7.042. $x(\lg 5-1)=\lg \left(2^{x}+1\right)-\lg 6$.

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7.042. $x(\lg 5-1)=\lg \left(2^{x}+1\right)-\lg 6$.
Solution.

$$
\begin{aligned}
& x(\lg 5-\lg 10)=\lg \left(2^{x}+1\right)-\lg 6, \quad x \lg \frac{5}{10}=\lg \frac{2^{x}+1}{6} \\
& \lg 2^{-x}=\lg \frac{2^{x}+1}{6}, 2^{-x}=\frac{2^{x}+1}{6}, 2^{2 x}+2^{x}-6=0
\end{aligned}
$$

Solving this equation as a quadratic in terms of $2^{x}$, we find $2^{x}=-3$ (not valid), $2^{x}=2$, from which we have $x=1$.

Answer: 1.