# RD Sharma Solutions Class 8 Powers Exercise 2.2

## RD Sharma Solutions Class 8 Chapter 2 Exercise 2.2

#### Exercise 2.2

Q1. Write each of the following in exponential form:

(i) $\left ( \frac{3}{2} \right )^{-1} \times \left ( \frac{3}{2} \right )^{-1} \times \left ( \frac{3}{2} \right )^{-1} \times \left ( \frac{3}{2} \right )^{-1}$

(ii) $\left ( \frac{2}{5} \right )^{-2} \times \left ( \frac{2}{5} \right )^{-2} \times \left ( \frac{2}{5} \right )^{-2}$

Solution:

(i) $\left ( \frac{3}{2} \right )^{-1} \times \left ( \frac{3}{2} \right )^{-1} \times \left ( \frac{3}{2} \right )^{-1} \times \left ( \frac{3}{2} \right )^{-1}$ = $\left (\frac{3}{2} \right )^{-1 + \left ( -1 \right ) + \left ( -1 \right ) + \left ( -1 \right )}$

Using law of exponents: $a^{m} \times a^{n} = a^{m + n}$ = $( \frac{3}{2} )^{-4}$

(ii) $\left (\frac{2}{5} \right )^{-2} \times \left (\frac{2}{5} \right )^{-2} \times \left (\frac{2}{5} \right )^{-2} = \left (\frac{2}{5} \right )^{-1 + \left ( -2 \right ) + \left ( -2 \right )}$

Using law of exponents: ${ a^{m} \times a^{n} = a^{m + n} }$ = $\left (\frac{2}{5} \right )^{-6}$

Q2. Evaluate:

(i) $5^{-2}$

(ii) $\left (-3 \right )^{-2}$

(iii) $\left ( \frac{1}{3} \right )^{-4}$

(iv) $\left ( \frac{-1}{2} \right )^{-1}$

Solution:

(i) $5^{-2} = \frac{1}{5^{2}}$ = $\frac{1}{25}$

(ii) $\left ( -3 \right )^{-2} = \frac{1}{\left (-3 \right )^{2}}$ = $\frac{1}{9}$

(iii) $\left ( \frac{1}{3} \right )^{-4} = \frac{1}{\left ( \frac{1}{3} \right )^{4}}$ = $\frac{1}{\frac{1}{81}}$ = 81

(iv) $\left ( \frac{-1}{2} \right )^{-1} \left ( \frac{1}{\frac{-1}{2}} \right )$ = -2

Q3. Express each of the following as a rational number in the form $\frac{p}{q}$:

(i) $6^{-1}$

(ii) $- 7^{-1}$

(iii) $\left ( \frac{1}{4} \right )^{-1}$

(iv) $\left ( -4 \right )^{-1} \times \left ( \frac{-3}{2} \right )^{-1}$

(v) $\left ( \frac{3}{5} \right )^{-1} \times \left ( \frac{5}{2} \right )^{-1}$

Solution:

(i) $6^{-1} = \frac{1}{6}$

(ii) $- 7^{-1} = \frac{1}{- 7}$ = $\frac{-1}{7}$

(iii) $\left ( \frac{1}{4} \right )^{-1} = \frac{1}{ \frac{1}{4} } = 4$

(iv) $\left ( -4 \right )^{-1} \times \left ( \frac{-3}{2} \right )^{-1} = \frac{1}{-4} \times \frac{1}{\frac{-3}{2}}$

= $\frac{1}{-4} \times \times \frac{2}{-3}$ = $\frac{1}{6}$

(v) $\left ( \frac{3}{5} \right )^{-1} \times \left ( \frac{5}{2} \right )^{-1} = \frac{1}{\frac{3}{5}} \times \frac{1}{\frac{5}{2}}$

= $\frac{5}{3} \times \frac{2}{5}$ = $\frac{2}{3}$

Q4. Simplify:

(i) $\left \{ 4^{-1} \times 3^{-1} \right \}^{2}$

(ii) $\left \{ 5^{-1} \div 6^{-1} \right \}^{3}$

(iii) $\left \{ 2^{-1} + 3^{-1} \right \}^{-1}$

(iv) $\left \{ 3^{-1} + 4^{-1} \right \}^{-1} \times 5^{-1}$

(v) $\left \{ 4^{-1} + 5^{-1} \right \}^{-1} + 3^{-1}$

Solution:

