2

If log10 8 = 0.90; find the value of :

(i) log10 4

(ii) log 32\sqrt{32}

(iii) log 0.125

Answer

Given,

⇒ log10 8 = 0.90

⇒ log10 23 = 0.90

⇒ 3log10 2 = 0.90

⇒ log10 2 = 0.903\dfrac{0.90}{3}

⇒ log10 2 = 0.30 .........(1)

(i) Given,

⇒ log10 4

⇒ log10 22

⇒ 2 log10 2

Substituting value of log10 2 from equation (1) in above equation, we get :

⇒ 2 × 0.30

⇒ 0.60

Hence, log10 4 = 0.60

(ii) Given,

⇒log 32⇒log 42⇒log 4+log 2⇒log 22+log 212⇒2×log 2+12×log 2\Rightarrow \text{log } \sqrt{32} \\[1em] \Rightarrow \text{log } 4\sqrt{2} \\[1em] \Rightarrow \text{log } 4 + \text{log } \sqrt{2} \\[1em] \Rightarrow \text{log } 2^2 + \text{log } 2^{\dfrac{1}{2}} \\[1em] \Rightarrow 2 \times \text{log 2} + \dfrac{1}{2} \times \text{log 2}

Substituting value of log 2 from equation (1), in above equation, we get :

⇒2×0.30+12×0.30⇒0.60+0.15⇒0.75\Rightarrow 2 \times 0.30 + \dfrac{1}{2} \times 0.30 \\[1em] \Rightarrow 0.60 + 0.15 \\[1em] \Rightarrow 0.75

Hence, log 32\sqrt{32} = 0.75

(iii) Given,

⇒log 0.125⇒log 1251000⇒log 18⇒log 123⇒log 2−3⇒−3×log 2⇒−3×0.30⇒−0.90\Rightarrow \text{log } 0.125 \\[1em] \Rightarrow \text{log } \dfrac{125}{1000} \\[1em] \Rightarrow \text{log } \dfrac{1}{8} \\[1em] \Rightarrow \text{log } \dfrac{1}{2^3} \\[1em] \Rightarrow \text{log } 2^{-3} \\[1em] \Rightarrow -3 \times \text{log 2} \\[1em] \Rightarrow -3 \times 0.30 \\[1em] \Rightarrow -0.90

Hence, log 0.125 = -0.90

Question 3

If log 27 = 1.431, find the value of :

(i) log 9

(ii) log 300

Answer

Given,

⇒ log 27 = 1.431

⇒ log 33 = 1.431

⇒ 3 log 3 = 1.431

⇒ log 3 = 1.4313\dfrac{1.431}{3} = 0.477 .......(1)

(i) Solving,

⇒ log 9

⇒ log 32

⇒ 2 log 3

Substituting value of log 3 from equation (1) in above equation, we get :

⇒ 2 × 0.477

⇒ 0.954

Hence, log 9 = 0.954

(ii) Solving,

⇒ log 300

⇒ log 3 × 100

⇒ log 3 + log 100

⇒ 0.477 + log 102

⇒ 0.477 + 2 log 10

⇒ 0.477 + 2 × 1

⇒ 0.477 + 2

⇒ 2.477

Hence, log 300 = 2.477

Question 4

If log10 a = b, find 103b - 2 in terms of a.

Answer

Given,

⇒ log10 a = b

⇒ 10b = a

Cubing both sides, we get :

⇒ (10b)3 = a3

⇒ 103b = a3

Dividing both sides by 102, we get :

⇒103b102=a3102⇒103b−2=a3100.\Rightarrow \dfrac{10^{3b}}{10^2} = \dfrac{a^3}{10^2} \\[1em] \Rightarrow 10^{3b - 2} = \dfrac{a^3}{100}.

Hence, 103b - 2 = a3100.\dfrac{a^3}{100}.

Question 5

If log5 x = y, find 52y + 3 in terms of x.

Answer

Given,

⇒ log5 x = y

⇒ x = 5y ......(1)

Solving, equation 52y + 3, we get :

⇒ 52y.53

⇒ (5y)2 × 125

Substituting value of 5y from equation (1) in above equation, we get :

⇒ 125x2.

Hence, 52y + 3 = 125x2.

Question 6

Given :

log3 m = x and log3 n = y

(i) Express 32x - 3 in terms of m.

(ii) Write down 31 - 2y + 3x in terms of m and n.

(iii) If 2 log3 A = 5x - 3y; find A in terms of m and n.

Answer

Given,

1st equation :

⇒ log3 m = x

⇒ m = 3x ......(1)

2nd equation :

⇒ log3 n = y

⇒ n = 3y ......(2)

(i) Given,

32x - 3

⇒ 32x.3-3

⇒ (3x)2.3-3

Substituting value of 3x from equation (1) in above equation, we get :

⇒ m2.3-3

⇒ m233\dfrac{m^2}{3^3}

⇒ m227\dfrac{m^2}{27}.

Hence, 32x - 3 = m227\dfrac{m^2}{27}.

(ii) Simplifying the expression,

⇒ 31 - 2y + 3x

⇒ 31.3-2y.33x

⇒ 3.(3y)-2.(3x)3

Substituting value of 3x and 3y from equation (1) and (2) in above equation, we get :

⇒ 3.n-2.m3

⇒ 3m3n2\dfrac{3m^3}{n^2}.

Hence, 31 - 2y + 3x = 3m3n2\dfrac{3m^3}{n^2}.

(iii) Given,

⇒ 2log3 A = 5x - 3y

⇒ log3 A2 = 5x - 3y

⇒ A2 = 35x - 3y

⇒ A2 = 35x.3-3y

⇒ A2 = (3x)5.(3y)-3

Substituting value of 3x and 3y from equation (1) and (2) in above equation, we get :

⇒ A2 = m5.n-3

⇒ A2 = m5n3\dfrac{m^5}{n^3}

⇒ A = m5n3\sqrt{\dfrac{m^5}{n^3}}.

Hence, A = m5n3\sqrt{\dfrac{m^5}{n^3}}.

Question 7(i)

Simplify :

log (a)3 - log a

Answer

Simplifying the expression,

⇒ log (a)3 - log a

⇒ 3log a - log a

⇒ 2log a

⇒ log a2

⇒ 2 log a.

Hence, log (a)3 - log a = 2 log a.

Question 7(ii)

Simplify :

log (a)3 ÷ log a

Answer

Simplifying the expression,

⇒ log (a)3 ÷ log a

⇒ 3log a ÷ log a

⇒ 3 log alog a\dfrac{\text{3 log a}}{\text{log a}}

⇒ 3.

Hence, log (a)3 ÷ log a = 3.

Question 8

If log (a + b) = log a + log b, find a in terms of b.

Answer

Given,

⇒ log (a + b) = log a + log b

⇒ log (a + b) = log ab

⇒ a + b = ab

⇒ ab - a = b

⇒ a(b - 1) = b

⇒ a = bb−1\dfrac{b}{b - 1}.

Hence, a = bb−1\dfrac{b}{b - 1}.

