Rd Sharma 2022 for Class 9 Maths Chapter 2 - Exponents Of Real Numbers

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Page No 42:

Question 1:

Simplify the following:

(i) $3 {(a^{4} b^{3})}^{10} \times 5 {(a^{2} b^{2})}^{3}$

(ii) ${(2 x^{- 2} y^{3})}^{3}$

(iii) $\frac{(4 \times 10^{7}) (6 \times 10^{- 5})}{8 \times 10^{4}}$

(iv) $\frac{4 a b^{2} (- 5 a b^{3})}{10 a^{2} b^{2}}$

(v) ${(\frac{x^{2} y^{2}}{a^{2} b^{3}})}^{n}$

(vi) $\frac{{(a^{3 n - 9})}^{6}}{a^{2 n - 4}}$

Answer:

(i)
$3 {(a^{4} b^{3})}^{10} \times 5 {(a^{2} b^{2})}^{3} = 3 \times a^{40} \times b^{30} \times 5 \times a^{6} \times b^{6} = 15 \times a^{40} \times a^{6} \times b^{30} \times b^{6} = 15 \times a^{40 + 6} \times b^{30 + 6} [a^{m} \times a^{n} = a^{m + n}] = 15 a^{46} b^{36}$

(ii)
${(2 x^{- 2} y^{3})}^{3} = 2^{3} \times {(x^{- 2})}^{3} \times {(y^{3})}^{3} = 8 \times x^{- 6} \times y^{9} = 8 x^{- 6} y^{9}$

(iii)
$\frac{(4 \times 10^{7}) (6 \times 10^{- 5})}{8 \times 10^{4}} = \frac{4 \times 10^{7} \times 6 \times 10^{- 5}}{8 \times 10^{4}} = \frac{24 \times 10^{7 + (- 5)}}{8 \times 10^{4}} = \frac{24 \times 10^{2}}{8 \times 10^{4}}$
$= \frac{24 \times 10^{2} \times 10^{- 4}}{8} = 3 \times 10^{2 + (- 4)} = 3 \times 10^{- 2} = \frac{3}{100}$

(iv)
$\frac{4 a b^{2} (- 5 a b^{3})}{10 a^{2} b^{2}} = \frac{4 \times a \times b^{2} \times (- 5) \times a \times b^{3}}{10 a^{2} b^{2}} = \frac{- 20 \times a^{1} \times a^{1} \times b^{2} \times b^{3}}{10 a^{2} b^{2}}$
$= \frac{- 20 \times a^{1 + 1} \times b^{2 + 3}}{10 a^{2} b^{2}} = - 2 \times a^{2} \times b^{5} \times a^{- 2} \times b^{- 2} = - 2 \times a^{2 + (- 2)} \times b^{5 + (- 2)} = - 2 \times a^{0} \times b^{3} = - 2 b^{3}$

(v)
${(\frac{x^{2} y^{2}}{a^{2} b^{3}})}^{n} = \frac{{(x^{2})}^{n} {(y^{2})}^{n}}{{(a^{2})}^{n} {(b^{3})}^{n}} = \frac{x^{2 n} y^{2 n}}{a^{2 n} b^{3 n}}$

(vi)
$\frac{{(a^{3 n - 9})}^{6}}{a^{2 n - 4}} = \frac{a^{6 (3 n - 9)}}{a^{2 n - 4}} = \frac{a^{(18 n - 54)}}{a^{2 n - 4}}$
$= a^{(18 n - 54)} \times a^{- (2 n - 4)} = a^{18 n - 54} \times a^{- 2 n + 4} = a^{18 n - 54 - 2 n + 4} = a^{16 n - 50}$

Page No 42:

Question 2:

If $a = 3$ and $b = - 2$ , find the values of:
(i) $a^{a} + b^{b}$
(ii) $a^{b} + b^{a}$
(iii) ${(a + b)}^{a b}$

Answer:

(i) $a^{a} + b^{b}$
Here, $a = 3$ and $b = - 2$ .
Put the values in the expression $a^{a} + b^{b}$ .
$3^{3} + {(- 2)}^{- 2} = 27 + \frac{1}{{(- 2)}^{2}} = 27 + \frac{1}{4} = \frac{108 + 1}{4} = \frac{109}{4}$

(ii) $a^{b} + b^{a}$
Here, $a = 3$ and $b = - 2$ .
Put the values in the expression $a^{b} + b^{a}$ .
$3^{- 2} + {(- 2)}^{3} = {(\frac{1}{3})}^{2} + (- 8) = \frac{1}{9} - 8 = \frac{1 - 72}{9} = - \frac{71}{9}$

(iii) ${(a + b)}^{a b}$
Here, $a = 3$ and $b = - 2$ .
Put the values in the expression ${(a + b)}^{a b}$ .
${[3 + (- 2)]}^{3 (- 2)} = {(1)}^{- 6} = 1$

Page No 42:

Question 3:

If x, y, a, b are positive real numbers, prove that:

(i) ${(\frac{x^{a}}{x^{b}})}^{c} \times {(\frac{x^{b}}{x^{c}})}^{a} \times {(\frac{x^{c}}{x^{a}})}^{b} = 1$

(ii) $\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1$

Answer:

(i)
${(\frac{x^{a}}{x^{b}})}^{c} \times {(\frac{x^{b}}{x^{c}})}^{a} \times {(\frac{x^{c}}{x^{a}})}^{b} = {(x^{a - b})}^{c} \times {(x^{b - c})}^{a} \times {(x^{c - a})}^{b} [∵ \frac{a^{m}}{a^{n}} = a^{m - n}, a \neq 0] = x^{(a - b) c} \times x^{(b - c) a} \times x^{(c - a) b} [∵ {(a^{m})}^{n} = a^{m n}] = x^{a c - b c} \times x^{a b - a c} \times x^{b c - a b} = x^{a c - b c + a b - a c + b c - a b} [∵ a^{m} \times a^{n} = a^{m + n}] = x^{0} = 1 [∵ a^{0} = 1, a \neq 0]$

(ii)
$\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = \frac{1}{1 + \frac{x^{a}}{x^{b}}} + \frac{1}{1 + \frac{x^{b}}{x^{a}}} [∵ \frac{a^{m}}{a^{n}} = a^{m - n}, a \neq 0] = \frac{1}{\frac{x^{b} + x^{a}}{x^{b}}} + \frac{1}{\frac{x^{a} + x^{b}}{x^{a}}} = \frac{x^{b}}{x^{a} + x^{b}} + \frac{x^{a}}{x^{a} + x^{b}} = \frac{x^{a} + x^{b}}{x^{a} + x^{b}} = 1$

Page No 42:

Question 4:

Solve the following equations for x:

(i) $7^{2 x + 3} = 1$

(ii) $2^{x + 1} = 4^{x - 3}$

(iii) $2^{5 x + 3} = 8^{x + 3}$

(iv) $4^{2 x} = \frac{1}{32}$

(v) $2^{3 x - 7} = 256$

Answer:

(i)
$7^{2 x + 3} = 1 \Rightarrow 7^{2 x + 3} = 7^{0} \Rightarrow 2 x + 3 = 0 \Rightarrow 2 x = - 3 \Rightarrow x = - \frac{3}{2}$

(ii)
$2^{x + 1} = 4^{x - 3} \Rightarrow 2^{x + 1} = {(2^{2})}^{x - 3} \Rightarrow 2^{x + 1} = (2^{2 x - 6}) \Rightarrow x + 1 = 2 x - 6 \Rightarrow x = 7$

(iii)
$2^{5 x + 3} = 8^{x + 3} \Rightarrow 2^{5 x + 3} = {(2^{3})}^{x + 3} \Rightarrow 2^{5 x + 3} = 2^{3 x + 9} \Rightarrow 5 x + 3 = 3 x + 9 \Rightarrow 2 x = 6 \Rightarrow x = 3$

(iv)
$4^{2 x} = \frac{1}{32} \Rightarrow {(2^{2})}^{2 x} = \frac{1}{2^{5}} \Rightarrow 2^{4 x} \times 2^{5} = 1 \Rightarrow 2^{4 x + 5} = 2^{0} \Rightarrow 4 x + 5 = 0 \Rightarrow x = - \frac{5}{4}$

(v)
$2^{3 x - 7} = 256 \Rightarrow 2^{3 x - 7} = 2^{8} \Rightarrow 3 x - 7 = 8 \Rightarrow 3 x = 15 \Rightarrow x = 5$

Page No 42:

Question 5:

Prove that:

(i) $\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = a b c$

(ii) ${(a^{- 1} + b^{- 1})}^{- 1} = \frac{a b}{a + b}$

(iii) ${(\frac{x^{a}}{x^{b}})}^{a^{2} + a b + b^{2}} \times {(\frac{x^{b}}{x^{c}})}^{b^{2} + b c + c^{2}} \times {(\frac{x^{c}}{x^{a}})}^{c^{2} + c a + a^{2}} = 1$

Answer:

(i) Consider the left hand side:
$\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = \frac{a + b + c}{\frac{1}{a b} + \frac{1}{b c} + \frac{1}{c a}} = \frac{a + b + c}{\frac{c + a + b}{a b c}} = (a + b + c) \times (\frac{a b c}{a + b + c}) = a b c$
Therefore left hand side is equal to the right hand side. Hence proved.

(ii)
Consider the left hand side:
${(a^{- 1} + b^{- 1})}^{- 1} = \frac{1}{a^{- 1} + b^{- 1}} = \frac{1}{\frac{1}{a} + \frac{1}{b}} = \frac{1}{\frac{b + a}{a b}} = \frac{a b}{a + b}$
Therefore left hand side is equal to the right hand side.
Hence proved.

(iii)
${(\frac{x^{a}}{x^{b}})}^{a^{2} + a b + b^{2}} \times {(\frac{x^{b}}{x^{c}})}^{b^{2} + b c + c^{2}} \times {(\frac{x^{c}}{x^{a}})}^{c^{2} + c a + a^{2}} = {(x^{a - b})}^{a^{2} + a b + b^{2}} \times {(x^{b - c})}^{b^{2} + b c + b^{2}} \times {(x^{c - a})}^{c^{2} + c a + a^{2}} = x^{(a - b) (a^{2} + a b + b^{2})} \times x^{(b - c) (b^{2} + b c + c^{2})} \times x^{(c - a) (c^{2} + c a + a^{2})} = x^{a^{3} - b^{3} + b^{3} - c^{3} + c^{3} - a^{3}} = x^{0} = 1$

Hence proved.

