Solve the equation: $\;$ $\log_{x - 1} 3 = 2$
Given equation: $\;\;$ $\log_{x - 1} 3 = 2$
i.e. $\;$ $\left(x - 1\right)^2 = 3$
i.e. $\;$ $x^2 - 2x + 1 = 3$
i.e. $\;$ $x^2 - 2x - 2 = 0$
i.e. $\;$ $x = \dfrac{2 \pm \sqrt{4 + 8}}{2} = \dfrac{2 \pm \sqrt{12}}{2} = \dfrac{2 \pm 2 \sqrt{3}}{2}$
i.e. $\;$ $x = 1 \pm \sqrt{3}$
When $\;$ $x = 1 - \sqrt{3}$, $\;$ then we have
$\log_{1 - \sqrt{3} - 1} 3 = 2$ $\implies$ $\log_{-\sqrt{3}} 3 = 2$
But base of a logarithm cannot be negative.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{1 + \sqrt{3} \right\}$