(i) ( − 243 32 ) − 3 5 \Big(-\dfrac{243}{32}\Big)^{-\dfrac{3}{5}} ( − 32 243 ) − 5 3
(ii) ( 5 23 64 ) − 2 3 \Big(5\dfrac{23}{64}\Big)^{-\dfrac{2}{3}} ( 5 64 23 ) − 3 2
Answer
Given,
⇒ ( − 243 32 ) − 3 5 ⇒ ( − 3 5 2 5 ) − 3 5 ⇒ [ ( − 3 2 ) 5 ] − 3 5 ⇒ ( − 3 2 ) − 5 × 3 5 ⇒ ( − 3 2 ) − 3 ⇒ ( − 2 3 ) 3 ⇒ − 8 27 . \Rightarrow \Big(-\dfrac{243}{32}\Big)^{-\dfrac{3}{5}}\\[1em] \Rightarrow \Big(-\dfrac{3^5}{2^5}\Big)^{-\dfrac{3}{5}}\\[1em] \Rightarrow \Big[\Big(-\dfrac{3}{2}\Big)^5\Big]^{-\dfrac{3}{5}}\\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^{-\dfrac{5 \times 3}{5}}\\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^{-3}\\[1em] \Rightarrow \Big(-\dfrac{2}{3}\Big)^3\\[1em] \Rightarrow -\dfrac{8}{27}. ⇒ ( − 32 243 ) − 5 3 ⇒ ( − 2 5 3 5 ) − 5 3 ⇒ [ ( − 2 3 ) 5 ] − 5 3 ⇒ ( − 2 3 ) − 5 5 × 3 ⇒ ( − 2 3 ) − 3 ⇒ ( − 3 2 ) 3 ⇒ − 27 8 .
Hence, ( − 243 32 ) − 3 5 = − 8 27 \Big(-\dfrac{243}{32}\Big)^{-\dfrac{3}{5}} = -\dfrac{8}{27} ( − 32 243 ) − 5 3 = − 27 8
(ii) Given,
⇒ ( 5 23 64 ) − 2 3 ⇒ ( 343 64 ) − 2 3 ⇒ ( 7 3 4 3 ) − 2 3 ⇒ [ ( 7 4 ) 3 ] − 2 3 ⇒ ( 7 4 ) − 2 × 3 3 ⇒ ( 7 4 ) − 2 ⇒ ( 4 7 ) 2 ⇒ 16 49 . \Rightarrow \Big(5\dfrac{23}{64}\Big)^{-\dfrac{2}{3}}\\[1em] \Rightarrow \Big(\dfrac{343}{64}\Big)^{-\dfrac{2}{3}}\\[1em] \Rightarrow \Big(\dfrac{7^3}{4^3}\Big)^{-\dfrac{2}{3}}\\[1em] \Rightarrow \Big[\Big(\dfrac{7}{4}\Big)^3\Big]^{-\dfrac{2}{3}}\\[1em] \Rightarrow \Big(\dfrac{7}{4}\Big)^{-\dfrac{2 \times 3}{3}}\\[1em] \Rightarrow \Big(\dfrac{7}{4}\Big)^{-2}\\[1em] \Rightarrow \Big(\dfrac{4}{7}\Big)^2\\[1em] \Rightarrow \dfrac{16}{49}. ⇒ ( 5 64 23 ) − 3 2 ⇒ ( 64 343 ) − 3 2 ⇒ ( 4 3 7 3 ) − 3 2 ⇒ [ ( 4 7 ) 3 ] − 3 2 ⇒ ( 4 7 ) − 3 2 × 3 ⇒ ( 4 7 ) − 2 ⇒ ( 7 4 ) 2 ⇒ 49 16 .
Hence, ( 5 23 64 ) − 2 3 = 16 49 \Big(5\dfrac{23}{64}\Big)^{-\dfrac{2}{3}} = \dfrac{16}{49} ( 5 64 23 ) − 3 2 = 49 16
Simplify the following:
(2a-3 b2 )3
Answer
Given,
⇒ (2a-3 b2 )3 = (2)3 (a-3 )3 (b2 )3
= 8a-9 b6 .
Hence, (2a-3 b2 )3 = 8a-9 b6 .
Simplify the following:
a − 1 + b − 1 ( a b ) − 1 \dfrac{a^{-1} + b^{-1}}{(ab)^{-1}} ( ab ) − 1 a − 1 + b − 1
Answer
Given,
⇒ a − 1 + b − 1 ( a b ) − 1 = ( a − 1 + b − 1 ) × 1 ( a b ) − 1 = ( 1 a + 1 b ) × a b = ( a + b a b ) × a b = a + b . \Rightarrow \dfrac{a^{-1} + b^{-1}}{(ab)^{-1}} = (a^{-1} + b^{-1}) \times \dfrac{1}{(ab)^{-1}} \\[1em] = \Big(\dfrac{1}{a} + \dfrac{1}{b}\Big) \times ab \\[1em] = \Big(\dfrac{a + b}{ab}\Big) \times ab \\[1em] = a + b. ⇒ ( ab ) − 1 a − 1 + b − 1 = ( a − 1 + b − 1 ) × ( ab ) − 1 1 = ( a 1 + b 1 ) × ab = ( ab a + b ) × ab = a + b .
Hence, a − 1 + b − 1 ( a b ) − 1 \dfrac{a^{-1} + b^{-1}}{(ab)^{-1}} ( ab ) − 1 a − 1 + b − 1
Simplify the following:
x − 1 y − 1 x − 1 + y − 1 \dfrac{x^{-1}y^{-1}}{x^{-1} + y^{-1}} x − 1 + y − 1 x − 1 y − 1
Answer
Given,
⇒ x − 1 y − 1 x − 1 + y − 1 = ( x y ) − 1 × 1 x − 1 + y − 1 = 1 x y × 1 1 x + 1 y = 1 x y × 1 x + y x y = 1 x y × x y x + y = 1 x + y . \Rightarrow \dfrac{x^{-1}y^{-1}}{x^{-1} + y^{-1}} = (xy)^{-1} \times \dfrac{1}{x^{-1} + y^{-1}} \\[1em] = \dfrac{1}{xy} \times \dfrac{1}{\dfrac{1}{x} + \dfrac{1}{y}} = \dfrac{1}{xy} \times \dfrac{1}{\dfrac{x + y}{xy}} \\[1em] = \dfrac{1}{xy} \times \dfrac{xy}{x + y} = \dfrac{1}{x + y}. ⇒ x − 1 + y − 1 x − 1 y − 1 = ( x y ) − 1 × x − 1 + y − 1 1 = x y 1 × x 1 + y 1 1 = x y 1 × x y x + y 1 = x y 1 × x + y x y = x + y 1 .
Hence, x − 1 y − 1 x − 1 + y − 1 = 1 x + y \dfrac{x^{-1}y^{-1}}{x^{-1} + y^{-1}} = \dfrac{1}{x + y} x − 1 + y − 1 x − 1 y − 1 = x + y 1
Simplify the following:
( 4 × 10 7 ) ( 6 × 10 − 5 ) 8 × 10 10 \dfrac{(4 \times 10^7)(6 \times 10^{-5})}{8 \times 10^{10}} 8 × 1 0 10 ( 4 × 1 0 7 ) ( 6 × 1 0 − 5 )
Answer
Given,
⇒ ( 4 × 10 7 ) ( 6 × 10 − 5 ) 8 × 10 10 = 4 × 6 × 10 7 × 10 − 5 8 × 10 10 = 24 × 10 7 + ( − 5 ) 8 × 10 10 = 24 × 10 2 8 × 10 10 = 3 × 10 2 × 10 − 10 = 3 × 10 2 + ( − 10 ) = 3 × 10 − 8 . \Rightarrow \dfrac{(4 \times 10^7)(6 \times 10^{-5})}{8 \times 10^{10}} = \dfrac{4 \times 6 \times 10^7 \times 10^{-5}}{8 \times 10^{10}} \\[1em] = \dfrac{24 \times 10^{7 + (-5)}}{8 \times 10^{10}} = \dfrac{24 \times 10^2}{8 \times 10^{10}} \\[1em] = 3 \times 10^2 \times 10^{-10} = 3 \times 10^{2 + (-10)} \\[1em] = 3 \times 10^{-8}. ⇒ 8 × 1 0 10 ( 4 × 1 0 7 ) ( 6 × 1 0 − 5 ) = 8 × 1 0 10 4 × 6 × 1 0 7 × 1 0 − 5 = 8 × 1 0 10 24 × 1 0 7 + ( − 5 ) = 8 × 1 0 10 24 × 1 0 2 = 3 × 1 0 2 × 1 0 − 10 = 3 × 1 0 2 + ( − 10 ) = 3 × 1 0 − 8 .
Hence, ( 4 × 10 7 ) ( 6 × 10 − 5 ) 8 × 10 10 = 3 × 10 − 8 . \dfrac{(4 \times 10^7)(6 \times 10^{-5})}{8 \times 10^{10}} = 3 \times 10^{-8}. 8 × 1 0 10 ( 4 × 1 0 7 ) ( 6 × 1 0 − 5 ) = 3 × 1 0 − 8 .
Simplify the following:
3 a b − 1 + 2 b a − 1 \dfrac{3a}{b^{-1}} + \dfrac{2b}{a^{-1}} b − 1 3 a + a − 1 2 b
Answer
Given,
⇒ 3 a b − 1 + 2 b a − 1 = 3 a 1 b + 2 b 1 a = 3 a b + 2 a b = 5 a b . \Rightarrow \dfrac{3a}{b^{-1}} + \dfrac{2b}{a^{-1}} = \dfrac{3a}{\dfrac{1}{b}} + \dfrac{2b}{\dfrac{1}{a}}\\[1em] = 3ab + 2ab = 5ab. ⇒ b − 1 3 a + a − 1 2 b = b 1 3 a + a 1 2 b = 3 ab + 2 ab = 5 ab .
Hence, 3 a b − 1 + 2 b a − 1 \dfrac{3a}{b^{-1}} + \dfrac{2b}{a^{-1}} b − 1 3 a + a − 1 2 b
Simplify the following:
50 × 4-1 + 8 1 3 8^{\dfrac{1}{3}} 8 3 1
Answer
Given,
⇒ 5 0 × 4 − 1 + 8 1 3 = 1 × 4 − 1 + ( 2 3 ) 1 3 = 1 × 1 4 + 2 = 1 + 8 4 = 9 4 = 2 1 4 . \Rightarrow 5^0 \times 4^{-1} + 8^{\dfrac{1}{3}} = 1 \times 4^{-1} + (2^3)^{\dfrac{1}{3}} \\[1em] = 1 \times \dfrac{1}{4} + 2 \\[1em] = \dfrac{1 + 8}{4} \\[1em] = \dfrac{9}{4} \\[1em] = 2\dfrac{1}{4}. ⇒ 5 0 × 4 − 1 + 8 3 1 = 1 × 4 − 1 + ( 2 3 ) 3 1 = 1 × 4 1 + 2 = 4 1 + 8 = 4 9 = 2 4 1 .
Hence, 5 0 × 4 − 1 + 8 1 3 = 2 1 4 5^0 \times 4^{-1} + 8^{\dfrac{1}{3}} = 2\dfrac{1}{4} 5 0 × 4 − 1 + 8 3 1 = 2 4 1
Simplify the following:
( 8 125 ) − 1 3 \Big(\dfrac{8}{125}\Big)^{-\dfrac{1}{3}} ( 125 8 ) − 3 1
Answer
Given,
⇒ ( 8 125 ) − 1 3 = ( 125 8 ) 1 3 = ( 5 3 2 3 ) 1 3 = 5 3 × 1 3 2 3 × 1 3 = 5 2 = 2 1 2 . \Rightarrow \Big(\dfrac{8}{125}\Big)^{-\dfrac{1}{3}} = \Big(\dfrac{125}{8}\Big)^{\dfrac{1}{3}} \\[1em] = \Big(\dfrac{5^3}{2^3}\Big)^{\dfrac{1}{3}} = \dfrac{5^{3 \times \dfrac{1}{3}}}{2^{3 \times \dfrac{1}{3}}}\\[1em] = \dfrac{5}{2} = 2\dfrac{1}{2}. ⇒ ( 125 8 ) − 3 1 = ( 8 125 ) 3 1 = ( 2 3 5 3 ) 3 1 = 2 3 × 3 1 5 3 × 3 1 = 2 5 = 2 2 1 .
Hence, ( 8 125 ) − 1 3 = 2 1 2 . \Big(\dfrac{8}{125}\Big)^{-\dfrac{1}{3}} = 2\dfrac{1}{2}. ( 125 8 ) − 3 1 = 2 2 1 .
Simplify the following:
( 0.027 ) − 1 3 (0.027)^{-\dfrac{1}{3}} ( 0.027 ) − 3 1
Answer
Given,
⇒ ( 0.027 ) − 1 3 = [ ( 0.3 ) 3 ] − 1 3 = ( 0.3 ) − 1 = 1 0.3 = 10 3 = 3 1 3 . \Rightarrow (0.027)^{-\dfrac{1}{3}} = [(0.3)^3]^{-\dfrac{1}{3}} \\[1em] = (0.3)^{-1} = \dfrac{1}{0.3} \\[1em] = \dfrac{10}{3} = 3\dfrac{1}{3}. \\[1em] ⇒ ( 0.027 ) − 3 1 = [( 0.3 ) 3 ] − 3 1 = ( 0.3 ) − 1 = 0.3 1 = 3 10 = 3 3 1 .
Hence,( 0.027 ) − 1 3 = 3 1 3 (0.027)^{-\dfrac{1}{3}} = 3\dfrac{1}{3} ( 0.027 ) − 3 1 = 3 3 1
Simplify the following:
( − 1 27 ) − 2 3 \Big(-\dfrac{1}{27}\Big)^{-\dfrac{2}{3}} ( − 27 1 ) − 3 2
Answer
Given,
⇒ ( − 1 27 ) − 2 3 = ( − 27 ) 2 3 = ( − 3 ) 3 × 2 3 = ( − 3 ) 2 = 9. \Rightarrow \Big(-\dfrac{1}{27}\Big)^{-\dfrac{2}{3}} = (-27)^{\dfrac{2}{3}} \\[1em] = (-3)^{3 \times \dfrac{2}{3}} = (-3)^2 \\[1em] = 9. ⇒ ( − 27 1 ) − 3 2 = ( − 27 ) 3 2 = ( − 3 ) 3 × 3 2 = ( − 3 ) 2 = 9.
Hence, ( − 1 27 ) − 2 3 \Big(-\dfrac{1}{27}\Big)^{-\dfrac{2}{3}} ( − 27 1 ) − 3 2
Simplify the following:
( 64 ) − 2 3 ÷ ( 9 ) − 3 2 (64)^{-\dfrac{2}{3}} ÷ (9)^{-\dfrac{3}{2}} ( 64 ) − 3 2 ÷ ( 9 ) − 2 3
Answer
Given,
⇒ ( 64 ) − 2 3 ÷ ( 9 ) − 3 2 = ( 1 64 ) 2 3 ÷ ( 1 9 ) 3 2 = ( 1 2 6 ) 2 3 ÷ ( 1 3 2 ) 3 2 = ( 1 ) 2 3 2 6 × 2 3 ÷ ( 1 ) 3 2 3 2 × 3 2 = 1 2 4 ÷ 1 3 3 = 1 2 4 × 3 3 = 27 16 = 1 11 16 . \Rightarrow (64)^{-\dfrac{2}{3}} ÷ (9)^{-\dfrac{3}{2}} = \Big(\dfrac{1}{64}\Big)^{\dfrac{2}{3}} ÷ \Big(\dfrac{1}{9}\Big)^{\dfrac{3}{2}} \\[1em] = \Big(\dfrac{1}{2^6}\Big)^{\dfrac{2}{3}} ÷ \Big(\dfrac{1}{3^2}\Big)^{\dfrac{3}{2}} \\[1em] = \dfrac{(1)^{\dfrac{2}{3}}}{2^{6 \times \dfrac{2}{3}}} ÷ \dfrac{(1)^{\dfrac{3}{2}}}{3^{2 \times \dfrac{3}{2}}} \\[1em] = \dfrac{1}{2^4} ÷ \dfrac{1}{3^3} \\[1em] = \dfrac{1}{2^4} \times 3^3 \\[1em] = \dfrac{27}{16} = 1\dfrac{11}{16}. ⇒ ( 64 ) − 3 2 ÷ ( 9 ) − 2 3 = ( 64 1 ) 3 2 ÷ ( 9 1 ) 2 3 = ( 2 6 1 ) 3 2 ÷ ( 3 2 1 ) 2 3 = 2 6 × 3 2 ( 1 ) 3 2 ÷ 3 2 × 2 3 ( 1 ) 2 3 = 2 4 1 ÷ 3 3 1 = 2 4 1 × 3 3 = 16 27 = 1 16 11 .
Hence, ( 64 ) − 2 3 ÷ ( 9 ) − 3 2 = 1 11 16 (64)^{-\dfrac{2}{3}} ÷ (9)^{-\dfrac{3}{2}} = 1\dfrac{11}{16} ( 64 ) − 3 2 ÷ ( 9 ) − 2 3 = 1 16 11
Simplify the following:
( 27 ) 2 n 3 × ( 8 ) − n 6 ( 18 ) − n 2 \dfrac{(27)^{\dfrac{2n}{3}} \times (8)^{-\dfrac{n}{6}}}{(18)^{-\dfrac{n}{2}}} ( 18 ) − 2 n ( 27 ) 3 2 n × ( 8 ) − 6 n
Answer
Given,
⇒ ( 27 ) 2 n 3 × ( 8 ) − n 6 ( 18 ) − n 2 = ( 3 3 ) 2 n 3 × ( 2 3 ) − n 6 ( 2 × 3 2 ) − n 2 = 3 2 n × ( 2 ) − n 2 ( 2 ) − n 2 × ( 3 2 ) − n 2 = 3 2 n 3 − n = 3 2 n 1 3 n = 3 2 n × 3 n = 3 ( 2 n + n ) = 3 3 n . \Rightarrow \dfrac{(27)^{\dfrac{2n}{3}} \times (8)^{-\dfrac{n}{6}}}{(18)^{-\dfrac{n}{2}}} \\[1em] = \dfrac{(3^3)^{\dfrac{2n}{3}} \times (2^3)^{-\dfrac{n}{6}}}{(2 \times 3^2)^{-\dfrac{n}{2}}} \\[1em] = \dfrac{3^{2n} \times (2)^{-\dfrac{n}{2}}}{(2)^{-\dfrac{n}{2}} \times (3^2)^{-\dfrac{n}{2}}} \\[1em] = \dfrac{3^{2n}}{3^{-n}} = \dfrac{3^{2n}}{\dfrac{1}{3^n}} \\[1em] = 3^{2n} \times 3^n \\[1em] = 3^{(2n + n)} = 3^{3n}. \\[1em] ⇒ ( 18 ) − 2 n ( 27 ) 3 2 n × ( 8 ) − 6 n = ( 2 × 3 2 ) − 2 n ( 3 3 ) 3 2 n × ( 2 3 ) − 6 n = ( 2 ) − 2 n × ( 3 2 ) − 2 n 3 2 n × ( 2 ) − 2 n = 3 − n 3 2 n = 3 n 1 3 2 n = 3 2 n × 3 n = 3 ( 2 n + n ) = 3 3 n .
