We have,
$$
\begin{aligned}
&\log _{2}\left(2^{x-1}+6\right)+\log _{2}\left(4^{x-1}\right)=5 \\
&\left.\Rightarrow 2^{x-1}+6\right)\left(2^{2 x-2}\right)=2^{5}
\end{aligned}
$$
Let
$$
y=2^{x-1}
$$
$$
\Rightarrow \quad \cdot(y+6) y^{2}=2^{5}
$$
$$
\begin{aligned}
&\Rightarrow \quad y^{3}+6 y^{2}-32=0 \\
&\Rightarrow \quad(y-2)\left(y^{2}+8 y+16\right)=0
\end{aligned}
$$
So,
$$
\begin{aligned}
y &=2 \\
2^{x-1} &=2^{1}
\end{aligned}
$$
On comparing both sides, we get
$$
\begin{array}{r}
x-1=1 \\
x=2
\end{array}
$$