Page No 42:

Question 6:

Simplify the following:

(i) $\frac{3^{n} \times 9^{n + 1}}{3^{n - 1} \times 9^{n - 1}}$

(ii) $\frac{5 \times 25^{n + 1} - 25 \times 5^{2 n}}{5 \times 5^{2 n + 3} - 25^{n + 1}}$

(iii) $\frac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^{x} - 2^{2} \times 5^{n}}$

(iv) $\frac{6 {(8)}^{n + 1} + 16 {(2)}^{3 n - 2}}{10 {(2)}^{3 n + 1} - 7 {(8)}^{n}}$

Answer:

(i)
$\frac{3^{n} \times 9^{n + 1}}{3^{n - 1} \times 9^{n - 1}} = \frac{3^{n} \times {(3^{2})}^{(n + 1)}}{3^{n - 1} \times {(3^{2})}^{n - 1}} = \frac{3^{n} \times 3^{2 n + 2}}{3^{n - 1} \times 3^{2 n - 2}}$
$= \frac{3^{n + 2 n + 2}}{3^{n - 1 + 2 n - 2}} = \frac{3^{3 n + 2}}{3^{3 n - 3}} = 3^{3 n + 2 - 3 n + 3} = 3^{5} = 243$

(ii)
$\frac{5 \times 25^{n + 1} - 25 \times 5^{2 n}}{5 \times 5^{2 n + 3} - 25^{n + 1}} = \frac{5 \times {(5^{2})}^{n + 1} - (5^{2}) \times 5^{2 n}}{5 \times 5^{2 n + 3} - {(5^{2})}^{n + 1}} = \frac{5 \times (5^{2 n + 2}) - (5^{2}) \times 5^{2 n}}{5 \times 5^{2 n + 3} - (5^{2 n + 2})} = \frac{5^{1 + 2 n + 2} - 5^{2 + 2 n}}{5^{1 + 2 n + 3} - 5^{2 n + 2}}$
$= \frac{5^{2 n + 3} - 5^{2 + 2 n}}{5^{2 n + 4} - 5^{2 n + 2}} = \frac{5^{2 + 2 n} (5 - 1)}{5^{2 n + 2} (5^{2} - 1)} = \frac{4}{24} = \frac{1}{6}$

(iii)
$\frac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^{n} - 2^{2} \times 5^{n}} = \frac{5^{n + 1} (5^{2} - 6)}{5^{n} (9 - 2^{2})} = \frac{5^{n} \times 5 \times (25 - 6)}{5^{n} (9 - 4)} = \frac{5 \times 19}{5} = 19$

(iv)
$\frac{6 {(8)}^{n + 1} + 16 {(2)}^{3 n - 2}}{10 {(2)}^{3 n + 1} - 7 {(8)}^{n}} = \frac{6 {(2^{3})}^{n + 1} + 16 {(2)}^{3 n - 2}}{10 {(2)}^{3 n + 1} - 7 {(2^{3})}^{n}} = \frac{6 (2^{3 n + 3}) + 16 {(2)}^{3 n - 2}}{10 {(2)}^{3 n + 1} - 7 (2^{3 n})}$
$= \frac{6 \times 2^{3 n} (2^{3}) + 16 {(2)}^{3 n} 2^{- 2}}{10 {(2)}^{3 n} (2^{1}) - 7 (2^{3 n})} = \frac{2^{3 n} (48 + 4)}{2^{3 n} (20 - 7)} = \frac{52}{13} = 4$

Page No 42:

Question 7:

Solve the following equations for x:

(i) $2^{2 x} - 2^{x + 3} + 2^{4} = 0$

(ii) $3^{2 x + 4} + 1 = 2 . 3^{x + 2}$

(iii) $4^{x - 1} \times {(0.5)}^{3 - 2 x} = {(\frac{1}{8})}^{x}$

Answer:

(i)
$2^{2 x} - 2^{x + 3} + 2^{4} = 0 \Rightarrow {(2^{x})}^{2} - (2^{x} \times 2^{3}) + {(2^{2})}^{2} = 0 \Rightarrow {(2^{x})}^{2} - 2 \times 2^{x} \times 2^{2} + {(2^{2})}^{2} = 0 \Rightarrow {(2^{x} - 2^{2})}^{2} = 0 \Rightarrow 2^{x} - 2^{2} = 0 \Rightarrow 2^{x} = 2^{2} \Rightarrow x = 2$

(ii)
$3^{2 x + 4} + 1 = 2 . 3^{x + 2} \Rightarrow {(3^{x + 2})}^{2} - 2 . 3^{x + 2} + 1 = 0 \Rightarrow {(3^{x + 2} - 1)}^{2} = 0 \Rightarrow 3^{x + 2} - 1 = 0 \Rightarrow 3^{x + 2} = 1 = 3^{0} \Rightarrow x + 2 = 0 \Rightarrow x = - 2$

(iii)
$4^{x - 1} \times {(0.5)}^{3 - 2 x} = {(\frac{1}{8})}^{x} \Rightarrow {(2^{2})}^{x - 1} \times {(\frac{1}{2})}^{3 - 2 x} = {[{(\frac{1}{2})}^{3}]}^{x} \Rightarrow 2^{2 x - 2} \times 2^{2 x - 3} = 2^{- 3 x} \Rightarrow 2^{2 x - 2 + 2 x - 3} = 2^{- 3 x} \Rightarrow 2^{4 x - 5} = 2^{- 3 x} \Rightarrow 4 x - 5 = - 3 x \Rightarrow x = \frac{5}{7}$

Page No 42:

Question 8:

If $49392 = a^{4} b^{2} c^{3}$ , find the values of a, b and c, where a, b and c are different positive primes.

Answer:

First find out the prime factorisation of 49392.

$\begin{matrix} 2 & 49392 \\ 2 & 24696 \\ 2 & 12348 \\ 2 & 6174 \\ 3 & 3087 \\ 3 & 1029 \\ 7 & 343 \\ 7 & 49 \\ 7 & 7 \\ 1 \end{matrix}$

It can be observed that 49392 can be written as $2^{4} \times 3^{2} \times 7^{3}$ , where 2, 3 and 7 are positive primes.
$∴ 49392 = 2^{4} 3^{2} 7^{3} = a^{4} b^{2} c^{3} \Rightarrow a = 2, b = 3, c = 7$

Page No 42:

Question 9:

If $1176 = 2^{a} 3^{b} 7^{c}$ , find a, b and c.

Answer:

First find out the prime factorisation of 1176.
$\begin{matrix} 2 & 1176 \\ 2 & 588 \\ 2 & 294 \\ 3 & 147 \\ 7 & 49 \\ 7 & 7 \\ 1 \end{matrix}$

It can be observed that 1176 can be written as $2^{3} \times 3^{1} \times 7^{2}$ .
$1176 = 2^{3} 3^{1} 7^{2} = 2^{a} 3^{b} 7^{c}$
Hence, a = 3, b = 1 and c = 2.

Page No 42:

Question 10:

Given $4725 = 3^{a} 5^{b} 7^{c}$ , find
(i) the integral values of a, b and c
(ii) the value of $2^{- a} 3^{b} 7^{c}$

Answer:

(i) Given $4725 = 3^{a} 5^{b} 7^{c}$
First find out the prime factorisation of 4725.
$\begin{matrix} 3 & 4725 \\ 3 & 1575 \\ 3 & 525 \\ 5 & 175 \\ 5 & 35 \\ 7 & 7 \\ 1 \end{matrix}$
It can be observed that 4725 can be written as $3^{3} \times 5^{2} \times 7^{1}$ .
$∴ 4725 = 3^{a} 5^{b} 7^{c} = 3^{3} 5^{2} 7^{1}$
Hence, a = 3, b = 2 and c = 1.

(ii)
When a = 3, b = 2 and c = 1,
$2^{- a} 3^{b} 7^{c} = 2^{- 3} \times 3^{2} \times 7^{1} = \frac{1}{8} \times 9 \times 7 = \frac{63}{8}$
Hence, the value of $2^{- a} 3^{b} 7^{c}$ is $\frac{63}{8}$ .

Page No 42:

Question 11:

If $a = x y^{p - 1}, b = x y^{q - 1} and c = x y^{r - 1}$ , prove that $a^{q - r} b^{r - p} c^{p - q} = 1$ .

Answer:

It is given that $a = x y^{p - 1}, b = x y^{q - 1} and c = x y^{r - 1}$ .
$∴ a^{q - r} b^{r - p} c^{p - q} = {(x y^{p - 1})}^{q - r} {(x y^{q - 1})}^{r - p} {(x y^{r - 1})}^{p - q} = x^{(q - r)} y^{(p - 1) (q - r)} x^{(r - p)} y^{(r - p) (q - 1)} x^{(p - q)} y^{(p - q) (r - 1)} = x^{(q - r)} x^{(r - p)} x^{(p - q)} y^{(p - 1) (q - r)} y^{(r - p) (q - 1)} y^{(p - q) (r - 1)} = x^{(q - r) + (r - p) + (p - q)} y^{(p - 1) (q - r) + (r - p) (q - 1) + (p - q) (r - 1)} = x^{q - r + r - p + p - q} y^{p q - q - p r + r + r q - r - p q + p + p r - p - q r + q} = x^{0} y^{0} = 1$

Page No 53:

Question 1:

Simplify:

(i) $(16^{- 1 / 5})^{5 / 2}$

(ii) $\sqrt[5]{{(32)}^{- 3}}$

(iii) $\sqrt[3]{(343)^{- 2}}$

(iv) $(0.001)^{1 / 3}$

(v) $\begin{array}{rcl} {[5 {(8^{\frac{1}{3}} + 27^{\frac{1}{3}})}^{3}]}^{^{\frac{1}{4}}} \end{array}$

(vi) ${(\frac{\sqrt{2}}{5})}^{8} \div {(\frac{\sqrt{2}}{5})}^{13}$

(vii) ${(\frac{5^{- 1} \times 7^{2}}{5^{2} \times 7^{- 4}})}^{7 / 2} \times {(\frac{5^{- 2} \times 7^{3}}{5^{3} \times 7^{- 5}})}^{- 5 / 2}$

(viii) $64^{\frac{- 1}{3}} (64^{\frac{1}{3}} - 64^{\frac{2}{3}})$

(ix) ${(1^{3} + 2^{3} + 3^{3})}^{\frac{1}{2}}$

Answer:

(i) Given ${(16^{- \frac{1}{5}})}^{\frac{5}{2}}$

By using law of rational exponents we have

Hence, the value of is.
(ii) $\sqrt[5]{{(32)}^{- 3}}$

$= \sqrt[5]{{(\frac{1}{32})}^{3}} = {(\frac{1}{32})}^{\frac{3}{5}} = {(\frac{1}{{(2)}^{5}})}^{\frac{3}{5}} = {(\frac{1}{2})}^{5 \times \frac{3}{5}}$
$= {(\frac{1}{2})}^{3} = (\frac{1}{8})$

(iii) Given $\sqrt[3]{(343)^{- 2}}$

Hence, the value of is.