Hence, ( 27 ) 2 n 3 × ( 8 ) − n 6 ( 18 ) − n 2 = 3 3 n \dfrac{(27)^{\dfrac{2n}{3}} \times (8)^{-\dfrac{n}{6}}}{(18)^{-\dfrac{n}{2}}} = 3^{3n} ( 18 ) − 2 n ( 27 ) 3 2 n × ( 8 ) − 6 n = 3 3 n
Simplify the following:
5. ( 25 ) n + 1 − 25. ( 5 ) 2 n 5. ( 5 ) 2 n + 3 − ( 25 ) n + 1 \dfrac{5.(25)^{n + 1} - 25.(5)^{2n}}{5.(5)^{2n + 3} - (25)^{n + 1}} 5. ( 5 ) 2 n + 3 − ( 25 ) n + 1 5. ( 25 ) n + 1 − 25. ( 5 ) 2 n
Answer
Given,
⇒ 5. ( 25 ) n + 1 − 25. ( 5 ) 2 n 5. ( 5 ) 2 n + 3 − ( 25 ) n + 1 = 5. ( 5 2 ) n + 1 − ( 5 2 ) ( 5 ) 2 n 5. ( 5 ) 2 n + 3 − ( 5 2 ) n + 1 = 5.5 2 n + 2 − 5 2 + 2 n 5 2 n + 4 − 5 2 n + 2 = 5 2 n + 3 − 5 2 n + 2 5 2 n + 4 − 5 2 n + 2 = 5 2 n .5 3 − 5 2 n .5 2 5 2 n .5 4 − 5 2 n .5 2 = 5 2 n ( 5 3 − 5 2 ) 5 2 n ( 5 4 − 5 2 ) = 125 − 25 625 − 25 = 100 600 = 1 6 . \Rightarrow \dfrac{5.(25)^{n + 1} - 25.(5)^{2n}}{5.(5)^{2n + 3} - (25)^{n + 1}} \\[1em] = \dfrac{5.(5^2)^{n + 1} - (5^2)(5)^{2n}}{5.(5)^{2n + 3} - (5^2)^{n + 1}} \\[1em] = \dfrac{5.5^{2n + 2} - 5^{2 + 2n}}{5^{2n + 4} - 5^{2n + 2}} \\[1em] = \dfrac{5^{2n + 3} - 5^{2n + 2}}{5^{2n + 4} - 5^{2n + 2}} \\[1em] = \dfrac{5^{2n}.5^3 - 5^{2n}.5^2}{5^{2n}.5^4 - 5^{2n}.5^2} \\[1em] = \dfrac{5^{2n}(5^3 - 5^2)}{5^{2n}(5^4 - 5^2)} \\[1em] = \dfrac{125 - 25}{625 - 25} \\[1em] = \dfrac{100}{600} = \dfrac{1}{6}. ⇒ 5. ( 5 ) 2 n + 3 − ( 25 ) n + 1 5. ( 25 ) n + 1 − 25. ( 5 ) 2 n = 5. ( 5 ) 2 n + 3 − ( 5 2 ) n + 1 5. ( 5 2 ) n + 1 − ( 5 2 ) ( 5 ) 2 n = 5 2 n + 4 − 5 2 n + 2 5. 5 2 n + 2 − 5 2 + 2 n = 5 2 n + 4 − 5 2 n + 2 5 2 n + 3 − 5 2 n + 2 = 5 2 n . 5 4 − 5 2 n . 5 2 5 2 n . 5 3 − 5 2 n . 5 2 = 5 2 n ( 5 4 − 5 2 ) 5 2 n ( 5 3 − 5 2 ) = 625 − 25 125 − 25 = 600 100 = 6 1 .
Hence, 5. ( 25 ) n + 1 − 25. ( 5 ) 2 n 5. ( 5 ) 2 n + 3 − ( 25 ) n + 1 = 1 6 \dfrac{5.(25)^{n + 1} - 25.(5)^{2n}}{5.(5)^{2n + 3} - (25)^{n + 1}} = \dfrac{1}{6} 5. ( 5 ) 2 n + 3 − ( 25 ) n + 1 5. ( 25 ) n + 1 − 25. ( 5 ) 2 n = 6 1
Simplify the following:
[ 8 − 4 3 ÷ 2 − 2 ] 1 2 \Big[8^{-\dfrac{4}{3}} ÷ 2^{-2}\Big]^{\dfrac{1}{2}} [ 8 − 3 4 ÷ 2 − 2 ] 2 1
Answer
Given,
⇒ [ 8 − 4 3 ÷ 2 − 2 ] 1 2 = [ ( 1 8 ) 4 3 ÷ ( 1 2 ) 2 ] 1 2 = [ ( 1 2 3 ) 4 3 ÷ ( 1 2 ) 2 ] 1 2 = [ 1 2 3 × 4 3 ÷ ( 1 2 2 ) ] 1 2 = [ 1 2 4 × 2 2 ] 1 2 = ( 1 2 2 ) 1 2 = 1 2 . \Rightarrow \Big[8^{-\dfrac{4}{3}} ÷ 2^{-2}\Big]^{\dfrac{1}{2}} = \Big[\Big(\dfrac{1}{8}\Big)^{\frac{4}{3}} ÷ \Big(\dfrac{1}{2}\Big)^2 \Big]^{\dfrac{1}{2}} \\[1em] = \Big[\Big(\dfrac{1}{2^3}\Big)^{\dfrac{4}{3}} ÷ \Big(\dfrac{1}{2}\Big)^2 \Big]^{\dfrac{1}{2}} \\[1em] = \Big[\dfrac{1}{2^{3 \times \dfrac{4}{3}}} ÷ \Big(\dfrac{1}{2^2}\Big)\Big]^{\dfrac{1}{2}} \\[1em] = \Big[\dfrac{1}{2^4} \times 2^2\Big]^{\dfrac{1}{2}} \\[1em] = \Big(\dfrac{1}{2^2}\Big)^{\dfrac{1}{2}} \\[1em] = \dfrac{1}{2}. ⇒ [ 8 − 3 4 ÷ 2 − 2 ] 2 1 = [ ( 8 1 ) 3 4 ÷ ( 2 1 ) 2 ] 2 1 = [ ( 2 3 1 ) 3 4 ÷ ( 2 1 ) 2 ] 2 1 = [ 2 3 × 3 4 1 ÷ ( 2 2 1 ) ] 2 1 = [ 2 4 1 × 2 2 ] 2 1 = ( 2 2 1 ) 2 1 = 2 1 .
Hence, [ 8 − 4 3 ÷ 2 − 2 ] 1 2 = 1 2 \Big[8^{-\dfrac{4}{3}} ÷ 2^{-2}\Big]^{\dfrac{1}{2}} = \dfrac{1}{2} [ 8 − 3 4 ÷ 2 − 2 ] 2 1 = 2 1
Simplify the following:
( 27 8 ) 2 3 − ( 1 4 ) − 2 + 5 0 \Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0 ( 8 27 ) 3 2 − ( 4 1 ) − 2 + 5 0
Answer
Given,
⇒ ( 27 8 ) 2 3 − ( 1 4 ) − 2 + 5 0 = ( 3 3 2 3 ) 2 3 − 4 2 + 1 = 3 3 × 2 3 2 3 × 2 3 − 16 + 1 = 3 2 2 2 − 15 = 9 4 − 15 = 9 − 60 4 = − 51 4 = − 12 3 4 . \Rightarrow \Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0 = \Big(\dfrac{3^3}{2^3}\Big)^{\dfrac{2}{3}} - 4^2 + 1 \\[1em] = \dfrac{3^{3 \times \dfrac{2}{3}}}{2^{3 \times \dfrac{2}{3}}} - 16 + 1 \\[1em] = \dfrac{3^2}{2^2} - 15 \\[1em] = \dfrac{9}{4} - 15 \\[1em] = \dfrac{9 - 60}{4} \\[1em] = -\dfrac{51}{4} = -12\dfrac{3}{4}. ⇒ ( 8 27 ) 3 2 − ( 4 1 ) − 2 + 5 0 = ( 2 3 3 3 ) 3 2 − 4 2 + 1 = 2 3 × 3 2 3 3 × 3 2 − 16 + 1 = 2 2 3 2 − 15 = 4 9 − 15 = 4 9 − 60 = − 4 51 = − 12 4 3 .
Hence, ( 27 8 ) 2 3 − ( 1 4 ) − 2 + 5 0 = − 12 3 4 . \Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0 = -12\dfrac{3}{4}. ( 8 27 ) 3 2 − ( 4 1 ) − 2 + 5 0 = − 12 4 3 .
Simplify the following:
( 3 x 2 ) − 3 × ( x 9 ) 2 3 (3x^2)^{-3} \times (x^9)^{\dfrac{2}{3}} ( 3 x 2 ) − 3 × ( x 9 ) 3 2
Answer
Given,
⇒ ( 3 x 2 ) − 3 × ( x 9 ) 2 3 = ( 1 3 x 2 ) 3 × [ ( x 3 ) 3 ] 2 3 = 1 27 x 6 × ( x 3 ) 3 × 2 3 = 1 27 x 6 × x 6 = 1 27 . \Rightarrow (3x^2)^{-3} \times (x^9)^{\dfrac{2}{3}} = \Big(\dfrac{1}{3x^2}\Big)^3 \times [(x^3)^3]^{\dfrac{2}{3}} \\[1em] = \dfrac{1}{27x^6} \times (x^3)^{3 \times \dfrac{2}{3}} \\[1em] = \dfrac{1}{27x^6} \times x^6 = \dfrac{1}{27}.\\[1em] ⇒ ( 3 x 2 ) − 3 × ( x 9 ) 3 2 = ( 3 x 2 1 ) 3 × [( x 3 ) 3 ] 3 2 = 27 x 6 1 × ( x 3 ) 3 × 3 2 = 27 x 6 1 × x 6 = 27 1 .
Hence, ( 3 x 2 ) − 3 × ( x 9 ) 2 3 = 1 27 . (3x^2)^{-3} \times (x^9)^{\dfrac{2}{3}}= \dfrac{1}{27}. ( 3 x 2 ) − 3 × ( x 9 ) 3 2 = 27 1 .
Simplify the following:
( 8 x 4 ) 1 3 ÷ x 1 3 (8x^4)^{\dfrac{1}{3}} ÷ x^{\dfrac{1}{3}} ( 8 x 4 ) 3 1 ÷ x 3 1
Answer
Given,
⇒ ( 8 x 4 ) 1 3 ÷ x 1 3 = ( 8 ) 1 3 ( x 4 ) 1 3 × 1 ( x ) 1 3 = ( 2 3 ) 1 3 . ( x ) 4 3 . ( x ) − 1 3 = 2. ( x ) 4 3 − 1 3 = 2. ( x ) 3 3 = 2 x . \Rightarrow (8x^4)^{\dfrac{1}{3}} ÷ x^{\dfrac{1}{3}} = (8)^{\dfrac{1}{3}} (x^4)^{\dfrac{1}{3}} \times \dfrac{1}{(x)^{\dfrac{1}{3}}} \\[1em] = (2^3)^{\dfrac{1}{3}}.(x)^{\dfrac{4}{3}}.(x)^{-\dfrac{1}{3}} \\[1em] = 2.(x)^{\dfrac{4}{3} - \dfrac{1}{3}} = 2.(x)^{\dfrac{3}{3}} = 2x. ⇒ ( 8 x 4 ) 3 1 ÷ x 3 1 = ( 8 ) 3 1 ( x 4 ) 3 1 × ( x ) 3 1 1 = ( 2 3 ) 3 1 . ( x ) 3 4 . ( x ) − 3 1 = 2. ( x ) 3 4 − 3 1 = 2. ( x ) 3 3 = 2 x .
Hence, ( 8 x 4 ) 1 3 ÷ x 1 3 (8x^4)^{\dfrac{1}{3}} ÷ x^{\dfrac{1}{3}} ( 8 x 4 ) 3 1 ÷ x 3 1 .
Simplify the following:
( 3 2 ) 0 + 3 − 4 × 3 6 + ( 1 3 ) − 2 (3^2)^0 + 3^{-4} \times 3^6 + \Big(\dfrac{1}{3}\Big)^{-2} ( 3 2 ) 0 + 3 − 4 × 3 6 + ( 3 1 ) − 2
Answer
Given,
⇒ ( 3 2 ) 0 + 3 − 4 × 3 6 + ( 1 3 ) − 2 = 9 0 + 1 3 4 × 3 6 + ( 3 ) 2 = 1 + 1 3 4 × 3 6 + 3 2 = 1 + 3 2 + 3 2 = 1 + 9 + 9 = 19. \Rightarrow (3^2)^0 + 3^{-4} \times 3^6 + \Big(\dfrac{1}{3}\Big)^{-2} \\[1em] = 9^0 + \dfrac{1}{3^4} \times 3^6 + (3)^2 \\[1em] = 1 + \dfrac{1}{3^4} \times 3^6 + 3^2 \\[1em] = 1 + 3^2 + 3^2 \\[1em] = 1 + 9 + 9 = 19. ⇒ ( 3 2 ) 0 + 3 − 4 × 3 6 + ( 3 1 ) − 2 = 9 0 + 3 4 1 × 3 6 + ( 3 ) 2 = 1 + 3 4 1 × 3 6 + 3 2 = 1 + 3 2 + 3 2 = 1 + 9 + 9 = 19.
Hence, ( 3 2 ) 0 + 3 − 4 × 3 6 ÷ ( 1 3 ) − 2 (3^2)^0 + 3^{-4} \times 3^6 ÷ \Big(\dfrac{1}{3}\Big)^{-2} ( 3 2 ) 0 + 3 − 4 × 3 6 ÷ ( 3 1 ) − 2
Simplify the following:
( 9 ) 5 2 − 3. ( 5 ) 0 − ( 1 81 ) − 1 2 (9)^{\dfrac{5}{2}} - 3.(5)^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}} ( 9 ) 2 5 − 3. ( 5 ) 0 − ( 81 1 ) − 2 1
Answer
Given,
⇒ ( 3 2 ) 5 2 − 3.1 − ( 1 9 2 ) − 1 2 = 3 2 × 5 2 − 3 − ( 9 2 ) 1 2 = 3 5 − 3 − 9 = 243 − 12 = 231. \Rightarrow (3^2)^{\dfrac{5}{2}} - 3.1 - \Big(\dfrac{1}{9^2}\Big)^{-\dfrac{1}{2}} \\[1em] = 3^{2 \times \dfrac{5}{2}} - 3 - (9^2)^{\dfrac{1}{2}} \\[1em] = 3^5 - 3 - 9 \\[1em] = 243 - 12 \\[1em] = 231. ⇒ ( 3 2 ) 2 5 − 3.1 − ( 9 2 1 ) − 2 1 = 3 2 × 2 5 − 3 − ( 9 2 ) 2 1 = 3 5 − 3 − 9 = 243 − 12 = 231.
Hence, ( 9 ) 5 2 − 3. ( 5 ) 0 − ( 1 81 ) − 1 2 (9)^{\dfrac{5}{2}} - 3.(5)^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}} ( 9 ) 2 5 − 3. ( 5 ) 0 − ( 81 1 ) − 2 1
Simplify the following:
16 3 4 + 2 ( 1 2 ) − 1 ( 3 ) 0 16^{\dfrac{3}{4}} + 2\Big(\dfrac{1}{2}\Big)^{-1}(3)^0 1 6 4 3 + 2 ( 2 1 ) − 1 ( 3 ) 0
Answer
Given,
⇒ 16 3 4 + 2 ( 1 2 ) − 1 ( 3 ) 0 = ( 2 4 ) 3 4 + 2. ( 2 ) 1 .1 = 2 3 + 4 = 8 + 4 = 12. \Rightarrow 16^{\dfrac{3}{4}} + 2\Big(\dfrac{1}{2}\Big)^{-1}(3)^0 = (2^4)^{\dfrac{3}{4}} + 2.(2)^1 .1 \\[1em] = 2^3 + 4 = 8 + 4 = 12. ⇒ 1 6 4 3 + 2 ( 2 1 ) − 1 ( 3 ) 0 = ( 2 4 ) 4 3 + 2. ( 2 ) 1 .1 = 2 3 + 4 = 8 + 4 = 12.
Hence, 16 3 4 + 2 ( 1 2 ) − 1 ( 3 ) 0 16^{\dfrac{3}{4}} + 2\Big(\dfrac{1}{2}\Big)^{-1}(3)^0 1 6 4 3 + 2 ( 2 1 ) − 1 ( 3 ) 0
Simplify the following:
( 81 ) 3 4 − ( 1 32 ) − 2 5 + ( 8 ) 1 3 ( 1 2 ) − 1 ( 2 ) 0 (81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}}\Big(\dfrac{1}{2}\Big)^{-1}(2)^0 ( 81 ) 4 3 − ( 32 1 ) − 5 2 + ( 8 ) 3 1 ( 2 1 ) − 1 ( 2 ) 0
Answer
Given,
⇒ ( 81 ) 3 4 − ( 1 32 ) − 2 5 + ( 8 ) 1 3 ( 1 2 ) − 1 ( 2 ) 0 = ( 3 4 ) 3 4 − ( 32 ) 2 5 + ( 2 3 ) 1 3 ( 2 ) 1 .1 = 3 3 − ( 2 5 ) 2 5 + 2.2 = 27 − 2 2 + 4 = 27 − 4 + 4 = 27. \Rightarrow (81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}}\Big(\dfrac{1}{2}\Big)^{-1}(2)^0 = (3^4)^{\dfrac{3}{4}} - (32)^{\dfrac{2}{5}} + (2^3)^{\dfrac{1}{3}}(2)^1.1 \\[1em] = 3^3 - (2^5)^{\dfrac{2}{5}} + 2.2 \\[1em] = 27 - 2^2 + 4 \\[1em] = 27 - 4 + 4 = 27. ⇒ ( 81 ) 4 3 − ( 32 1 ) − 5 2 + ( 8 ) 3 1 ( 2 1 ) − 1 ( 2 ) 0 = ( 3 4 ) 4 3 − ( 32 ) 5 2 + ( 2 3 ) 3 1 ( 2 ) 1 .1 = 3 3 − ( 2 5 ) 5 2 + 2.2 = 27 − 2 2 + 4 = 27 − 4 + 4 = 27.
Hence, ( 81 ) 3 4 − ( 1 32 ) − 2 5 + ( 8 ) 1 3 ( 1 2 ) − 1 ( 2 ) 0 (81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}}\Big(\dfrac{1}{2}\Big)^{-1}(2)^0 ( 81 ) 4 3 − ( 32 1 ) − 5 2 + ( 8 ) 3 1 ( 2 1 ) − 1 ( 2 ) 0
Simplify the following:
( 64 125 ) − 2 3 ÷ 1 ( 256 625 ) 1 4 + ( 25 64 3 ) 0 \Big(\dfrac{64}{125}\Big)^{-\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}}} + \Big(\dfrac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0 ( 125 64 ) − 3 2 ÷ ( 625 256 ) 4 1 1 + ( 3 64 25 ) 0
Answer
Given,
⇒ ( 64 125 ) − 2 3 ÷ 1 ( 256 625 ) 1 4 + ( 25 64 3 ) 0 = ( 125 64 ) 2 3 × ( 256 625 ) 1 4 + 1 = [ ( 5 4 ) 3 ] 2 3 × [ ( 4 5 ) 4 ] 1 4 + 1 = ( 5 4 ) 2 × 4 5 + 1 = 5 4 + 1 = 9 4 = 2 1 4 . \Rightarrow \Big(\dfrac{64}{125}\Big)^{-\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}}} + \Big(\dfrac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0 \\[1em] = \Big(\dfrac{125}{64}\Big)^{\dfrac{2}{3}} \times \Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}} + 1 \\[1em] = \Big[\Big(\dfrac{5}{4}\Big)^3\Big]^{\dfrac{2}{3}} \times \Big[\Big(\dfrac{4}{5}\Big)^4\Big]^{\dfrac{1}{4}} + 1 \\[1em] = \Big(\dfrac{5}{4}\Big)^2 \times \dfrac{4}{5} + 1 \\[1em] = \dfrac{5}{4} + 1 \\[1em] = \dfrac{9}{4} = 2\dfrac{1}{4}. ⇒ ( 125 64 ) − 3 2 ÷ ( 625 256 ) 4 1 1 + ( 3 64 25 ) 0 = ( 64 125 ) 3 2 × ( 625 256 ) 4 1 + 1 = [ ( 4 5 ) 3 ] 3 2 × [ ( 5 4 ) 4 ] 4 1 + 1 = ( 4 5 ) 2 × 5 4 + 1 = 4 5 + 1 = 4 9 = 2 4 1 .
