7.254. $\log _{2} \sqrt[3]{4}+\log _{8}\left(9^{x+1}-1\right)=1+\log _{8}\left(3^{x+1}+1\right)$.
## Solution.

Domain of definition: $9^{x+1}-1>0 \Leftrightarrow x>-1$.

Since $\log _{2} \sqrt[3]{4}=\frac{2}{3}$, we have

$$
\begin{aligned}
& \log _{8}\left(9 \cdot 3^{2 x}-1\right)-\log _{8}\left(3 \cdot 3^{x}+1\right)=\frac{1}{3} \Leftrightarrow \log _{8} \frac{9 \cdot 3^{2 x}-1}{3 \cdot 3^{x}+1}=\frac{1}{3} \Leftrightarrow \\
& \Leftrightarrow \frac{9 \cdot 3^{2 x}-1}{3 \cdot 3^{x}+1}=8^{\frac{1}{3}}=2 \Leftrightarrow 9 \cdot 3^{2 x}-6 \cdot 3^{x}-3=0
\end{aligned}
$$

Solving this equation as a quadratic in terms of $3^{x}$, we get $3^{x}=-\frac{1}{3}$ (no solutions), or $3^{x}=3^{0} \Rightarrow x=0$.

Answer: 0.