(iv) Given $(0.001)^{\frac{1}{3}}$

Hence, the value of $(0.001)^{\frac{1}{3}}$ is.

(v) Given $\begin{array}{rcl} {[5 {(8^{\frac{1}{3}} + 27^{\frac{1}{3}})}^{3}]}^{^{\frac{1}{4}}} \end{array}$

$\begin{array}{rcl} {[5 {(8^{\frac{1}{3}} + 27^{\frac{1}{3}})}^{3}]}^{^{\frac{1}{4}}} & = & {[5 {({(2^{3})}^{\frac{1}{3}} + {(3^{3})}^{\frac{1}{3}})}^{3}]}^{^{\frac{1}{4}}} \\ = & {[5 {(2 + 3)}^{3}]}^{^{\frac{1}{4}}} \\ = & {[5 {(5)}^{3}]}^{^{\frac{1}{4}}} \\ = & {[5^{4}]}^{^{\frac{1}{4}}} \\ = & 5^{4 \times \frac{1}{4}} \\ = & 5 \end{array}$

Hence the value of $\begin{array}{rcl} {[5 {(8^{\frac{1}{3}} + 27^{\frac{1}{3}})}^{3}]}^{^{\frac{1}{4}}} \end{array}$ is 5.

(vi) Given ${(\frac{\sqrt{2}}{5})}^{8} \div {(\frac{\sqrt{2}}{5})}^{13}$ .

By using the law of rational exponents

Hence the value of is

(vii) Given ${(\frac{5^{- 1} \times 7^{2}}{5^{2} \times 7^{- 4}})}^{7 / 2} \times {(\frac{5^{- 2} \times 7^{3}}{5^{3} \times 7^{- 5}})}^{- 5 / 2}$ .

Hence, the value of ${(\frac{5^{- 1} \times 7^{2}}{5^{2} \times 7^{- 4}})}^{7 / 2} \times {(\frac{5^{- 2} \times 7^{3}}{5^{3} \times 7^{- 5}})}^{- 5 / 2}$ is .

(viii)
$\begin{array}{rcl} 6 4^{\frac{- 1}{3}} [64^{\frac{1}{3}} - 64^{\frac{2}{3}}] & = & {(4^{3})}^{- \frac{1}{3}} [{(4^{3})}^{\frac{1}{3}} - {(4^{3})}^{\frac{2}{3}}] \\ = & (4^{3 \times \frac{- 1}{3}}) [4^{3 \times \frac{1}{3}} - 4^{3 \times \frac{2}{3}}] \\ = & \frac{1}{4} [4 - 16] = \frac{- 12}{4} = - 3 \end{array}$

(ix)
$\begin{array}{rcl} {(1^{3} + 2^{3} + 3^{3})}^{\frac{1}{2}} & = & \sqrt{(1 + 8 + 27)} \\ = & \sqrt{36} \\ = & 6^{2 \times \frac{1}{2}} [∵ {(a^{m})}^{n} = a^{m n}] \\ = & 6 \end{array}$

Page No 53:

Question 2:

Show that:

(i) ${(x^{a - b})}^{a + b} {(x^{b - c})}^{b + c} {(x^{c - a})}^{c + a} = 1$

(ii) ${(\frac{3^{a}}{3^{b}})}^{a + b} {(\frac{3^{b}}{3^{c}})}^{b + c} {(\frac{3^{c}}{3^{a}})}^{c + a} = 1$

Answer:

(i)
$\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}}$
$= \frac{1}{1 + \frac{x^{a}}{x^{b}}} + \frac{1}{1 + \frac{x^{b}}{x^{a}}} = \frac{x^{b}}{x^{b} + x^{a}} + \frac{x^{a}}{x^{a} + x^{b}} = \frac{x^{b} + x^{a}}{x^{a} + x^{b}} = 1$

(ii)
${(\frac{3^{a}}{3^{b}})}^{a + b} {(\frac{3^{b}}{3^{c}})}^{b + c} {(\frac{3^{c}}{3^{a}})}^{c + a} = {(3^{a - b})}^{a + b} {(3^{b - c})}^{b + c} {(3^{c - a})}^{c + a} = 3^{a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}} = 3^{0} = 1$

Page No 53:

Question 3:

If $27^{x} = \frac{9}{3^{x}},$ find x.

Answer:

We are given. We have to find the value of

Since

By using the law of exponents we get,

On equating the exponents we get,

Hence,

Page No 53:

Question 4:

Find the values of x in each of the following:

(i) $2^{5 x} \div 2 x = \sqrt[5]{2^{20}}$

(ii) $(2^{3})^{4} = (2^{2})^{x}$

(iii) ${(\frac{3}{5})}^{x} {(\frac{5}{3})}^{2 x} = \frac{125}{27}$

(iv) $5^{x - 2} \times 3^{2 x - 3} = 135$

(v) $2^{x - 7} \times 5^{x - 4} = 1250$

(vi) ${(\sqrt[3]{4})}^{2 x + \frac{1}{2}} = \frac{1}{32}$

(vii) $5^{2 x + 3} = 1$

(viii) ${(13)}^{\sqrt{x}} = 4^{4} - 3^{4} - 6$

(ix) ${(\sqrt{\frac{3}{5}})}^{x + 1} = \frac{125}{27}$

Answer:

From the following we have to find the value of x

(i) Given

By using rational exponents

On equating the exponents we get,

The value of x is

(ii) Given

On equating the exponents

Hence the value of x is

(iii) Given

Comparing exponents we have,

Hence the value of x is

(iv) Given

On equating the exponents of 5 and 3 we get,

And,

The value of x is

(v) Given

On equating the exponents we get

And,

Hence the value of x is

(vi) ${(\sqrt[3]{4})}^{2 x + \frac{1}{2}} = \frac{1}{32}$

${(2^{2})}^{\frac{1}{3} (\frac{4 x + 1}{2})} = {(\frac{1}{2})}^{5} \Rightarrow 2^{(\frac{4 x + 1}{3})} = 2^{- 5}$

On comparing we get,
$\frac{4 x + 1}{3} = - 5 \Rightarrow 4 x + 1 = - 15 \Rightarrow 4 x = - 16 \Rightarrow x = - 4$

(vii) $5^{2 x + 3} = 1$

$5^{2 x + 3} = 5^{0} \Rightarrow 2 x + 3 = 0 \Rightarrow x = \frac{- 3}{2}$

(viii) ${(13)}^{\sqrt{x}} = 4^{4} - 3^{4} - 6$

${(13)}^{\sqrt{x}} = {(2^{2})}^{4} - 3^{4} - 6 \Rightarrow {(13)}^{\sqrt{x}} = 2^{8} - 3^{4} - 6 \Rightarrow {(13)}^{\sqrt{x}} = 256 - 81 - 6 \Rightarrow {(13)}^{\sqrt{x}} = 169 \Rightarrow {(13)}^{\sqrt{x}} = {(13)}^{2}$

On comparing we get,
$\sqrt{x} = 2 on squaring both sides we get, x = 4$

(ix) ${(\sqrt{\frac{3}{5}})}^{x + 1} = \frac{125}{27}$
${(\sqrt{\frac{3}{5}})}^{x + 1} = {(\frac{5}{3})}^{3} \Rightarrow {(\frac{3}{5})}^{\frac{x + 1}{2}} = {(\frac{3}{5})}^{- 3}$

On comparing we get,

$\frac{x + 1}{2} = - 3 \Rightarrow x + 1 = - 6 \Rightarrow x = - 7$

Page No 53:

Question 5:

If $x = 2^{\frac{1}{3}} + 2^{\frac{2}{3}}$ , show that x³ – 6x = 6.

Answer:

$x = 2^{\frac{1}{3}} + 2^{\frac{2}{3}}$
Cubing on both sides, we get
$x^{3} = {(2^{\frac{1}{3}} + 2^{\frac{2}{3}})}^{3} \Rightarrow x^{3} = {(2^{\frac{1}{3}})}^{3} + {(2^{\frac{2}{3}})}^{3} + 3 \times 2^{\frac{1}{3}} \times 2^{\frac{2}{3}} (2^{\frac{1}{3}} + 2^{\frac{2}{3}}) \Rightarrow x^{3} = 2 + 2^{2} + 3 \times 2^{\frac{1 + 2}{3}} (x) \Rightarrow x^{3} = 2 + 4 + 3 \times 2 x \Rightarrow x^{3} - 6 x = 6$

Page No 53:

Question 6:

Determine ${(8 x)}^{x}, if 9^{x + 2} = 240 + 9^{x}$ .

Answer:

$9^{x + 2} = 240 + 9^{x}$
$\Rightarrow 9^{x} \times 9^{2} = 240 + 9^{x} \Rightarrow 9^{x} (9^{2} - 1) = 240 \Rightarrow 9^{x} (81 - 1) = 240 \Rightarrow 9^{x} \times 80 = 240 \Rightarrow 9^{x} = 3 \Rightarrow 3^{2 x} = 3^{1} \Rightarrow 2 x = 1 \Rightarrow x = \frac{1}{2}$
$∴ {(8 x)}^{x} = {(8 \times \frac{1}{2})}^{\frac{1}{2}} = {(4)}^{\frac{1}{2}} = 2$

Page No 53:

Question 7:

If $3^{x + 1} = 9^{x - 2}$ , find the value of 2^1+x.

Answer:

$3^{x + 1} = 9^{x - 2}$
$\Rightarrow 3^{x} \times 3 = \frac{9^{x}}{9^{2}} \Rightarrow 3^{x} \times 3 = \frac{{(3^{2})}^{x}}{{(3^{2})}^{2}} = \frac{3^{2 x}}{3^{4}} \Rightarrow 3^{4} \times 3 = \frac{3^{2 x}}{3^{x}} \Rightarrow 3^{5} = 3^{2 x - x}$
$\Rightarrow 3^{5} = 3^{x}$
Comparing both sides, we get
x = 5
So, $2^{1 + x} = 2^{1 + 5} = 2^{6} = 64$

Page No 53:

Question 8:

If $3^{4 x} = {(81)}^{- 1}$ and $10^{\frac{1}{y}} = 0.0001$ , find the value of $2^{- x + 4 y}$ .