(i)$\left \{ 4^{-1} \times 3^{-1} \right \}^{2} = \left ( \frac{1}{4} \times \frac{1}{3} \right )^{2}$

= $\left ( \frac{1}{12} \right )^{2}$ = $\left ( \frac{1}{144} \right )$

(ii) $\left ( 5^{-1} \div 6^{-1} \right )^{3} = \left ( \frac{1}{5} \div \frac{1}{6} \right )^{3}$

= $\left ( \frac{6}{5} \right )^{3}$ = $\left ( \frac{216}{125} \right )$

(iii) $\left \{ 2^{-1} + 3^{-1} \right \}^{-1} = \left ( \frac{1}{2} + \frac{1}{3} \right )^{-1}$

= $\left ( \frac{5}{6} \right )^{-1}$ = $\left ( \frac{6}{5} \right )$

(iv) $\left \{ 3^{-1} + 4^{-1} \right \}^{-1} \times 5^{-1} = \left ( \frac{1}{3} \times \frac{1}{4} \right )^{-1} \times \frac{1}{5}$

= $\left ( \frac{1}{12} \right )^{-1} \times \frac{1}{5}$

= $12 \times \frac{1}{5}$ = $\frac{12}{5}$

(v) $\left \{ 4^{-1} + 5^{-1} \right \}^{-1} + 3^{-1} = \left ( \frac{1}{4} – \frac{1}{5} \right ) \div \frac{1}{3}$

= $\left ( \frac{5 – 4}{20} \right ) \times 3$

= $\frac{1}{20} \times 3$ = $\frac{3}{20}$

Q5. Express each of the following rational numbers with a negative exponent:

(i) $\left ( \frac{1}{4} \right )^{3}$

(ii) $\left ( 3 \right )^{5}$

(iii) $\left ( \frac{3 }{5}\right )^{4}$

(iv) $\left \{ \left ( \frac{3}{2} \right )^{4} \right \}^{-3}$

(v) $\left \{ \left ( \frac{7}{4} \right )^{4 } \right \}^{-3}$

Solution:

(i) $\left ( \frac{1}{4} \right )^{3}$

= $\left ( \frac{4}{1} \right )^{-3}$

(ii) $\left ( 3 \right )^{5}$

= $\left ( \frac{1}{3} \right )^{-5}$

(iii) $\left ( \frac{3 }{5}\right )^{4}$

= $\left ( \frac{5 }{3}\right )^{-4}$

(iv) $\left \{ \left ( \frac{3}{2} \right )^{4} \right \}^{-3}$

= $\left ( \frac{3}{2} \right )^{-12}$

(v) $\left \{ \left ( \frac{7}{4} \right )^{4 } \right \}^{-3}$

= $\left ( \frac{7}{4} \right )^{-12}$

Q6. Express each of the following rational numbers with a positive exponent.

(i) $\left ( \frac{3}{4} \right )^{-2}$

(ii) $\left ( \frac{5}{4} \right )^{-3}$

(iii) $4^{3} \times 4^{-9}$

(iv) $\left \{ \left ( \frac{4}{3} \right )^{-3} \right \}^{-4}$

(v) $\left \{ \left ( \frac{3}{2} \right )^{4} \right \}^{-2}$

Solution:

(i) $\left ( \frac{3}{4} \right )^{-2}$

= $\left ( \frac{4}{3} \right )^{2}$

(ii) $\left ( \frac{5}{4} \right )^{-3}$

= $\left ( \frac{4}{5} \right )^{3}$

(iii) $4^{3} \times 4^{-9}$

= $4^{3 – 9} = 4^{-6}$

= $\left (\frac{1}{4} \right )^{6}$

(iv) $\left \{ \left ( \frac{4}{3} \right )^{-3} \right \}^{-4}$

= $\left ( \frac{4}{3} \right )^{-4 \times -3}$

= $\left ( \frac{4}{3} \right )^{12}$

(v) $\left \{ \left ( \frac{3}{2} \right )^{4} \right \}^{-2}$

= $\left ( \frac{3}{2} \right )^{4 \times -2}$

= $\left ( \frac{3}{2} \right )^{-8}$

= $\left ( \frac{2}{3} \right )^{8}$

Q7. Simplify:

