Find the value of $x$ if $8^{x - 1} = 2^{x + 3}$ .

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Solution

The correct option is D $3$
$8^{x - 1} = 2^{x + 3}$
${(2^{3})}^{x - 1} = 2^{x + 3}$
$2^{3 x - 3} = 2^{x + 3}$
As the bases are equal, so exponents must be equal
So $3 x - 3 = x + 3$
$3 x - x = 3 + 3$
$2 x = 6$
$x = \frac{6}{2} = 3$

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