Answer:

It is given that $3^{4 x} = {(81)}^{- 1}$ and $10^{\frac{1}{y}} = 0.0001$ .
Now,
$3^{4 x} = {(81)}^{- 1} \Rightarrow 3^{4 x} = {(3^{4})}^{- 1} \Rightarrow {(3^{x})}^{4} = {(3^{- 1})}^{4} \Rightarrow x = - 1$
And,
$10^{\frac{1}{y}} = 0.0001 \Rightarrow 10^{\frac{1}{y}} = \frac{1}{10000} \Rightarrow 10^{\frac{1}{y}} = {(\frac{1}{10})}^{4} \Rightarrow 10^{\frac{1}{y}} = {(10)}^{- 4} \Rightarrow \frac{1}{y} = - 4 \Rightarrow y = - \frac{1}{4}$
Therefore, the value of $2^{- x + 4 y}$ is $2^{1 + 4 (- \frac{1}{4})} = 2^{0} = 1$ .

Page No 53:

Question 9:

If $5^{3 x} = 125 and 10^{y} = 0.001$ , find x and y.

Answer:

It is given that $5^{3 x} = 125 and 10^{y} = 0.001$ .
Now,
$5^{3 x} = 125 \Rightarrow 5^{3 x} = 5^{3} \Rightarrow 3 x = 3 \Rightarrow x = 1$
And,
$10^{y} = 0.001 \Rightarrow 10^{y} = \frac{1}{1000} \Rightarrow 10^{y} = 10^{- 3} \Rightarrow y = - 3$
Hence, the values of x and y are 1 and −3, respectively.

Page No 53:

Question 10:

Solve the following equations:
(i) $3^{x + 1} = 27 \times 3^{4}$
(ii) $4^{2 x} = {(\sqrt[3]{16})}^{- \frac{6}{y}} = {(\sqrt{8})}^{2}$
(iii) $3^{x - 1} \times 5^{2 y - 3} = 225$
(iv) $8^{x + 1} = 16^{y + 2} and, {(\frac{1}{2})}^{3 + x} = {(\frac{1}{4})}^{3 y}$
(v) $4^{x - 1} \times {(0.5)}^{3 - 2 x} = {(\frac{1}{8})}^{x}$

(vi) $\sqrt{\frac{a}{b}} = {(\frac{b}{a})}^{1 - 2 x}$ , where a and b are distinct primes

Answer:

(i)
$3^{x + 1} = 27 \times 3^{4} \Rightarrow 3^{x + 1} = 3^{3} \times 3^{4} \Rightarrow 3^{x + 1} = 3^{3 + 4} \Rightarrow 3^{x + 1} = 3^{7} \Rightarrow x + 1 = 7 \Rightarrow x = 6$

(ii)
$4^{2 x} = {(\sqrt[3]{16})}^{- \frac{6}{y}} = {(\sqrt{8})}^{2} \Rightarrow 4^{2 x} = {(\sqrt{8})}^{2} and {(\sqrt[3]{16})}^{- \frac{6}{y}} = {(\sqrt{8})}^{2} \Rightarrow 4^{2 x} = (8^{\frac{1}{2} \times 2}) and (16^{\frac{1}{3} \times - \frac{6}{y}}) = (8^{\frac{1}{2} \times 2}) \Rightarrow 4^{2 x} = 8 and (16^{- \frac{2}{y}}) = 8 \Rightarrow 2^{4 x} = 2^{3} and (2^{- \frac{8}{y}}) = 2^{3} \Rightarrow 4 x = 3 and - \frac{8}{y} = 3 \Rightarrow x = \frac{3}{4} and y = - \frac{8}{3}$

(iii)
$3^{x - 1} \times 5^{2 y - 3} = 225 \Rightarrow 3^{x - 1} \times 5^{2 y - 3} = 3 \times 3 \times 5 \times 5 \Rightarrow 3^{x - 1} \times 5^{2 y - 3} = 3^{2} \times 5^{2} \Rightarrow x - 1 = 2 and 2 y - 3 = 2 \Rightarrow x = 3 and y = \frac{5}{2}$

(iv)
$8^{x + 1} = 16^{y + 2} and, {(\frac{1}{2})}^{3 + x} = {(\frac{1}{4})}^{3 y} \Rightarrow {(2^{3})}^{x + 1} = {(2^{4})}^{y + 2} and {(\frac{1}{2})}^{3 + x} = {(\frac{1}{2^{2}})}^{3 y} \Rightarrow {(2)}^{3 x + 3} = {(2)}^{4 y + 8} and {(\frac{1}{2})}^{3 + x} = {(\frac{1}{2})}^{6 y} \Rightarrow 3 x + 3 = 4 y + 8 and 3 + x = 6 y$
Now,
$3 + x = 6 y \Rightarrow x = 6 y - 3$
Putting x = 6y − 3 in $3 x - 4 y = 5$ , we get
$3 (6 y - 3) - 4 y = 5 \Rightarrow 18 y - 9 - 4 y = 5 \Rightarrow 14 y = 14 \Rightarrow y = 1$
Putting y = 1 in $x = 6 y - 3$ , we get
$x = 6 \times 1 - 3 = 3$

(v)
$4^{x - 1} \times {(0.5)}^{3 - 2 x} = {(\frac{1}{8})}^{x} \Rightarrow {(2^{2})}^{x - 1} \times {(\frac{1}{2})}^{3 - 2 x} = {(\frac{1}{2^{3}})}^{x} \Rightarrow {(2)}^{2 x - 2} \times {(2)}^{- (3 - 2 x)} = {(2)}^{- 3 x} \Rightarrow {(2)}^{2 x - 2 - 3 + 2 x} = {(2)}^{- 3 x} \Rightarrow 4 x - 5 = - 3 x \Rightarrow 7 x = 5 \Rightarrow x = \frac{5}{7}$

(vi)
$\sqrt{\frac{a}{b}} = {(\frac{b}{a})}^{1 - 2 x} \Rightarrow {(\frac{a}{b})}^{\frac{1}{2}} = {(\frac{a}{b})}^{- (1 - 2 x)} \Rightarrow \frac{1}{2} = - (1 - 2 x) \Rightarrow \frac{1}{2} = 2 x - 1 \Rightarrow \frac{3}{2} = 2 x \Rightarrow x = \frac{3}{4}$

Page No 53:

Question 11:

If $2^{x} \times 3^{y} \times 5^{z} = 2160$ , find x , y and z. Hence, compute the value of $3^{x} \times 2^{- y} \times 5^{- z}$ .

Answer:

Given: $2^{x} \times 3^{y} \times 5^{z} = 2160$
First, find out the prime factorisation of 2160.
$\begin{matrix} 2 & 2160 \\ 2 & 1080 \\ 2 & 540 \\ 2 & 270 \\ 3 & 135 \\ 3 & 45 \\ 3 & 15 \\ 5 & 5 \\ 1 \end{matrix}$
It can be observed that 2160 can be written as $2^{4} \times 3^{3} \times 5^{1}$ .
Also,
$2^{x} \times 3^{y} \times 5^{z} = 2^{4} \times 3^{3} \times 5^{1} \Rightarrow x = 4, y = 3, z = 1$
Therefore, the value of $3^{x} \times 2^{- y} \times 5^{- z}$ is $3^{4} \times 2^{- 3} \times 5^{- 1} = 81 \times \frac{1}{8} \times \frac{1}{5} = \frac{81}{40}$ .

Page No 53:

Question 12:

If $1176 = 2^{a} \times 3^{b} \times 7^{c}$ , find the values of a, b and c. Hence, compute the value of $2^{a} \times 3^{b} \times 7^{- c}$ as a fraction.

Answer:

First find the prime factorisation of 1176.
$\begin{matrix} 2 & 1176 \\ 2 & 588 \\ 2 & 294 \\ 3 & 147 \\ 7 & 49 \\ 7 & 7 \\ 1 \end{matrix}$
It can be observed that 1176 can be written as $2^{3} \times 3^{1} \times 7^{2}$ .
$1176 = 2^{a} 3^{b} 7^{c} = 2^{3} 3^{1} 7^{2}$
So, a = 3, b = 1 and c = 2.
Therefore, the value of $2^{a} \times 3^{b} \times 7^{- c}$ is $2^{3} \times 3^{1} \times 7^{- 2} = 8 \times 3 \times \frac{1}{49} = \frac{24}{49}$

Page No 54:

Question 13:

Assuming that x, y, z are positive real numbers, simplify each of the following:

(i) ${(\sqrt{x^{- 3}})}^{5}$

(ii) $\sqrt{x^{3} y^{- 2}}$

(iii) $\sqrt[5]{243 x^{10} y^{5} z^{10}}$

(iv) ${(\frac{x^{- 4}}{y^{- 10}})}^{5 / 4}$

(v) ${(\frac{\sqrt{2}}{\sqrt{3}})}^{5} {(\frac{6}{7})}^{2}$

Answer:

We have to simplify the following, assuming thatare positive real numbers

(i) Given

As x is positive real number then we have

Hence the simplified value of is.

(ii) Given

As x and y are positive real numbers then we can write

By using law of rational exponents we have

Hence the simplified value of is .

(iii) Given $\sqrt[5]{243 x^{10} y^{5} z^{10}}$

$\begin{array}{rcl} \sqrt[5]{243 x^{10} y^{5} z^{10}} & = & {(243 x^{10} y^{5} z^{10})}^{\frac{1}{5}} \\ = & {(3^{5} x^{10} y^{5} z^{10})}^{\frac{1}{5}} \\ = & {(3^{5})}^{\frac{1}{5}} {(x^{10})}^{\frac{1}{5}} {(y^{5})}^{\frac{1}{5}} {(z^{10})}^{\frac{1}{5}} \\ = & 3^{5 \times \frac{1}{5}} \times x^{10 \times \frac{1}{5}} \times y^{5 \times \frac{1}{5}} \times z^{10 \times \frac{1}{5}} \\ = & 3 x^{2} y z^{2} \end{array}$

Hence, the value of $\sqrt[5]{243 x^{10} y^{5} z^{10}}$ is $\begin{array}{rcl} 3 x^{2} y z^{2} \end{array}$ .