Question 9(i)

Prove that :

(log a)2 - (log b)2 = log (ab)\Big(\dfrac{a}{b}\Big). log (ab)

Answer

To prove:

(log a)2 - (log b)2 = log (ab)\Big(\dfrac{a}{b}\Big). log (ab)

Solving L.H.S. of the above equation, we get :

⇒ (log a)2 - (log b)2

⇒ (log a + log b)(log a - log b)

⇒ log ab. log (ab)\Big(\dfrac{a}{b}\Big)

Hence, proved that (log a)2 - (log b)2 = log (ab)\Big(\dfrac{a}{b}\Big). log (ab)

Question 9(ii)

Prove that :

If a log b + b log a - 1 = 0, then ba.ab = 10

Answer

Given,

⇒ a log b + b log a - 1 = 0

⇒ log ba + log ab = 1

⇒ log ba + log ab = log 10

⇒ log ba.ab = log 10

⇒ ba.ab = 10.

Hence, proved that ba.ab = 10.

Question 10(i)

If log (a + 1) = log (4a - 3) - log 3; find a.

Answer

Given,

⇒ log (a + 1) = log (4a - 3) - log 3

⇒ log (a + 1) = log 4a−33\dfrac{4a - 3}{3}

⇒ (a + 1) = 4a−33\dfrac{4a - 3}{3}

⇒ 4a - 3 = 3(a + 1)

⇒ 4a - 3 = 3a + 3

⇒ 4a - 3a = 3 + 3

⇒ a = 6.

Hence, a = 6.

Question 10(ii)

If 2 log y - log x - 3 = 0, express x in terms of y.

Answer

Given,

⇒ 2 log y - log x - 3 = 0

⇒ 2 log y - log x = 3

⇒ log y2 - log x = 3

⇒ log y2x\dfrac{y^2}{x} = 3

⇒ y2x=103\dfrac{y^2}{x} = 10^3

⇒ y2 = x.103

⇒ x = y2103=y21000\dfrac{y^2}{10^3} = \dfrac{y^2}{1000}.

Hence, y = y21000\dfrac{y^2}{1000}.

Question 10(iii)

Prove that :

log10 125 = 3(1 - log10 2)

Answer

To prove:

log10 125 = 3(1 - log10 2)

Solving R.H.S. of the equation, we get :

⇒ 3(1 - log10 2)

⇒ 3(log10 10 - log10 2)

⇒ 3(log10 102\dfrac{10}{2})

⇒ 3 log10 5

⇒ log10 53

⇒ log10 125

Since, L.H.S. = R.H.S.

Hence, proved that log10 125 = 3(1 - log10 2).

Exercise 8(D)

Question 1(a)

The value of (log 8 - log 2) ÷ log 32 is :

  1. 25\dfrac{2}{5}

  2. 52\dfrac{5}{2}

  3. 2

  4. 4

Answer

Simplifying the expression :

⇒(log 8 - log 2) ÷ log 32⇒log 8 - log 2log 32⇒log 23−log 2log 25⇒3 log 2 - log 25 log 2⇒2 log 25 log 2⇒25.\Rightarrow \text{(log 8 - log 2) ÷ log 32} \\[1em] \Rightarrow \dfrac{\text{log 8 - log 2}}{\text{log 32}} \\[1em] \Rightarrow \dfrac{\text{log 2}^3 - \text{log 2}}{\text{log 2}^5} \\[1em] \Rightarrow \dfrac{\text{3 log 2 - log 2}}{\text{5 log 2}} \\[1em] \Rightarrow \dfrac{\text{2 log 2}}{\text{5 log 2}} \\[1em] \Rightarrow \dfrac{2}{5}.

Hence, Option 1 is the correct option.

Question 1(b)

If log3 (x + 1) = 2, the value of x is :

  1. 9

  2. 8

  3. 3

  4. 13\dfrac{1}{3}

Answer

Given,

⇒ log3 (x + 1) = 2

⇒ x + 1 = 32

⇒ x + 1 = 9

⇒ x = 9 - 1 = 8.

Hence, Option 2 is the correct option.

Question 1(c)

If 2 log x = log 250 - 1, the value of x is :

  1. 17

  2. -5

  3. 5

  4. 10

Answer

Given,

⇒ 2 log x = log 250 - 1

⇒ 2 log x = log 250 - log 10

⇒ log x2 = log 25010\dfrac{250}{10}

⇒ x2 = 25

⇒ x = 25\sqrt{25} = 5.

Hence, Option 3 is the correct option.

Question 1(d)

If log3 x - log3 2 - 1 = 0; the value of x is :

  1. 3

  2. -3

  3. -6

  4. 6

Answer

Given,

⇒ log3 x - log3 2 - 1 = 0

⇒ log3 x = log3 2 + 1

⇒ log3 x = log3 2 + log3 3

⇒ log3 x = log3 (2 × 3)

⇒ log3 x = log3 6

⇒ x = 6.

Hence, Option 4 is the correct option.

Question 1(e)

The value of 2 log 3 - 13\dfrac{1}{3} log 64 + log 12 is :

  1. log 27

  2. 27

  3. -27

  4. -log 27

Answer

Given,

⇒2 log 3−13log 64 + log 12⇒log 32−log 6413+log 12⇒log 32−(log 43)13+log 12⇒log 9−(log 4)3×13+log 12⇒log 9 - log 4 + log 12⇒log 9 + log 12 - log 4⇒log 9×124⇒log 1084⇒log 27.\Rightarrow \text{2 log 3} - \dfrac{1}{3} \text{log 64 + log 12} \\[1em] \Rightarrow \text{log 3}^2 - \text{log 64}^{\dfrac{1}{3}} + \text{log 12} \\[1em] \Rightarrow \text{log 3}^2 - (\text{log 4}^3)^{\dfrac{1}{3}} + \text{log 12} \\[1em] \Rightarrow \text{log 9} - (\text{log 4})^{3 \times \dfrac{1}{3}} + \text{log 12} \\[1em] \Rightarrow \text{log 9 - log 4 + log 12} \\[1em] \Rightarrow \text{log 9 + log 12 - log 4} \\[1em] \Rightarrow \text{log } \dfrac{9 \times 12}{4} \\[1em] \Rightarrow \text{log } \dfrac{108}{4} \\[1em] \Rightarrow \text{log } 27.

Hence, Option 1 is the correct option.

Question 2

If 32 log a+23 log b −1=0\dfrac{3}{2} \text{ log a} + \dfrac{2}{3} \text{ log b } - 1 = 0, find the value of a9.b4

Answer

Given,

⇒32 log a+23 log b −1=0⇒log a32+log b23=1⇒log a32+log b23=log 10⇒log (a32×b23)=log 10⇒(a32×b23)=10\Rightarrow \dfrac{3}{2} \text{ log a} + \dfrac{2}{3} \text{ log b } - 1 = 0 \\[1em] \Rightarrow \text{log a}^{\dfrac{3}{2}} + \text{log b}^{\dfrac{2}{3}} = 1 \\[1em] \Rightarrow \text{log a}^{\dfrac{3}{2}} + \text{log b}^{\dfrac{2}{3}} = \text{log 10} \\[1em] \Rightarrow \text{log } (a^{\dfrac{3}{2}} \times b^{\dfrac{2}{3}}) = \text{log 10} \\[1em] \Rightarrow (a^{\dfrac{3}{2}} \times b^{\dfrac{2}{3}}) = 10

Cubing and then squaring both sides, we get :

⇒[(a32×b23)3]2=[(10)3]2⇒(a32×b23)6=106⇒a32×6.b23×6=106⇒a182.b123=106⇒a9.b4=106.\Rightarrow [(a^{\dfrac{3}{2}} \times b^{\dfrac{2}{3}})^3]^2 = [(10)^3]^2 \\[1em] \Rightarrow (a^{\dfrac{3}{2}} \times b^{\dfrac{2}{3}})^6 = 10^6 \\[1em] \Rightarrow a^{\dfrac{3}{2} \times 6}.b^{\dfrac{2}{3} \times 6} = 10^6 \\[1em] \Rightarrow a^{\dfrac{18}{2}}.b^{\dfrac{12}{3}} = 10^6 \\[1em] \Rightarrow a^9.b^4 = 10^6.

