If $64^{x} = 4^{x^{2} - 4}$ , then find the value of $x$ .

x = 4

x = - 1

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x = - 4

x = 1

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x = 10

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x = \sqrt{20}

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x = 3

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Solution

The correct option is A $x = 4$ or $x = - 1$
$64^{x} = 4^{3 x} = 4^{x^{2} - 4}$ , by equating powers , we get,
$x^{2} - 4 = 3 x$ , which implies $x^{2} - 3 x - 4 = 0$
$\Rightarrow x^{2} - 4 x + x - 4 = 0$
$\Rightarrow x (x - 4) + 1 (x - 4) = 0$
$\Rightarrow (x - 4) (x + 1) = 0$
The roots are $x = 4, - 1$

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