Chapter 8: Exponents and Powers

Ans: Option (b) is correct.

Use the law of exponent ${\left( {{a^m}} \right)^n} = {\left( a \right)^{mn}}$, where a is a non zero integer.

Thus, 

${\left( {{z^{ - 2}}} \right)^3} = {z^{ - 2 \times 3}} = {z^{ - 6}}$

Ans: Option (c) is correct.

For (-a)m, if m is odd, then (-a)m is negative.

${\left( { - \dfrac{1}{2}} \right)^3} = \left( { - \dfrac{1}{2}} \right) \times \left( { - \dfrac{1}{2}} \right) \times \left( { - \dfrac{1}{2}} \right) = \left( { - \dfrac{1}{8}} \right)$

Ans: Option (b) is correct.

Reciprocal of a is $\dfrac{{\text{1}}}{{\text{a}}}.$

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

$\Rightarrow \dfrac{{{3^4}}}{{{2^4}}} \ne {\left( {\dfrac{3}{2}} \right)^{ - 4}}$

In questions 34 to 65, fill in the blanks to make the statements true.

Ans: 10-10

Multiplicative inverse is the reciprocal of a number. Thus,

${10^{10}} = \dfrac{1}{{{{10}^{10}}}} = {10^{ - 10}}$

Ans: ${a^3} \times {a^{ - 10}} = {a^{ - 7}}$

$\Rightarrow {a^m} \times {a^n} = {a^{m + n}}$

$\Rightarrow {a^3} \times {a^{ - 10}} = {a^{3 - 10}} = {a^{ - 7}}$

Ans: 1

As, a0 =1, so 50 = 1.

Ans: -1

As, xm × xn = (x)m+n

${5^5} \times {5^{ - 5}} = {\left( 5 \right)^{5 - 5}} = {\left( 5 \right)^0} = 1$

Ans: $\left( {\dfrac{1}{{{2^6}}}} \right)$

As, (xm) n = (x)mn

${\left( {\dfrac{1}{{{2^3}}}} \right)^2} = {\left( {\dfrac{{{1^3}}}{{{2^3}}}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^{3 \times 2}} = {\left( {\dfrac{1}{2}} \right)^6} = \dfrac{1}{{{2^6}}}$

Ans: (2)-6

In 8-2, 8 is the base and -2 is the power.

${8^{ - 2}} = \dfrac{1}{{{8^2}}} = \dfrac{1}{{64}}$

${2^6} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$

$\Rightarrow \dfrac{1}{{64}} = \dfrac{1}{{{2^6}}} = {2^{ - 6}}$

Ans: negative.