Hence, a9.b4 = 106.

Question 3

If x = 1 + log 2 - log 5, y = 2 log 3 and z = log a - log 5; find the value of a, if x + y = 2z.

Answer

Given,

⇒ x + y = 2z

⇒ 1 + log 2 - log 5 + 2 log 3 = 2 (log a - log 5)

⇒ log 10 + log 2 - log 5 + log 32 = 2 log a5\dfrac{a}{5}

⇒ log 10 + log 2 + log 9 - log 5 = log (a5)2\Big(\dfrac{a}{5}\Big)^2

⇒ log 10×2×95\dfrac{10 \times 2 \times 9}{5} = log (a5)2\Big(\dfrac{a}{5}\Big)^2

⇒ 2 × 2 × 9 = (a5)2\Big(\dfrac{a}{5}\Big)^2

⇒ a2 = 52 × 2 × 2 × 9

⇒ a2 = 900

⇒ a = 900\sqrt{900} = 30.

Hence, a = 30.

Question 4

If x = log 0.6; y = log 1.25 and z = log 3 - 2 log 2, find the values of :

(i) x + y - z

(ii) 5x + y - z

Answer

(i) Substituting values of x, y and z in equation x + y - z, we get :

⇒ log 0.6 + log 1.25 - (log 3 - 2 log 2)

⇒ log 0.6 + log 1.25 - log 3 + 2 log 2

⇒ log 0.6 + log 1.25 + log 22 - log 3

⇒ log 0.6×1.25×223\dfrac{0.6 \times 1.25 \times 2^2}{3}

⇒ log 0.2×1.25×40.2 \times 1.25 \times 4

⇒ log 1

⇒ 0.

Hence, x + y - z = 0.

(ii) Substituting value of x + y - z from part (i) in 5x + y - z, we get :

⇒ 50

⇒ 1.

Hence, 5x + y - z = 1.

Question 5

If a2 = log x, b3 = log y and 3a2 - 2b3 = 6 log z, express y in terms of x and z.

Answer

⇒ 3a2 - 2b3 = 6 log z

⇒ 3log x - 2log y = 6 log z

⇒ log x3 - log y2 = log z6

⇒ log x3y2\dfrac{x^3}{y^2} = log z6

⇒ x3y2=z6\dfrac{x^3}{y^2} = z^6

⇒ y2=x3z6y^2 = \dfrac{x^3}{z^6}

⇒ y=x3z6y = \dfrac{\sqrt{x^3}}{\sqrt{z^6}}

⇒ y = x32÷z3x^{\dfrac{3}{2}} ÷ z^3.

Hence, y=x32÷z3y = x^{\dfrac{3}{2}} ÷ z^3.

Question 6

If log a−b2=12\dfrac{a - b}{2} = \dfrac{1}{2} (log a + log b), show that :

a2 + b2 = 6ab

Answer

Given,

⇒log a−b2=12(log a + log b)⇒log a−b2=12(log ab)⇒log a−b2=log (ab)12⇒a−b2=(ab)12\Rightarrow \text{log } \dfrac{a - b}{2} = \dfrac{1}{2} \text{(log a + log b)} \\[1em] \Rightarrow \text{log } \dfrac{a - b}{2} = \dfrac{1}{2} (\text{log ab}) \\[1em] \Rightarrow \text{log } \dfrac{a - b}{2} = \text{log (ab)} ^{\dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{a - b}{2} = (ab)^{\dfrac{1}{2}}

Squaring both sides, we get :

⇒(a−b2)2=(ab)12×2⇒a2+b2−2ab4=ab⇒a2+b2−2ab=4ab⇒a2+b2=4ab+2ab⇒a2+b2=6ab.\Rightarrow \Big(\dfrac{a - b}{2}\Big)^2 = (ab)^{\dfrac{1}{2} \times 2} \\[1em] \Rightarrow \dfrac{a^2 + b^2 - 2ab}{4} = ab \\[1em] \Rightarrow a^2 + b^2 - 2ab = 4ab \\[1em] \Rightarrow a^2 + b^2 = 4ab + 2ab \\[1em] \Rightarrow a^2 + b^2 = 6ab.

Hence, proved that a2 + b2 = 6ab.

Question 7

If a2 + b2 = 23ab, show that :

log a+b5=12\dfrac{a + b}{5} = \dfrac{1}{2} (log a + log b).

Answer

Given,

⇒ a2 + b2 = 23ab

⇒ a2 + b2 + 2ab = 23ab + 2ab

⇒ (a + b)2 = 25ab

⇒ (a+b)225\dfrac{(a + b)^2}{25} = ab

⇒ (a+b5)2\Big(\dfrac{a + b}{5}\Big)^2 = ab

Taking log on both sides, we get :

⇒ log (a+b5)2\Big(\dfrac{a + b}{5}\Big)^2 = log ab

⇒ 2 log a+b5\dfrac{a + b}{5} = log a + log b

⇒ log a+b5\dfrac{a + b}{5} = 12\dfrac{1}{2} (log a + log b).

Hence, proved that log a+b5\dfrac{a + b}{5} = 12\dfrac{1}{2} (log a + log b).

Question 8

If m = log 20 and n = log 25, find the value of x, so that : 2 log (x - 4) = 2m - n

Answer

Given,

⇒ 2 log (x - 4) = 2m - n

Substituting value of m and n in above equation, we get :

⇒ 2 log (x - 4) = 2 log 20 - log 25

⇒ log (x - 4)2 = log 202 - log 25

⇒ log (x - 4)2 = log 400 - log 25

⇒ log (x - 4)2 = log 40025\dfrac{400}{25}

⇒ log (x - 4)2 = log 16

⇒ (x - 4)2 = 16

⇒ x2 + 16 - 8x = 16

⇒ x2 - 8x + 16 - 16 = 0

⇒ x2 - 8x = 0

⇒ x(x - 8) = 0

⇒ x = 0 or x - 8 = 0

⇒ x = 0 or x = 8.

x cannot be zero as then (x - 4) will be negative.

Hence, x = 8.

Question 9

Solve for x and y; if x > 0 and y > 0 :

log xy = log xy\dfrac{x}{y} + 2 log 2 = 2.

Answer

Given,

⇒ log xy = log xy\dfrac{x}{y} + 2 log 2 = 2 ........(1)

Solving L.H.S. of the equation :

⇒log xy=log xy+2 log 2⇒log xy=log xy+log 22⇒log xy=log xy+log 4⇒log xy=log(xy×4)⇒xy=4xy⇒y2=4xx⇒y2=4⇒y=4=2.\Rightarrow \text{log xy} = \text{log }\dfrac{x}{y} + \text{2 log 2} \\[1em] \Rightarrow \text{log xy} = \text{log }\dfrac{x}{y} + \text{log 2}^2 \\[1em] \Rightarrow \text{log xy} = \text{log }\dfrac{x}{y} + \text{log 4} \\[1em] \Rightarrow \text{log xy} = \text{log} \Big(\dfrac{x}{y} \times 4\Big) \\[1em] \Rightarrow xy = \dfrac{4x}{y} \\[1em] \Rightarrow y^2 = \dfrac{4x}{x} \\[1em] \Rightarrow y^2 = 4 \\[1em] \Rightarrow y = \sqrt{4} = 2.