(i) $\left \{ \left ( \frac{1}{3} \right )^{-3} – \left ( \frac{1}{2} \right )^{-3} \right \} \div \left ( \frac{1}{4} \right )^{-3}$

(ii) $\left ( 3^{2} – 2^{2} \right ) \times \left ( \frac{2}{3} \right )^{-3}$

(iii) $\left \{ \left ( \frac{1}{2} \right )^{-1} \times \left ( -4 \right )^{-1} \right \}^{-1}$

(iv) $\left [ \left \{ \left ( \frac{-1}{4} \right )^{2} \right \}^{-2} \right ]^{-1}$

(v) $\left \{ \left ( \frac{2}{3} \right )^{2} \right \}^{3} \times \left ( \frac{1}{3} \right )^{-4} \times 3^{-1} \times 6^{-1}$

Solution:

(i) $\left \{ \left ( \frac{1}{3} \right )^{-3} – \left ( \frac{1}{2} \right )^{-3} \right \} \div \left ( \frac{1}{4} \right )^{-3} = \left ( \frac{1}{\left ( 1/3 \right )^{3}} – \frac{1}{\left ( 1/2 \right )^{3}} \right ) \div \frac{1}{\left (1/4 \right )^{3}} \\ \\ = \left ( \frac{1}{\left (1/27 \right )} – \frac{1}{\left (1/8 \right )} \right ) \div \frac{1}{\left (1/64 \right )} \\ \\ = \left ( \frac{27}{1} – \frac{8}{1} \right ) \div 64 \\ \\ = \left ( 19 \right ) \times \frac{1}{64} \\ \\ = \frac{19}{64}$

(ii) $\left ( 3^{2} – 2^{2} \right ) \times \left ( \frac{2}{3} \right )^{-3} = \left ( 9 – 4 \right ) \times \frac{1}{ \left (2 / 3 \right )^{2}} \\ \\ = 5 \times \frac{27}{8} \\ \\ = \frac{135}{8}$

(iii) $\left ( \left ( \frac{1}{2} \right )^{-1} \times \left ( -4 \right )^{-1} \right )^{-1} = \left ( \left ( \frac{1}{1/2} \right ) \times \left ( \frac{1}{-4} \right ) \right )^{-1} \\ \\ = \left ( 2 \times \left ( \frac{1}{-4} \right ) \right )^{-1} \\ \\ = \left ( \frac{1}{-2} \right ) \\ \\ = \frac{1}{1/\left ( -2 \right )} \\ \\ = -2$

(iv) $\left(\left(\left(\frac{-1}{4}\right)^{2}\right)^{-2}\right)^{-1}=\left(\left(\frac{\left(-1\right)^{2}}{4^{2}}\right)^{-2}\right)^{-1}$

= $\left ( \left ( \frac{1}{16} \right )^{-2} \right )^{-1}$

= $\left ( \left ( \frac{1}{\left (1/16 \right )^{2}} \right ) \right )^{-1}$

= $\left ( \frac{1}{\left ( 1/256 \right )} \right ) ^{-1}$

= $256^{-1} = \frac{1}{256}$

(v) $\left \{ \left ( \frac{2}{3} \right )^{2} \right \}^{3} \times \left ( \frac{1}{3} \right )^{-4} \times 3^{-1} \times 6^{-1} \\ \\ = \left ( \frac{2^{2}}{3^{2}} \right )^{3} \times \frac{1}{\left (1/3 \right )^{4}} \times \frac{1}{3} \times \frac{1}{6} \\ \\ = \frac{4^{3}}{9^{3}} \times 81 \times \frac{1}{18} \\ \\ = \frac{64}{729} \times 81 \times \frac{1}{18} \\ \\ = \frac{64}{9} \times \frac{1}{18} \\ \\ = 64 \times \frac{1}{162} \\ \\ = \frac{64}{162} \\ \\ = \frac{32}{81}$

Q8. By what number should $5^{-1}$ be multiplies so that the product may be equal to $\left ( – 7 \right )^{-1}$ ?

