Mend the logs!

$\begin{aligned} \log_5 x + 6 \log_x 5 & = 5 \\ \log_5 x + \frac 6 {\log_5 x} & = 5 \\ \log_5^2 x - 5 \log_5 x + 6 & = 0 \\ (\log_5 x - 2)(\log_5 x -3) & = 0 \end{aligned}$

$\implies \begin{cases} \log_5 x = 2 & \implies x = 5^2 = 25 \\ \log_5 x = 3 & \implies x = 5^3 = 125 \end{cases}$

Therefore, the two possible values of $x$ are 25 and 125 ,

$\large \color{#D61F06} \log_{5}{x} + 6\log_{x}{5} = 5$

In this we will change the base of $\log_{x}{5}$ into $5$ . Using the change of base rule for logarithms:

$\large log_{x}{5} = \frac{\log_{5}{5}}{\log_{5}{x}}$ $\color{#3D99F6}(Note: \log_{5}{5} =1)$

$\large log_{x}{5} = \frac{1}{\log_{5}{x}}$

So $6\log_{x}{5} =\frac{6}{\log{5}{x}}$

$\large \therefore \log_{5}{x} + \frac{6}{\log_{5}{x}}$

$\large \color{#3D99F6} let \log_{5}{x} \ denote \ 'y'$

Then it becomes

$\large \color{#20A900} \Rightarrow \!\ y + \frac{6}{y} = 5$

$y^2 + 6 = 5y$ $\color{#3D99F6}(Multiply\ both\ sides\ by\ y)$

$\Rightarrow \!\ y^2 -\ 5y + 6\ =\ 0$ $\color{#3D99F6}(I\ have\ skipped\ the\ factorising\ part)$

$y=3\ or\ y=2$

$\large \color{magenta}\therefore \log_{5}{x}\ =3\ or\ 2$

$\therefore x\ =\ 5^3\ or\ 5^2$

$\large \color{#D61F06} x\ =\ 125\ or\ 25$