Hence, ( 64 125 ) − 2 3 ÷ 1 ( 256 625 ) 1 4 + ( 25 64 3 ) 0 = 2 1 4 \Big(\dfrac{64}{125}\Big)^{-\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}}} + \Big(\dfrac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0 = 2\dfrac{1}{4} ( 125 64 ) − 3 2 ÷ ( 625 256 ) 4 1 1 + ( 3 64 25 ) 0 = 2 4 1
Simplify the following:
5 n + 3 − 6 × 5 n + 1 9 × 5 n − 2 2 × 5 n \dfrac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 2^2 \times 5^n} 9 × 5 n − 2 2 × 5 n 5 n + 3 − 6 × 5 n + 1
Answer
Given,
⇒ 5 n + 3 − 6 × 5 n + 1 9 × 5 n − 2 2 × 5 n = 5 n .5 3 − 6.5 n .5 1 5 n ( 9 − 2 2 ) = 5 n . ( 5 3 − 30 ) 5 n ( 9 − 4 ) = 125 − 30 5 = 95 5 = 19. \Rightarrow \dfrac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 2^2 \times 5^n} = \dfrac{5^n.5^3 - 6.5^n.5^1}{5^n(9 - 2^2)} \\[1em] = \dfrac{5^n.(5^3 - 30)}{5^n(9 - 4)} \\[1em] = \dfrac{125 - 30}{5} \\[1em] = \dfrac{95}{5} = 19. ⇒ 9 × 5 n − 2 2 × 5 n 5 n + 3 − 6 × 5 n + 1 = 5 n ( 9 − 2 2 ) 5 n . 5 3 − 6. 5 n . 5 1 = 5 n ( 9 − 4 ) 5 n . ( 5 3 − 30 ) = 5 125 − 30 = 5 95 = 19.
Hence, 5 n + 3 − 6 × 5 n + 1 9 × 5 n − 2 2 × 5 n \dfrac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 2^2 \times 5^n} 9 × 5 n − 2 2 × 5 n 5 n + 3 − 6 × 5 n + 1
Simplify the following:
[ ( 64 ) 2 3 .2 − 2 ÷ 8 0 ] − 1 2 \Big[(64)^{\dfrac{2}{3}}.2^{-2} ÷ 8^0\Big]^{-\dfrac{1}{2}} [ ( 64 ) 3 2 . 2 − 2 ÷ 8 0 ] − 2 1
Answer
Given,
⇒ [ ( 64 ) 2 3 .2 − 2 ÷ 8 0 ] − 1 2 = [ [ ( 4 ) 3 ] 2 3 × ( 1 2 ) 2 ÷ 1 ] − 1 2 = [ ( 4 2 ) × ( 1 4 ) ] − 1 2 = [ 16 × 1 4 ] − 1 2 = [ 4 ] − 1 2 = ( 1 4 ) 1 2 = 1 2 = 1 2 . \Rightarrow \Big[(64)^{\dfrac{2}{3}}.2^{-2} ÷ 8^0\Big]^{-\dfrac{1}{2}} = \Big[[(4)^3]^{\dfrac{2}{3}} \times \Big(\dfrac{1}{2}\Big)^2 ÷ 1 \Big]^{-\dfrac{1}{2}} \\[1em] = \Big[(4^2) \times \Big(\dfrac{1}{4}\Big)\Big]^{-\dfrac{1}{2}} \\[1em] = \Big[16 \times \dfrac{1}{4}\Big]^{-\dfrac{1}{2}} \\[1em] = [4]^{-\dfrac{1}{2}} \\[1em] = \Big(\dfrac{1}{4}\Big)^{\dfrac{1}{2}} = \dfrac{1}{2} \\[1em] = \dfrac{1}{2}. ⇒ [ ( 64 ) 3 2 . 2 − 2 ÷ 8 0 ] − 2 1 = [ [( 4 ) 3 ] 3 2 × ( 2 1 ) 2 ÷ 1 ] − 2 1 = [ ( 4 2 ) × ( 4 1 ) ] − 2 1 = [ 16 × 4 1 ] − 2 1 = [ 4 ] − 2 1 = ( 4 1 ) 2 1 = 2 1 = 2 1 .
Hence, [ ( 64 ) − 2 3 .2 − 2 ÷ 8 0 ] − 1 2 = 1 2 \Big[(64)^{-\dfrac{2}{3}}.2^{-2} ÷ 8^0\Big]^{-\dfrac{1}{2}} = \dfrac{1}{2} [ ( 64 ) − 3 2 . 2 − 2 ÷ 8 0 ] − 2 1 = 2 1
Simplify the following:
3 n × 9 n + 1 ÷ ( 3 n − 1 × 9 n − 1 ) 3^n \times 9^{n + 1} ÷ (3^{n - 1} \times 9^{n - 1}) 3 n × 9 n + 1 ÷ ( 3 n − 1 × 9 n − 1 )
Answer
Given,
⇒ 3 n × 9 n + 1 ÷ ( 3 n − 1 × 9 n − 1 ) = 3 n × 9 n + 1 × 1 3 n − 1 × 9 n − 1 = 3 n × 9 n .9 1 3 n .3 − 1 × 9 n .9 − 1 = 9 3 − 1 × 9 − 1 = 9 × 3 × 9 = 243. \Rightarrow 3^n \times 9^{n + 1} ÷ (3^{n - 1} \times 9^{n - 1}) \\[1em] = 3^n \times 9^{n + 1} \times \dfrac{1}{3^{n - 1} \times 9^{n - 1}} \\[1em] = \dfrac{3^n \times 9^n.9^1}{3^n.3^{-1} \times 9^n.9^{-1}} \\[1em] = \dfrac{9}{3^{-1} \times 9^{-1}} \\[1em] = 9 \times 3 \times 9 \\[1em] = 243. ⇒ 3 n × 9 n + 1 ÷ ( 3 n − 1 × 9 n − 1 ) = 3 n × 9 n + 1 × 3 n − 1 × 9 n − 1 1 = 3 n . 3 − 1 × 9 n . 9 − 1 3 n × 9 n . 9 1 = 3 − 1 × 9 − 1 9 = 9 × 3 × 9 = 243.
Hence, 3 n × 9 n + 1 ÷ ( 3 n − 1 × 9 n − 1 ) 3^n \times 9^{n + 1} ÷ (3^{n - 1} \times 9^{n - 1}) 3 n × 9 n + 1 ÷ ( 3 n − 1 × 9 n − 1 )
Simplify the following:
2 2 × 256 4 64 3 − ( 1 2 ) − 2 \dfrac{\sqrt{2^2} \times \sqrt[4]{256}}{\sqrt[3]{64}} - \Big(\dfrac{1}{2}\Big)^{-2} 3 64 2 2 × 4 256 − ( 2 1 ) − 2
Answer
Given,
⇒ 2 2 × 256 4 64 3 − ( 1 2 ) − 2 ⇒ 2 2 × 2 8 4 2 6 3 − ( 2 ) 2 ⇒ ( 2 2 ) 1 2 × ( 2 8 ) 1 4 ( 2 6 ) 1 3 − 4 ⇒ ( 2 ) 2 × 1 2 × ( 2 ) 8 × 1 4 ( 2 ) 6 × 1 3 − 4 ⇒ 2 × 2 2 2 2 − 4 ⇒ 2 − 4 ⇒ − 2. \Rightarrow \dfrac{\sqrt{2^2} \times \sqrt[4]{256}}{\sqrt[3]{64}} - \Big(\dfrac{1}{2}\Big)^{-2} \\[1em] \Rightarrow \dfrac{\sqrt{2^2} \times \sqrt[4]{2^8}}{\sqrt[3]{2^6}} - (2)^{2} \\[1em] \Rightarrow \dfrac{(2^2)^{\dfrac{1}{2}} \times (2^8)^{\dfrac{1}{4}}}{(2^6)^{\dfrac{1}{3}}} - 4 \\[1em] \Rightarrow \dfrac{(2)^{2 \times \dfrac{1}{2}} \times (2)^{8 \times \dfrac{1}{4}}}{(2)^{6 \times \dfrac{1}{3}}} - 4 \\[1em] \Rightarrow \dfrac{2 \times 2^2}{2^2} - 4 \\[1em] \Rightarrow 2 - 4 \\[1em] \Rightarrow -2. ⇒ 3 64 2 2 × 4 256 − ( 2 1 ) − 2 ⇒ 3 2 6 2 2 × 4 2 8 − ( 2 ) 2 ⇒ ( 2 6 ) 3 1 ( 2 2 ) 2 1 × ( 2 8 ) 4 1 − 4 ⇒ ( 2 ) 6 × 3 1 ( 2 ) 2 × 2 1 × ( 2 ) 8 × 4 1 − 4 ⇒ 2 2 2 × 2 2 − 4 ⇒ 2 − 4 ⇒ − 2.
Hence, 2 2 × 256 4 64 3 − ( 1 2 ) − 2 \dfrac{\sqrt{2^2} \times \sqrt[4]{256}}{\sqrt[3]{64}} - \Big(\dfrac{1}{2}\Big)^{-2} 3 64 2 2 × 4 256 − ( 2 1 ) − 2
Simplify the following:
3 − 6 7 × 4 − 3 7 × 9 3 7 × 2 6 7 2 2 + 2 0 + 2 − 2 \dfrac{3^{-\dfrac{6}{7}} \times 4^{-\dfrac{3}{7}} \times 9^{\dfrac{3}{7}} \times 2^{\dfrac{6}{7}}}{2^2 + 2^0 + 2^{-2}} 2 2 + 2 0 + 2 − 2 3 − 7 6 × 4 − 7 3 × 9 7 3 × 2 7 6
Answer
Given,
⇒ 3 − 6 7 × 4 − 3 7 × 9 3 7 × 2 6 7 2 2 + 2 0 + 2 − 2 ⇒ 3 − 6 7 × ( 2 2 ) − 3 7 × ( 3 2 ) 3 7 × 2 6 7 4 + 1 + ( 1 2 ) 2 ⇒ 3 − 6 7 × ( 2 ) − 6 7 × ( 3 ) 6 7 × 2 6 7 4 + 1 + ( 1 4 ) ⇒ 3 − 6 7 + 6 7 × ( 2 ) − 6 7 + 6 7 ( 16 + 4 + 1 4 ) ⇒ 3 0 × 2 0 21 4 ⇒ 1 21 4 ⇒ 4 21 . \Rightarrow \dfrac{3^{-\dfrac{6}{7}} \times 4^{-\dfrac{3}{7}} \times 9^{\dfrac{3}{7}} \times 2^{\dfrac{6}{7}}}{2^2 + 2^0 + 2^{-2}} \\[1em] \Rightarrow \dfrac{3^{-\dfrac{6}{7}} \times (2^2)^{-\dfrac{3}{7}} \times (3^2)^{\dfrac{3}{7}} \times 2^{\dfrac{6}{7}}}{4 + 1 + \Big(\dfrac{1}{2}\Big)^{2}} \\[1em] \Rightarrow \dfrac{3^{-\dfrac{6}{7}} \times (2)^{-\dfrac{6}{7}} \times (3)^{\dfrac{6}{7}} \times 2^{\dfrac{6}{7}}}{4 + 1 + \Big(\dfrac{1}{4}\Big)} \\[1em] \Rightarrow \dfrac{3^{-\dfrac{6}{7} + \dfrac{6}{7}} \times (2)^{-\dfrac{6}{7} + \dfrac{6}{7}}}{\Big(\dfrac{16 + 4 + 1}{4}\Big)} \\[1em] \Rightarrow \dfrac{3^0 \times 2^0}{\dfrac{21}{4}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{21}{4}} \\[1em] \Rightarrow \dfrac{4}{21}. ⇒ 2 2 + 2 0 + 2 − 2 3 − 7 6 × 4 − 7 3 × 9 7 3 × 2 7 6 ⇒ 4 + 1 + ( 2 1 ) 2 3 − 7 6 × ( 2 2 ) − 7 3 × ( 3 2 ) 7 3 × 2 7 6 ⇒ 4 + 1 + ( 4 1 ) 3 − 7 6 × ( 2 ) − 7 6 × ( 3 ) 7 6 × 2 7 6 ⇒ ( 4 16 + 4 + 1 ) 3 − 7 6 + 7 6 × ( 2 ) − 7 6 + 7 6 ⇒ 4 21 3 0 × 2 0 ⇒ 4 21 1 ⇒ 21 4 .
Hence, 3 − 6 7 × 4 − 3 7 × 9 3 7 × 2 6 7 2 2 + 2 0 + 2 − 2 = 4 21 \dfrac{3^{-\dfrac{6}{7}} \times 4^{-\dfrac{3}{7}} \times 9^{\dfrac{3}{7}} \times 2^{\dfrac{6}{7}}}{2^2 + 2^0 + 2^{-2}} = \dfrac{4}{21} 2 2 + 2 0 + 2 − 2 3 − 7 6 × 4 − 7 3 × 9 7 3 × 2 7 6 = 21 4 .
Simplify the following:
( 32 ) 2 5 × ( 4 ) − 1 2 × ( 8 ) 1 3 2 − 2 ÷ ( 64 ) − 1 3 \dfrac{(32)^{\dfrac{2}{5}} \times (4)^{-\dfrac{1}{2}} \times (8)^{\dfrac{1}{3}}}{2^{-2} ÷ (64)^{-\dfrac{1}{3}}} 2 − 2 ÷ ( 64 ) − 3 1 ( 32 ) 5 2 × ( 4 ) − 2 1 × ( 8 ) 3 1
Answer
Given,
⇒ ( 32 ) 2 5 × ( 4 ) − 1 2 × ( 8 ) 1 3 2 − 2 ÷ ( 64 ) − 1 3 ⇒ ( 2 5 ) 2 5 × ( 2 2 ) − 1 2 × ( 2 3 ) 1 3 2 − 2 ÷ ( 2 6 ) − 1 3 ⇒ 2 2 × 2 − 1 × 2 1 2 − 2 ÷ 2 − 2 ⇒ 2 2 − 1 + 1 2 − 2 2 − 2 ⇒ 2 2 1 = 2 2 = 4. \Rightarrow \dfrac{(32)^{\dfrac{2}{5}} \times (4)^{-\dfrac{1}{2}} \times (8)^{\dfrac{1}{3}}}{2^{-2} ÷ (64)^{-\dfrac{1}{3}}} \\[1em] \Rightarrow \dfrac{(2^5)^{\dfrac{2}{5}} \times (2^2)^{-\dfrac{1}{2}} \times (2^3)^{\dfrac{1}{3}}}{2^{-2} ÷ (2^6)^{-\dfrac{1}{3}}} \\[1em] \Rightarrow \dfrac{2^2 \times 2^{-1} \times 2^1}{2^{-2} ÷ 2^{-2}} \\[1em] \Rightarrow \dfrac{2^{2 - 1 + 1}}{\dfrac{2^{-2}}{2^{-2}}} \\[1em] \Rightarrow \dfrac{2^2}{1} = 2^2 = 4. ⇒ 2 − 2 ÷ ( 64 ) − 3 1 ( 32 ) 5 2 × ( 4 ) − 2 1 × ( 8 ) 3 1 ⇒ 2 − 2 ÷ ( 2 6 ) − 3 1 ( 2 5 ) 5 2 × ( 2 2 ) − 2 1 × ( 2 3 ) 3 1 ⇒ 2 − 2 ÷ 2 − 2 2 2 × 2 − 1 × 2 1 ⇒ 2 − 2 2 − 2 2 2 − 1 + 1 ⇒ 1 2 2 = 2 2 = 4.
Hence, ( 32 ) 2 5 × ( 4 ) − 1 2 × ( 8 ) 1 3 2 − 2 ÷ ( 64 ) − 1 3 \dfrac{(32)^{\dfrac{2}{5}} \times (4)^{-\dfrac{1}{2}} \times (8)^{\dfrac{1}{3}}}{2^{-2} ÷ (64)^{-\dfrac{1}{3}}} 2 − 2 ÷ ( 64 ) − 3 1 ( 32 ) 5 2 × ( 4 ) − 2 1 × ( 8 ) 3 1
Simplify the following:
5 2 ( x + 6 ) × ( 25 ) − 7 + 2 x ( 125 ) 2 x \dfrac{5^{2(x + 6)} \times (25)^{-7 + 2x}}{(125)^{2x}} ( 125 ) 2 x 5 2 ( x + 6 ) × ( 25 ) − 7 + 2 x
Answer
Given,
⇒ 5 2 ( x + 6 ) × ( 25 ) − 7 + 2 x ( 125 ) 2 x ⇒ 5 2 x + 12 × ( 5 2 ) − 7 + 2 x ( 5 3 ) 2 x ⇒ 5 2 x + 12 × 5 − 14 + 4 x 5 6 x ⇒ 5 2 x + 12 + ( − 14 + 4 x ) 5 6 x ⇒ 5 6 x − 2 .5 − 6 x ⇒ 5 6 x − 2 + ( − 6 x ) ⇒ 5 − 2 = 1 5 2 = 1 25 . \Rightarrow \dfrac{5^{2(x + 6)} \times (25)^{-7 + 2x}}{(125)^{2x}} \\[1em] \Rightarrow \dfrac{5^{2x + 12} \times (5^2)^{-7 + 2x}}{(5^3)^{2x}} \\[1em] \Rightarrow \dfrac{5^{2x + 12} \times 5^{-14 + 4x}}{5^{6x}} \\[1em] \Rightarrow \dfrac{5^{2x + 12 + (-14 + 4x)}}{5^{6x}} \\[1em] \Rightarrow 5^{6x - 2}.5^{-6x} \\[1em] \Rightarrow 5^{6x - 2 + (-6x)} \\[1em] \Rightarrow 5^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25}. ⇒ ( 125 ) 2 x 5 2 ( x + 6 ) × ( 25 ) − 7 + 2 x ⇒ ( 5 3 ) 2 x 5 2 x + 12 × ( 5 2 ) − 7 + 2 x ⇒ 5 6 x 5 2 x + 12 × 5 − 14 + 4 x ⇒ 5 6 x 5 2 x + 12 + ( − 14 + 4 x ) ⇒ 5 6 x − 2 . 5 − 6 x ⇒ 5 6 x − 2 + ( − 6 x ) ⇒ 5 − 2 = 5 2 1 = 25 1 .
