If $4^{x} = 8^{y}$ , find $\frac{x}{y} - 1$ .

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Solution

Given,
$4^{x} = 8^{y}$
or, $2^{2 x} = 2^{3 y}$
or, $2 x = 3 y$
or, $\frac{x}{y} = \frac{3}{2}$ .
Now,
$\frac{x}{y} - 1$
$= \frac{3}{2} - 1$
$= \frac{1}{2}$

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