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Laws of Exponents for Real Numbers

Simplify: 2-...

Question

Simplify:
$(2^{- 1} + 3^{- 1})^{- 1} \div 4^{- 1}$

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Solution

$(2^{- 1} + 3^{- 1})^{- 1} \div 4^{- 1}$
we know $a^{- 1} =$ $\frac{1}{a} .$
$(\frac{1}{2} + \frac{1}{3})^{- 1} \div \frac{1}{4}$
$(\frac{5}{6})^{- 1} \div \frac{1}{4}$ $\Rightarrow \frac{6 / 5}{1 / 4}$ = $\frac{24}{5}$

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