1B. Solve the equation

$$
\log _{2}\left(9^{x-1}+7\right)=\log _{2}\left(3^{x-1}+1\right)+2
$$
Solution. First, let's determine the domain. From the properties of the exponential function, $9^{x-1}>0,3^{x-1}>0$ for the entire set of real numbers, so $D=\mathbb{R}$. Using properties of logarithmic functions, from $2=\log _{2} 4$ and $\log _{2} a+\log _{2} b=\log _{2} a b$, the equation takes the form

$$
\log _{2}\left(9^{x-1}+7\right)=\log _{2}\left(3^{x-1}+1\right)+\log _{2} 4 \Leftrightarrow \log _{2}\left(9^{x-1}+7\right)=\log _{2} 4\left(3^{x-1}+1\right)
$$

Then

$$
\begin{gathered}
9^{x-1}+7=4\left(3^{x-1}+1\right) \Leftrightarrow, 9^{x-1}+7=4\left(3^{x-1}+1\right) \Leftrightarrow \\
3^{2(x-1)}-4 \cdot 3^{x-1}+7-4=0 \Leftrightarrow 3^{2(x-1)}-4 \cdot 3^{x-1}+3=0
\end{gathered}
$$

By introducing the substitution $3^{x-1}=t$, the equation is reduced to the quadratic equation $t^{2}-4 t+3=0$, with solutions $t_{1}=1, t_{2}=3$. Returning to the substitution $3^{x-1}=1=3^{0}$ and $3^{x-1}=3=3^{1}$, we get linear equations $x-1=0$, or $x-1=1$. Finally, the solutions for the variable $x$ are $x \in\{1,2\}$.