Express the following in exponential form , using laws :
$[{(- 2^{0})}^{- 4} \div {(4^{0})}^{- 3}] + {(7^{2})}^{- 3}$

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Solution

Given that , $[{(- 2^{0})}^{- 4} \div {(4^{0})}^{- 3}] + {(7^{2})}^{- 3}$
$= \frac{1}{{(- 2^{0})}^{4}} \div \frac{1}{{(4^{0})}^{3}} + \frac{1}{{(7^{2})}^{3}}$
$= \frac{1}{{(- 1)}^{4} \times 2^{0 \times 4}} \div \frac{1}{4^{0 \times 3}} + \frac{1}{7^{2 \times 3}}$
$= \frac{1}{1 \times 2^{0}} \div \frac{1}{4^{0}} + \frac{1}{7^{6}}$
$= \frac{1}{1} \div \frac{1}{1} + \frac{1}{7^{6}}$
$= 1 + 7^{- 6}$
Hence, the required value is $1 + 7^{- 6}$ .

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