Hence, 5 2 ( x + 6 ) × ( 25 ) − 7 + 2 x ( 125 ) 2 x = 1 25 \dfrac{5^{2(x + 6)} \times (25)^{-7 + 2x}}{(125)^{2x}} = \dfrac{1}{25} ( 125 ) 2 x 5 2 ( x + 6 ) × ( 25 ) − 7 + 2 x = 25 1
Simplify the following:
7 2 n + 3 − ( 49 ) n + 2 ( ( 343 ) n + 1 ) 2 3 \dfrac{7^{2n + 3} - (49)^{n + 2}}{((343)^{n + 1})^{\dfrac{2}{3}}} (( 343 ) n + 1 ) 3 2 7 2 n + 3 − ( 49 ) n + 2
Answer
Given,
⇒ 7 2 n + 3 − ( 49 ) n + 2 ( ( 343 ) n + 1 ) 2 3 ⇒ 7 2 n + 3 − ( 7 2 ) n + 2 ( ( 7 3 ) n + 1 ) 2 3 ⇒ 7 2 n + 3 − 7 2 n + 4 7 3 × ( n + 1 ) × 2 3 ⇒ 7 2 n .7 3 − 7 2 n .7 4 7 2 ( n + 1 ) ⇒ 343.7 2 n − 2401.7 2 n 7 2 n + 2 ⇒ 7 2 n ( 343 − 2401 ) 7 2 n .7 2 ⇒ − 2058 49 = − 42. \Rightarrow \dfrac{7^{2n + 3} - (49)^{n + 2}}{((343)^{n + 1})^{\dfrac{2}{3}}} \\[1em] \Rightarrow \dfrac{7^{2n + 3} - (7^2)^{n + 2}}{((7^3)^{n + 1})^{\dfrac{2}{3}}} \\[1em] \Rightarrow \dfrac{7^{2n + 3} - 7^{2n + 4}}{7^{3 \times (n + 1) \times \dfrac{2}{3}}} \\[1em] \Rightarrow \dfrac{7^{2n}.7^3 - 7^{2n}.7^{4}}{7^{2(n + 1)}} \\[1em] \Rightarrow \dfrac{343.7^{2n} - 2401.7^{2n}}{7^{2n + 2}} \\[1em] \Rightarrow \dfrac{7^{2n}(343 - 2401)}{7^{2n}.7^2} \\[1em] \Rightarrow \dfrac{-2058}{49} = -42. ⇒ (( 343 ) n + 1 ) 3 2 7 2 n + 3 − ( 49 ) n + 2 ⇒ (( 7 3 ) n + 1 ) 3 2 7 2 n + 3 − ( 7 2 ) n + 2 ⇒ 7 3 × ( n + 1 ) × 3 2 7 2 n + 3 − 7 2 n + 4 ⇒ 7 2 ( n + 1 ) 7 2 n . 7 3 − 7 2 n . 7 4 ⇒ 7 2 n + 2 343. 7 2 n − 2401. 7 2 n ⇒ 7 2 n . 7 2 7 2 n ( 343 − 2401 ) ⇒ 49 − 2058 = − 42.
Hence, 7 2 n + 3 − ( 49 ) n + 2 ( ( 343 ) n + 1 ) 2 3 \dfrac{7^{2n + 3} - (49)^{n + 2}}{((343)^{n + 1})^{\dfrac{2}{3}}} (( 343 ) n + 1 ) 3 2 7 2 n + 3 − ( 49 ) n + 2 .
Simplify the following:
( 27 ) 4 3 + ( 32 ) 0.8 + ( 0.8 ) − 1 (27)^{\dfrac{4}{3}} + (32)^{0.8} + (0.8)^{-1} ( 27 ) 3 4 + ( 32 ) 0.8 + ( 0.8 ) − 1
Answer
Given,
⇒ ( 27 ) 4 3 + ( 32 ) 0.8 + ( 0.8 ) − 1 ⇒ ( 3 3 ) 4 3 + ( 2 5 ) 8 10 + ( 1 0.8 ) 1 ⇒ ( 3 ) 3 × 4 3 + ( 2 ) 5 × 8 10 + ( 10 8 ) ⇒ ( 3 ) 4 + 2 4 + 5 4 ⇒ 81 + 16 + 5 4 ⇒ 97 + 5 4 ⇒ 388 + 5 4 ⇒ 393 4 ⇒ 98.25 \Rightarrow (27)^{\dfrac{4}{3}} + (32)^{0.8} + (0.8)^{-1} \\[1em] \Rightarrow (3^3)^{\dfrac{4}{3}} + (2^5)^{\dfrac{8}{10}} + \Big(\dfrac{1}{0.8}\Big)^{1} \\[1em] \Rightarrow (3)^{3 \times \dfrac{4}{3}} + (2)^{5 \times \dfrac{8}{10}} + \Big(\dfrac{10}{8}\Big) \\[1em] \Rightarrow (3)^4 + 2^4 + \dfrac{5}{4} \\[1em] \Rightarrow 81 + 16 + \dfrac{5}{4} \\[1em] \Rightarrow 97 + \dfrac{5}{4} \\[1em] \Rightarrow \dfrac{388 + 5}{4} \\[1em] \Rightarrow \dfrac{393}{4} \\[1em] \Rightarrow 98.25 ⇒ ( 27 ) 3 4 + ( 32 ) 0.8 + ( 0.8 ) − 1 ⇒ ( 3 3 ) 3 4 + ( 2 5 ) 10 8 + ( 0.8 1 ) 1 ⇒ ( 3 ) 3 × 3 4 + ( 2 ) 5 × 10 8 + ( 8 10 ) ⇒ ( 3 ) 4 + 2 4 + 4 5 ⇒ 81 + 16 + 4 5 ⇒ 97 + 4 5 ⇒ 4 388 + 5 ⇒ 4 393 ⇒ 98.25
Hence, ( 27 ) 4 3 + ( 32 ) 0.8 + ( 0.8 ) − 1 (27)^{\dfrac{4}{3}} + (32)^{0.8} + (0.8)^{-1} ( 27 ) 3 4 + ( 32 ) 0.8 + ( 0.8 ) − 1
Simplify the following:
( 32 − 5 ) 1 3 ( 32 + 5 ) 1 3 (\sqrt{32} - \sqrt{5})^{\dfrac{1}{3}}(\sqrt{32} + \sqrt{5})^{\dfrac{1}{3}} ( 32 − 5 ) 3 1 ( 32 + 5 ) 3 1
Answer
Given,
⇒ ( 32 − 5 ) 1 3 ( 32 + 5 ) 1 3 ⇒ [ ( 32 − 5 ) ( 32 + 5 ) ] 1 3 \Rightarrow (\sqrt{32} - \sqrt{5})^{\dfrac{1}{3}}(\sqrt{32} + \sqrt{5})^{\dfrac{1}{3}} \\[1em] \Rightarrow [(\sqrt{32} - \sqrt{5})(\sqrt{32} + \sqrt{5})]^{\dfrac{1}{3}} ⇒ ( 32 − 5 ) 3 1 ( 32 + 5 ) 3 1 ⇒ [( 32 − 5 ) ( 32 + 5 ) ] 3 1
As (a - b)(a + b) = a2 - b2 we get,
⇒ [ ( 32 ) 2 − ( 5 2 ) ] 1 3 ⇒ [ 32 − 5 ] 1 3 ⇒ ( 27 ) 1 3 ⇒ ( 3 3 ) 1 3 ⇒ 3. \Rightarrow [(\sqrt{32})^2 - (\sqrt{5}^2)]^{\dfrac{1}{3}} \\[1em] \Rightarrow [32 - 5]^{\dfrac{1}{3}} \\[1em] \Rightarrow (27)^{\dfrac{1}{3}} \\[1em] \Rightarrow (3^3)^{\dfrac{1}{3}} \\[1em] \Rightarrow 3. ⇒ [( 32 ) 2 − ( 5 2 ) ] 3 1 ⇒ [ 32 − 5 ] 3 1 ⇒ ( 27 ) 3 1 ⇒ ( 3 3 ) 3 1 ⇒ 3.
Hence, ( 32 − 5 ) 1 3 ( 32 + 5 ) 1 3 (\sqrt{32} - \sqrt{5})^{\dfrac{1}{3}}(\sqrt{32} + \sqrt{5})^{\dfrac{1}{3}} ( 32 − 5 ) 3 1 ( 32 + 5 ) 3 1
Simplify the following:
( x 1 3 − x − 1 3 ) ( x 2 3 + 1 + x − 2 3 ) \Big(x^{\dfrac{1}{3}} - x^{-\dfrac{1}{3}}\Big)\Big(x^{\dfrac{2}{3}} + 1 + x^{-\dfrac{2}{3}}\Big) ( x 3 1 − x − 3 1 ) ( x 3 2 + 1 + x − 3 2 )
Answer
The above equation can be written as,
⇒ ( x 1 3 − x − 1 3 ) [ ( x 1 3 ) 2 + x 1 3 . x − 1 3 + ( x − 1 3 ) 2 ) \Rightarrow \Big(x^{\dfrac{1}{3}} - x^{-\dfrac{1}{3}}\Big)\Big[\Big(x^{\dfrac{1}{3}}\Big)^2 + x^{\dfrac{1}{3}}.x^{-\dfrac{1}{3}} + \Big(x^{-\dfrac{1}{3}}\Big)^2\Big) ⇒ ( x 3 1 − x − 3 1 ) [ ( x 3 1 ) 2 + x 3 1 . x − 3 1 + ( x − 3 1 ) 2 )
We know that,
a3 - b3 = (a - b)(a2 + ab + b2 )
∴ ( x 1 3 − x − 1 3 ) [ ( x 1 3 ) 2 + x 1 3 . x − 1 3 + ( x − 1 3 ) 2 ] = ( x 1 3 ) 3 − ( x − 1 3 ) 3 ⇒ ( x 3 × 1 3 ) − ( x 3 × − 1 3 ) ⇒ x 1 − x − 1 ⇒ x − 1 x . \therefore \Big(x^{\dfrac{1}{3}} - x^{-\dfrac{1}{3}}\Big)\Big[\Big(x^{\dfrac{1}{3}}\Big)^2 + x^{\dfrac{1}{3}}.x^{-\dfrac{1}{3}} + \Big(x^{-\dfrac{1}{3}}\Big)^2\Big] = \Big(x^{\dfrac{1}{3}}\Big)^3 - \Big(x^{-\dfrac{1}{3}}\Big)^3 \\[1em] \Rightarrow \Big(x^{3 \times \dfrac{1}{3}}\Big) - \Big(x^{3 \times -\dfrac{1}{3}}\Big) \\[1em] \Rightarrow x^1 - x^{-1} \\[1em] \Rightarrow x - \dfrac{1}{x}. ∴ ( x 3 1 − x − 3 1 ) [ ( x 3 1 ) 2 + x 3 1 . x − 3 1 + ( x − 3 1 ) 2 ] = ( x 3 1 ) 3 − ( x − 3 1 ) 3 ⇒ ( x 3 × 3 1 ) − ( x 3 ×− 3 1 ) ⇒ x 1 − x − 1 ⇒ x − x 1 .
Hence, ( x 1 3 − x − 1 3 ) ( x 2 3 + 1 + x − 2 3 ) = x − 1 x \Big(x^{\dfrac{1}{3}} - x^{-\dfrac{1}{3}}\Big)\Big(x^{\dfrac{2}{3}} + 1 + x^{-\dfrac{2}{3}}\Big) = x - \dfrac{1}{x} ( x 3 1 − x − 3 1 ) ( x 3 2 + 1 + x − 3 2 ) = x − x 1
Simplify the following:
( x m x n ) l . ( x n x l ) m . ( x l x m ) n \Big(\dfrac{x^m}{x^n}\Big)^l.\Big(\dfrac{x^n}{x^l}\Big)^m.\Big(\dfrac{x^l}{x^m}\Big)^n ( x n x m ) l . ( x l x n ) m . ( x m x l ) n
Answer
Given,
⇒ ( x m x n ) l . ( x n x l ) m . ( x l x m ) n = ( x m . x − n ) l . ( x n . x − l ) m . ( x l . x − m ) n = ( x m − n ) l . ( x n − l ) m . ( x l − m ) n = x m l − n l . x n m − l m . x l n − m n = x m l − n l + n m − l m + l n − m n = x 0 = 1. \Rightarrow \Big(\dfrac{x^m}{x^n}\Big)^l.\Big(\dfrac{x^n}{x^l}\Big)^m.\Big(\dfrac{x^l}{x^m}\Big)^n = (x^m.x^{-n})^l.(x^n.x^{-l})^m.(x^l.x^{-m})^n \\[1em] = (x^{m - n})^l.(x^{n - l})^m.(x^{l - m})^n \\[1em] = x^{ml - nl}.x^{nm - lm}.x^{ln - mn} \\[1em] = x^{ml - nl + nm - lm + ln - mn} \\[1em] = x^0 = 1. ⇒ ( x n x m ) l . ( x l x n ) m . ( x m x l ) n = ( x m . x − n ) l . ( x n . x − l ) m . ( x l . x − m ) n = ( x m − n ) l . ( x n − l ) m . ( x l − m ) n = x m l − n l . x nm − l m . x l n − mn = x m l − n l + nm − l m + l n − mn = x 0 = 1.
Hence, ( x m x n ) l . ( x n x l ) m . ( x l x m ) n \Big(\dfrac{x^m}{x^n}\Big)^l.\Big(\dfrac{x^n}{x^l}\Big)^m.\Big(\dfrac{x^l}{x^m}\Big)^n ( x n x m ) l . ( x l x n ) m . ( x m x l ) n .
Simplify the following:
( x a + b x c ) a − b . ( x b + c x a ) b − c . ( x c + a x b ) c − a \Big(\dfrac{x^{a + b}}{x^c}\Big)^{a - b}.\Big(\dfrac{x^{b + c}}{x^a}\Big)^{b - c}.\Big(\dfrac{x^{c + a}}{x^b}\Big)^{c - a} ( x c x a + b ) a − b . ( x a x b + c ) b − c . ( x b x c + a ) c − a
Answer
Given,
⇒ ( x a + b x c ) a − b . ( x b + c x a ) b − c . ( x c + a x b ) c − a = ( x a + b − c ) a − b . ( x b + c − a ) b − c . ( x c + a − b ) c − a = x ( a 2 − a b + b a − b 2 − c a + c b ) . x ( b 2 − b c + c b − c 2 − a b + a c ) . x c 2 − c a + a c − a 2 − b c + b a = x a 2 − a b + b a − b 2 − c a + c b + b 2 − b c + c b − c 2 − a b + a c + c 2 − c a + a c − a 2 − b c + b a = x a 2 − a 2 − a b + b a − b 2 + b 2 − c a + a c + c b − b c + c b − b c − a b + b a − c a + a c − c 2 + c 2 = x 0 = 1. \Rightarrow \Big(\dfrac{x^{a + b}}{x^c}\Big)^{a - b}.\Big(\dfrac{x^{b + c}}{x^a}\Big)^{b - c}.\Big(\dfrac{x^{c + a}}{x^b}\Big)^{c - a} \\[1em] = (x^{a + b - c})^{a - b}.(x^{b + c - a})^{b - c}.(x^{c + a - b})^{c - a} \\[1em] = x^{(a^2 - ab + ba - b^2 - ca + cb)}.x^{(b^2 - bc + cb - c^2 - ab + ac)}.x^{c^2 - ca + ac - a^2 - bc + ba} \\[1em] = x^{a^2 - ab + ba - b^2 - ca + cb + b^2 - bc + cb - c^2 - ab + ac + c^2 - ca + ac - a^2 - bc + ba} \\[1em] = x^{a^2 - a^2 - ab + ba - b^2 + b^2 - ca + ac + cb - bc + cb - bc - ab + ba - ca + ac - c^2 + c^2} \\[1em] = x^0 = 1. ⇒ ( x c x a + b ) a − b . ( x a x b + c ) b − c . ( x b x c + a ) c − a = ( x a + b − c ) a − b . ( x b + c − a ) b − c . ( x c + a − b ) c − a = x ( a 2 − ab + ba − b 2 − c a + c b ) . x ( b 2 − b c + c b − c 2 − ab + a c ) . x c 2 − c a + a c − a 2 − b c + ba = x a 2 − ab + ba − b 2 − c a + c b + b 2 − b c + c b − c 2 − ab + a c + c 2 − c a + a c − a 2 − b c + ba = x a 2 − a 2 − ab + ba − b 2 + b 2 − c a + a c + c b − b c + c b − b c − ab + ba − c a + a c − c 2 + c 2 = x 0 = 1.
Hence, ( x a + b x c ) a − b . ( x b + c x a ) b − c . ( x c + a x b ) c − a \Big(\dfrac{x^{a + b}}{x^c}\Big)^{a - b}.\Big(\dfrac{x^{b + c}}{x^a}\Big)^{b - c}.\Big(\dfrac{x^{c + a}}{x^b}\Big)^{c - a} ( x c x a + b ) a − b . ( x a x b + c ) b − c . ( x b x c + a ) c − a
Simplify the following:
x l x m l m . x m x n m n . x n x l n l \sqrt[lm]{\dfrac{x^l}{x^m}}.\sqrt[mn]{\dfrac{x^m}{x^n}}.\sqrt[nl]{\dfrac{x^n}{x^l}} l m x m x l . mn x n x m . n l x l x n
Answer
Given,
⇒ x l x m l m . x m x n m n . x n x l n l = x l − m l m . x m − n m n . x n − l n l = ( x ) l − m l m . ( x ) m − n m n . ( x ) n − l n l = ( x ) l − m l m + m − n m n + n − l n l = ( x ) ( l − m ) n + ( m − n ) l + ( n − l ) m l m n = ( x ) l n − m n + m l − n l + n m − l m l m n = ( x ) 0 = 1. \Rightarrow \sqrt[lm]{\dfrac{x^l}{x^m}}.\sqrt[mn]{\dfrac{x^m}{x^n}}.\sqrt[nl]{\dfrac{x^n}{x^l}} \\[1em] = \sqrt[lm]{x^{l - m}}.\sqrt[mn]{x^{m - n}}.\sqrt[nl]{x^{n - l}} \\[1em] = (x)^{\dfrac{l - m}{lm}}.(x )^{\dfrac{m - n}{mn}}.(x)^{\dfrac{n - l}{nl}} \\[1em] = (x)^{\dfrac{l - m}{lm} + \dfrac{m - n}{mn} + \dfrac{n - l}{nl}} \\[1em] = (x)^{\dfrac{(l - m)n + (m - n)l + (n - l)m}{lmn}} \\[1em] = (x)^{\dfrac{ln - mn + ml - nl + nm - lm}{lmn}} \\[1em] = (x)^0 \\[1em] = 1. ⇒ l m x m x l . mn x n x m . n l x l x n = l m x l − m . mn x m − n . n l x n − l = ( x ) l m l − m . ( x ) mn m − n . ( x ) n l n − l = ( x ) l m l − m + mn m − n + n l n − l = ( x ) l mn ( l − m ) n + ( m − n ) l + ( n − l ) m = ( x ) l mn l n − mn + m l − n l + nm − l m = ( x ) 0 = 1.
Hence, x l x m l m . x m x n m n . x n x l n l \sqrt[lm]{\dfrac{x^l}{x^m}}.\sqrt[mn]{\dfrac{x^m}{x^n}}.\sqrt[nl]{\dfrac{x^n}{x^l}} l m x m x l . mn x n x m . n l x l x n
Simplify the following:
( x a x b ) a 2 + a b + b 2 . ( x b x c ) b 2 + b c + c 2 . ( x c x a ) c 2 + a c + a 2 \Big(\dfrac{x^a}{x^b}\Big)^{a^2 + ab + b^2}.\Big(\dfrac{x^b}{x^c}\Big)^{b^2 + bc + c^2}.\Big(\dfrac{x^c}{x^a}\Big)^{c^2 + ac + a^2} ( x b x a ) a 2 + ab + b 2 . ( x c x b ) b 2 + b c + c 2 . ( x a x c ) c 2 + a c + a 2
Answer
Given,
⇒ ( x a x b ) a 2 + a b + b 2 . ( x b x c ) b 2 + b c + c 2 . ( x c x a ) c 2 + a c + a 2 ⇒ ( x ( a − b ) ) ( a 2 + a b + b 2 ) . ( x ( b − c ) ) ( b 2 + b c + c 2 ) . ( x ( c − a ) ) ( c 2 + c a + a 2 ) \Rightarrow \Big(\dfrac{x^a}{x^b}\Big)^{a^2 + ab + b^2}.\Big(\dfrac{x^b}{x^c}\Big)^{b^2 + bc + c^2}.\Big(\dfrac{x^c}{x^a}\Big)^{c^2 + ac + a^2} \\[1em] \Rightarrow (x^{(a - b)})^{(a^2 + ab + b^2)}.(x^{(b - c)})^{(b^2 + bc + c^2)}.(x^{(c - a)})^{(c^2 + ca + a^2)} ⇒ ( x b x a ) a 2 + ab + b 2 . ( x c x b ) b 2 + b c + c 2 . ( x a x c ) c 2 + a c + a 2 ⇒ ( x ( a − b ) ) ( a 2 + ab + b 2 ) . ( x ( b − c ) ) ( b 2 + b c + c 2 ) . ( x ( c − a ) ) ( c 2 + c a + a 2 )
By formula,
(x3 - y3 ) = (x - y)(x2 + y2 + xy) we get,
⇒ ( x ) a 3 − b 3 . ( x ) b 3 − c 3 . ( x ) c 3 − a 3 ⇒ x a 3 − b 3 + b 3 − c 3 + c 3 − a 3 ⇒ x 0 = 1. \Rightarrow (x)^{a^3 - b^3}.(x)^{b^3 - c^3}.(x)^{c^3 - a^3} \\[1em] \Rightarrow x^{a^3 - b^3 + b^3 - c^3 + c^3 - a^3} \\[1em] \Rightarrow x^{0} = 1. ⇒ ( x ) a 3 − b 3 . ( x ) b 3 − c 3 . ( x ) c 3 − a 3 ⇒ x a 3 − b 3 + b 3 − c 3 + c 3 − a 3 ⇒ x 0 = 1.
