Find the value of $x$ , if $8^{x - 3} = 16^{4 - x} .$

4

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\frac{11}{3}

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\frac{25}{7}

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\frac{7}{2}

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Solution

The correct option is C $\frac{25}{7}$
$8^{x - 3} = {(2^{3})}^{x - 3} = 2^{3 (x - 3)} = 2^{(3 x - 9)} - - (i)$

$16^{4 - x} = {(2^{4})}^{4 - x} = 2^{4 (4 - x)} = 2^{(16 - 4 x)} - - (i i)$

Equating (i) and (ii), we get

$2^{(3 x - 9)} = 2^{(16 - 4 x)}$

Since the bases are same, we can equate the powers.
$\Rightarrow 3 x - 9 = 16 - 4 x$
$\Rightarrow 7 x = 25$
$x = \frac{25}{7}$

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