$7.23 \quad 7^{\lg x}-5^{\lg x+1}=3 \cdot 5^{\lg x-1}-13 \cdot 7^{\lg x-1}$
7.23 Group the powers with base 7 on the left side of the equation, and the powers with base 5 on the right side: $7^{\lg x}+13 \cdot 7^{\lg x-1}=3 \cdot 5^{\lg x-1}+5^{\lg x+1}$.

By factoring out the power with the smaller exponent and performing further transformations, we get

$$
\begin{aligned}
& 7^{\lg x-1}(7+13)=5^{\lg x-1}\left(3+5^{2}\right) ; \frac{7^{\lg x-1} \cdot 20}{5^{\lg x-1} \cdot 28}=1 ; \\
& \left(\frac{7}{5}\right)^{\lg x-2}=\left(\frac{7}{5}\right)^{0}
\end{aligned}
$$

From this, $\lg x-2=0$, i.e., $x=100$.

Answer: $x=100$.