7.104. $\frac{\lg x^{2}}{\lg (6 x-5)}=1$.
Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ 6 x-5>0,\end{array}, \frac{5}{6}<x \neq 1\right.$.

We have $\lg x^{2}=\lg (6 x-5)$, from which $x^{2}=6 x-5, x^{2}-6 x+5=0$, hence $x_{1}=5$ and $x_{2}=1 ; x_{2}=1$ does not satisfy the domain of definition.

Answer: 5.