7.121. $\log _{2}\left(4^{x}+4\right)=x+\log _{2}\left(2^{x+1}-3\right)$

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7.121. $\log _{2}\left(4^{x}+4\right)=x+\log _{2}\left(2^{x+1}-3\right)$
## Solution.

Domain of definition: $2^{x+1}-3>0$.

Rewrite the equation as

$$
\begin{aligned}
& \log _{2}\left(2^{2 x}+4\right)-\log _{2}\left(2 \cdot 2^{x}-3\right)=x, \quad \log _{2} \frac{2^{2 x}+4}{2 \cdot 2^{x}-3}=x \\
& \frac{2^{2 x}+4}{2 \cdot 2^{x}-3}=2^{x}, 2^{2 x}-3 \cdot 2^{x}-4=0 .
\end{aligned}
$$

Solving this equation as a quadratic in terms of $2^{x}$, we get $2^{x}=-1, \varnothing$; or $2^{x}=4$, hence $x=2$.

Answer: 2.