7.124. $\log _{5}\left(4^{x}+144\right)-4 \log _{5} 2=1+\log _{5}\left(2^{x-2}+1\right)$
## Solution.

We have

$$
\begin{aligned}
& \log _{5} \frac{2^{2 x}+144}{16}=\log _{5} 5\left(\frac{2^{x}}{4}+1\right) \frac{2^{2 x}+144}{16}=\frac{5\left(2^{x}+4\right)}{4} \\
& 2^{2 x}-20 \cdot 2^{x}+64=0
\end{aligned}
$$

- Solving this equation as a quadratic in terms of $2^{x}$, we get $\left(2^{x}\right)_{1}=4$ or $\left(2^{x}\right)_{2}=16$, hence $x_{1}=2, x_{2}=4$.

Answer: $2 ; 4$.