7.034. $\lg \left(3^{x}-2^{4-x}\right)=2+0.25 \lg 16-0.5 x \lg 4$.
Solution.

Domain of definition: $3^{x}-2^{4-x}>0$.

From the condition

$$
\begin{aligned}
& \lg \left(3^{x}-2^{4-x}\right)=\lg 100+\lg 2-\lg 2^{x} \Rightarrow \lg \left(3^{x}-2^{4-x}\right)=\lg \frac{100 \cdot 2}{2^{x}} \\
& 3^{x}-2^{4-x}=\frac{200}{2^{x}}
\end{aligned}
$$

From here $6^{x}=216$, hence $x=3$.

Answer: 3.