If $8^{x} = 16^{x - 1}$ , find $x$ .

\frac{1}{8}

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\frac{1}{2}

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2

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4

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Solution

The correct option is D $4$
Given, $8^{x} = (16)^{x - 1}$
Left hand side, $8^{x} = (2^{3})^{x} = 2^{3 x}$
Right hand side $16^{x - 1} = (2^{4})^{x - 1} = 2^{4 (x - 1)} = 2^{(4 x - 4)}$
Equating both sides, we get
$2^{3 x} = 2^{4 x - 4}$
As bases are equal, then powers must be equal, so
$\Rightarrow 3 x = 4 x - 4$
$\Rightarrow x = 4$

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