(iv) Given ${(\frac{x^{- 4}}{y^{- 10}})}^{5 / 4}$

$\begin{array}{rcl} {(\frac{x^{- 4}}{y^{- 10}})}^{5 / 4} & = & {(\frac{y^{10}}{x^{4}})}^{5 / 4} \\ = & \frac{{(y^{10})}^{5 / 4}}{{(x^{4})}^{5 / 4}} \\ = & \frac{y^{10 \times \frac{5}{4}}}{x^{4 \times \frac{5}{4}}} \\ = & \frac{y^{\frac{25}{2}}}{x^{5}} \end{array}$

Hence, the value of ${(\frac{x^{- 4}}{y^{- 10}})}^{5 / 4}$ is $\begin{array}{rcl} \frac{y^{\frac{25}{2}}}{x^{5}} \end{array}$ .

(v) ${(\frac{\sqrt{2}}{\sqrt{3}})}^{5} {(\frac{6}{7})}^{2}$

${(\frac{\sqrt{2}}{\sqrt{3}})}^{5} {(\frac{6}{7})}^{2} = {(\frac{\sqrt{2}}{\sqrt{3}})}^{2 + 2 + 1} {(\frac{6}{7})}^{2} = {(\frac{\sqrt{2}}{\sqrt{3}})}^{2} \times {(\frac{\sqrt{2}}{\sqrt{3}})}^{2} \times {(\frac{\sqrt{2}}{\sqrt{3}})}^{1} \times {(\frac{6}{7})}^{2} = (\frac{2}{3}) \times (\frac{2}{3}) \times {(\frac{\sqrt{2}}{\sqrt{3}})}^{1} \times {(\frac{6}{7})}^{2}$
$= (\frac{2}{3}) \times (\frac{2}{3}) \times {(\frac{\sqrt{2}}{\sqrt{3}})}^{1} \times {(\frac{6}{7})}^{2} = \frac{16 \sqrt{2}}{49 \sqrt{3}} = \sqrt{\frac{512}{7203}} = {(\frac{512}{7203})}^{\frac{1}{2}}$

Page No 54:

Question 14:

Prove that:

(i) $\sqrt{3 \times 5^{- 3}} \div \sqrt[3]{3^{- 1}} \sqrt{5} \times \sqrt[6]{3 \times 5^{6}} = \frac{3}{5}$

(ii) $9^{3 / 2} - 3 \times 5^{0} - {(\frac{1}{81})}^{-}^{1 / 2} = 15$

(iii) ${(\frac{1}{4})}^{- 2} - 3 \times 8^{2 / 3} \times 4^{0} + {(\frac{9}{16})}^{- 1 / 2} = \frac{16}{3}$

(iv) $\frac{2^{1 / 2} \times 3^{1 / 3} \times 4^{1 / 4}}{10^{- 1 / 5} \times 5^{3 / 5}} \div \frac{3^{4 / 3} \times 5^{- 7 / 5}}{4^{- 3 / 5} \times 6} = 10$

(v) $\sqrt{\frac{1}{4} + (0.01)^{- 1 / 2}} - (27)^{2 / 3} = \frac{3}{2}$

(vi) $\frac{2^{n} + 2^{n - 1}}{2^{n + 1} - 2^{n}} = \frac{3}{2}$

(vii) ${(\frac{64}{125})}^{- 2 / 3} + \frac{1}{{(\frac{256}{625})}^{1 / 4}} + (\frac{\sqrt{25}}{\sqrt[3]{64}}) = \frac{65}{16}$

(viii) $\frac{3^{- 3} \times 6^{2} \times \sqrt{98}}{5^{2} \times \sqrt[3]{1 / 25} \times (15) -^{4 / 3} \times 3^{1 / 3}} = 28 \sqrt{2}$

(ix) $\frac{(0.6)^{0} - (0.1) - 1}{{(\frac{3}{8})}^{- 1} {(\frac{3}{2})}^{3} + {(- \frac{1}{3})}^{- 1}} = - \frac{3}{2}$

Answer:

(i) We have to prove that

By using rational exponent we get,

Hence,

(ii) We have to prove that. So,

Hence,

(iii) We have to prove that

Now,

Hence,

(iv) We have to prove that. So,

Let

Hence,

(v) We have to prove that

Let

Hence,

(vi) We have to prove that . So,

Let

Hence,

(vii) We have to prove that. So let

By taking least common factor we get

Hence,

(viii) We have to prove that. So,

Let

Hence,

(ix) We have to prove that. So,

Let

Hence,

Page No 54:

Question 15:

If x, y, z are positive real numbers, prove that:

(i) $\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1$

(ii) ${[\{\frac{x^{a (a - b)}}{x^{a (a + b)}}\} \div \{\frac{x^{b (b - a)}}{x^{b (b + a)}}\}]}^{a + b} = 1$

(iii) ${(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = 1$

(iv) ${\{{(x^{a - a^{- 1}})}^{\frac{1}{a - 1}}\}}^{\frac{a}{a + 1}} = x$

(v) ${(\frac{a^{x + 1}}{a^{y + 1}})}^{x + y} {(\frac{a^{y + 2}}{a^{z + 2}})}^{y + z} {(\frac{a^{z + 3}}{a^{x + 3}})}^{z + x} = 1$

(vi) ${(\frac{x^{a^{2} + b^{2}}}{x^{a b}})}^{a + b} {(\frac{x^{b^{2} + c^{2}}}{x^{b c}})}^{b + c} {(\frac{x^{c^{2} + a^{2}}}{x^{a c}})}^{a + c} = x^{2 (a^{3} + b^{3} + c^{3})}$

Answer:

(i)
$\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = \frac{1}{1 + \frac{x^{a}}{x^{b}}} + \frac{1}{1 + \frac{x^{b}}{x^{a}}} [∵ \frac{a^{m}}{a^{n}} = a^{m - n}, a \neq 0] = \frac{1}{\frac{x^{b} + x^{a}}{x^{b}}} + \frac{1}{\frac{x^{a} + x^{b}}{x^{a}}} = \frac{x^{b}}{x^{a} + x^{b}} + \frac{x^{a}}{x^{a} + x^{b}} = \frac{x^{a} + x^{b}}{x^{a} + x^{b}} = 1$

(ii)
${[\{\frac{x^{a (a - b)}}{x^{a (a + b)}}\} \div \{\frac{x^{b (b - a)}}{x^{b (b + a)}}\}]}^{a + b} = {[\{\frac{x^{a (a - b)}}{x^{a (a + b)}}\} \times \{\frac{x^{b (b + a)}}{x^{b (b - a)}}\}]}^{a + b} = {[\{\frac{x^{a^{2} - a b}}{x^{a^{2} + a b}}\} \times \{\frac{x^{b^{2} + a b}}{x^{b^{2} - a b}}\}]}^{a + b}$
$= {[\{x^{a^{2} - a b - a^{2} - a b}\} \times \{x^{b^{2} + a b - b^{2} + a b}\}]}^{a + b} = {[x^{- 2 a b} \times x^{2 a b}]}^{a + b} = {[x^{- 2 a b + 2 a b}]}^{a + b} = {[x^{0}]}^{a + b} = 1$

(iii)
${(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = {(x)}^{\frac{1}{a - b} \times \frac{1}{a - c}} {(x)}^{\frac{1}{b - c} \times}^{\frac{1}{b - a}} {(x)}^{\frac{1}{c - a} \times}^{\frac{1}{c - b}} = {(x)}^{\frac{1}{a - b} \times \frac{1}{a - c} + \frac{1}{b - c} \times \frac{1}{b - a} + \frac{1}{c - a} \times \frac{1}{c - b}} = {(x)}^{\frac{(b - c) - (a - c) + (a - b)}{(a - b) (a - c) (b - c)}} = x^{0} = 1$

(iv)
${\{{(x^{a - a^{- 1}})}^{\frac{1}{a - 1}}\}}^{\frac{a}{a + 1}} = \{{(x^{a - \frac{1}{a}})}^{\frac{1}{a - 1} \times \frac{a}{a + 1}}\} = {\{x^{\frac{a^{2} - 1}{a}}\}}^{\frac{a}{a^{2} - 1}} = x^{\frac{a^{2} - 1}{a} \times \frac{a}{a^{2} - 1}} = x^{1} = x$

(v)
${(\frac{a^{x + 1}}{a^{y + 1}})}^{x + y} {(\frac{a^{y + 2}}{a^{z + 2}})}^{y + z} {(\frac{a^{z + 3}}{a^{x + 3}})}^{z + x} = {(a^{x + 1 - y - 1})}^{x + y} {(a^{y + 2 - z - 2})}^{y + z} {(a^{z + 3 - x - 3})}^{z + x} = {(a^{x - y})}^{x + y} {(a^{y - z})}^{y + z} {(a^{z - x})}^{z + x} = a^{x^{2} - y^{2} + y^{2} - z^{2} + z^{2} - x^{2}} = a^{0} = 1$

(vi)
${(\frac{x^{a^{2} + b^{2}}}{x^{a b}})}^{a + b} {(\frac{x^{b^{2} + c^{2}}}{x^{b c}})}^{b + c} {(\frac{x^{c^{2} + a^{2}}}{x^{a c}})}^{a + c} = {(x^{a^{2} + b^{2} - a b})}^{a + b} {(x^{b^{2} + c^{2} - b c})}^{b + c} {(x^{c^{2} + a^{2} - a c})}^{a + c} = [x^{(a + b) (a^{2} + b^{2} - a b)}] [x^{(b + c) (b^{2} + c^{2} - b c)}] [x^{(a + c) (c^{2} + a^{2} - a c)}] = (x^{a^{3} + b^{3}}) (x^{b^{3} + c^{3}}) (x^{a^{3} + c^{3}}) = x^{2 (a^{3} + b^{3} + c^{3})}$

Page No 54:

Question 16:

If $2^{x} = 3^{y} = 12^{z}, show that \frac{1}{z} = \frac{1}{y} + \frac{2}{x}$ .

Answer:

Let $2^{x} = 3^{y} = 12^{z} = k$
$\Rightarrow 2 = k^{\frac{1}{x}}, 3 = k^{\frac{1}{y}}, 12 = k^{\frac{1}{z}}$
Now,
$12 = k^{\frac{1}{z}} \Rightarrow 2^{2} \times 3 = k^{\frac{1}{z}} \Rightarrow {(k^{\frac{1}{x}})}^{2} \times k^{\frac{1}{y}} = k^{\frac{1}{z}} \Rightarrow k^{\frac{2}{x} + \frac{1}{y}} = k^{\frac{1}{z}} \Rightarrow \frac{2}{x} + \frac{1}{y} = \frac{1}{z}$

Page No 54:

Question 17:

If $2^{x} = 3^{y} = 6^{- z}, show that \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ .

Answer:

Let $2^{x} = 3^{y} = 6^{- z} = k$
$\Rightarrow 2 = k^{\frac{1}{x}}, 3 = k^{\frac{1}{y}}, 6 = k^{\frac{1}{- z}}$
Now,
$6 = 2 \times 3 = k^{\frac{1}{- z}} \Rightarrow k^{\frac{1}{x}} \times k^{\frac{1}{y}} = k^{\frac{1}{- z}} \Rightarrow k^{\frac{1}{x} + \frac{1}{y}} = k^{\frac{1}{- z}} \Rightarrow \frac{1}{x} + \frac{1}{y} = \frac{1}{- z} \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

Page No 54:

Question 18:

If a^x= b^y= c^zand b² = ac, then show that $y = \frac{2 z x}{z + x}$ .

Answer:

Let $a^{x} = b^{y} = c^{z} = k$
So, $a = k^{\frac{1}{x}}, b = k^{\frac{1}{y}}, c = k^{\frac{1}{z}}$
Thus,
$b^{2} = a c \Rightarrow {(k^{\frac{1}{y}})}^{2} = (k^{\frac{1}{x}}) (k^{\frac{1}{z}}) \Rightarrow k^{\frac{2}{y}} = k^{\frac{1}{x} + \frac{1}{z}} \Rightarrow \frac{2}{y} = \frac{1}{x} + \frac{1}{z}$
$\Rightarrow \frac{2}{y} = \frac{z + x}{x z} \Rightarrow y = \frac{2 z x}{z + x}$

Page No 54:

Question 19:

If 3^x = 5^y = (75)^z, show that $z = \frac{x y}{2 x + y}$ .

Answer:

Let $3^{x} = 5^{y} = {(75)}^{z} = k$
$\Rightarrow 3 = k^{\frac{1}{x}}, 5 = k^{\frac{1}{y}}, 75 = k^{\frac{1}{z}} \Rightarrow 5^{2} \times 3 = k^{\frac{1}{z}} \Rightarrow {(k^{\frac{1}{y}})}^{2} \times k^{\frac{1}{x}} = k^{\frac{1}{z}} \Rightarrow k^{\frac{2}{y}} \times k^{\frac{1}{x}} = k^{\frac{1}{z}}$
$\Rightarrow k^{\frac{2}{y} + \frac{1}{x}} = k^{\frac{1}{z}} \Rightarrow \frac{2}{y} + \frac{1}{x} = \frac{1}{z} \Rightarrow \frac{2 x + y}{x y} = \frac{1}{z} \Rightarrow z = \frac{x y}{2 x + y}$

Page No 55:

Question 20:

If a and b are distinct primes such that $\sqrt[3]{a^{6} b^{- 4}} = a^{x} b^{2 y}$ , find x and y.

Answer:

Given: $\sqrt[3]{a^{6} b^{- 4}} = a^{x} b^{2 y}$
$\sqrt[3]{a^{6} b^{- 4}} = a^{x} b^{2 y} \Rightarrow {(a^{6} b^{- 4})}^{\frac{1}{3}} = a^{x} b^{2 y} \Rightarrow a^{6 \times \frac{1}{3}} b^{- 4 \times \frac{1}{3}} = a^{x} b^{2 y} \Rightarrow a^{2} b^{- \frac{4}{3}} = a^{x} b^{2 y} \Rightarrow x = 2 and y = - \frac{2}{3}$

Page No 55:

Question 21:

If a and b are different positive primes such that

(i) ${(\frac{a^{- 1} b^{2}}{a^{2} b^{- 4}})}^{7} \div (\frac{a^{3} b^{- 5}}{a^{- 2} b^{3}}) = a^{x} b^{y}$ , find x and y.

(ii) ${(a + b)}^{- 1} (a^{- 1} + b^{- 1}) = a^{x} b^{y}$ , find x + y + 2.

Answer:

(i)
${(\frac{a^{- 1} b^{2}}{a^{2} b^{- 4}})}^{7} \div (\frac{a^{3} b^{- 5}}{a^{- 2} b^{3}}) = a^{x} b^{y} \Rightarrow (\frac{a^{- 7} b^{14}}{a^{14} b^{- 28}}) \div (\frac{a^{3} b^{- 5}}{a^{- 2} b^{3}}) = a^{x} b^{y} \Rightarrow (a^{- 7 - 14} b^{14 + 28}) \div (a^{3 + 2} b^{- 5 - 3}) = a^{x} b^{y} \Rightarrow (a^{- 21} b^{42}) \div (a^{5} b^{- 8}) = a^{x} b^{y} \Rightarrow a^{- 21 - 5} b^{42 + 8} = a^{x} b^{y} \Rightarrow a^{- 26} b^{50} = a^{x} b^{y} \Rightarrow x = - 26 and y = 50$

(ii)
${(a + b)}^{- 1} (a^{- 1} + b^{- 1}) = a^{x} b^{y} \Rightarrow \frac{1}{(a + b)} (\frac{1}{a} + \frac{1}{b}) = a^{x} b^{y} \Rightarrow \frac{1}{(a + b)} (\frac{a + b}{a b}) = a^{x} b^{y} \Rightarrow (\frac{1}{a b}) = a^{x} b^{y} \Rightarrow {(a b)}^{- 1} = a^{x} b^{y} \Rightarrow a^{- 1} b^{- 1} = a^{x} b^{y} \Rightarrow x = - 1 and y = - 1$
Therefore, the value of x + y + 2 is −1 −1 + 2 = 0.

Page No 55:

Question 22:

Simplify:

(i) ${(\frac{x^{a + b}}{x^{c}})}^{a - b} {(\frac{x^{b + c}}{x^{a}})}^{b - c} {(\frac{x^{c + a}}{x^{b}})}^{c - a}$

(ii) $l m \sqrt{\frac{x^{l}}{x^{m}}} \times m n \sqrt{\frac{x^{m}}{x^{n}}} \times n l \sqrt{\frac{x^{n}}{x^{l}}}$

Answer:

(i)
${(\frac{x^{a + b}}{x^{c}})}^{a - b} {(\frac{x^{b + c}}{x^{a}})}^{b - c} {(\frac{x^{c + a}}{x^{b}})}^{c - a} = (\frac{x^{(a + b) (a - b)}}{x^{c (a - b)}}) (\frac{x^{(b + c) (b - c)}}{x^{a (b - c)}}) (\frac{x^{(c + a) (c - a)}}{x^{b (c - a)}}) = (\frac{x^{(a + b) (a - b)}}{x^{c a - b c}}) (\frac{x^{(b + c) (b - c)}}{x^{a b - a c}}) (\frac{x^{(c + a) (c - a)}}{x^{b c - a b}}) = (\frac{x^{(a^{2} - b^{2})} x^{(b^{2} - c^{2})} x^{(c^{2} - a^{2})}}{x^{c a - b c + a b - a c + b c - a b}}) = \frac{x^{a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}}}{x^{c a - b c + a b - a c + b c - a b}} = \frac{x^{0}}{x^{0}} = 1$

(ii)
$\sqrt[l m]{\frac{x^{l}}{x^{m}}} \times \sqrt[m n]{\frac{x^{m}}{x^{n}}} \times \sqrt[n l]{\frac{x^{n}}{x^{l}}} = {(\frac{x^{l}}{x^{m}})}^{\frac{1}{m l}} \times {(\frac{x^{m}}{x^{n}})}^{\frac{1}{m n}} \times {(\frac{x^{n}}{x^{l}})}^{\frac{1}{n l}} = {(x^{l - m})}^{\frac{1}{m l}} \times {(x^{m - n})}^{\frac{1}{m n}} \times {(x^{n - l})}^{\frac{1}{n l}} = x^{\frac{l - m}{m l}} \times x^{\frac{m - n}{m n}} \times x^{\frac{n - l}{n l}} = x^{\frac{l - m}{m l} + \frac{m - n}{m n} + \frac{n - l}{n l}} = x^{\frac{l n - m n + l m - n l + n m - l m}{n m l}} = x^{0} = 1$

Page No 55:

Question 23:

Show that: $\frac{{(a + \frac{1}{b})}^{m} \times {(a - \frac{1}{b})}^{n}}{{(b + \frac{1}{a})}^{m} \times {(b - \frac{1}{a})}^{n}} = {(\frac{a}{b})}^{m + n}$

Answer:

$\frac{{(a + \frac{1}{b})}^{m} \times {(a - \frac{1}{b})}^{n}}{{(b + \frac{1}{a})}^{m} \times {(b - \frac{1}{a})}^{n}} = \frac{{(\frac{a b + 1}{b})}^{m} \times {(\frac{a b - 1}{b})}^{n}}{{(\frac{a b + 1}{a})}^{m} \times {(\frac{a b - 1}{a})}^{n}} = {(\frac{\frac{a b + 1}{b}}{\frac{a b + 1}{a}})}^{m} \times {(\frac{\frac{a b - 1}{b}}{\frac{a b - 1}{a}})}^{n} = {(\frac{a b + 1}{b} \times \frac{a}{a b + 1})}^{m} \times {(\frac{a b - 1}{b} \times \frac{a}{a b - 1})}^{n} = {(\frac{a}{b})}^{m} \times {(\frac{a}{b})}^{n} = {(\frac{a}{b})}^{m} \times {(\frac{a}{b})}^{n} = {(\frac{a}{b})}^{m + n}$

Page No 55:

Question 24:

(i) If $a = x^{m + n} y^{l}, b = x^{n + l} y^{m} and c = x^{l + m} y^{n}$ , prove that $a^{m - n} b^{n - l} c^{l - m} = 1$ .

(ii) If $x = a^{m + n}, y = a^{n + l} and z = a^{l + m}$ , prove that $x^{m} y^{n} z^{l} = x^{n} y^{l} z^{m}$ .