From equation (1), we get :

⇒ log xy = 2

⇒ log x(2) = 2

⇒ log 2x = 2

⇒ log 2x = 2log 10

⇒ log 2x = log 102

⇒ 2x = 102

⇒ 2x = 100

⇒ x = 1002\dfrac{100}{2} = 50.

Hence, x = 50 and y = 2.

Question 10(i)

Find x, if :

logx 625 = -4

Answer

Given,

⇒ logx 625 = -4

⇒ x-4 = 625

⇒ 1x4\dfrac{1}{x^4} = 625

⇒ x4=1625x^4 = \dfrac{1}{625}

⇒ x4=(15)4x^4 = \Big(\dfrac{1}{5}\Big)^4

⇒ x = 15\dfrac{1}{5} = 0.2

Hence, x = 0.2

Question 10(ii)

Find x, if :

logx (5x - 6) = 2

Answer

Given,

⇒ logx (5x - 6) = 2

⇒ (5x - 6) = x2

⇒ x2 - 5x + 6 = 0

⇒ x2 - 3x - 2x + 6 = 0

⇒ x(x - 3) - 2(x - 3) = 0

⇒ (x - 2)(x - 3) = 0

⇒ x - 2 = 0 or x - 3 = 0

⇒ x = 2 or x = 3.

Hence, x = 2 or 3.

Question 11

If p = log 20 and q = log 25, find the value of x, if 2 log (x + 1) = 2p - q.

Answer

Given,

⇒ 2 log (x + 1) = 2p - q

Substituting value of p and q in above equation, we get :

⇒ 2 log (x + 1) = 2 log 20 - log 25

⇒ log (x + 1)2 = log 202 - log 25

⇒ log (x + 1)2 = log 400 - log 25

⇒ log (x + 1)2 = log 40025\dfrac{400}{25}

⇒ log (x + 1)2 = log 16

⇒ (x + 1)2 = 16

⇒ x2 + 1 + 2x = 16

⇒ x2 + 2x + 1 - 16 = 0

⇒ x2 + 2x - 15 = 0

⇒ x2 + 5x - 3x - 15 = 0

⇒ x(x + 5) - 3(x + 5) = 0

⇒ (x - 3)(x + 5) = 0

⇒ x - 3 = 0 or x + 5 = 0

⇒ x = 3 or x = -5.

But x cannot be negative, then x + 1 will be negative which is not possible.

Hence, x = 3.

Question 12

If log2 (x + y) = log3 (x - y) = log 25log 0.2\dfrac{\text{log 25}}{\text{log 0.2}}, find the values of x and y.

Answer

Given,

log2 (x + y) = log3 (x - y) = log 25log 0.2\dfrac{\text{log 25}}{\text{log 0.2}}

Solving, equation :

⇒log2 (x+y)=log 25log 0.2⇒log2 (x+y)=log0.2 25⇒log2 (x+y)=log210 25⇒log2 (x+y)=log15 52⇒log2 (x+y)=log5−1 52⇒log2 (x+y)=2−1log5 5⇒log2 (x+y)=−2×1⇒log2 (x+y)=−2⇒x+y=2−2⇒x+y=122⇒x+y=14 ......(1)\Rightarrow \text{log}_2 \space (x + y) = \dfrac{\text{log 25}}{\text{log 0.2}} \\[1em] \Rightarrow \text{log}_2 \space (x + y) = \text{log}_{0.2} \space 25 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = \text{log}_{\dfrac{2}{10}} \space 25 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = \text{log}_{\dfrac{1}{5}} \space 5^2 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = \text{log}_{5^{-1}} \space 5^2 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = \dfrac{2}{-1}\text{log}_{5} \space 5 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = -2 \times 1 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = -2 \\[1em] \Rightarrow x + y = 2^{-2} \\[1em] \Rightarrow x + y = \dfrac{1}{2^2} \\[1em] \Rightarrow x + y = \dfrac{1}{4}\text{ ......(1)}

Solving, equation :

⇒log3 (x−y)=log 25log 0.2⇒log3 (x−y)=log0.2 25⇒log3 (x−y)=log210 25⇒log3 (x−y)=log15 52⇒log3 (x−y)=log5−1 52⇒log3 (x−y)=2−1log55⇒log3 (x−y)=−2×1⇒log3 (x−y)=−2⇒x−y=3−2⇒x−y=132⇒x−y=19 ......(2)\Rightarrow \text{log}_3 \space (x - y) = \dfrac{\text{log 25}}{\text{log 0.2}} \\[1em] \Rightarrow \text{log}_3 \space (x - y) = \text{log}_{0.2} \space 25 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = \text{log}_{\dfrac{2}{10}} \space 25 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = \text{log}_{\dfrac{1}{5}} \space 5^2 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = \text{log}_{5^{-1}} \space 5^2 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = \dfrac{2}{-1}\text{log}_{5}5 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = -2 \times 1 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = -2 \\[1em] \Rightarrow x - y = 3^{-2} \\[1em] \Rightarrow x - y = \dfrac{1}{3^2} \\[1em] \Rightarrow x - y = \dfrac{1}{9}\text{ ......(2)}

Adding equation (1) and (2), we get :

⇒(x+y)+(x−y)=14+19⇒x+x+y−y=9+436⇒2x=1336⇒x=1336×2=1372.\Rightarrow (x + y) + (x - y) = \dfrac{1}{4} + \dfrac{1}{9} \\[1em] \Rightarrow x + x + y - y = \dfrac{9 + 4}{36} \\[1em] \Rightarrow 2x = \dfrac{13}{36} \\[1em] \Rightarrow x = \dfrac{13}{36 \times 2} = \dfrac{13}{72}.

Substituting value of x in equation (1), we get :

⇒x+y=14⇒1372+y=14⇒y=14−1372⇒y=18−1372⇒y=572.\Rightarrow x + y = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{13}{72} + y = \dfrac{1}{4} \\[1em] \Rightarrow y = \dfrac{1}{4} - \dfrac{13}{72} \\[1em] \Rightarrow y = \dfrac{18 - 13}{72} \\[1em] \Rightarrow y = \dfrac{5}{72}.

Hence, x = 1372 and y=572\dfrac{13}{72}\text{ and y} = \dfrac{5}{72}.

Question 13

Given :

log xlog y=32\dfrac{\text{log x}}{\text{log y}} = \dfrac{3}{2} and log (xy) = 5; find the values of x and y.

Answer

Solving, equation :

⇒log xlog y=32⇒2 log x = 3 log y⇒log x2=log y3⇒x2=y3⇒x=y3 ........(1)\Rightarrow \dfrac{\text{log x}}{\text{log y}} = \dfrac{3}{2} \\[1em] \Rightarrow \text{2 log x = 3 log y} \\[1em] \Rightarrow \text{log } x^2 = \text{log } y^3 \\[1em] \Rightarrow x^2 = y^3 \\[1em] \Rightarrow x = \sqrt{y^3}\text{ ........(1)}

Substituting value of x from equation (1) in log (xy) = 5, we get :

⇒log (y3×y)=5⇒log (y32×y)=5⇒log y32+1=5⇒y52=105⇒(y12)5=105⇒(y)5=105⇒y=10\Rightarrow \text{log } (\sqrt{y^3} \times y) = 5 \\[1em] \Rightarrow \text{log } (y^{\dfrac{3}{2}}\times y) = 5 \\[1em] \Rightarrow \text{log } y^{\dfrac{3}{2} + 1} = 5 \\[1em] \Rightarrow y^{\dfrac{5}{2}} = 10^5 \\[1em] \Rightarrow (y^{\dfrac{1}{2}})^5 =10^5 \\[1em] \Rightarrow (\sqrt{y})^5 = 10^5 \\[1em] \Rightarrow \sqrt{y} = 10

Squaring both sides, we get :

⇒y=102\Rightarrow y = 10^2 = 100.

