If $2 {log}_{4} x = 1 + {log}_{4} (x - 1)$ find the value of x

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Solution

Given $\Rightarrow {log}_{4} x^{2} - {log}_{4} (x - 1) = 1$
$\Rightarrow {log}_{4} \frac{x^{2}}{x - 1} = 1 \Rightarrow 4^{1} = \frac{x^{2}}{x - 1}$
$\Rightarrow x^{2} = 4 x - 4 \Rightarrow x^{2} - 4 x + 4 = 0$
$\Rightarrow (x - 2)^{2} = 0 \Rightarrow x = 2$

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