Answer:

(i) Given: $a = x^{m + n} y^{l}, b = x^{n + l} y^{m} and c = x^{l + m} y^{n}$
Putting the values of a, b and c in $a^{m - n} b^{n - l} c^{l - m}$ , we get
$a^{m - n} b^{n - l} c^{l - m} = {(x^{m + n} y^{l})}^{m - n} {(x^{n + l} y^{m})}^{n - l} {(x^{l + m} y^{n})}^{l - m} = [x^{(m + n) (m - n)} y^{l (m - n)}] [x^{(n + l) (n - l)} y^{m (n - l)}] [x^{(l + m) (l - m)} y^{n (l - m)}] = x^{(m^{2} - n^{2})} x^{(n^{2} - l^{2})} x^{(l^{2} - m^{2})} y^{l m - l n} y^{m n - m l} y^{n l - n m} = x^{m^{2} - n^{2} + n^{2} - l^{2} + l^{2} - m^{2}} y^{l m - l n + m n - m l + n l - n m} = x^{0} y^{0} = 1$

(ii) Given: $x = a^{m + n}, y = a^{n + l} and z = a^{l + m}$
Putting the values of x, y and z in $x^{m} y^{n} z^{l}$ , we get
$x^{m} y^{n} z^{l} = {(a^{m + n})}^{m} {(a^{n + l})}^{n} {(a^{l + m})}^{l} = (a^{m^{2} + n m}) (a^{n^{2} + l n}) (a^{l^{2} + l m}) = a^{m^{2} + n^{2} + l^{2} + n m + l n + l m}$
Putting the values of x, y and z in $x^{n} y^{l} z^{m}$ , we get
$x^{n} y^{l} z^{m} = {(a^{m + n})}^{n} {(a^{n + l})}^{l} {(a^{l + m})}^{m} = (a^{m n + n^{2}}) (a^{n l + l^{2}}) (a^{l m + m^{2}}) = a^{m n + n^{2} + n l + l^{2} + l m + m^{2}}$
So, $x^{m} y^{n} z^{l}$ = $x^{n} y^{l} z^{m}$

Page No 56:

Question 1:

Write ${(625)}^{- 1 / 4}$ in decimal form.

Answer:

We have to writein decimal form. So,

${(625)}^{\frac{- 1}{4}} = \frac{1}{625^{\frac{1}{4}}} = {(\frac{1}{(5^{4})})}^{\frac{1}{4}}$

Hence the decimal form of is

Page No 56:

Question 2:

State the product law of exponents.

Answer:

State the product law of exponents.

If is any real number and , are positive integers, then

By definition, we have

(factor) ( factor)

to factors

Thus the exponent "product rule" tells us that, when multiplying two powers that have the same base, we can add the exponents.

Page No 56:

Question 3:

State the quotient law of exponents.

Answer:

The quotient rule tells us that we can divide two powers with the same base by subtracting the exponents. If a is a non-zero real number and m, n are positive integers, then

We shall divide the proof into three parts

(i) when

(ii) when

(iii) when

Case 1

When

We have

$\frac{a^{m}}{a^{n}} = \frac{a \times a \times a . . . . to m factors}{a \times a \times a . . . . t o n f a c t o r s} \frac{a^{m}}{a^{n}} = a \times a \times a . . . . t o (m - n) f a c t o r s \frac{a^{m}}{a^{n}} = a^{m - n}$

Case 2

When

We get

Cancelling common factors in numerator and denominator we get,

By definition we can write 1 as

Case 3

When

In this case, we have

Hence, whether, or,

Page No 56:

Question 4:

State the power law of exponents.

Answer:

The "power rule" tell us that to raise a power to a power, just multiply the exponents.

If a is any real number and m, n are positive integers, then

We have,

factors

Hence,

Page No 56:

Question 5:

If 2⁴ × 4² =16^x, then find the value of x.

Answer:

We have to find the value of x provided

So,

By equating the exponents we get

Hence the value of x is .

Page No 56:

Question 6:

If 3^x^-1 = 9 and 4^y⁺² = 64, what is the value of $\frac{x}{y}$ ?

Answer:

We have to find the value of for

So,

By equating the exponent we get

Let’s take

By equating the exponent we get

By substituting in we get

Hence the value of is

Page No 56:

Question 7:

Write the value of $\sqrt[3]{7} \times \sqrt[3]{49} .$

Answer:

We have to find the value of . So,

By using law rational exponents we get,

Hence the value of is

Page No 56:

Question 8:

Write ${(\frac{1}{9})}^{- 1 / 2} \times (64)^{- 1 / 3}$ as a rational number.

Answer:

We have to find the value of . So,

Hence the value of the value of is .

Page No 56:

Question 9:

Write the value of $\sqrt[3]{125 \times 27}$ .

Answer:

We have to find the value of $\sqrt[3]{125 \times 27}$ . So,

$\sqrt[3]{125 \times 27} = \sqrt[3]{5^{3} \times 3^{3}} = 5 \times 3 = 15$

Hence the value of the value of is .

Page No 57:

Question 10:

Write the value of ${\{5 (8^{1 / 3} + 27^{1 / 3})^{3}\}}^{1 / 4} .$

Answer:

We have to find the value of. So,

By using rational exponents we get

Hence the simplified value of is

Page No 57:

Question 11:

Simplify ${[{\{{(625)}^{- 1 / 2}\}}^{- 1 / 4}]}^{2}$

Answer:

We have to simplify. So,

Hence, the value of is

Page No 57:

Question 12:

If (x − 1)³ = 8, What is the value of (x + 1)² ?

Answer:

We have to find the value of , where

Consider

By equating the base, we get

By substituting in

Hence the value of is .

Page No 57:

Question 1:

(21² – 15²)^2/3 is equal to ________.

Answer:

${(21^{2} - 15^{2})}^{\frac{2}{3}} = {(441 - 225)}^{\frac{2}{3}} = {(216)}^{\frac{2}{3}} = {(6^{3})}^{\frac{2}{3}} = {(6)}^{3 \times \frac{2}{3}} = {(6)}^{2} = 36$

Hence, (21² – 15²)^2/3 is equal to 36.

Page No 57:

Question 2:

81^1/4 × 9^3/2 × 27^–4/3 is equal to _________.

Answer:

$81^{\frac{1}{4}} \times 9^{\frac{3}{2}} \times 27^{- \frac{4}{3}} = {(3^{4})}^{\frac{1}{4}} \times {(3^{2})}^{\frac{3}{2}} \times \frac{1}{27^{\frac{4}{3}}} = {(3)}^{4 \times \frac{1}{4}} \times {(3)}^{2 \times \frac{3}{2}} \times \frac{1}{{(3^{3})}^{\frac{4}{3}}} = {(3)}^{1} \times {(3)}^{3} \times \frac{1}{{(3)}^{3 \times \frac{4}{3}}} = {(3)}^{1 + 3} \times \frac{1}{3^{4}} = {(3)}^{4} \times \frac{1}{3^{4}} = 1$

Hence, 81^1/4 × 9^3/2 × 27^–4/3 is equal to 1.

Page No 57:

Question 3:

${\{\frac{{(169)}^{- 3}}{{(196)}^{- 8}}\}}^{\frac{1}{48}} =$ __________.

Answer:

${\{\frac{{(169)}^{- 3}}{{(196)}^{- 8}}\}}^{\frac{1}{48}} = {\{\frac{{(13^{2})}^{- 3}}{{(14^{2})}^{- 8}}\}}^{\frac{1}{48}} = {\{\frac{{(13)}^{2 \times (- 3)}}{{(14)}^{2 \times (- 8)}}\}}^{\frac{1}{48}} = {\{\frac{{(13)}^{- 6}}{{(14)}^{- 16}}\}}^{\frac{1}{48}} = \frac{{(13)}^{- 6 \times \frac{1}{48}}}{{(14)}^{- 16 \times \frac{1}{48}}} = \frac{{(13)}^{- \frac{1}{8}}}{{(14)}^{- \frac{1}{3}}} = \frac{\frac{1}{13^{\frac{1}{8}}}}{\frac{1}{14^{\frac{1}{3}}}} = \frac{{(14)}^{\frac{1}{3}}}{{(13)}^{\frac{1}{8}}}$

Hence, ${\{\frac{{(169)}^{- 3}}{{(196)}^{- 8}}\}}^{\frac{1}{48}} =$ $\frac{{(14)}^{\frac{1}{3}}}{{(13)}^{\frac{1}{8}}}$ .

Page No 57:

Question 4:

If x = 8^2/3 × 32^–2/5, then x^–5= ________.

Answer:

$Let x = 8^{\frac{2}{3}} \times 32^{- \frac{2}{5}} \Rightarrow x = {(2^{3})}^{\frac{2}{3}} \times {(2^{5})}^{- \frac{2}{5}} \Rightarrow x = {(2)}^{3 \times \frac{2}{3}} \times {(2)}^{5 \times (- \frac{2}{5})} \Rightarrow x = {(2)}^{2} \times {(2)}^{- 2} \Rightarrow x = 2^{2} \times \frac{1}{2^{2}} \Rightarrow x = 1 Now, x^{- 5} = \frac{1}{x^{5}} = \frac{1}{1^{5}} = 1$

Hence, if x = 8^2/3 × 32^–2/5, then x^–5= 1.

Page No 57:

Question 5:

If 6ⁿ = 1296, then 6^n–3 = _________.

Answer:

$Let 6^{n} = 1296 \Rightarrow 6^{n} = 6^{4} \Rightarrow n = 4 Now, 6^{n - 3} = 6^{4 - 3} = 6^{1} = 6$

Hence, if 6ⁿ = 1296, then 6^n–3 = 6.

Page No 57:

Question 6:

The value of 4 × (256)^–1/4 ÷ (243)^1/5 is ________.

Answer:

$Let x = 4 \times {(256)}^{- \frac{1}{4}} \div {(243)}^{\frac{1}{5}} \Rightarrow x = 4 \times {(4^{4})}^{- \frac{1}{4}} \div {(3^{5})}^{\frac{1}{5}} \Rightarrow x = 4 \times {(4)}^{4 \times (- \frac{1}{4})} \div {(3)}^{5 \times \frac{1}{5}} \Rightarrow x = 4 \times {(4)}^{- 1} \div {(3)}^{1} \Rightarrow x = 4 \times \frac{1}{4} \div {(3)}^{1} \Rightarrow x = 1 \div 3 \Rightarrow x = \frac{1}{3}$

Hence, the value of 4 × (256)^–1/4 ÷ (243)^1/5 is $\frac{1}{3}$ .

Page No 57:

Question 7:

If ${(6 x)}^{6} = 6^{2^{3}}, then x =________.$

Answer:

$Let {(6 x)}^{6} = 6^{2^{3}} \Rightarrow {(6 x)}^{6} = 6^{8} \Rightarrow 6^{6} x^{6} = 6^{8} \Rightarrow x^{6} = \frac{6^{8}}{6^{6}} \Rightarrow x^{6} = 6^{8 - 6} \Rightarrow x^{6} = 6^{2} \Rightarrow x = {(6^{2})}^{\frac{1}{6}} \Rightarrow x = 6^{2 \times \frac{1}{6}} \Rightarrow x = 6^{\frac{1}{3}}$

Hence, if ${(6 x)}^{6} = 6^{2^{3}}, then x = 6^{\frac{1}{3}} .$

Page No 57:

Question 8:

${\{{(\frac{a}{b})}^{\sqrt{99} - \sqrt{97}}\}}^{\sqrt{99} + \sqrt{97}}$ = ___________.