Substituting value of y in equation (1), we get :

⇒x=(102)3=106=103=1000.\Rightarrow x = \sqrt{(10^2)^3} \\[1em] = \sqrt{10^6} \\[1em] = 10^3 \\[1em] = 1000.

Hence, x = 1000 and y = 100.

Question 14

Given log10 x = 2a and log10 y = b2\dfrac{b}{2}.

(i) Write 10a in terms of x.

(ii) Write 102b + 1 in terms of y.

(iii) If log10P = 3a - 2b, express P in terms of x and y.

Answer

(i) Given,

⇒ log10 x = 2a

⇒ x = 102a

⇒ x = (10a)2

Taking square root on both sides, we get :

⇒x=10a\Rightarrow \sqrt{x} = 10^a

Hence, 10a = x\sqrt{x}.

(ii) Given,

⇒ 102b + 1

⇒ 102b.101 .........(1)

Given,

⇒log10y=b2⇒y=10b2⇒y4=(10b2)4⇒y4=102b ......(2)\Rightarrow \text{log}_{10}y = \dfrac{b}{2} \\[1em] \Rightarrow y = 10^{\dfrac{b}{2}} \\[1em] \Rightarrow y^4 = (10^{\dfrac{b}{2}})^{4} \\[1em] \Rightarrow y^4 = 10^{2b} \text{ ......(2)}

Substituting value of 102b from equation (2) in equation (1), we get :

⇒ y4.101

⇒ 10y4.

Hence, 102b + 1 = 10y4.

(iii) Given,

⇒ log10 P = 3a - 2b

⇒ P = 103a - 2b

⇒ P = 103a.10-2b

⇒ P = 103a102b\dfrac{10^{3a}}{10^{2b}}

We know that: (shown above)

y4 = 102b

and

x\sqrt{x} = 10a

∴ P = 103a102b\dfrac{10^{3a}}{10^{2b}} = (10a)3y4=(x)3y4=x32y4\dfrac{(10^{a})^3}{y^4} = \dfrac{(\sqrt{x})^3}{y^4} = \dfrac{x^{\dfrac{3}{2}}}{y^4}.

Hence, P = x32y4\dfrac{x^{\dfrac{3}{2}}}{y^4}

Question 15

Solve :

log5 (x + 1) - 1 = 1 + log5 (x - 1).

Answer

Given,

⇒ log5 (x + 1) - 1 = 1 + log5 (x - 1)

⇒ log5 (x + 1) - log5 (x - 1) = 1 + 1

⇒ log5 x+1x−1\dfrac{x + 1}{x - 1} = 2

⇒ x+1x−1=52\dfrac{x + 1}{x - 1} = 5^2

⇒ x + 1 = 25(x - 1)

⇒ x + 1 = 25x - 25

⇒ 25x - x = 1 + 25

⇒ 24x = 26

⇒ x = 2624=1312=1112\dfrac{26}{24} = \dfrac{13}{12} = 1\dfrac{1}{12}.

Hence, x = 11121\dfrac{1}{12}.

Test Yourself

Question 1(a)

The value of log3 81 is:

  1. 4

  2. -4

  3. 14\dfrac{1}{4}

  4. -14\dfrac{1}{4}

Answer

Given,

⇒ log3 81

⇒ log3 (3)4

⇒ 4.log3 3

⇒ 4 x 1

⇒ 4.

Hence, option 1 is the correct option.

Question 1(b)

The value of log16 2 is:

  1. 4

  2. -4

  3. 14\dfrac{1}{4}

  4. -14\dfrac{1}{4}

Answer

Let, log16 2 = x

⇒ 16x = 2

⇒ (24)x = 2

⇒ 24x = 21

⇒ 4x = 1

⇒ x = 14\dfrac{1}{4}.

Hence, option 3 is the correct option.

Question 1(c)

If log(3x - 2) = 2, then the value of x is:

  1. 34

  2. 30

  3. 17

  4. none of these

Answer

Given, log(3x - 2) = 2

⇒ log10(3x - 2) = 2

⇒ 3x - 2 = 102

⇒ 3x - 2 = 100

⇒ 3x = 100 + 2

⇒ 3x = 102

⇒ x = 1023\dfrac{102}{3}

⇒ x = 34.

Hence, option 1 is the correct option.

Question 1(d)

2 + 12\dfrac{1}{2} log 9 - 2 log 5 is equal to:

  1. -log 12

  2. log 24

  3. log 12

  4. log 1825\dfrac{18}{25}

Answer

Given,

⇒ 2 + 12\dfrac{1}{2} log 9 - 2 log 5

⇒ 2log 10 + log 912\text{log 9}^\dfrac{1}{2} - log 52

⇒ log 102 + log 9\sqrt{9} - log 52

⇒ log 100 + log 3 - log 25

⇒ log (100 x 3) - log 25

⇒ log 300 - log 25

⇒ log 30025\dfrac{300}{25}

⇒ log 12.

Hence, option 3 is the correct option.

Question 1(e)

3 + log 10-2 is equal to:

  1. 5

  2. 1

  3. -5

  4. -1

Answer

Given,

⇒ 3 + log 10-2

⇒ 3 + (-2) log 10

⇒ 3 + (-2) × 1

⇒ 3 - 2

⇒ 1.

Hence, option 2 is the correct option.

Question 1(f)

Statement 1: log2 (x2 - 4) = 5 ⇒ x = 6

Statement 2: x2 - 4 = 25

⇒ x2 = 36 and x = ±\pm 6

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

Given,

⇒ log2 (x2 - 4) = 5

⇒ x2 - 4 = 25

⇒ x2 - 4 = 32

⇒ x2 = 32 + 4

⇒ x2 = 36

⇒ x = 36\sqrt{36}

⇒ x = ±6\pm 6

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Question 1(g)

Statement 1: log3x = a, then 9a = 1x2\dfrac{1}{x^2}

Statement 2: log3x = a ⇒ 3a = x

∴ 9a = (32)a = (3a)2 = x2

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

Let, log3x = a

⇒ 3a = x

⇒ (3a)2 = x2

⇒ (32)a = x2

⇒ 9a = x2

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Question 1(h)

Assertion (A): log3x=4\text{log}_{\sqrt{3}}x = 4

⇒ x = 9

Reason (R): x = (3)4(\sqrt{3})^4 = 32 = 9

  1. A is true, but R is false.

  2. A is false, but R is true.

  3. Both A and R are true and R is the correct reason for A.

  4. Both A and R are true and R is the incorrect reason for A.

Answer

Given,

⇒log3x=4⇒(3)4=x⇒x=9.\Rightarrow \text{log}_{\sqrt{3}}x = 4 \\[1em] \Rightarrow (\sqrt{3})^4 = x \\[1em] \Rightarrow x = 9.

