RD Sharma Solutions Class 8 Powers Exercise 2.2

RD Sharma Class 8 Solutions Chapter 2 Ex 2.2 PDF Free Download

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

Answer:

(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 \(x^{-1}\).

Answer:

(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.

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