Question

Find the value of

1. (30 × 4−1) × 22
2. (2−1 × 4−1) ÷ 2−2
3. ($\frac{1}{2}$)−2 + ($\frac{1}{3}$)−2 + ($\frac{1}{4}$)−2
4. (3−1 +4−1 + 5−1)0
5. {($\frac{−2}{3}$)−2}2

Answer 1

(i) (30 × 4−1) × 22
a0 = 1 and a−m = $\frac{1}{a^m}$
(30 × 4−1) × 22
= (1 + $\frac{1}{4}$) × 22

= $[\frac{(4 + 1)}{4}] ×$ 22

$= (\frac{5}{4}) ×$ 22

= $\frac{5}{2}$2 × 22 [Since 4 = 2 × 2 = 22]
= 5

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(ii) (2−1 × 4−1) ÷ 2−2
(am)n = amn, a−m = $\frac{1}{a^m}$
(2−1 × 4−1) ÷ 2−2
= [2−1 × {(2)2}−1] ÷ 2−2 [Since 4 = 2 × 2 = 22]
= (2−1 × 2−2) ÷ 2−2 [∵(am)n = amn]
= 2−3 ÷ 2−2 [∵am × an = am + n]
= 2−3−(−2) [∵am ÷ an = am−n]
= 2−3 + 2
= 2−1
= $\frac{1}{2}$ [∵a−m = $\frac{1}{a^m}$]

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(iii) $(\frac{1}{2})^{−2} \+ (\frac{1}{3})^{−2} \+ (\frac{1}{4})^{−2}$

$(\frac{a}{b})^{−m} = (\frac{b}{a})^m$

$(\frac{1}{2})^{−2} \+ (\frac{1}{3})^{−2} \+ (\frac{1}{4})^{−2}$

$= (\frac{2}{1})^2 \+ (\frac{3}{1})^2 \+ (\frac{4}{1})^2$

= (2)2 + (3)2 + (4)2
= 4 + 9 + 16 = 29

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(iv)(3−1 +4−1 + 5−1)0
a0 = 1 and a−m = $\frac{1}{a^m}$
(3−1 + 4−1 + 5−1)0
$= (\frac{1}{3} + \frac{1}{4} + \frac{1}{5})^0$ [Since a−m = $\frac{1}{a^m}$]
= 1 [Since a0 = 1]

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(v) {$(\frac{−2}{3})^{−2}$}2
a−m = $\frac{1}{a^m}$ and $(\frac{a}{b})^m = \frac{a^m}{b^m}$

{$(\frac{−2}{3})^{−2}$}2

= {$(\frac{3}{−2})^2$}2 [Since a−m = $\frac{1}{a^m}$]

={$\frac{3^2}{(−2)^2}$}2 [Since $(\frac{a}{b})^m = \frac{a^m}{b^m}$]

$=(\frac{9}{4})^2 = \frac{81}{16}$