We determine $x$ such that $2^x\cdot 8^{2x+2}=\dfrac{2\cdot 2^{x^2+15}}{(4^x)^2}$.
$$\begin{align}
2^x\cdot 8^{2x+2}=\frac{2\cdot 2^{x^2+15}}{(4^x)^2} &\Leftrightarrow 2^x\cdot (2^3)^{2x+2}=\frac{2^1\cdot 2^{x^2+15}}{4^{2x}}\\
&\Leftrightarrow 2^x\cdot 2^{6x+6}=\frac{2^{x^2+16}}{(2^2)^{2x}}\\
&\Leftrightarrow 2^{7x+6}=\frac{2^{x^2+16}}{2^{4x}}\\
&\Leftrightarrow 2^{7x+6}=2^{x^2-4x+16}\\
&\Leftrightarrow 7x+6=x^2-4x+16\\
&\Leftrightarrow x^2-11x+10=0\\
&\Leftrightarrow(x-10)(x-1)=0\\
&\Leftrightarrow x=1 \textrm{ or } x=10.
\end{align}$$