Answer:

${\{{(\frac{a}{b})}^{\sqrt{99} - \sqrt{97}}\}}^{\sqrt{99} + \sqrt{97}} = {(\frac{a}{b})}^{(\sqrt{99} - \sqrt{97}) \times (\sqrt{99} + \sqrt{97})} = {(\frac{a}{b})}^{{(\sqrt{99})}^{2} - {(\sqrt{97})}^{2}} (using the identity : a^{2} - b^{2} = (a + b) (a - b)) = {(\frac{a}{b})}^{99 - 97} = {(\frac{a}{b})}^{2} = \frac{a^{2}}{b^{2}}$

Hence, ${\{{(\frac{a}{b})}^{\sqrt{99} - \sqrt{97}}\}}^{\sqrt{99} + \sqrt{97}}$ = $\frac{a^{2}}{b^{2}} .$

Page No 57:

Question 9:

If $\frac{{(p + \frac{1}{q})}^{p - q} {(p - \frac{1}{q})}^{p + q}}{{(q + \frac{1}{p})}^{p - q} {(q - \frac{1}{p})}^{p + q}} = {(\frac{p}{q})}^{x}, then x =_________.$

Answer:

$\frac{{(p + \frac{1}{q})}^{p - q} {(p - \frac{1}{q})}^{p + q}}{{(q + \frac{1}{p})}^{p - q} {(q - \frac{1}{p})}^{p + q}} = {(\frac{p}{q})}^{x} \Rightarrow \frac{{(\frac{q p + 1}{q})}^{p - q} {(\frac{p q - 1}{q})}^{p + q}}{{(\frac{p q + 1}{p})}^{p - q} {(\frac{p q - 1}{p})}^{p + q}} = {(\frac{p}{q})}^{x} \Rightarrow \frac{\frac{{(q p + 1)}^{p - q}}{{(q)}^{p - q}} \times \frac{{(p q - 1)}^{p + q}}{{(q)}^{p + q}}}{\frac{{(p q + 1)}^{p - q}}{{(p)}^{p - q}} \times \frac{{(p q - 1)}^{p + q}}{{(p)}^{p + q}}} = {(\frac{p}{q})}^{x} \Rightarrow \frac{{(q p + 1)}^{p - q}}{{(q)}^{p - q}} \times \frac{{(p q - 1)}^{p + q}}{{(q)}^{p + q}} \times \frac{{(p)}^{p - q}}{{(p q + 1)}^{p - q}} \times \frac{{(p)}^{p + q}}{{(p q - 1)}^{p + q}} = {(\frac{p}{q})}^{x} \Rightarrow \frac{{(p)}^{p - q}}{{(q)}^{p - q}} \times \frac{{(p)}^{p + q}}{{(q)}^{p + q}} = {(\frac{p}{q})}^{x} \Rightarrow \frac{{(p)}^{p - q + p + q}}{{(q)}^{p - q + p + q}} = {(\frac{p}{q})}^{x} \Rightarrow \frac{{(p)}^{2 p}}{{(q)}^{2 p}} = {(\frac{p}{q})}^{x} \Rightarrow {(\frac{p}{q})}^{2 p} = {(\frac{p}{q})}^{x} \Rightarrow 2 p = x \Rightarrow x = 2 p$

Hence, if $\frac{{(p + \frac{1}{q})}^{p - q} {(p - \frac{1}{q})}^{p + q}}{{(q + \frac{1}{p})}^{p - q} {(q - \frac{1}{p})}^{p + q}} = {(\frac{p}{q})}^{x}, then x = 2 p .$

Page No 57:

Question 10:

If 5ⁿ⁺² = 625, then (12n + 3)^1/3 = _________.

Answer:

$Let 5^{n + 2} = 625 \Rightarrow 5^{n + 2} = 5^{4} \Rightarrow n + 2 = 4 \Rightarrow n = 2 Now, {(12 n + 3)}^{\frac{1}{3}} = {(12 (2) + 3)}^{\frac{1}{3}} = {(24 + 3)}^{\frac{1}{3}} = {(27)}^{\frac{1}{3}} = {(3^{3})}^{\frac{1}{3}} = {(3)}^{3 \times \frac{1}{3}} = 3$

Hence, if 5ⁿ⁺² = 625, then (12n + 3)^1/3 = 3.

Page No 57:

Question 11:

If ${(\frac{a^{3 - 7 a} \times a^{6 - 2 a}}{a^{2 a} \times a^{9 - 2 a}})}^{\frac{1}{9}} =$ __________.

Answer:

${(\frac{a^{3 - 7 a} \times a^{6 - 2 a}}{a^{2 a} \times a^{9 - 2 a}})}^{\frac{1}{9}} = {(\frac{a^{3 - 7 a + 6 - 2 a}}{a^{2 a + 9 - 2 a}})}^{\frac{1}{9}} = {(\frac{a^{9 - 9 a}}{a^{9}})}^{\frac{1}{9}} = {(a^{9 - 9 a - 9})}^{\frac{1}{9}} = {(a^{- 9 a})}^{\frac{1}{9}} = {(a)}^{- 9 a \times \frac{1}{9}} = {(a)}^{- a} = \frac{1}{a^{a}}$

Hence, ${(\frac{a^{3 - 7 a} \times a^{6 - 2 a}}{a^{2 a} \times a^{9 - 2 a}})}^{\frac{1}{9}} = \frac{1}{a^{a}} .$

Page No 57:

Question 12:

If $\frac{1}{{(243)}^{x}} = {(729)}^{y} = 3^{3}$ , then 5x + 6y = __________.

Answer:

$Given : \frac{1}{{(243)}^{x}} = {(729)}^{y} = 3^{3} \frac{1}{{(243)}^{x}} = 3^{3} \Rightarrow \frac{1}{{(3^{5})}^{x}} = 3^{3} \Rightarrow \frac{1}{3^{5 x}} = 3^{3} \Rightarrow 3^{- 5 x} = 3^{3} \Rightarrow - 5 x = 3 \Rightarrow 5 x = - 3 . . . (1) {(729)}^{y} = 3^{3} \Rightarrow {(3^{6})}^{y} = 3^{3} \Rightarrow 3^{6 y} = 3^{3} \Rightarrow 6 y = 3 . . . (2) Adding (1) and (2), we get 5 x + 6 y = - 3 + 3 = 0$

Hence, 5x + 6y = 0.

Page No 57:

Question 13:

If 6^x–y = 36 and 3^x+y= 729, then x² – y² = _________.

Answer:

$Given : 6^{x - y} = 36 and 3^{x + y} = 729 6^{x - y} = 36 \Rightarrow 6^{x - y} = 6^{2} \Rightarrow x - y = 2 . . . (1) 3^{x + y} = 729 \Rightarrow 3^{x + y} = 3^{6} \Rightarrow x + y = 6 . . . (2) Adding (1) and (2), we get 2 x = 8 \Rightarrow x = 4 Substituting the value of x in (2), we get 4 + y = 6 \Rightarrow y = 2 Now, x^{2} - y^{2} = 4^{2} - 2^{2} = 16 - 4 = 12$

Hence, x² – y² = 12.

Page No 57:

Question 14:

$\sqrt[4]{\sqrt[3]{2^{2}}}$ equals __________.

Answer:

$\sqrt[4]{\sqrt[3]{2^{2}}} = {\{{(2^{2})}^{\frac{1}{3}}\}}^{\frac{1}{4}} = {(2^{2})}^{\frac{1}{3} \times \frac{1}{4}} = {(2^{2})}^{\frac{1}{12}} = {(2)}^{2 \times \frac{1}{12}} = {(2)}^{\frac{1}{6}}$

Hence, $\sqrt[4]{\sqrt[3]{2^{2}}}$ equals ${(2)}^{\frac{1}{6}} .$

Page No 57:

Question 15:

The product $\sqrt[3]{2} . \sqrt[4]{2} . \sqrt[12]{32}$ is equal to ________.

Answer:

$\sqrt[3]{2} . \sqrt[4]{2} . \sqrt[12]{32} = {(2)}^{\frac{1}{3}} . {(2)}^{\frac{1}{4}} . {(32)}^{\frac{1}{12}} = {(2)}^{\frac{1}{3}} . {(2)}^{\frac{1}{4}} . {(2^{5})}^{\frac{1}{12}} = {(2)}^{\frac{1}{3}} . {(2)}^{\frac{1}{4}} . {(2)}^{5 \times \frac{1}{12}} = {(2)}^{\frac{1}{3}} . {(2)}^{\frac{1}{4}} . {(2)}^{\frac{5}{12}} = {(2)}^{\frac{1}{3} + \frac{1}{4} + \frac{5}{12}} = {(2)}^{\frac{4 + 3 + 5}{12}} = {(2)}^{\frac{12}{12}} = {(2)}^{1} = 2$

Hence, the product $\sqrt[3]{2} . \sqrt[4]{2} . \sqrt[12]{32}$ is equal to 2.

Page No 57:

Question 16:

$\sqrt[4]{{(81)}^{- 2}}$ is equal to _________.

Answer:

$\sqrt[4]{{(81)}^{- 2}} = {\{{(81)}^{- 2}\}}^{\frac{1}{4}} = {(81)}^{- 2 \times \frac{1}{4}} = {(81)}^{- \frac{1}{2}} = {(3^{4})}^{- \frac{1}{2}} = {(3)}^{4 \times (- \frac{1}{2})} = {(3)}^{- 2} = \frac{1}{3^{2}} = \frac{1}{9}$

Hence, $\sqrt[4]{{(81)}^{- 2}}$ is equal to $\frac{1}{9}$ .

Page No 57:

Question 17:

The value of (256)^0.16 × (256)^0.09is ________.

Answer:

${(256)}^{0.16} \times {(256)}^{0.09} = {(256)}^{0.16 + 0.09} = {(256)}^{0.25} = {(256)}^{\frac{25}{100}} = {(256)}^{\frac{1}{4}} = {(2^{8})}^{\frac{1}{4}} = {(2)}^{8 \times \frac{1}{4}} = {(2)}^{2} = 4$

Hence, the value of (256)^0.16 × (256)^0.09is 4.

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