Hence, ( x a x b ) a 2 + a b + b 2 . ( x b x c ) b 2 + b c + c 2 . ( x c x a ) c 2 + a c + a 2 \Big(\dfrac{x^a}{x^b}\Big)^{a^2 + ab + b^2}.\Big(\dfrac{x^b}{x^c}\Big)^{b^2 + bc + c^2}.\Big(\dfrac{x^c}{x^a}\Big)^{c^2 + ac + a^2} ( x b x a ) a 2 + ab + b 2 . ( x c x b ) b 2 + b c + c 2 . ( x a x c ) c 2 + a c + a 2
Simplify the following:
( x a x − b ) a 2 − a b + b 2 . ( x b x − c ) b 2 − b c + c 2 . ( x c x − a ) c 2 − c a + a 2 \Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2}.\Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2}.\Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2} ( x − b x a ) a 2 − ab + b 2 . ( x − c x b ) b 2 − b c + c 2 . ( x − a x c ) c 2 − c a + a 2
Answer
Given,
⇒ ( x a x − b ) a 2 − a b + b 2 . ( x b x − c ) b 2 − b c + c 2 . ( x c x − a ) c 2 − c a + a 2 ⇒ ( x ( a − ( − b ) ) ) ( a 2 − a b + b 2 ) . ( x ( b − ( − c ) ) ) ( b 2 − b c + c 2 ) . ( x ( c − ( − a ) ) ) ( c 2 − c a + a 2 ) ⇒ ( x ( a + b ) ) ( a 2 − a b + b 2 ) . ( x ( b + c ) ) ( b 2 − b c + c 2 ) . ( x ( c + a ) ) ( c 2 − c a + a 2 ) \Rightarrow \Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2}.\Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2}.\Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2} \\[1em] \Rightarrow (x^{(a - (-b))})^{(a^2 - ab + b^2)}.(x^{(b - (-c))})^{(b^2 - bc + c^2)}.(x^{(c - (-a))})^{(c^2 - ca + a^2)} \\[1em] \Rightarrow (x^{(a + b)})^{(a^2 - ab + b^2)}.(x^{(b + c)})^{(b^2 - bc + c^2)}.(x^{(c + a)})^{(c^2 - ca + a^2)} ⇒ ( x − b x a ) a 2 − ab + b 2 . ( x − c x b ) b 2 − b c + c 2 . ( x − a x c ) c 2 − c a + a 2 ⇒ ( x ( a − ( − b )) ) ( a 2 − ab + b 2 ) . ( x ( b − ( − c )) ) ( b 2 − b c + c 2 ) . ( x ( c − ( − a )) ) ( c 2 − c a + a 2 ) ⇒ ( x ( a + b ) ) ( a 2 − ab + b 2 ) . ( x ( b + c ) ) ( b 2 − b c + c 2 ) . ( x ( c + a ) ) ( c 2 − c a + a 2 )
By formula,
(x3 + y3 ) = (x + y)(x2 + y2 - xy) we get,
⇒ ( x ) a 3 + b 3 . ( x ) b 3 + c 3 . ( x ) c 3 + a 3 ⇒ x a 3 + b 3 + b 3 + c 3 + c 3 + a 3 ⇒ x 2 ( a 3 + b 3 + c 3 ) . \Rightarrow (x)^{a^3 + b^3}.(x)^{b^3 + c^3}.(x)^{c^3 + a^3} \\[1em] \Rightarrow x^{a^3 + b^3 + b^3 + c^3 + c^3 + a^3} \\[1em] \Rightarrow x^{2(a^3 + b^3 + c^3)}. ⇒ ( x ) a 3 + b 3 . ( x ) b 3 + c 3 . ( x ) c 3 + a 3 ⇒ x a 3 + b 3 + b 3 + c 3 + c 3 + a 3 ⇒ x 2 ( a 3 + b 3 + c 3 ) .
Hence, ( x a x − b ) a 2 − a b + b 2 . ( x b x − c ) b 2 − b c + c 2 . ( x c x − a ) c 2 − c a + a 2 = x 2 ( a 3 + b 3 + c 3 ) \Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2}.\Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2}.\Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2} = x^{2(a^3 + b^3+ c^3)} ( x − b x a ) a 2 − ab + b 2 . ( x − c x b ) b 2 − b c + c 2 . ( x − a x c ) c 2 − c a + a 2 = x 2 ( a 3 + b 3 + c 3 )
Simplify the following:
(a-1 + b-1 ) ÷ (a-2 - b-2 )
Answer
Given,
⇒ ( a − 1 + b − 1 ) ÷ ( a − 2 − b − 2 ) ⇒ ( 1 a + 1 b ) ÷ ( 1 a 2 − 1 b 2 ) ⇒ ( b + a a b ) ÷ ( b 2 − a 2 a 2 b 2 ) ⇒ ( b + a a b ) × ( a 2 b 2 b 2 − a 2 ) ⇒ a b ( b + a ) b 2 − a 2 ⇒ a b ( b + a ) ( b − a ) ( b + a ) ⇒ a b b − a . \Rightarrow (a^{-1} + b^{-1}) ÷ (a^{-2} - b^{-2}) \\[1em] \Rightarrow \Big(\dfrac{1}{a} + \dfrac{1}{b}\Big) ÷ \Big(\dfrac{1}{a^2} - \dfrac{1}{b^2}\Big) \\[1em] \Rightarrow \Big(\dfrac{b + a}{ab}\Big) ÷ \Big(\dfrac{b^2 - a^2}{a^2b^2}\Big) \\[1em] \Rightarrow \Big(\dfrac{b + a}{ab}\Big) \times \Big(\dfrac{a^2b^2}{b^2 - a^2}\Big) \\[1em] \Rightarrow \dfrac{ab(b + a)}{b^2 - a^2} \\[1em] \Rightarrow \dfrac{ab(b + a)}{(b - a)(b + a)} \\[1em] \Rightarrow \dfrac{ab}{b - a}. ⇒ ( a − 1 + b − 1 ) ÷ ( a − 2 − b − 2 ) ⇒ ( a 1 + b 1 ) ÷ ( a 2 1 − b 2 1 ) ⇒ ( ab b + a ) ÷ ( a 2 b 2 b 2 − a 2 ) ⇒ ( ab b + a ) × ( b 2 − a 2 a 2 b 2 ) ⇒ b 2 − a 2 ab ( b + a ) ⇒ ( b − a ) ( b + a ) ab ( b + a ) ⇒ b − a ab .
Hence, (a-1 + b-1 ) ÷ (a-2 - b-2 ) = a b b − a . \dfrac{ab}{b - a}. b − a ab .
Simplify the following:
1 1 + a m − n + 1 1 + a n − m \dfrac{1}{1 + a^{m - n}} + \dfrac{1}{1 + a^{n - m}} 1 + a m − n 1 + 1 + a n − m 1
Answer
Given,
⇒ 1 1 + a m − n + 1 1 + a n − m ⇒ 1 1 + a m . a − n + 1 1 + a n . a − m ⇒ 1 1 + a m a n + 1 1 + a n a m ⇒ 1 a n + a m a n + 1 a m + a n a m ⇒ a n a n + a m + a m a m + a n ⇒ a n + a m a n + a m ⇒ 1. \Rightarrow \dfrac{1}{1 + a^{m - n}} + \dfrac{1}{1 + a^{n - m}} \\[1em] \Rightarrow \dfrac{1}{1 + a^m.a^{-n}} + \dfrac{1}{1 + a^n.a^{-m}} \\[1em] \Rightarrow \dfrac{1}{1 + \dfrac{a^m}{a^n}} + \dfrac{1}{1 + \dfrac{a^n}{a^m}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{a^n + a^m}{a^n}} + \dfrac{1}{\dfrac{a^m + a^n}{a^m}} \\[1em] \Rightarrow \dfrac{a^n}{a^n + a^m} + \dfrac{a^m}{a^m + a^n} \\[1em] \Rightarrow \dfrac{a^n + a^m}{a^n + a^m} \\[1em] \Rightarrow 1. ⇒ 1 + a m − n 1 + 1 + a n − m 1 ⇒ 1 + a m . a − n 1 + 1 + a n . a − m 1 ⇒ 1 + a n a m 1 + 1 + a m a n 1 ⇒ a n a n + a m 1 + a m a m + a n 1 ⇒ a n + a m a n + a m + a n a m ⇒ a n + a m a n + a m ⇒ 1.
Hence, 1 1 + a m − n + 1 1 + a n − m \dfrac{1}{1 + a^{m - n}} + \dfrac{1}{1 + a^{n - m}} 1 + a m − n 1 + 1 + a n − m 1
Prove the following:
(a + b)-1 (a-1 + b-1 ) = 1 a b \dfrac{1}{ab} ab 1
Answer
Given,
⇒ ( a + b ) − 1 ( a − 1 + b − 1 ) = 1 a b \Rightarrow (a + b)^{-1}(a^{-1} + b^{-1}) = \dfrac{1}{ab} ⇒ ( a + b ) − 1 ( a − 1 + b − 1 ) = ab 1
Solving L.H.S. of above equation,
⇒ 1 a + b × ( 1 a + 1 b ) ⇒ 1 a + b × ( b + a a b ) ⇒ 1 a b . \Rightarrow \dfrac{1}{a + b} \times \Big(\dfrac{1}{a} + \dfrac{1}{b}\Big) \\[1em] \Rightarrow \dfrac{1}{a + b} \times \Big(\dfrac{b + a}{ab}\Big) \\[1em] \Rightarrow \dfrac{1}{ab}. ⇒ a + b 1 × ( a 1 + b 1 ) ⇒ a + b 1 × ( ab b + a ) ⇒ ab 1 .
Since, L.H.S. = R.H.S.
Hence, proved that (a + b)-1 (a-1 + b-1 ) = 1 a b \dfrac{1}{ab} ab 1
Prove the following:
x + y + z x − 1 y − 1 + y − 1 z − 1 + z − 1 x − 1 = x y z \dfrac{x + y + z}{x^{-1}y^{-1} + y^{-1}z^{-1} + z^{-1}x^{-1}} = xyz x − 1 y − 1 + y − 1 z − 1 + z − 1 x − 1 x + y + z = x yz
Answer
Given,
x + y + z x − 1 y − 1 + y − 1 z − 1 + z − 1 x − 1 = x y z \dfrac{x + y + z}{x^{-1}y^{-1} + y^{-1}z^{-1} + z^{-1}x^{-1}} = xyz x − 1 y − 1 + y − 1 z − 1 + z − 1 x − 1 x + y + z = x yz
Solving L.H.S. of above equation,
⇒ x + y + z 1 x y + 1 y z + 1 z x ⇒ x + y + z z + x + y x y z ⇒ x y z ( x + y + z ) x + y + z ⇒ x y z . \Rightarrow \dfrac{x + y + z}{\dfrac{1}{xy} + \dfrac{1}{yz} + \dfrac{1}{zx}} \\[1em] \Rightarrow \dfrac{x + y + z}{\dfrac{z + x + y}{xyz}} \\[1em] \Rightarrow \dfrac{xyz(x + y + z)}{x + y + z} \\[1em] \Rightarrow xyz. ⇒ x y 1 + yz 1 + z x 1 x + y + z ⇒ x yz z + x + y x + y + z ⇒ x + y + z x yz ( x + y + z ) ⇒ x yz .
Hence, proved that x + y + z x − 1 y − 1 + y − 1 z − 1 + z − 1 x − 1 \dfrac{x + y + z}{x^{-1}y^{-1} + y^{-1}z^{-1} + z^{-1}x^{-1}} x − 1 y − 1 + y − 1 z − 1 + z − 1 x − 1 x + y + z
If a = cz , b = ax and c = by , prove that xyz = 1.
Answer
Given,
a = cz , b = ax and c = by
Substituting value of c in a = cz we get,
⇒ a = (by )z = byz
Substituting value of b in above equation we get,
⇒ a = (ax )yz xyz
∴ xyz = 1.
Hence, proved that xyz = 1.
If a = xyp - 1 , b = xyq - 1 and c = xyr - 1 , prove that
aq - r .br - p .cp - q = 1.
Answer
Given,
a = xyp - 1 , b = xyq - 1 and c = xyr - 1 .
Substituting value of a, b and c in L.H.S. of aq - r .br - p .cp - q = 1 we get,
⇒ (xyp - 1 )q - r .(xyq - 1 )r - p .(xyr - 1 )p - q
⇒ (xy)(p - 1)(q - r) .(xy)(q - 1)(r - p) .(xy)(r - 1)(p - q)
⇒ (xy)(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)
⇒ (xy)pq - pr - q + r + qr - qp - r + p + rp - rq - p + q
⇒ (xy)p - p - q + q + r - r + pq - qp - pr + rp + qr - rq
⇒ (xy)0
⇒ 1.
Hence,prove that aq - r .br - p .cp - q = 1.
If 2x = 3y = 6-z , prove that 1 x + 1 y + 1 z = 0. \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0. x 1 + y 1 + z 1 = 0.
Answer
Let 2x = 3y = 6-z = k.
∴ 2 x = k or 2 = k 1 x \therefore 2^x = k \text{ or } 2 = k^{\dfrac{1}{x}} ∴ 2 x = k or 2 = k x 1
∴ 3 y = k or 3 = k 1 y \therefore 3^y = k \text{ or } 3 = k^{\dfrac{1}{y}} ∴ 3 y = k or 3 = k y 1
∴ 6 − z = k or 6 = k − 1 z \therefore 6^{-z} = k \text{ or } 6 = k^{-\dfrac{1}{z}} ∴ 6 − z = k or 6 = k − z 1
We know that,
2 × 3 = 6
Substituting value of 2, 3 and 6 from (i), (ii) and (iii) in above equation we get,
⇒ k 1 x × k 1 y = k − 1 z ⇒ k 1 x + 1 y = k − 1 z ∴ 1 x + 1 y = − 1 z ⇒ 1 x + 1 y + 1 z = 0. \Rightarrow k^{\dfrac{1}{x}} \times k^{\dfrac{1}{y}} = k^{-\dfrac{1}{z}} \\[1em] \Rightarrow k^{\dfrac{1}{x} + \dfrac{1}{y}} = k^{-\dfrac{1}{z}} \\[1em] \therefore \dfrac{1}{x} + \dfrac{1}{y} = -\dfrac{1}{z} \\[1em] \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0. ⇒ k x 1 × k y 1 = k − z 1 ⇒ k x 1 + y 1 = k − z 1 ∴ x 1 + y 1 = − z 1 ⇒ x 1 + y 1 + z 1 = 0.
Hence, proved that 1 x + 1 y + 1 z = 0 \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0 x 1 + y 1 + z 1 = 0
If 2x = 3y = 12z , prove that x = 2 y z y − z . \dfrac{2yz}{y - z}. y − z 2 yz .
Answer
Let 2x = 3y = 12z = k.
∴ 2 x = k or 2 = k 1 x \therefore 2^x = k \text{ or } 2 = k^{\dfrac{1}{x}} ∴ 2 x = k or 2 = k x 1
∴ 3 y = k or 3 = k 1 y \therefore 3^y = k \text{ or } 3 = k^{\dfrac{1}{y}} ∴ 3 y = k or 3 = k y 1
∴ 12 z = k or 12 = k 1 z \therefore 12^{z} = k \text{ or } 12 = k^{\dfrac{1}{z}} ∴ 1 2 z = k or 12 = k z 1
We know that,
22 × 3 = 12
Substituting value of 2, 3 and 12 from (i), (ii) and (iii) in above equation we get,
⇒ ( k 1 x ) 2 × k 1 y = k 1 z ⇒ k 2 x × k 1 y = k 1 z ⇒ k 2 x + 1 y = k 1 z ∴ 2 x + 1 y = 1 z ⇒ 2 x = 1 z − 1 y ⇒ 2 x = y − z y z ⇒ x 2 = y z y − z ⇒ x = 2 y z y − z . \Rightarrow (k^{\dfrac{1}{x}})^2 \times k^{\dfrac{1}{y}} = k^{\dfrac{1}{z}} \\[1em] \Rightarrow k^{\dfrac{2}{x}} \times k^{\dfrac{1}{y}} = k^{\dfrac{1}{z}} \\[1em] \Rightarrow k^{\dfrac{2}{x} + \dfrac{1}{y}} = k^{\dfrac{1}{z}} \\[1em] \therefore \dfrac{2}{x} + \dfrac{1}{y} = \dfrac{1}{z} \\[1em] \Rightarrow \dfrac{2}{x} = \dfrac{1}{z} - \dfrac{1}{y} \\[1em] \Rightarrow \dfrac{2}{x} = \dfrac{y - z}{yz} \\[1em] \Rightarrow \dfrac{x}{2} = \dfrac{yz}{y - z} \\[1em] \Rightarrow x = \dfrac{2yz}{y - z}. ⇒ ( k x 1 ) 2 × k y 1 = k z 1 ⇒ k x 2 × k y 1 = k z 1 ⇒ k x 2 + y 1 = k z 1 ∴ x 2 + y 1 = z 1 ⇒ x 2 = z 1 − y 1 ⇒ x 2 = yz y − z ⇒ 2 x = y − z yz ⇒ x = y − z 2 yz .
Hence, proved that x = 2 y z y − z . x = \dfrac{2yz}{y - z}. x = y − z 2 yz .
Simplify and express with positive exponents:
( 3 x 2 ) 0 , ( x y ) − 2 , ( − 27 a 9 ) 2 3 (3x^2)^0, (xy)^{-2}, (-27a^9)^{\dfrac{2}{3}} ( 3 x 2 ) 0 , ( x y ) − 2 , ( − 27 a 9 ) 3 2
Answer
(3x2 )0 = 1,
( x y ) − 2 = ( 1 x y ) 2 = 1 x 2 y 2 (xy)^{-2} = \Big(\dfrac{1}{xy}\Big)^2 = \dfrac{1}{x^2y^2} ( x y ) − 2 = ( x y 1 ) 2 = x 2 y 2 1
( − 27 a 9 ) 2 3 = [ ( − 3 a 3 ) 3 ] 2 3 = ( − 3 a 3 ) 3 × 2 3 = ( − 3 a 3 ) 2 = 9 a 6 . (-27a^9)^{\dfrac{2}{3}} = [(-3a^3)^3]^{\dfrac{2}{3}} \\[1em] = (-3a^3)^{3 \times \dfrac{2}{3}} \\[1em] = (-3a^3)^2 \\[1em] = 9a^6. ( − 27 a 9 ) 3 2 = [( − 3 a 3 ) 3 ] 3 2 = ( − 3 a 3 ) 3 × 3 2 = ( − 3 a 3 ) 2 = 9 a 6 .
If a = 3 and b = -2, find the values of:
(i) aa + bb
(ii) ab + ba
Answer
(i) Given, a = 3 and b = -2.
