Chapter 8: Exponents and Powers

Q11Written answer

The multiplicative inverse of ${\left( { - \dfrac{5}{9}} \right)^{ - 99}}$ is

(a) ${\left( { - \dfrac{5}{9}} \right)^{99}}$

(b) $\left( {{{\dfrac{5}{9}}^{99}}} \right)$

(c) ${\left( {\dfrac{9}{{ - 5}}} \right)^{99}}$

(d) ${\left( {\dfrac{9}{5}} \right)^{99}}$

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Ans: Option (a) is correct.

For multiplicative inverse, a is called multiplicative inverse of b, if a x b = 1.

Thus,

$\Rightarrow a \times {\left( {\dfrac{{ - 5}}{9}} \right)^{ - 99}} = 1$

$\Rightarrow a = \dfrac{1}{{{{\left( {\dfrac{{ - 5}}{9}} \right)}^{ - 99}}}} = {\left( { - \dfrac{5}{9}} \right)^{99}}$

Q12Written answer

If x be any non-zero integer and m, n be negative integers, then xm $\times$ xn is equal to

(a) xm

(b) xm+n

(c) xn

(d) xm–n

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Ans: Option (b) is correct.

Using law of exponents, am × an = (a)m+n (Where, a is a non-zero integer)

Similarly, xm$\times$xn = (x)m+n

Q13Written answer

If y be any non-zero integer, then y0 is equal to

(a) 1

(b) 0

(c) – 1

(c) Not defined

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Ans: Option (a) is correct.

Using law of exponents, a0 = 1

Similarly, y0 = 1.

Q14Written answer

If x be any non-zero integer, then x–1 is equal to

(a) x

(b) $\dfrac{1}{x}$

(c) -x

(d) $\dfrac{{ - 1}}{x}$

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Ans: Option (b) is correct.

Use the law of exponent ${a^{ - m}} = \dfrac{1}{{{a^m}}}$, where a is a non zero integer.

Thus, ${x^{ - 1}} = \dfrac{1}{{{x^1}}} = \dfrac{1}{x}$

Q15Written answer

If x be any integer different from zero and m be any positive integer, then x–m is equal to

(a) xm

(b) -xm

(c) $\dfrac{1}{{{x^m}}}$

(d) $\dfrac{{ - 1}}{{{x^m}}}$

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Ans: Option (c) is correct.

Use the law of exponent ${a^{ - m}} = \dfrac{1}{{{a^m}}}$, where a is a non zero integer.

Thus, ${x^{ - m}} = \dfrac{1}{{{x^m}}}$

Q16Written answer

If x be any integer different from zero and m, n be any integers, then

(xm)n is equal to

(a) xm+n

(b) xmn

(c) ${x^{\dfrac{m}{n}}}$

(d) ${x^{m - n}}$

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Ans: Option (b) is correct.

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

Thus, ${\left( {{x^m}} \right)^n} = {\left( x \right)^{m \times n}} = {x^{mn}}$.

Q17Written answer

Which of the following is equal to ${\left( { - \dfrac{3}{4}} \right)^{ - 3}}$?

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

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

(c) ${\left( {\dfrac{4}{3}} \right)^3}$

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

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Ans: Option (d) is correct.

Use the law of exponent ${a^{ - m}} = \dfrac{1}{{{a^m}}}$, where a is a non zero integer.

Thus, ${\left( {\dfrac{{ - 3}}{4}} \right)^{ - 3}} = \dfrac{1}{{{{\left( {\dfrac{{ - 3}}{4}} \right)}^3}}} = {\left( { - \dfrac{4}{3}} \right)^3}$

Q18Written answer

${\left( { - \dfrac{5}{7}} \right)^{ - 5}}$ is equal to

(a) ${\left( {\dfrac{5}{7}} \right)^{ - 5}}$

(b) ${\left( {\dfrac{7}{5}} \right)^5}$

(c) ${\left( {\dfrac{7}{5}} \right)^5}$

(d) ${\left( { - \dfrac{7}{5}} \right)^5}$

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Ans: Option (d) is correct.

Use the law of exponent ${a^{ - m}} = \dfrac{1}{{{a^m}}}$, where a is a non zero integer.

Thus, ${\left( {\dfrac{{ - 5}}{7}} \right)^{ - 5}} = \dfrac{1}{{{{\left( {\dfrac{{ - 5}}{7}} \right)}^5}}} = {\left( { - \dfrac{7}{5}} \right)^5}$

Q19Written answer

${\left( {\dfrac{{ - 7}}{5}} \right)^{ - 1}}$ is equal to

(a) $\dfrac{5}{7}$

(b) $ - \dfrac{5}{7}$

(c) $\dfrac{7}{5}$

(d) $\dfrac{{ - 7}}{5}$

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Ans: Option (b) is correct.

Use the law of exponent ${a^{ - m}} = \dfrac{1}{{{a^m}}}$, where a is a non zero integer.

Thus, ${\left( {\dfrac{{ - 7}}{5}} \right)^{ - 1}} = \dfrac{1}{{{{\left( {\dfrac{{ - 7}}{5}} \right)}^1}}} = \left( { - \dfrac{5}{7}} \right)$

Q20Written answer

(–9)3 ÷ (–9)8 is equal to

(a) (9)5

(b) (9)–5

(c) (– 9)5

(d) (– 9)–5

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Ans: Option (d) is correct.

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

Thus, $\dfrac{{{{\left( { - 9} \right)}^3}}}{{{{\left( { - 9} \right)}^8}}} = {\left( { - 9} \right)^{3 - 8}} = {\left( { - 9} \right)^{ - 5}}$

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