Solution:

Write numbers into simplest form:

$5^{-1} = \frac{1}{5}$ and $\left ( -7 \right )^{-1} = \frac{1}{-7}$

Let x be the resultant number, so according to the statement

$\frac{1}{5}x = \frac{-1}{7}$

Multiplying both side by 5, we get:

x = $-\frac{5}{7}$

Q9. By what number should $\left ( \frac{1}{2} \right )^{-1}$ be multiplies so that the product may be equal to $\left ( \frac{-4}{7} \right )^{-1}$?

Solution:

Express fractions into simplest form

$\left ( \frac{1}{2} \right )^{-1} = 2$,

And $\left ( \frac{-4}{7} \right )^{-1} = -\frac{7}{4}$

Let x be the required number

$2x = -\frac{7}{4}$

Dividing both sides by 2, we get

$x = -\frac{7}{8}$

Q10. By what number should $\left ( -15 \right )^{-1}$ be divided so that the quotient may be equal to $\left ( -5 \right )^{-1}$

Solution:

Expressing in fraction form, we get:

$\left (-15 \right )^{-1} = – \frac{1}{15}$ (using $a^{-1 } = \frac{1}{a}$)

And

$\left (- 5 \right )^{-1} = – \frac{1}{5}$ (using $a^{-1 } = \frac{1}{a}$)

We have to find a number x such that

$- \frac{1}{15} \div x = -\frac{1}{5}$

Solving this equation, we get:

$-\frac{1}{15} \times \frac{1}{x} = – \frac{1}{5}$

$\frac{ – 5}{ – 15} = x$

$x = \frac{1}{3}$

Q11. By what number should $\left (\frac{5}{3} \right )^{-2}$ be multiplies so that the product may be $\left (\frac{7}{3} \right )^{-1}$?

Solution:

Expressing as a positive exponent, we have:

$\left ( \frac{5}{3} \right )^{-2} = \frac{1}{\left (5/3 \right )^{2}} \\ \\ = \frac{1}{25/9} \\ \\ = \frac{9}{25} \\ \\ and \\ \\ = \left (\frac{7}{3} \right )^{-1} = \frac{3}{7} \\ \\ We \; have\; to \; find \; a \; number \; x \; such \; that \\ \\ \frac{9}{25} \times x = \frac{3}{7}$

Multiplying both sides by 25/9, we get:

$x = \frac{3}{7} \times \frac{25}{9} = \frac{1}{7} \times \frac{25}{3} = \frac{25}{21}$

Hence, $\left (\frac{5}{3} \right )^{-2}$ should be multiplied by $\frac{25}{21}$ top obtain $\left (\frac{7}{3} \right )^{-1}$.

Q12. Find x, if:

(i) $\left ( \frac{1}{4} \right )^{-4} \times \left ( \frac{1}{4} \right )^{-8} = \left ( \frac{1}{4} \right )^{-4x}$

(ii) $\left ( \frac{-1}{2} \right )^{-19} \times \left ( \frac{-1}{2} \right )^{8} = \left ( \frac{-1}{2} \right )^{-2x + 1}$

(iii) $\left ( \frac{3}{2} \right )^{-3} \times \left ( \frac{3}{2} \right )^{5} = \left ( \frac{3}{2} \right )^{2x + 1}$

(iv) $\left (\frac{2}{5} \right )^{-3} \times \left (\frac{2}{5} \right )^{15} = \left (\frac{2}{5} \right )^{2 + 3x}$

(v) $\left ( \frac{5}{4} \right )^{-x} \div \left ( \frac{5}{4} \right )^{-4} = \left ( \frac{5}{4} \right )^{5}$

(vi) $\left ( \frac{8}{3} \right )^{2x + 1} \times \left ( \frac{8}{3} \right )^{5} = \left ( \frac{8}{3} \right )^{x + 2}$

(i) We have:

$\\ \left ( \frac{1}{4} \right )^{-4} \times \left ( \frac{1}{4} \right )^{-8} = \left ( \frac{1}{4} \right )^{-4x} \\ \\ \left ( \frac{1}{4} \right )^{-12} = \left ( \frac{1}{4} \right )^{-4x} \\ \\ -12 = -4x \\ 3 = x$

