If $3^{x} = 4^{x - 1}$ , then $x =$

\frac{2 {log}_{3} 2}{2 {log}_{3} 2 - 1}

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\frac{2}{2 - {log}_{2} 3}

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\frac{1}{1 - {log}_{4} 3}

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\frac{2 {log}_{2} 3}{2 {log}_{2} 3 - 1}

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Solution

The correct option is A $\frac{2 {log}_{3} 2}{2 {log}_{3} 2 - 1}$
$3^{x} = 4^{x - 1}$
apply logarithm with base 3
$x = (2 x - 2) {log}_{3} 2$
$x (1 - 2 {log}_{3} 2) = - 2 {log}_{3} 2 ⟹ x = \frac{2 {log}_{3} 2}{2 {log}_{3} 2 - 1}$

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