If $4^{n - 2} + 4^{2} = 32$ , then what is the value of $n$ ?

2

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4

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6

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Solution

The correct option is A $4$
Given, $4^{n - 2} + 4^{2} = 32$
$4^{n - 2} = 32 - 4^{2}$
$4^{n - 2} = 32 - 16$
$4^{n - 2} = 16$
$4^{n - 2} = 4^{2}$
$n - 2 = 2$ ..... (as the bases are equal)
$n = 4$

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