∴ Both A and R are true and R is the correct reason for A.

Hence, option 3 is the correct option.

Question 1(i)

Assertion (A): log 2 = a and log 3 = b

⇒ 1 + log 12 = 2a + b

Reason (R): 1 + log 12

= 1 + log 2 x 2 x 3

= 1 + 2log 2 + log 3

= 1 + 2a + b

  1. A is true, but R is false.

  2. A is false, but R is true.

  3. Both A and R are true and R is the correct reason for A.

  4. Both A and R are true and R is the incorrect reason for A.

Answer

Given, log 2 = a and log 3 = b

⇒ 1 + log 12

⇒ 1 + log (4 x 3)

⇒ 1 + log (2 x 2 x 3)

⇒ 1 + log 4 + log 3

⇒ 1 + log 22 + log 3

= 1 + 2log 2 + log 3

= 1 + 2a + b.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Question 2

If log2 x = a and log3 y = a, write 72a in terms of x and y.

Answer

Given,

⇒ log2 x = a

⇒ x = 2a ........(1)

⇒ log3 y = a

⇒ y = 3a ........(2)

Simplifying (72)a, we get :

⇒ 72a

⇒ (23 × 32)a

⇒ (23)a × (32)a

⇒ (2a)3 × (3a)2

Substituting value of 2a and 3a from equation (1) and (2), in above equation, we get :

⇒ x3.y2

Hence, 72a = x3.y2

Question 3

Solve for x :

log (x - 1) + log (x + 1) = log2 1

Answer

Given,

⇒ log (x - 1) + log (x + 1) = log2 1

⇒ log (x - 1)(x + 1) = 0

⇒ log (x2 + x - x - 1) = 0

⇒ log (x2 - 1) = 0

⇒ x2 - 1 = 100

⇒ x2 - 1 = 1

⇒ x2 = 1 + 1

⇒ x2 = 2

⇒ x = 2\sqrt{2}.

Hence, x = 2\sqrt{2}.

Question 4

If log (x2 - 21) = 2, show that x = ±11\pm 11.

Answer

Given,

⇒ log (x2 - 21) = 2

⇒ x2 - 21 = 102

⇒ x2 - 21 = 100

⇒ x2 = 100 + 21

⇒ x2 = 121

⇒ x = 121\sqrt{121}

⇒ x = ±11\pm 11.

Hence, proved that x = ±11\pm 11.

Question 5

If x = (100)a, y = (10000)b and z = (10)c, find log 10yx2z3\dfrac{10\sqrt{y}}{x^2z^3} in terms of a, b and c.

Answer

Given,

⇒ x = (100)a

⇒ x = (102)a

⇒ x = 102a ..........(1)

Given,

⇒ y = (10000)b

⇒ y = (104)b

⇒ y = 104b ..........(2)

Given,

⇒ z = 10c ..........(3)

Substituting value of x, y and z from equations (1), (2) and (3) respectively in log 10yx2z3\dfrac{10\sqrt{y}}{x^2z^3}, we get :

⇒log 10yx2z3⇒log 10×104b(102a)2×(10c)3⇒log 10×(10b)4(104a)×(103c)⇒log 10×102b104a+3c⇒log 102b+1104a+3c⇒log 102b+1−log 104a+3c⇒(2b+1)×log 10−(4a+3c)×log 10⇒(2b+1)×1−(4a+3c)×1⇒(2b+1)−(4a+3c)⇒2b+1−4a−3c.\Rightarrow \text{log } \dfrac{10\sqrt{y}}{x^2z^3} \\[1em] \Rightarrow \text{log } \dfrac{10 \times \sqrt{10^{4b}}}{(10^{2a})^2 \times (10^c)^3} \\[1em] \Rightarrow \text{log } \dfrac{10 \times \sqrt{(10^{b})^4}}{(10^{4a}) \times (10^{3c})} \\[1em] \Rightarrow \text{log } \dfrac{10 \times 10^{2b}}{10^{4a + 3c}} \\[1em] \Rightarrow \text{log } \dfrac{10^{2b + 1}}{10^{4a + 3c}} \\[1em] \Rightarrow \text{log } 10^{2b + 1} - \text{log } 10^{4a + 3c} \\[1em] \Rightarrow (2b + 1) \times \text{log 10} - (4a + 3c) \times \text{log 10} \\[1em] \Rightarrow (2b + 1) \times 1 - (4a + 3c) \times 1 \\[1em] \Rightarrow (2b + 1) - (4a + 3c) \\[1em] \Rightarrow 2b + 1 - 4a - 3c.

Hence, log 10yx2z3\dfrac{10\sqrt{y}}{x^2z^3} = 2b + 1 - 4a - 3c.

Question 6

If 3(log 5 - log 3) - (log 5 - 2 log 6) = 2 - log x, find x.

Answer

Given,

⇒ 3(log 5 - log 3) - (log 5 - 2 log 6) = 2 - log x

⇒ 3log 5 - 3log 3 - log 5 + 2 log 6 = 2 - log x

⇒ 2log 5 - 3log 3 + 2log 6 = 2 - log x

⇒ 2log 5 - log 33 + log 62 = 2 - log x

⇒ log 52 - log 27 + log 36 + log x = 2

⇒ log 25 + log 36 + log x - log 27 = 2

⇒ log 25×36×x27\dfrac{25 \times 36 \times x}{27} = 2

⇒ 100x3=102\dfrac{100x}{3} = 10^2

⇒ 100x = 3 × 102

⇒ 100x = 300

⇒ x = 300100\dfrac{300}{100} = 3.

Hence, x = 3.

Question 7

Given log x = 2m - n, log y = n - 2m and log z = 3m - 2n, find in terms of m and n, the value of log x2y3z4\dfrac{x^2y^3}{z^4}.

Answer

Given,

1st equation :

⇒ log x = 2m - n

⇒ x = 102m - n .......(1)

2nd equation :

⇒ log y = n - 2m

⇒ y = 10n - 2m .......(2)

3rd equation :

⇒ log z = 3m - 2n

⇒ z = 103m - 2n .......(3)

Substituting value of x, y and z in log x2y3z4\dfrac{x^2y^3}{z^4}, we get :

⇒log (102m−n)2(10n−2m)3(103m−2n)4⇒log 102(2m−n).103(n−2m)104(3m−2n)⇒log 104m−2n.103n−6m1012m−8n⇒log 104m−2n+3n−6m−(12m−8n)⇒log 10n−2m−12m+8n⇒log 109n−14m⇒(9n−14m) log 10⇒(9n−14m)×1⇒9n−14m.\Rightarrow \text{log } \dfrac{(10^{2m - n})^2(10^{n - 2m})^3}{(10^{3m - 2n})^4} \\[1em] \Rightarrow \text{log } \dfrac{10^{2(2m - n)}.10^{3(n - 2m)}}{10^{4(3m - 2n)}} \\[1em] \Rightarrow \text{log } \dfrac{10^{4m - 2n}.10^{3n - 6m}}{10^{12m - 8n}} \\[1em] \Rightarrow \text{log } 10^{4m - 2n + 3n - 6m - (12m - 8n)} \\[1em] \Rightarrow \text{log } 10^{n - 2m - 12m + 8n} \\[1em] \Rightarrow \text{log } 10^{9n - 14m} \\[1em] \Rightarrow (9n - 14m) \text{ log 10} \\[1em] \Rightarrow (9n - 14m) \times 1 \\[1em] \Rightarrow 9n - 14m.