(ii) We have:

$\\ \\ \left ( \frac{-1}{2} \right )^{-19} \times \left ( \frac{-1}{2} \right )^{8} = \left ( \frac{-1}{2} \right )^{-2x + 1} \\ \left ( \frac{-1}{2} \right )^{-11} = \left ( \frac{-1}{2} \right )^{-2x + 1} \\ \\ -11 = -2x + 1 \\ \\ -12 = -2x \\ \\ 6 = x$

Therefore, x = 6

(iii) We have:

$\\ \left ( \frac{3}{2} \right )^{-3} \times \left ( \frac{3}{2} \right )^{5} = \left ( \frac{3}{2} \right )^{2x + 1} \\ \\ \left ( \frac{3}{2} \right )^{2} = \left ( \frac{3}{2} \right )^{2x + 1} \\ \\ 2 = 2x + 1 \\ \\ 1 = 2x \\ \\ \frac{1}{2} = x$

Therefore, $x= \frac{1}{2}$

(iv) We have:

$\\ \left (\frac{2}{5} \right )^{-3} \times \left (\frac{2}{5} \right )^{15} = \left (\frac{2}{5} \right )^{2 + 3x} \\ \\ \left (\frac{2}{5} \right )^{12} = \left (\frac{2}{5} \right )^{2x + 1} \\ \\ 12 = 2 + 3x \\ \\ 10 = 3x \\ \\ \frac{10}{3} = x$

Therefore, $x= \frac{10}{3}$

(v) We have:

$\\ \left ( \frac{5}{4} \right )^{-x} \div \left ( \frac{5}{4} \right )^{-4} = \left ( \frac{5}{4} \right )^{5} \\ \\ \left ( \frac{5}{4} \right )^{-x + 4} = \left ( \frac{5}{4} \right )^{5} \\ \\ -x + 4 = 5 \\ -x = 1 \\ \\ x = -1$

Therefore, x = -1

(vi) We have:

$\\ \left ( \frac{8}{3} \right )^{2x + 1} \times \left ( \frac{8}{3} \right )^{5} = \left ( \frac{8}{3} \right )^{x + 2} \\ \\ \left ( \frac{8}{3} \right )^{2x + 6} = \left ( \frac{8}{3} \right )^{x + 2} \\ \\ 2x + 6 = x + 2 \\ x = -4$

Therefore, x = -4

Q13.

(i) if $x = \left ( \frac{3}{2} \right )^{2} \times \left ( \frac{2}{3} \right )^{-4}$, find the value of $x^{-2}$.

(ii) If $x = \left ( \frac{4}{5} \right )^{-2} \div \left ( \frac{1}{4} \right )^{2}$, find the value of $x^{-1}$.

(i) First, we have to find x.

$\\ x = \left ( \frac{3}{2} \right )^{2} \times \left ( \frac{2}{3} \right )^{-4} \\ \\ = \left (\frac{3}{2} \right )^{2} \times \left ( \frac{3}{2} \right )^{4} \\ \\ = \left ( \frac{3}{2} \right )^{6}$

Hence, $x^{-2}$ is:

$x^{-2} = \left ( \left ( \frac{3}{2} \right )^{6} \right )^{-2} \\ \\ = \left ( \frac{3}{2} \right )^{-12} \\ \\ = \left ( \frac{2}{3} \right )^{12}$

(ii) First we will have to find x.

$x = \left ( \frac{4}{5} \right )^{-2} \div \left ( \frac{1}{4} \right )^{2} \\ \\ = \left ( \frac{4^{-2}}{5^{-2}} \right ) \times 4^{2} \\ \\ = \frac{4^{0}}{5^{-2}} \\ \left ( 5^{2} \right )^{-1} \\ \\ = \frac{1}{5^{2}}$

Q14. Find the value of x for which $5^{2x} \div 5^{-3} = 5^{5}$.

Sol:

$\\ 5^{2x} \div 5^{-3} = 5^{5} \\ 5^{2x + 3} = 5^{5} \\ 2x + 3 = 5 \\ 2x = 2 \\ x = 1$

The value of x is 1.