Substituting values of a and b in aa + bb we get,
= 33 + (-2)-2
= 27 + ( − 1 2 ) 2 \Big(-\dfrac{1}{2}\Big)^2 ( − 2 1 ) 2
= 27 + 1 4 = 27 1 4 \dfrac{1}{4} = 27\dfrac{1}{4} 4 1 = 27 4 1
Hence, a a + b b = 27 1 4 . a^a + b^b = 27\dfrac{1}{4}. a a + b b = 27 4 1 .
(ii) Given, a = 3 and b = -2.
Substituting values of a and b in ab + ba we get,
= ( 3 ) − 2 + ( − 2 ) 3 = ( 1 3 ) 2 + ( − 8 ) = 1 9 − 8 = 1 − 72 9 = − 71 9 = − 7 8 9 . = (3)^{-2} + (-2)^3 \\[1em] = \Big(\dfrac{1}{3}\Big)^2 + (-8) \\[1em] = \dfrac{1}{9} - 8 \\[1em] = \dfrac{1 - 72}{9} \\[1em] = -\dfrac{71}{9} = -7\dfrac{8}{9}. = ( 3 ) − 2 + ( − 2 ) 3 = ( 3 1 ) 2 + ( − 8 ) = 9 1 − 8 = 9 1 − 72 = − 9 71 = − 7 9 8 .
Hence, a a + b b = − 7 8 9 . a^a + b^b = -7\dfrac{8}{9}. a a + b b = − 7 9 8 .
If x = 103 × 0.0099, y = 10-2 × 110, find the value of x y \sqrt{\dfrac{x}{y}} y x
Answer
Substituting value of x and y in x y \sqrt{\dfrac{x}{y}} y x
⇒ x y = 10 3 × 0.0099 10 − 2 × 110 = 9.9 10 − 2 × 110 = 9.9 × 10 2 110 = 990 110 = 9 = 3. \Rightarrow \sqrt{\dfrac{x}{y}} = \sqrt{\dfrac{10^3 \times 0.0099}{10^{-2} \times 110}} \\[1em] = \sqrt{\dfrac{9.9}{10^{-2} \times 110}} \\[1em] = \sqrt{\dfrac{9.9 \times 10^2}{110}} \\[1em] = \sqrt{\dfrac{990}{110}} \\[1em] = \sqrt{9} = 3. ⇒ y x = 1 0 − 2 × 110 1 0 3 × 0.0099 = 1 0 − 2 × 110 9.9 = 110 9.9 × 1 0 2 = 110 990 = 9 = 3.
Hence, x y = 3. \sqrt{\dfrac{x}{y}} = 3. y x = 3.
Evaluate x 1 2 . y − 1 . z 2 3 x^{\dfrac{1}{2}}.y^{-1}.z^{\dfrac{2}{3}} x 2 1 . y − 1 . z 3 2
Answer
Substituting value of x, y and z in x 1 2 . y − 1 . z 2 3 x^{\dfrac{1}{2}}.y^{-1}.z^{\dfrac{2}{3}} x 2 1 . y − 1 . z 3 2
⇒ x 1 2 . y − 1 . z 2 3 = ( 9 ) 1 2 . ( 2 ) − 1 . ( 8 ) 2 3 = ( 3 2 ) 1 2 × 1 2 × ( 2 3 ) 2 3 = 3 × 1 2 × 2 2 = 12 2 = 6. \Rightarrow x^{\dfrac{1}{2}}.y^{-1}.z^{\dfrac{2}{3}} = (9)^{\dfrac{1}{2}}.(2)^{-1}.(8)^{\dfrac{2}{3}} \\[1em] = (3^2)^{\dfrac{1}{2}} \times \dfrac{1}{2} \times (2^3)^{\dfrac{2}{3}} \\[1em] = 3 \times \dfrac{1}{2} \times 2^2 \\[1em] = \dfrac{12}{2} = 6. ⇒ x 2 1 . y − 1 . z 3 2 = ( 9 ) 2 1 . ( 2 ) − 1 . ( 8 ) 3 2 = ( 3 2 ) 2 1 × 2 1 × ( 2 3 ) 3 2 = 3 × 2 1 × 2 2 = 2 12 = 6.
Hence, x 1 2 . y − 1 . z 2 3 x^{\dfrac{1}{2}}.y^{-1}.z^{\dfrac{2}{3}} x 2 1 . y − 1 . z 3 2
If x4 y2 z3 = 49392, find the values of x, y and z, where x, y and z are different positive primes.
Answer
On prime factorization of 49392 we get,
49392 = 24 32 73
∴ x4 y2 z3 = 24 32 73
On comparing we get,
x = 2, y = 3 and z = 7.
Hence, x = 2, y = 3 and z = 7.
If a 6 b − 4 3 = a x . b 2 y \sqrt[3]{a^6b^{-4}} = a^x.b^{2y} 3 a 6 b − 4 = a x . b 2 y
Answer
Given,
⇒ a 6 b − 4 3 = a x . b 2 y ⇒ ( a 6 b − 4 ) 1 3 = a x . b 2 y ⇒ a 6 3 . b − 4 3 = a x . b 2 y ⇒ a 2 . b − 4 3 = a x . b 2 y ∴ x = 2 and 2 y = − 4 3 ⇒ x = 2 and y = − 2 3 . \Rightarrow \sqrt[3]{a^6b^{-4}} = a^x.b^{2y} \\[1em] \Rightarrow (a^6b^{-4})^{\dfrac{1}{3}} = a^x.b^{2y} \\[1em] \Rightarrow a^{\dfrac{6}{3}}.b^{-\dfrac{4}{3}} = a^x.b^{2y} \\[1em] \Rightarrow a^2.b^{-\dfrac{4}{3}} = a^x.b^{2y} \\[1em] \therefore x = 2 \text{ and } 2y = -\dfrac{4}{3} \\[1em] \Rightarrow x = 2 \text{ and } y = -\dfrac{2}{3}. ⇒ 3 a 6 b − 4 = a x . b 2 y ⇒ ( a 6 b − 4 ) 3 1 = a x . b 2 y ⇒ a 3 6 . b − 3 4 = a x . b 2 y ⇒ a 2 . b − 3 4 = a x . b 2 y ∴ x = 2 and 2 y = − 3 4 ⇒ x = 2 and y = − 3 2 .
Hence, x = 2 and y = − 2 3 -\dfrac{2}{3} − 3 2
If (p + q)-1 (p-1 + q-1 ) = pa qb , prove that a + b +2 = 0, where p and q are different positive primes.
Answer
Given,
⇒ ( p + q ) − 1 ( p − 1 + q − 1 ) = p a q b ⇒ 1 ( p + q ) ( 1 p + 1 q ) = p a q b ⇒ 1 ( p + q ) × p + q p q = p a q b ⇒ 1 p q = p a q b ⇒ p − 1 q − 1 = p a q b ∴ a = − 1 and b = − 1. \Rightarrow (p + q)^{-1}(p^{-1} + q^{-1}) = p^aq^b \\[1em] \Rightarrow \dfrac{1}{(p + q)}\Big(\dfrac{1}{p} + \dfrac{1}{q}\Big) = p^aq^b \\[1em] \Rightarrow \dfrac{1}{(p + q)} \times \dfrac{p + q}{pq} = p^aq^b \\[1em] \Rightarrow \dfrac{1}{pq} = p^aq^b \\[1em] \Rightarrow p^{-1}q^{-1} = p^aq^b \\[1em] \therefore a = -1 \text{ and } b = -1. ⇒ ( p + q ) − 1 ( p − 1 + q − 1 ) = p a q b ⇒ ( p + q ) 1 ( p 1 + q 1 ) = p a q b ⇒ ( p + q ) 1 × pq p + q = p a q b ⇒ pq 1 = p a q b ⇒ p − 1 q − 1 = p a q b ∴ a = − 1 and b = − 1.
Substituting values of a and b in L.H.S. of a + b + 2 = 0 we get,
a + b + 2 = -1 + (-1) + 2 = -2 + 2 = 0.
Hence, proved that a + b + 2 = 0.
If ( p − 1 q 2 p 2 q − 4 ) 7 ÷ ( p 3 q − 5 p − 2 q 3 ) − 5 = p x q y \Big(\dfrac{p^{-1}q^{2}}{p^2q^{-4}}\Big)^7 ÷ \Big(\dfrac{p^3q^{-5}}{p^{-2}q^{3}}\Big)^{-5} = p^xq^y ( p 2 q − 4 p − 1 q 2 ) 7 ÷ ( p − 2 q 3 p 3 q − 5 ) − 5 = p x q y
Answer
Given,
⇒ ( p − 1 q 2 p 2 q − 4 ) 7 ÷ ( p 3 q − 5 p − 2 q 3 ) − 5 = p x q y ⇒ ( p − 1 q 2 p 2 q − 4 ) 7 ÷ ( p − 2 q 3 p 3 q − 5 ) 5 = p x q y ⇒ p − 7 q 14 p 14 q − 28 × ( p 3 q − 5 p − 2 q 3 ) 5 = p x q y ⇒ q 14 . q 28 p 14 . p 7 × p 15 q − 25 p − 10 q 15 = p x q y ⇒ q 14 + 28 p 14 + 7 × p 15 . p 10 q 25 . q 15 = p x q y ⇒ q 42 p 21 × p 25 q 40 = p x q y ⇒ p 4 q 2 = p x q y ∴ x = 4 and y = 2. \Rightarrow \Big(\dfrac{p^{-1}q^{2}}{p^2q^{-4}}\Big)^7 ÷ \Big(\dfrac{p^3q^{-5}}{p^{-2}q^{3}}\Big)^{-5} = p^xq^y \\[1em] \Rightarrow \Big(\dfrac{p^{-1}q^{2}}{p^2q^{-4}}\Big)^7 ÷ \Big(\dfrac{p^{-2}q^3}{p^3q^{-5}}\Big)^5 = p^xq^y \\[1em] \Rightarrow \dfrac{p^{-7}q^{14}}{p^{14}q^{-28}} \times \Big(\dfrac{p^3q^{-5}}{p^{-2}q^3}\Big)^5 = p^xq^y \\[1em] \Rightarrow \dfrac{q^{14}.q^{28}}{p^{14}.p^7} \times \dfrac{p^{15}q^{-25}}{p^{-10}q^{15}} = p^xq^y \\[1em] \Rightarrow \dfrac{q^{14 +28}}{p^{14 + 7}} \times \dfrac{p^{15}.p^{10}}{q^{25}.q^{15}} = p^xq^y \\[1em] \Rightarrow \dfrac{q^{42}}{p^{21}} \times \dfrac{p^{25}}{q^{40}} = p^xq^y \\[1em] \Rightarrow p^{4}q^{2} = p^xq^y \\[1em] \therefore x = 4 \text{ and } y = 2. ⇒ ( p 2 q − 4 p − 1 q 2 ) 7 ÷ ( p − 2 q 3 p 3 q − 5 ) − 5 = p x q y ⇒ ( p 2 q − 4 p − 1 q 2 ) 7 ÷ ( p 3 q − 5 p − 2 q 3 ) 5 = p x q y ⇒ p 14 q − 28 p − 7 q 14 × ( p − 2 q 3 p 3 q − 5 ) 5 = p x q y ⇒ p 14 . p 7 q 14 . q 28 × p − 10 q 15 p 15 q − 25 = p x q y ⇒ p 14 + 7 q 14 + 28 × q 25 . q 15 p 15 . p 10 = p x q y ⇒ p 21 q 42 × q 40 p 25 = p x q y ⇒ p 4 q 2 = p x q y ∴ x = 4 and y = 2.
x + y = 4 + 2 = 6.
Hence, x + y = 6.
Solve the following equation for x:
52x + 3 = 1
Answer
Given,
⇒ 52x + 3 = 1
⇒ 52x + 3 = 50
⇒ 2x + 3 = 0
⇒ 2x = -3
⇒ x = -3 2 \dfrac{3}{2} 2 3
Hence, x = -3 2 \dfrac{3}{2} 2 3
Solve the following equation for x:
( 13 ) x = 4 4 − 3 4 − 6 (13)^{\sqrt{x}} = 4^4 - 3^4 - 6 ( 13 ) x = 4 4 − 3 4 − 6
Answer
Given,
⇒ ( 13 ) x = 4 4 − 3 4 − 6 ⇒ ( 13 ) x = 256 − 81 − 6 ⇒ ( 13 ) x = 169 ⇒ ( 13 ) x = 13 2 ∴ x = 2 ⇒ x = 4. \Rightarrow (13)^{\sqrt{x}} = 4^4 - 3^4 - 6 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 256 - 81 - 6 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 169 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 13^2 \\[1em] \therefore \sqrt{x} = 2 \\[1em] \Rightarrow x = 4. ⇒ ( 13 ) x = 4 4 − 3 4 − 6 ⇒ ( 13 ) x = 256 − 81 − 6 ⇒ ( 13 ) x = 169 ⇒ ( 13 ) x = 1 3 2 ∴ x = 2 ⇒ x = 4.
Hence, x = 4.
Solve the following equation for x:
( 3 5 ) x + 1 = 125 27 \Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27} ( 5 3 ) x + 1 = 27 125
Answer
Given,
⇒ ( 3 5 ) x + 1 = 125 27 ⇒ ( 3 5 ) x + 1 2 = ( 5 3 ) 3 ⇒ ( 3 5 ) x + 1 2 = ( 3 5 ) − 3 ∴ x + 1 2 = − 3 ⇒ x + 1 = − 6 ⇒ x = − 7. \Rightarrow \Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27} \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x + 1}{2}} = \Big(\dfrac{5}{3}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x + 1}{2}} = \Big(\dfrac{3}{5}\Big)^{-3} \\[1em] \therefore \dfrac{x + 1}{2} = -3 \\[1em] \Rightarrow x + 1 = -6 \\[1em] \Rightarrow x = -7. ⇒ ( 5 3 ) x + 1 = 27 125 ⇒ ( 5 3 ) 2 x + 1 = ( 3 5 ) 3 ⇒ ( 5 3 ) 2 x + 1 = ( 5 3 ) − 3 ∴ 2 x + 1 = − 3 ⇒ x + 1 = − 6 ⇒ x = − 7.
Hence, x = -7.
Solve the following equation for x:
( 4 3 ) 2 x + 1 2 = 1 32 (\sqrt[3]{4})^{2x + \dfrac{1}{2}} = \dfrac{1}{32} ( 3 4 ) 2 x + 2 1 = 32 1
Answer
Given,
⇒ ( 4 3 ) 2 x + 1 2 = 1 32 ⇒ [ ( 2 2 ) 1 3 ] 2 x + 1 2 = 1 2 5 ⇒ ( 2 2 ) 2 x 3 + 1 6 = 2 − 5 ⇒ ( 2 ) 4 x 3 + 1 3 = 2 − 5 ⇒ 4 x + 1 3 = − 5 ⇒ 4 x + 1 = − 15 ⇒ 4 x = − 16 ⇒ x = − 4. \Rightarrow (\sqrt[3]{4})^{2x + \dfrac{1}{2}} = \dfrac{1}{32} \\[1em] \Rightarrow [(2^2)^{\dfrac{1}{3}}]^{2x + \dfrac{1}{2}} = \dfrac{1}{2^5} \\[1em] \Rightarrow (2^2)^{\dfrac{2x}{3} + \dfrac{1}{6}} = 2^{-5} \\[1em] \Rightarrow (2)^{\dfrac{4x}{3} + \dfrac{1}{3}} = 2^{-5} \\[1em] \Rightarrow \dfrac{4x + 1}{3} = -5 \\[1em] \Rightarrow 4x + 1 = -15 \\[1em] \Rightarrow 4x = -16 \\[1em] \Rightarrow x = -4. ⇒ ( 3 4 ) 2 x + 2 1 = 32 1 ⇒ [( 2 2 ) 3 1 ] 2 x + 2 1 = 2 5 1 ⇒ ( 2 2 ) 3 2 x + 6 1 = 2 − 5 ⇒ ( 2 ) 3 4 x + 3 1 = 2 − 5 ⇒ 3 4 x + 1 = − 5 ⇒ 4 x + 1 = − 15 ⇒ 4 x = − 16 ⇒ x = − 4.
Hence, x = -4.
Solve the following equation for x:
p q = ( q p ) 1 − 2 x \sqrt{\dfrac{p}{q}} = \Big(\dfrac{q}{p}\Big)^{1 - 2x} q p = ( p q ) 1 − 2 x
Answer
Given,
⇒ p q = ( q p ) 1 − 2 x ⇒ ( p q ) 1 2 = ( p q ) − ( 1 − 2 x ) ⇒ ( p q ) 1 2 = ( p q ) ( 2 x − 1 ) ⇒ 1 2 = 2 x − 1 ⇒ 2 x = 1 + 1 2 ⇒ 2 x = 3 2 ⇒ x = 3 4 . \Rightarrow \sqrt{\dfrac{p}{q}} = \Big(\dfrac{q}{p}\Big)^{1 - 2x} \\[1em] \Rightarrow \Big(\dfrac{p}{q}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{p}{q}\Big)^{-(1 - 2x)} \\[1em] \Rightarrow \Big(\dfrac{p}{q}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{p}{q}\Big)^{(2x - 1)} \\[1em] \Rightarrow \dfrac{1}{2} = 2x - 1 \\[1em] \Rightarrow 2x = 1 + \dfrac{1}{2} \\[1em] \Rightarrow 2x = \dfrac{3}{2} \\[1em] \Rightarrow x = \dfrac{3}{4}. ⇒ q p = ( p q ) 1 − 2 x ⇒ ( q p ) 2 1 = ( q p ) − ( 1 − 2 x ) ⇒ ( q p ) 2 1 = ( q p ) ( 2 x − 1 ) ⇒ 2 1 = 2 x − 1 ⇒ 2 x = 1 + 2 1 ⇒ 2 x = 2 3 ⇒ x = 4 3 .
Hence, x = 3 4 \dfrac{3}{4} 4 3
Solve the following equation for x:
4 x − 1 × ( 0.5 ) 3 − 2 x = ( 1 8 ) x 4^{x - 1} \times (0.5)^{3 - 2x} = \Big(\dfrac{1}{8}\Big)^x 4 x − 1 × ( 0.5 ) 3 − 2 x = ( 8 1 ) x
Answer
Given,
⇒ 4 x − 1 × ( 0.5 ) 3 − 2 x = ( 1 8 ) x ⇒ ( 2 2 ) x − 1 × ( 5 10 ) 3 − 2 x = ( 1 2 3 ) x ⇒ 2 2 x − 2 × ( 1 2 ) 3 − 2 x = ( 2 − 3 ) x ⇒ 2 2 x − 2 × ( 2 − 1 ) 3 − 2 x = 2 − 3 x ⇒ 2 2 x − 2 × 2 2 x − 3 = 2 − 3 x ⇒ 2 2 x − 2 + 2 x − 3 = 2 − 3 x ⇒ 2 4 x − 5 = 2 − 3 x ⇒ 4 x − 5 = − 3 x ⇒ 4 x + 3 x = 5 ⇒ 7 x = 5 ⇒ x = 5 7 . \Rightarrow 4^{x - 1} \times (0.5)^{3 - 2x} = \Big(\dfrac{1}{8}\Big)^x \\[1em] \Rightarrow (2^2)^{x - 1} \times \Big(\dfrac{5}{10}\Big)^{3 - 2x} = \Big(\dfrac{1}{2^3}\Big)^x \\[1em] \Rightarrow 2^{2x - 2} \times \Big(\dfrac{1}{2}\Big)^{3 - 2x} = (2^{-3})^x \\[1em] \Rightarrow 2^{2x - 2} \times (2^{-1})^{3 - 2x} = 2^{-3x} \\[1em] \Rightarrow 2^{2x - 2} \times 2^{2x - 3} = 2^{-3x} \\[1em] \Rightarrow 2^{2x - 2 + 2x - 3} = 2^{-3x} \\[1em] \Rightarrow 2^{4x - 5} = 2^{-3x} \\[1em] \Rightarrow 4x - 5 = -3x \\[1em] \Rightarrow 4x + 3x = 5 \\[1em] \Rightarrow 7x = 5 \\[1em] \Rightarrow x = \dfrac{5}{7}. ⇒ 4 x − 1 × ( 0.5 ) 3 − 2 x = ( 8 1 ) x ⇒ ( 2 2 ) x − 1 × ( 10 5 ) 3 − 2 x = ( 2 3 1 ) x ⇒ 2 2 x − 2 × ( 2 1 ) 3 − 2 x = ( 2 − 3 ) x ⇒ 2 2 x − 2 × ( 2 − 1 ) 3 − 2 x = 2 − 3 x ⇒ 2 2 x − 2 × 2 2 x − 3 = 2 − 3 x ⇒ 2 2 x − 2 + 2 x − 3 = 2 − 3 x ⇒ 2 4 x − 5 = 2 − 3 x ⇒ 4 x − 5 = − 3 x ⇒ 4 x + 3 x = 5 ⇒ 7 x = 5 ⇒ x = 7 5 .