Hence, log x2y3z4\dfrac{x^2y^3}{z^4} = 9n - 14m.

Question 8

Given logx 25 - logx 5 = 2 - logx 1125\dfrac{1}{125}; find x.

Answer

Given,

⇒ logx 25 - logx 5 = 2 - logx 1125\dfrac{1}{125}

⇒ logx 25 - logx 5 + logx 1125\dfrac{1}{125} = 2

⇒ logx 25×11255\dfrac{25 \times \dfrac{1}{125}}{5} = 2

⇒ logx 25625\dfrac{25}{625} = 2

⇒ x2 = 25625\dfrac{25}{625}

⇒ x2=125x^2 = \dfrac{1}{25}

⇒ x = 125=15\sqrt{\dfrac{1}{25}} = \dfrac{1}{5}.

Hence, x = 15\dfrac{1}{5}.

Question 9

Solve for x, if :

logx 49 - logx 7 + logx 1343\dfrac{1}{343} + 2 = 0

Answer

Given,

⇒logx 49+logx 1343−logx7=−2⇒logx 49×13437=−2⇒logx 497×343=−2⇒logx 149=−2⇒x−2=149⇒x−2=172⇒x−2=7−2⇒x=7.\Rightarrow \text{log}_x \space {49} + \text{log}_x \space {\dfrac{1}{343}} - \text{log}_x7 = -2 \\[1em] \Rightarrow \text{log}_x \space {\dfrac{49 \times \dfrac{1}{343}}{7}} = -2 \\[1em] \Rightarrow \text{log}_x \space {\dfrac{49}{7 \times 343}} = -2 \\[1em] \Rightarrow \text{log}_x \space {\dfrac{1}{49}} = -2 \\[1em] \Rightarrow x^{-2} = \dfrac{1}{49} \\[1em] \Rightarrow x^{-2} = \dfrac{1}{7^2} \\[1em] \Rightarrow x^{-2} = 7^{-2} \\[1em] \Rightarrow x = 7.

Hence, x = 7.

Question 10

If a2 = log x, b3 = log y and a22−b33\dfrac{a^2}{2} - \dfrac{b^3}{3} = log c, find c in terms of x and y.

Answer

Given,

a2 = log x and b3 = log y

Substituting value of a2 and b3 in a22−b33\dfrac{a^2}{2} - \dfrac{b^3}{3} = log c, we get :

⇒a22−b33=log c⇒log x2−log y3=log c⇒3 log x - 2 log y6=log c⇒log x3−log y2=6 log c⇒log x3−log y2=log c6⇒log c6=logx3y2⇒c6=x3y2⇒c=x3y26.\Rightarrow \dfrac{a^2}{2} - \dfrac{b^3}{3} = \text{log c} \\[1em] \Rightarrow \dfrac{\text{log x}}{2} - \dfrac{\text{log y}}{3} = \text{log c} \\[1em] \Rightarrow \dfrac{\text{3 log x - 2 log y}}{6} = \text{log c} \\[1em] \Rightarrow \text{log x}^3 - \text{log y}^2 = \text{6 log c} \\[1em] \Rightarrow \text{log x}^3 - \text{log y}^2 = \text{log c}^6 \\[1em] \Rightarrow \text{log c}^6 = log \dfrac{x^3}{y^2} \\[1em] \Rightarrow c^6 = \dfrac{x^3}{y^2} \\[1em] \Rightarrow c = \sqrt[6]{\dfrac{x^3}{y^2}}.

Hence, c = x3y26.\sqrt[6]{\dfrac{x^3}{y^2}}.

Question 11

Given x = log10 12, y = log4 2 × log10 9 and z = log10 0.4, find :

(i) x - y - z

(ii) 13x - y - z

Answer

Given,

x = log10 12, y = log4 2 × log10 9 and z = log10 0.4

(i) Substituting value of x, y and z in equation x - y - z, we get :

⇒ x - y - z = log10 12 - log4 2 × log10 9 - log10 0.4

= log10 12 - log(22) 2 × log10 9 - log10 0.4

= log10 12 - 12\dfrac{1}{2} log 2 2 × log10 9 - log10 0.4

= log10 12 - 12\dfrac{1}{2} × 1 × log10 9 - log10 0.4

= log10 12 - 12\dfrac{1}{2} × log10 9 - log10 0.4

= log1012 - log10 9129^{\dfrac{1}{2}} - log10 0.4

= log1012 - log103 - log100.4

= log1012 - (log103 + log100.4)

= log10 123×0.4\dfrac{12}{3 \times 0.4}

= log10 121.2\dfrac{12}{1.2}

= log10 10

= 1.

Hence, x - y - z = 1.

(ii) Substituting value of x - y - z in 13x - y - z, we get :

⇒ 131 = 13.

Hence, 13x - y - z = 13.

Question 12

Solve for x, logx 155=2−logx 3515\sqrt{5} = 2 - \text{log}_x \space 3\sqrt{5}

Answer

Given,

⇒logx 155=2−logx 35⇒logx 155+logx 35=2⇒logx (155×35)=2⇒logx 225=2⇒x2=225⇒x=225=15.\Rightarrow \text{log}_x \space 15\sqrt{5} = 2 - \text{log}_x \space 3\sqrt{5} \\[1em] \Rightarrow \text{log}_x \space 15\sqrt{5} + \text{log}_x \space 3\sqrt{5} = 2 \\[1em] \Rightarrow \text{log}_x \space (15\sqrt{5} \times 3\sqrt{5}) = 2 \\[1em] \Rightarrow \text{log}_x \space 225 = 2 \\[1em] \Rightarrow x^2 = 225 \\[1em] \Rightarrow x = \sqrt{225} = 15.

Hence, x = 15.

Question 13(i)

Evaluate :

logb a × logc b × loga c

Answer

Simplifying the expression,

⇒logba×logcb×logac⇒log alog b×log blog c×log clog a⇒log a.log b.log clog b.log c.log a⇒1.\Rightarrow \text{log}_b a \times \text{log}_c b \times \text{log}_a 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 a}} \\[1em] \Rightarrow \dfrac{\text{log a.log b.log c}}{\text{log b.log c.log a}} \\[1em] \Rightarrow 1.

Hence, logb a × logc b × loga c = 1.

Question 13(ii)

Evaluate :

log3 8 ÷ log9 16

Answer

Simplifying the expression,

⇒log3 8÷log9 16⇒log 8log 3÷log 16log 9⇒log 8log 3×log 9log 16⇒log 23log 3×log 32log 24⇒3 log 2log 3×2 log 34 log 2⇒6 log 2.log 34 log 2.log 3⇒64⇒32⇒112.\Rightarrow \text{log}_3 \space 8 ÷ \text{log}_9 \space 16 \\[1em] \Rightarrow \dfrac{\text{log 8}}{\text{log 3}} ÷ \dfrac{\text{log 16}}{\text{log 9}} \\[1em] \Rightarrow \dfrac{\text{log 8}}{\text{log 3}} \times \dfrac{\text{log 9}}{\text{log 16}} \\[1em] \Rightarrow \dfrac{\text{log 2}^3}{\text{log 3}} \times \dfrac{\text{log 3}^2}{\text{log 2}^4} \\[1em] \Rightarrow \dfrac{\text{3 log 2}}{\text{log 3}} \times \dfrac{\text{2 log 3}}{\text{4 log 2}} \\[1em] \Rightarrow \dfrac{\text{6 log 2.log 3}}{\text{4 log 2.log 3}} \\[1em] \Rightarrow \dfrac{6}{4} \\[1em] \Rightarrow \dfrac{3}{2} \\[1em] \Rightarrow 1\dfrac{1}{2}.

