Logarithms #4

The equation is undefined when $x=0$ and $x=1$ , since $\log_x a$ is undefined.

For $x< 1$ , the $RHS$ is always $<0$ but the $LHS >0$ , therefore, there is no solution for $x \le 1$ .

For $x > 1$ , $|x-1| = x-1$ and there is a trivial solution, $x=2$ , then $LHS = RHS = 1$ , but is not an answer option.

The other solution is given below:

$\begin{aligned} |x-1|^{\log_3 x^2 - 2\log_x 9} & = (x-1)^7 \\ \Rightarrow (x-1)^{\log_3 x^2 - 2\log_x 9} & = (x-1)^7 \\ \Rightarrow \log_3 x^2 - 2\log_x 9 & = 7 \\ 2\log_3 x - \frac{2\log_3 9}{\log_3 x} & = 7 \\ 2 (\log_3 x)^2 - 7 \log_3 x - 4 & = 0 \\ (2\log_3 x + 1)(\log_3 x - 4) & = 0 \\ \log_3 x & = \begin{cases} - \frac{1}{2} \color{#D61F06}{< 1 \quad \text{rejected}} \\ \quad 4 \color{#3D99F6}{> 1 \quad \text{accepted}} \end{cases} \\ \Rightarrow x & = 3^4 = \boxed{81} \end{aligned}$