Hence, x = 5 7 \dfrac{5}{7} 7 5
If 53x = 125 and (10)y = 0.001, find x and y.
Answer
Given,
⇒ 53x = 125
⇒ 53x = 53
⇒ 3x = 3
⇒ x = 1.
Given,
⇒ 10 y = 0.001 ⇒ 10 y = 1 1000 ⇒ 10 y = 1 10 3 ⇒ 10 y = 10 − 3 ⇒ y = − 3. \Rightarrow 10^y = 0.001 \\[1em] \Rightarrow 10^y = \dfrac{1}{1000} \\[1em] \Rightarrow 10^y = \dfrac{1}{10^3} \\[1em] \Rightarrow 10^y = 10^{-3} \\[1em] \Rightarrow y = -3. ⇒ 1 0 y = 0.001 ⇒ 1 0 y = 1000 1 ⇒ 1 0 y = 1 0 3 1 ⇒ 1 0 y = 1 0 − 3 ⇒ y = − 3.
Hence, x = 1 and y = -3.
If 9 n .3 2 .3 n − ( 27 ) n 3 3 m .2 3 = 1 27 \dfrac{9^n.3^2.3^n - (27)^n}{3^{3m}.2^3} = \dfrac{1}{27} 3 3 m . 2 3 9 n . 3 2 . 3 n − ( 27 ) n = 27 1
Answer
Given,
⇒ 9 n .3 2 .3 n − ( 27 ) n 3 3 m .2 3 = 1 27 ⇒ ( 3 2 ) n .3 2 .3 n − ( 3 3 ) n 3 3 m .8 = 1 3 3 ⇒ 3 2 n + n .3 2 − 3 3 n 3 3 m .8 = 1 3 3 ⇒ 3 3 n .9 − 3 3 n 3 3 m .8 = 1 3 3 ⇒ 3 3 n ( 9 − 1 ) 3 3 m .8 = 1 3 3 ⇒ 3 3 n .8 3 3 m .8 = 3 − 3 ⇒ 3 3 n 3 3 m = 3 − 3 ⇒ 3 3 n = 3 3 m .3 − 3 ⇒ 3 3 n = 3 3 m + ( − 3 ) ⇒ 3 n = 3 m − 3 ⇒ 3 n = 3 ( m − 1 ) ⇒ n = m − 1 ⇒ m = 1 + n . \Rightarrow \dfrac{9^n.3^2.3^n - (27)^n}{3^{3m}.2^3} = \dfrac{1}{27} \\[1em] \Rightarrow \dfrac{(3^2)^n.3^2.3^n - (3^3)^n}{3^{3m}.8} = \dfrac{1}{3^3} \\[1em] \Rightarrow \dfrac{3^{2n + n}.3^2 - 3^{3n}}{3^{3m}.8} = \dfrac{1}{3^3} \\[1em] \Rightarrow \dfrac{3^{3n}.9 - 3^{3n}}{3^{3m}.8} = \dfrac{1}{3^3} \\[1em] \Rightarrow \dfrac{3^{3n}(9 - 1)}{3^{3m}.8} = \dfrac{1}{3^3} \\[1em] \Rightarrow \dfrac{3^{3n}.8}{3^{3m}.8} = 3^{-3} \\[1em] \Rightarrow \dfrac{3^{3n}}{3^{3m}} = 3^{-3} \\[1em] \Rightarrow 3^{3n} = 3^{3m}.3^{-3} \\[1em] \Rightarrow 3^{3n} = 3^{3m + (-3)} \\[1em] \Rightarrow 3n = 3m - 3 \\[1em] \Rightarrow 3n = 3(m - 1) \\[1em] \Rightarrow n = m - 1 \\[1em] \Rightarrow m = 1 + n. ⇒ 3 3 m . 2 3 9 n . 3 2 . 3 n − ( 27 ) n = 27 1 ⇒ 3 3 m .8 ( 3 2 ) n . 3 2 . 3 n − ( 3 3 ) n = 3 3 1 ⇒ 3 3 m .8 3 2 n + n . 3 2 − 3 3 n = 3 3 1 ⇒ 3 3 m .8 3 3 n .9 − 3 3 n = 3 3 1 ⇒ 3 3 m .8 3 3 n ( 9 − 1 ) = 3 3 1 ⇒ 3 3 m .8 3 3 n .8 = 3 − 3 ⇒ 3 3 m 3 3 n = 3 − 3 ⇒ 3 3 n = 3 3 m . 3 − 3 ⇒ 3 3 n = 3 3 m + ( − 3 ) ⇒ 3 n = 3 m − 3 ⇒ 3 n = 3 ( m − 1 ) ⇒ n = m − 1 ⇒ m = 1 + n .
Hence, proved that m = 1 + n.
If 34x = 81-1 and ( 10 ) 1 y (10)^{\dfrac{1}{y}} ( 10 ) y 1 -x .(16)y .
Answer
Given,
⇒ 34x = 81-1
⇒ 34x = (34 )-1
⇒ 34x = (3-4 )
⇒ 4x = -4
⇒ x = -1.
Given,
⇒ ( 10 ) 1 y = 0.0001 ⇒ ( 10 ) 1 y = 1 10 4 ⇒ ( 10 ) 1 y = 10 − 4 ⇒ 1 y = − 4 ⇒ y = − 1 4 . \Rightarrow (10)^{\dfrac{1}{y}} = 0.0001 \\[1em] \Rightarrow (10)^{\dfrac{1}{y}} = \dfrac{1}{10^4} \\[1em] \Rightarrow (10)^{\dfrac{1}{y}} = 10^{-4} \\[1em] \Rightarrow \dfrac{1}{y} = -4 \\[1em] \Rightarrow y = -\dfrac{1}{4}. ⇒ ( 10 ) y 1 = 0.0001 ⇒ ( 10 ) y 1 = 1 0 4 1 ⇒ ( 10 ) y 1 = 1 0 − 4 ⇒ y 1 = − 4 ⇒ y = − 4 1 .
Substituting value of x and y in 2-x .(16)y we get,
⇒ 2 − 1 . ( 16 ) − 1 4 ⇒ 2 × ( 2 4 ) − 1 4 ⇒ 2 × 2 4 × − 1 4 ⇒ 2 × 2 − 1 ⇒ 2 × 1 2 ⇒ 1. \Rightarrow 2^{-1}.(16)^{-\dfrac{1}{4}} \\[1em] \Rightarrow 2 \times (2^4)^{-\dfrac{1}{4}} \\[1em] \Rightarrow 2 \times 2^{4 \times -\dfrac{1}{4}} \\[1em] \Rightarrow 2 \times 2^{-1} \\[1em] \Rightarrow 2 \times \dfrac{1}{2} \\[1em] \Rightarrow 1. ⇒ 2 − 1 . ( 16 ) − 4 1 ⇒ 2 × ( 2 4 ) − 4 1 ⇒ 2 × 2 4 ×− 4 1 ⇒ 2 × 2 − 1 ⇒ 2 × 2 1 ⇒ 1.
Hence, 2-x .(16)y = 1.
If 3x + 1 = 9x - 2 , find the value of 21 + x .
Answer
Given,
⇒ 3x + 1 = 9x - 2
⇒ 3x + 1 = (32 )x - 2
⇒ 3x + 1 = 32(x - 2)
⇒ 3x + 1 = 32x - 4
⇒ x + 1 = 2x - 4
⇒ 2x - x = 1 + 4
⇒ x = 5.
Substituting value of x in 21 + x we get,
⇒ 21 + x = 21 + 5 = 26 = 64.
Hence, 21 + x = 64.
Solve the following equations:
(i) 3(2x + 1) - 2x + 2 + 5 = 0
(ii) 3x = 9.3y , 8.2y = 4x .
Answer
(i) Given,
⇒ 3(2x + 1) - 2x + 2 + 5 = 0
⇒ 3.2x + 3 - 2x .22 + 5 = 0
⇒ 3.2x - 4.2x + 3 + 5 = 0
⇒ -2x + 8 = 0
⇒ 2x = 8
⇒ 2x = 23
⇒ x = 3.
Hence, x = 3.
(ii) Given,
⇒ 3x = 9.3y
⇒ 3x = 32 .3y
⇒ 3x = 32 + y
⇒ x = y + 2 ......(i)
Given,
⇒ 8.2y = 4x
⇒ 23 .2y = (22 )x
⇒ 23 + y = 22x
⇒ 2x = y + 3 .......(ii)
Subtracting (i) from (ii) we get,
⇒ 2x - x = (y + 3) - (y + 2)
⇒ x = 1.
Substituting value of x in (i) we get,
⇒ 1 = y + 2
⇒ y = -1.
Hence, x = 1 and y = -1.
If 2x .3y .5z = 2160, find the values of x, y and z. Hence, compute the value of 3x .2-y .5-z .
Answer
2160 = 24 .33 .51 .
∴ 24 .33 .51 = 2x .3y .5z .
∴ x = 4, y = 3 and z = 1.
Substituting values of x, y and z in 3x .2-y .5-z
= 3 4 × 2 − 3 × 5 − 1 = 81 × 1 2 3 × 1 5 = 81 × 1 8 × 1 5 = 81 40 = 2 1 40 . = 3^4 \times 2^{-3} \times 5^{-1} \\[1em] = 81 \times \dfrac{1}{2^3} \times \dfrac{1}{5} \\[1em] = 81 \times \dfrac{1}{8} \times \dfrac{1}{5} \\[1em] = \dfrac{81}{40} = 2\dfrac{1}{40}. = 3 4 × 2 − 3 × 5 − 1 = 81 × 2 3 1 × 5 1 = 81 × 8 1 × 5 1 = 40 81 = 2 40 1 .
Hence, x = 4, y = 3 and z = 1 and 3x .2-y .5-z = 2 1 40 . 2\dfrac{1}{40}. 2 40 1 .
If x = 2 and y = -3, find the values of
(i) xx + yy
(ii) xy + yx
Answer
(i) xx + yy = 22 + (-3)-3
= 4 + ( − 1 3 ) 3 \Big(-\dfrac{1}{3}\Big)^3 ( − 3 1 ) 3
= 4 + − 1 27 -\dfrac{1}{27} − 27 1
= 4 - 1 27 \dfrac{1}{27} 27 1
= 108 − 1 27 = 107 27 = 3 26 27 \dfrac{108 - 1}{27} = \dfrac{107}{27} = 3\dfrac{26}{27} 27 108 − 1 = 27 107 = 3 27 26
Hence, xx + yy = 3 26 27 3\dfrac{26}{27} 3 27 26
(ii) xy + yx
x y + y x = 2 − 3 + ( − 3 ) 2 = ( 1 2 ) 3 + 9 = 1 8 + 9 = 1 + 72 8 = 73 8 = 9 1 8 . x^y + y^x = 2^{-3} + (-3)^{2} \\[1em] = \Big(\dfrac{1}{2}\Big)^3 + 9 \\[1em] = \dfrac{1}{8} + 9 \\[1em] = \dfrac{1 + 72}{8} \\[1em] = \dfrac{73}{8} = 9\dfrac{1}{8}. x y + y x = 2 − 3 + ( − 3 ) 2 = ( 2 1 ) 3 + 9 = 8 1 + 9 = 8 1 + 72 = 8 73 = 9 8 1 .
Hence, xy + yx = 9 1 8 9\dfrac{1}{8} 9 8 1
If p = xm + n .yl , q = xn + l .ym and r = xl + m .yn , prove that
pm - n .qn - l .rl - m = 1
Answer
Substituting value of p, q and r in pm - n .qn - l .rl - m = 1.
= (xm + n .yl )m - n .(xn + l .ym )n - l .(xl + m .yn )l - m
= x(m + n)(m - n) .yl(m - n) .x(n + l)(n - l) .ym(n - l) .x(l + m)(l - m) .yn(l - m)
= xm2 - n2 .xn2 - l2 .xl2 - m2 .ylm - ln .ymn - ml .ynl - nm
= xm2 - n2 + n2 - l2 + l2 - m2 .ylm - ln + mn - ml + nl - nm
= x0 .y0
= 1.1
= 1.
Hence, proved that, pm - n .qn - l .rl - m = 1.
If x = am + n , y = an + l and z = al + m , prove that xm yn zl = xn yl zm .
Answer
Substituting values of x, y and z in L.H.S. of xm yn zl = xn yl zm we get,
xm yn zl = (am + n )m .(an + l )n .(al + m )l
= am2 + mn .an2 + ln .alm + l2
= am2 + n2 + l2 + mn + lm + ln .......(i)
Substituting values of x, y and z in R.H.S. of xm yn zl = xn yl zm we get,
xn yl zm = (am + n )n .(an + l )l .(al + m )m
= amn + n2 .anl + l2 .alm + m2
= am2 + n2 + l2 + mn + lm + ln .........(ii)
Since (i) = (ii),
Hence, proved that xm yn zl = xn yl zm .
Show that ( p + 1 q ) m × ( p − 1 q ) n ( q + 1 p ) m × ( q − 1 p ) n = ( p q ) m + n \dfrac{\Big(p + \dfrac{1}{q}\Big)^m \times \Big(p - \dfrac{1}{q}\Big)^n}{\Big(q + \dfrac{1}{p}\Big)^m \times \Big(q - \dfrac{1}{p}\Big)^n} = \Big(\dfrac{p}{q}\Big)^{m + n} ( q + p 1 ) m × ( q − p 1 ) n ( p + q 1 ) m × ( p − q 1 ) n = ( q p ) m + n
Answer
Given,
⇒ ( p + 1 q ) m × ( p − 1 q ) n ( q + 1 p ) m × ( q − 1 p ) n = ( p q + 1 q ) m × ( p q − 1 q ) n ( p q + 1 p ) m × ( p q − 1 p ) n = ( p q + 1 ) m q m × ( p q − 1 ) n q n ( p q + 1 ) m p m × ( p q − 1 ) n p n = ( p q + 1 ) m × ( p q − 1 ) n × p m × p n ( p q + 1 ) m × ( p q − 1 ) n × q m × q n = p m × p n q m × q n = p m + n q m + n = ( p q ) m + n . \Rightarrow \dfrac{\Big(p + \dfrac{1}{q}\Big)^m \times \Big(p - \dfrac{1}{q}\Big)^n}{\Big(q + \dfrac{1}{p}\Big)^m \times \Big(q - \dfrac{1}{p}\Big)^n} = \dfrac{\Big(\dfrac{pq + 1}{q}\Big)^m \times \Big(\dfrac{pq - 1}{q}\Big)^n}{\Big(\dfrac{pq + 1}{p}\Big)^m \times \Big(\dfrac{pq - 1}{p}\Big)^n} \\[1em] = \dfrac{\dfrac{(pq + 1)^m}{q^m} \times \dfrac{(pq - 1)^n}{q^n}}{\dfrac{(pq + 1)^m}{p^m} \times \dfrac{(pq - 1)^n}{p^n}} \\[1em] = \dfrac{(pq + 1)^m \times (pq - 1)^n \times p^m \times p^n}{(pq + 1)^m \times (pq - 1)^n \times q^m \times q^n} \\[1em] = \dfrac{p^m \times p^n}{q^m \times q^n} \\[1em] = \dfrac{p^{m + n}}{q^{m + n}} \\[1em] = \Big(\dfrac{p}{q}\Big)^{m + n}. ⇒ ( q + p 1 ) m × ( q − p 1 ) n ( p + q 1 ) m × ( p − q 1 ) n = ( p pq + 1 ) m × ( p pq − 1 ) n ( q pq + 1 ) m × ( q pq − 1 ) n = p m ( pq + 1 ) m × p n ( pq − 1 ) n q m ( pq + 1 ) m × q n ( pq − 1 ) n = ( pq + 1 ) m × ( pq − 1 ) n × q m × q n ( pq + 1 ) m × ( pq − 1 ) n × p m × p n = q m × q n p m × p n = q m + n p m + n = ( q p ) m + n .
Hence, proved that ( p + 1 q ) m × ( p − 1 q ) n ( q + 1 p ) m × ( q − 1 p ) n = ( p q ) m + n \dfrac{\Big(p + \dfrac{1}{q}\Big)^m \times \Big(p - \dfrac{1}{q}\Big)^n}{\Big(q + \dfrac{1}{p}\Big)^m \times \Big(q - \dfrac{1}{p}\Big)^n} = \Big(\dfrac{p}{q}\Big)^{m + n} ( q + p 1 ) m × ( q − p 1 ) n ( p + q 1 ) m × ( p − q 1 ) n = ( q p ) m + n
If x is a positive real number and exponents are rational numbers, then simplify the following :
( x ( a + b ) ) 2 ( x ( b + c ) ) 2 ( x ( c + a ) ) 2 ( x a x b x c ) 4 \dfrac{(x^{(a + b)})^2(x^{(b + c)})^2(x^{(c + a)})^2}{(x^ax^bx^c)^4} ( x a x b x c ) 4 ( x ( a + b ) ) 2 ( x ( b + c ) ) 2 ( x ( c + a ) ) 2
Answer
Given,
⇒ ( x ( a + b ) ) 2 ( x ( b + c ) ) 2 ( x ( c + a ) ) 2 ( x a x b x c ) 4 = x 2 a + 2 b . x 2 b + 2 c . x 2 c + 2 a x ( a + b + c ) 4 = x 2 a + 2 b + 2 b + 2 c + 2 c + 2 a x 4 a + 4 b + 4 c = x 4 a + 4 b + 4 c x 4 a + 4 b + 4 c = 1. \Rightarrow \dfrac{(x^{(a + b)})^2(x^{(b + c)})^2(x^{(c + a)})^2}{(x^ax^bx^c)^4} = \dfrac{x^{2a + 2b}.x^{2b +2c}.x^{2c + 2a}}{x^{(a + b + c)4}} \\[1em] = \dfrac{x^{2a + 2b + 2b + 2c + 2c + 2a}}{x^{4a + 4b + 4c}} \\[1em] = \dfrac{x^{4a + 4b + 4c}}{x^{4a + 4b + 4c}} \\[1em] = 1. ⇒ ( x a x b x c ) 4 ( x ( a + b ) ) 2 ( x ( b + c ) ) 2 ( x ( c + a ) ) 2 = x ( a + b + c ) 4 x 2 a + 2 b . x 2 b + 2 c . x 2 c + 2 a = x 4 a + 4 b + 4 c x 2 a + 2 b + 2 b + 2 c + 2 c + 2 a = x 4 a + 4 b + 4 c x 4 a + 4 b + 4 c = 1.
