Find the value of $7^{- 3}$ .

$343$

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$\frac{1}{343}$

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$49$

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$\frac{1}{49}$

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Solution

The correct option is B
$\frac{1}{343}$

$∵ a^{- n} = \frac{1}{n}$

$Hence, 7^{- 3} = \frac{1}{7^{3}}$
$= \frac{1}{7 \times 7 \times 7} = \frac{1}{343}$

Suggest Corrections

Similar questions

a

b

\frac{3 + \sqrt{7}}{3 - \sqrt{7}} = a + b \sqrt{7}

\sqrt{- \sqrt{3} + \sqrt{3 + 8 \sqrt{7 + 4 \sqrt{3}}}}

(\sqrt{7} - \sqrt{3})^{2} + (\sqrt{7} + \sqrt{3})^{2}

7^{2 x} \div 7^{3} = 7^{9}

x

\frac{3^{7}}{3^{4} \times 3^{3}}

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