Hence, log3 8 ÷ log9 16 = 1121\dfrac{1}{2}.

Question 13(iii)

Evaluate :

log5 8log 25 16×log 100 10\dfrac{\text{log}_5 \space 8}{\text{log }_{25} \space 16 \times \text{log }_{100} \space 10}

Answer

Simplifying the expression,
⇒log5 8log 25 16×log 100 10⇒log5 8log 52 16×log102 10⇒log5 812×log 5 16×12×log10 10⇒log5 814×log5 16⇒4log5 8log5 16⇒4×log 8log 5log 16log 5⇒4 log 8. log 5log 16. log 5⇒4 log 8log 16⇒4 log 23log 24⇒4×3×log 24×log 2⇒124⇒3.\Rightarrow \dfrac{\text{log}_5 \space 8}{\text{log }_{25} \space 16 \times \text{log }_{100} \space 10} \\[1em] \Rightarrow \dfrac{\text{log}_5 \space 8}{\text{log }_{5^2} \space 16 \times \text{log}_{10^2} \space 10} \\[1em] \Rightarrow \dfrac{\text{log}_5 \space 8}{\dfrac{1}{2} \times \text{log }_{5} \space 16 \times \dfrac{1}{2} \times \text{log}_{10} \space 10} \\[1em] \Rightarrow \dfrac{\text{log}_5 \space 8}{\dfrac{1}{4} \times \text{log}_{5} \space 16} \\[1em] \Rightarrow \dfrac{4\text{log}_5 \space 8}{\text{log}_{5} \space 16} \\[1em] \Rightarrow \dfrac{4 \times \dfrac{\text{log 8}}{\text{log 5}}}{\dfrac{\text{log 16}}{\text{log 5}}} \\[1em] \Rightarrow \dfrac{\text{4 log 8. log 5}}{\text{log 16. log 5}} \\[1em] \Rightarrow \dfrac{\text{4 log 8}}{\text{log 16}} \\[1em] \Rightarrow \dfrac{\text{4 log 2}^3}{\text{log 2}^4} \\[1em] \Rightarrow \dfrac{4 \times 3 \times \text{log 2}}{4 \times \text{log 2}} \\[1em] \Rightarrow \dfrac{12}{4} \\[1em] \Rightarrow 3.

Hence, log5 8log 25 16×log 100 10\dfrac{\text{log}_5 \space 8}{\text{log }_{25} \space 16 \times \text{log }_{100} \space 10} = 3.

Question 14

Show that :

loga m ÷ logab m = 1 + loga b

Answer

Simplifying L.H.S. of the given equation, we get :

⇒loga m÷logab m⇒log mlog a÷log mlog ab⇒log mlog a×log ablog m⇒log ablog a⇒loga ab⇒loga a+loga b⇒1+loga b.\Rightarrow \text{log}_a \space m ÷ \text{log}_{ab} \space m \\[1em] \Rightarrow \dfrac{\text{log m}}{\text{log a}} ÷ \dfrac{\text{log m}}{\text{log ab}}\\[1em] \Rightarrow \dfrac{\text{log m}}{\text{log a}} \times \dfrac{\text{log ab}}{\text{log m}}\\[1em] \Rightarrow \dfrac{\text{log ab}}{\text{log a}} \\[1em] \Rightarrow \text{log}_{a} \space ab \\[1em] \Rightarrow \text{log}_{a} \space a + \text{log}_{a} \space b \\[1em] \Rightarrow 1 + \text{log}_{a} \space b.

Hence, proved that loga m ÷ logab m = 1 + loga b.

Question 15

If log27 x=223\text{log}_{\sqrt{27}} \space x = 2\dfrac{2}{3}, find x.

Answer

Given,

⇒log27 x=223⇒log27 x=83⇒log33 x=83⇒log332 x=83⇒132 log3 x=83⇒23 log3 x=83⇒ log3 x=83×32⇒ log3 x=4⇒x=34=81.\Rightarrow \text{log}_{\sqrt{27}} \space x = 2\dfrac{2}{3} \\[1em] \Rightarrow \text{log}_{\sqrt{27}} \space x = \dfrac{8}{3} \\[1em] \Rightarrow \text{log}_{\sqrt{3^3}} \space x = \dfrac{8}{3} \\[1em] \Rightarrow \text{log}_{3^{\frac{3}{2}}} \space x = \dfrac{8}{3} \\[1em] \Rightarrow \dfrac{1}{\dfrac{3}{2}} \text{ log}_{3} \space x = \dfrac{8}{3} \\[1em] \Rightarrow \dfrac{2}{3}\text{ log}_{3} \space x = \dfrac{8}{3} \\[1em] \Rightarrow \text{ log}_{3} \space x = \dfrac{8}{3} \times \dfrac{3}{2} \\[1em] \Rightarrow \text{ log}_{3} \space x = 4 \\[1em] \Rightarrow x = 3^4 = 81.

Hence, x = 81.

Question 16

1loga bc+1+1logb ca+1+1logc ab+1\dfrac{1}{\text{log}_{a} \space bc + 1} + \dfrac{1}{\text{log}_{b} \space ca + 1} + \dfrac{1}{\text{log}_{c} \space ab + 1}

Answer

Evaluating,

⇒1loga bc+1+1logb ca+1+1logc ab+1⇒1log bclog a+1+1log calog b+1+1log ablog c+1⇒1log bc + log alog a+1log ca + log blog b+1log ab + log clog c⇒log alog bc + log a+log blog ca + log b+log clog ab + log c⇒log alog b + log c + log a+log blog c + log a + log b+log clog a + log b + log c⇒log a + log b + log clog a + log b + log c⇒1.\Rightarrow \dfrac{1}{\text{log}_{a} \space bc + 1} + \dfrac{1}{\text{log}_{b} \space ca + 1} + \dfrac{1}{\text{log}_{c} \space ab + 1} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{log bc}}{\text{log a}} + 1} + \dfrac{1}{\dfrac{\text{log ca}}{\text{log b}} + 1} + \dfrac{1}{\dfrac{\text{log ab}}{\text{log c}} + 1} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{log bc + log a}}{\text{log a}}} + \dfrac{1}{\dfrac{\text{log ca + log b}}{\text{log b}}} + \dfrac{1}{\dfrac{\text{log ab + log c}}{\text{log c}}} \\[1em] \Rightarrow \dfrac{\text{log a}}{\text{log bc + log a}} + \dfrac{\text{log b}}{\text{log ca + log b}} + \dfrac{\text{log c}}{\text{log ab + log c}} \\[1em] \Rightarrow \dfrac{\text{log a}}{\text{log b + log c + log a}} + \dfrac{\text{log b}}{\text{log c + log a + log b}} + \dfrac{\text{log c}}{\text{log a + log b + log c}} \\[1em] \Rightarrow \dfrac{\text{log a + log b + log c}}{\text{log a + log b + log c}} \\[1em] \Rightarrow 1.

Hence, 1loga bc+1+1logb ca+1+1logc ab+1\dfrac{1}{\text{log}_{a} \space bc + 1} + \dfrac{1}{\text{log}_{b} \space ca + 1} + \dfrac{1}{\text{log}_{c} \space ab + 1} = 1.

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