Hence, ( x ( a + b ) ) 2 ( x ( b + c ) ) 2 ( x ( c + a ) ) 2 ( x a x b x c ) 4 \dfrac{(x^{(a + b)})^2(x^{(b + c)})^2(x^{(c + a)})^2}{(x^ax^bx^c)^4} ( x a x b x c ) 4 ( x ( a + b ) ) 2 ( x ( b + c ) ) 2 ( x ( c + a ) ) 2
If x is a positive real number and exponents are rational numbers, then simplify the following :
( x a 2 x b 2 ) 1 a + b ( x b 2 x c 2 ) 1 b + c ( x c 2 x a 2 ) 1 c + a \Big(\dfrac{x^{a^2}}{x^{b^2}}\Big)^{\dfrac{1}{a + b}}\Big(\dfrac{x^{b^2}}{x^{c^2}}\Big)^{\dfrac{1}{b + c}}\Big(\dfrac{x^{c^2}}{x^{a^2}}\Big)^{\dfrac{1}{c + a}} ( x b 2 x a 2 ) a + b 1 ( x c 2 x b 2 ) b + c 1 ( x a 2 x c 2 ) c + a 1
Answer
Given,
⇒ ( x a 2 x b 2 ) 1 a + b ( x b 2 x c 2 ) 1 b + c ( x c 2 x a 2 ) 1 c + a ⇒ ( x a 2 − b 2 ) 1 a + b ( x b 2 − c 2 ) 1 b + c ( x c 2 − a 2 ) 1 c + a ⇒ ( x ) a 2 − b 2 a + b . ( x ) b 2 − c 2 b + c . ( x ) c 2 − a 2 c + a ⇒ ( x ) ( a − b ) ( a + b ) a + b . ( x ) ( b − c ) ( b + c ) b + c . ( x ) ( c − a ) ( c + a ) c + a ⇒ ( x ) a − b . ( x ) b − c . ( x ) c − a ⇒ x a − b + b − c + c − a ⇒ x 0 = 1. \Rightarrow \Big(\dfrac{x^{a^2}}{x^{b^2}}\Big)^{\dfrac{1}{a + b}}\Big(\dfrac{x^{b^2}}{x^{c^2}}\Big)^{\dfrac{1}{b + c}}\Big(\dfrac{x^{c^2}}{x^{a^2}}\Big)^{\dfrac{1}{c + a}} \\[1em] \Rightarrow (x^{a^2 - b^2})^{\dfrac{1}{a + b}}(x^{b^2 - c^2})^{\dfrac{1}{b + c}}(x^{c^2 - a^2})^{\dfrac{1}{c + a}} \\[1em] \Rightarrow (x)^{\dfrac{a^2 - b^2}{a + b}}.(x)^{\dfrac{b^2 - c^2}{b + c}}.(x)^{\dfrac{c^2 - a^2}{c + a}} \\[1em] \Rightarrow (x)^{\dfrac{(a - b)(a + b)}{a + b}}.(x)^{\dfrac{(b - c)(b + c)}{b + c}}.(x)^{\dfrac{(c - a)(c + a)}{c + a}} \\[1em] \Rightarrow (x)^{a - b}.(x)^{b - c}.(x)^{c - a} \\[1em] \Rightarrow x^{a - b + b - c + c - a} \\[1em] \Rightarrow x^{0} = 1. ⇒ ( x b 2 x a 2 ) a + b 1 ( x c 2 x b 2 ) b + c 1 ( x a 2 x c 2 ) c + a 1 ⇒ ( x a 2 − b 2 ) a + b 1 ( x b 2 − c 2 ) b + c 1 ( x c 2 − a 2 ) c + a 1 ⇒ ( x ) a + b a 2 − b 2 . ( x ) b + c b 2 − c 2 . ( x ) c + a c 2 − a 2 ⇒ ( x ) a + b ( a − b ) ( a + b ) . ( x ) b + c ( b − c ) ( b + c ) . ( x ) c + a ( c − a ) ( c + a ) ⇒ ( x ) a − b . ( x ) b − c . ( x ) c − a ⇒ x a − b + b − c + c − a ⇒ x 0 = 1.
Hence, ( x a 2 x b 2 ) 1 a + b ( x b 2 x c 2 ) 1 b + c ( x c 2 x a 2 ) 1 c + a \Big(\dfrac{x^{a^2}}{x^{b^2}}\Big)^{\dfrac{1}{a + b}}\Big(\dfrac{x^{b^2}}{x^{c^2}}\Big)^{\dfrac{1}{b + c}}\Big(\dfrac{x^{c^2}}{x^{a^2}}\Big)^{\dfrac{1}{c + a}} ( x b 2 x a 2 ) a + b 1 ( x c 2 x b 2 ) b + c 1 ( x a 2 x c 2 ) c + a 1
If x is a positive real number and exponents are rational numbers, then simplify the following :
( x b x c ) b + c − a ( x c x a ) c + a − b ( x a x b ) a + b − c \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a}\Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} ( x c x b ) b + c − a ( x a x c ) c + a − b ( x b x a ) a + b − c
Answer
Given,
⇒ ( x b x c ) b + c − a ( x c x a ) c + a − b ( x a x b ) a + b − c ⇒ ( x b − c ) b + c − a . ( x c − a ) c + a − b . ( x a − b ) a + b − c ⇒ x ( b − c ) ( b + c − a ) . ( x ) ( c − a ) ( c + a − b ) . ( x ) ( a − b ) ( a + b − c ) ⇒ x b 2 + b c − b a − c b − c 2 + c a . ( x ) c 2 + c a − c b − a c − a 2 + a b . ( x ) a 2 + a b − a c − b a − b 2 + b c ⇒ x b 2 − b 2 + b c − c b − c b + b c − b a + a b + a b − b a − c 2 + c 2 + c a + c a − a c − a c − a 2 + a 2 ⇒ x 0 = 1. \Rightarrow \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a}\Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} \\[1em] \Rightarrow (x^{b - c})^{b + c - a}.(x^{c - a})^{c + a - b}.(x^{a - b})^{a + b - c} \\[1em] \Rightarrow x^{(b - c)(b + c - a)}.(x)^{(c - a)(c + a - b)}.(x)^{(a - b)(a + b - c)} \\[1em] \Rightarrow x^{b^2 + bc - ba - cb - c^2 + ca}.(x)^{c^2 + ca - cb - ac - a^2 + ab}.(x)^{a^2 + ab - ac - ba - b^2 + bc} \\[1em] \Rightarrow x^{b^2 - b^2 + bc - cb - cb + bc - ba + ab + ab - ba - c^2 + c^2 + ca + ca - ac - ac - a^2 + a^2} \\[1em] \Rightarrow x^0 = 1. ⇒ ( x c x b ) b + c − a ( x a x c ) c + a − b ( x b x a ) a + b − c ⇒ ( x b − c ) b + c − a . ( x c − a ) c + a − b . ( x a − b ) a + b − c ⇒ x ( b − c ) ( b + c − a ) . ( x ) ( c − a ) ( c + a − b ) . ( x ) ( a − b ) ( a + b − c ) ⇒ x b 2 + b c − ba − c b − c 2 + c a . ( x ) c 2 + c a − c b − a c − a 2 + ab . ( x ) a 2 + ab − a c − ba − b 2 + b c ⇒ x b 2 − b 2 + b c − c b − c b + b c − ba + ab + ab − ba − c 2 + c 2 + c a + c a − a c − a c − a 2 + a 2 ⇒ x 0 = 1.
Hence, ( x b x c ) b + c − a ( x c x a ) c + a − b ( x a x b ) a + b − c \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a}\Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} ( x c x b ) b + c − a ( x a x c ) c + a − b ( x b x a ) a + b − c
Show that 1 1 + a y − x + a z − x + 1 1 + a z − y + a x − y + 1 1 + a x − z + a y − z \dfrac{1}{1 + a^{y - x} + a^{z - x}} + \dfrac{1}{1 + a^{z - y} + a^{x - y}} + \dfrac{1}{1 + a^{x - z} + a^{y - z}} 1 + a y − x + a z − x 1 + 1 + a z − y + a x − y 1 + 1 + a x − z + a y − z 1
Answer
Given,
1 1 + a y − x + a z − x + 1 1 + a z − y + a x − y + 1 1 + a x − z + a y − z \dfrac{1}{1 + a^{y - x} + a^{z - x}} + \dfrac{1}{1 + a^{z - y} + a^{x - y}} + \dfrac{1}{1 + a^{x - z} + a^{y - z}} 1 + a y − x + a z − x 1 + 1 + a z − y + a x − y 1 + 1 + a x − z + a y − z 1
Solving L.H.S. of the above equation,
⇒ 1 1 + a y − x + a z − x + 1 1 + a z − y + a x − y + 1 1 + a x − z + a y − z ⇒ 1 1 + a y a − x + a z . a − x + 1 1 + a z . a − y + a x . a − y + 1 1 + a x . a − z + a y . a − z ⇒ 1 1 + a y a x + a z a x + 1 1 + a z a y + a x a y + 1 1 + a x a z + a y a z ⇒ 1 a x + a y + a z a x + 1 a y + a z + a x a y + 1 a z + a x + a y a z ⇒ a x a x + a y + a z + a y a x + a y + a z + a z a x + a y + a z ⇒ a x + a y + a z a x + a y + a z ⇒ 1. \Rightarrow \dfrac{1}{1 + a^{y - x} + a^{z - x}} + \dfrac{1}{1 + a^{z - y} + a^{x - y}} + \dfrac{1}{1 + a^{x - z} + a^{y - z}} \\[1em] \Rightarrow \dfrac{1}{1 + a^ya^{-x} + a^{z}.a^{-x}} + \dfrac{1}{1 + a^z.a^{-y} + a^x.a^{-y}} + \dfrac{1}{1 + a^x.a^{-z} + a^y.a^{-z}} \\[1em] \Rightarrow \dfrac{1}{1 + \dfrac{a^y}{a^x} + \dfrac{a^z}{a^x}} + \dfrac{1}{1 + \dfrac{a^z}{a^y} + \dfrac{a^x}{a^y}} + \dfrac{1}{1 + \dfrac{a^x}{a^z} + \dfrac{a^y}{a^z}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{a^x + a^y + a^z}{a^x}} + \dfrac{1}{\dfrac{a^y + a^z + a^x}{a^y}} + \dfrac{1}{\dfrac{a^z + a^x + a^y}{a^z}} \\[1em] \Rightarrow \dfrac{a^x}{a^x + a^y + a^z} + \dfrac{a^y}{a^x + a^y + a^z} + \dfrac{a^z}{a^x + a^y + a^z} \\[1em] \Rightarrow \dfrac{a^x + a^y + a^z}{a^x + a^y + a^z} \\[1em] \Rightarrow 1. ⇒ 1 + a y − x + a z − x 1 + 1 + a z − y + a x − y 1 + 1 + a x − z + a y − z 1 ⇒ 1 + a y a − x + a z . a − x 1 + 1 + a z . a − y + a x . a − y 1 + 1 + a x . a − z + a y . a − z 1 ⇒ 1 + a x a y + a x a z 1 + 1 + a y a z + a y a x 1 + 1 + a z a x + a z a y 1 ⇒ a x a x + a y + a z 1 + a y a y + a z + a x 1 + a z a z + a x + a y 1 ⇒ a x + a y + a z a x + a x + a y + a z a y + a x + a y + a z a z ⇒ a x + a y + a z a x + a y + a z ⇒ 1.
Hence, proved that,
1 1 + a y − x + a z − x + 1 1 + a z − y + a x − y + 1 1 + a x − z + a y − z \dfrac{1}{1 + a^{y - x} + a^{z - x}} + \dfrac{1}{1 + a^{z - y} + a^{x - y}} + \dfrac{1}{1 + a^{x - z} + a^{y - z}} 1 + a y − x + a z − x 1 + 1 + a z − y + a x − y 1 + 1 + a x − z + a y − z 1
If (3)x = (5)y = (75)z , show that z = x y 2 x + y z = \dfrac{xy}{2x + y} z = 2 x + y x y
Answer
Let (3)x = (5)y = (75)z = k
∴ (3)x = k
⇒ 3 = k 1 x k^{\dfrac{1}{x}} k x 1
∴ (5)y = k
⇒ 5 = k 1 y k^{\dfrac{1}{y}} k y 1
∴ (75)z = k
⇒ 75 = k 1 z k^{\dfrac{1}{z}} k z 1
⇒ 3.52 = k 1 z k^{\dfrac{1}{z}} k z 1
Substituting value of x and y from (i) and (ii) in above equation we get,
⇒ k 1 x × ( k 1 y ) 2 = k 1 z ⇒ k 1 x × k 2 y = k 1 z ⇒ k 1 x + 2 y = k 1 z ⇒ 1 x + 2 y = 1 z ⇒ y + 2 x x y = 1 z ⇒ z = x y 2 x + y . \Rightarrow k^{\dfrac{1}{x}} \times (k^{\dfrac{1}{y}})^2 = k^{\dfrac{1}{z}} \\[1em] \Rightarrow k^{\dfrac{1}{x}} \times k^{\dfrac{2}{y}} = k^{\dfrac{1}{z}} \\[1em] \Rightarrow k^{\dfrac{1}{x} + \dfrac{2}{y}} = k^{\dfrac{1}{z}} \\[1em] \Rightarrow \dfrac{1}{x} + \dfrac{2}{y} = \dfrac{1}{z} \\[1em] \Rightarrow \dfrac{y + 2x}{xy} = \dfrac{1}{z} \\[1em] \Rightarrow z = \dfrac{xy}{2x + y}. ⇒ k x 1 × ( k y 1 ) 2 = k z 1 ⇒ k x 1 × k y 2 = k z 1 ⇒ k x 1 + y 2 = k z 1 ⇒ x 1 + y 2 = z 1 ⇒ x y y + 2 x = z 1 ⇒ z = 2 x + y x y .
Hence, proved that z = x y 2 x + y z = \dfrac{xy}{2x + y} z = 2 x + y x y
Solve the following equations for x :
3 x + 1 = 27 \sqrt{3^{x + 1}} = 27 3 x + 1 = 27
Answer
(i) Solving,
⇒ 3 x + 1 = 27 ⇒ 3 x + 1 = 3 3 ⇒ 3 x + 1 = 3 6 ⇒ x + 1 = 6 ⇒ x = 6 − 1 = 5. \Rightarrow \sqrt{3^{x + 1}} = 27 \\[1em] \Rightarrow \sqrt{3^{x + 1}} = 3^3 \\[1em] \Rightarrow \sqrt{3^{x + 1}} = \sqrt{3^6} \\[1em] \Rightarrow x + 1 = 6 \\[1em] \Rightarrow x = 6 - 1 = 5. ⇒ 3 x + 1 = 27 ⇒ 3 x + 1 = 3 3 ⇒ 3 x + 1 = 3 6 ⇒ x + 1 = 6 ⇒ x = 6 − 1 = 5.
Hence, x = 5.
Solve the following equation :
4 2 x = ( 16 3 ) − 6 y = ( 8 ) 2 4^{2x} = (\sqrt[3]{16})^{-\frac{6}{y}} = (\sqrt{8})^2 4 2 x = ( 3 16 ) − y 6 = ( 8 ) 2
Answer
Given,
⇒ 4 2 x = ( 16 3 ) − 6 y = ( 8 ) 2 ∴ 4 2 x = ( 8 ) 2 ⇒ ( 2 2 ) 2 x = 8 ⇒ ( 2 ) 4 x = ( 2 3 ) ⇒ 4 x = 3 ⇒ x = 3 4 . \Rightarrow 4^{2x} = (\sqrt[3]{16})^{-\frac{6}{y}} = (\sqrt{8})^2 \\[1em] \therefore 4^{2x} = (\sqrt{8})^2 \\[1em] \Rightarrow (2^2)^{2x} = 8 \\[1em] \Rightarrow (2)^{4x} = (2^3) \\[1em] \Rightarrow 4x = 3 \\[1em] \Rightarrow x = \dfrac{3}{4}. ⇒ 4 2 x = ( 3 16 ) − y 6 = ( 8 ) 2 ∴ 4 2 x = ( 8 ) 2 ⇒ ( 2 2 ) 2 x = 8 ⇒ ( 2 ) 4 x = ( 2 3 ) ⇒ 4 x = 3 ⇒ x = 4 3 .
Similarly,
⇒ ( 16 3 ) − 6 y = ( 8 ) 2 ⇒ ( 16 ) 1 3 × − 6 y = 8 ⇒ ( 16 ) − 2 y = 8 ⇒ ( 2 4 ) − 2 y = 2 3 ⇒ ( 2 ) − 8 y = 2 3 ⇒ − 8 y = 3 ⇒ y = − 8 3 . \Rightarrow (\sqrt[3]{16})^{-\frac{6}{y}} = (\sqrt{8})^2 \\[1em] \Rightarrow (16)^{\frac{1}{3} \times -\frac{6}{y}} = 8 \\[1em] \Rightarrow (16)^{-\frac{2}{y}} = 8 \\[1em] \Rightarrow (2^4)^{-\frac{2}{y}} = 2^3 \\[1em] \Rightarrow (2)^{-\frac{8}{y}} = 2^3 \\[1em] \Rightarrow -\dfrac{8}{y} = 3 \\[1em] \Rightarrow y = -\dfrac{8}{3}. ⇒ ( 3 16 ) − y 6 = ( 8 ) 2 ⇒ ( 16 ) 3 1 ×− y 6 = 8 ⇒ ( 16 ) − y 2 = 8 ⇒ ( 2 4 ) − y 2 = 2 3 ⇒ ( 2 ) − y 8 = 2 3 ⇒ − y 8 = 3 ⇒ y = − 3 8 .
Hence, x = 3 4 and y = − 8 3 \dfrac{3}{4} \text{ and } y = -\dfrac{8}{3} 4 3 and y = − 3 8
Solve the following equation :
3x - 1 × 52y - 3 = 225
Answer
Given,
⇒ 3x - 1 × 52y - 3 = 225
⇒ 3x - 1 × 52y - 3 = 32 .52
∴ x - 1 = 2 and 2y - 3 = 2
⇒ x = 2 + 1 and 2y = 2 + 3
⇒ x = 3 and y = 5 2 \dfrac{5}{2} 2 5
Hence, x = 3 and y = 5 2 \dfrac{5}{2} 2 5
Solve the following equation :
8x + 1 = 16y + 2 , ( 1 2 ) 3 + x = ( 1 4 ) 3 y \Big(\dfrac{1}{2}\Big)^{3 + x} = \Big(\dfrac{1}{4}\Big)^{3y} ( 2 1 ) 3 + x = ( 4 1 ) 3 y
Answer
Given,
⇒ 8x + 1 = 16y + 2
⇒ (23 )x + 1 = (24 )y + 2
⇒ 23x + 3 = 24y + 8
⇒ 3x + 3 = 4y + 8
⇒ 3x - 4y = 5 .......(i)
Given,
⇒ ( 1 2 ) 3 + x = ( 1 4 ) 3 y ⇒ ( 1 2 ) 3 + x = [ ( 1 2 ) 2 ] 3 y ⇒ ( 1 2 ) 3 + x = ( 1 2 ) 6 y ⇒ 3 + x = 6 y ⇒ 6 y − x = 3....... ( i i ) \Rightarrow \Big(\dfrac{1}{2}\Big)^{3 + x} = \Big(\dfrac{1}{4}\Big)^{3y} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^{3 + x} = \Big[\Big(\dfrac{1}{2}\Big)^2\Big]^{3y} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^{3 + x} = \Big(\dfrac{1}{2}\Big)^{6y} \\[1em] \Rightarrow 3 + x = 6y \\[1em] \Rightarrow 6y - x = 3 .......(ii) ⇒ ( 2 1 ) 3 + x = ( 4 1 ) 3 y ⇒ ( 2 1 ) 3 + x = [ ( 2 1 ) 2 ] 3 y ⇒ ( 2 1 ) 3 + x = ( 2 1 ) 6 y ⇒ 3 + x = 6 y ⇒ 6 y − x = 3....... ( ii )
Multiplying (ii) by 3 we get,
⇒ 18y - 3x = 9 .......(iii)
Adding (i) and (iii) we get,
⇒ 3x - 4y + (18y - 3x) = 5 + 9
Substituting value of y in (i) we get,
⇒ 3x - 4(1) = 5
Hence, x = 3 and y = 1.
