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
(a^m)ⁿ = a^mn, 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    [∵(a^m)ⁿ = a^mn]
= 2⁻³ ÷ 2⁻²                  [∵a^m × aⁿ = a^{m + n}]
= 2^{−3−(−2)}                     [∵a^m ÷ aⁿ = a^m−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 \((\frac{a}{b})^m = \frac{a^m}{b^m}\)]

\(=(\frac{9}{4})^2 = \frac{81}{16}\)