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Laws of Exponents for Real Numbers

Simplify: [ ...

Question

Simplify: ${[{(2)}^{- 1} + {(4)}^{- 1} + {(3)}^{- 1}]}^{- 1}$

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Solution

$[{(2)}^{- 1} + {(4)}^{- 1} + {(3)}^{- 1}]$
$= {[\frac{1}{2} + \frac{1}{4} + \frac{1}{3}]}^{- 1}$
$= {[\frac{1}{2} + \frac{1}{4} + \frac{1}{3}]}^{- 1}$
$= {(\frac{13}{12})}^{- 1}$
$= \frac{12}{13}$

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