Answer: (a) \(x\gt \frac13\). (b) \(x=\frac25\) or \(x=2\).

For \(\log_5(12x-4)\) to exist,

\(12x-4\gt 0.\)

Therefore

\(x\gt \frac13.\)

Now use change of base:

\(\log_x125=\frac{\log_5 125}{\log_5 x}=\frac{3}{\log_5 x}.\)

So

\(\frac{6}{\log_x125}=2\log_5 x.\)

The equation becomes

\(\log_5(12x-4)=2\log_5 x+1.\)

Since \(1=\log_5 5\),

\(2\log_5 x+1=\log_5(x^2)+\log_5 5=\log_5(5x^2).\)

Thus

\(\log_5(12x-4)=\log_5(5x^2),\)

so

\(12x-4=5x^2.\)

Hence

\(5x^2-12x+4=0.\)

Factorising,

\((5x-2)(x-2)=0.\)

Therefore

\(x=\frac25\quad\text{or}\quad x=2.\)

Both satisfy the required logarithm domains.