Question
$\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)$
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Answer
$\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{32}$ Solution: Here, $\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{16\times2}$ $\big[\because(\text{a}^{\text{m}})^{\text{n}}=\text{a}^{\text{mn}}\big]$ $=\Big(\frac{-1}{4}\Big)^{32}$ $\therefore\bigg[\Big(\frac{-1}{4}\Big)^{16}\bigg]^{2}=\Big(\frac{-1}{4}\Big)^{32}$
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