Find out the value of n if $3^{n} = 4^{n - 1}$

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Solution

$3^{x} = 4^{x - 1}$
$∴ x = {log}_{3} 4^{(x - 1)}$
$∴ x = (x - 1) {log}_{3} 4$
$∴ x (1 - {log}_{3} 4) = - 1$
$∴ x = \frac{1}{{log}_{3} 4}$
$x = 0.261$

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