If $2\left(2^{2 x}\right)=4^{x}+64$, what is the value of $x$?
(a) -2
(b) -1
(c) 1
(d) 2
(e) 3
The correct option is (e).

Solution 1: We observe that the terms on the right side of the given equation can be written as powers of 2. Indeed, $4^{x}=\left(2^{2}\right)^{x}=2^{2 x}$ and $64=2^{6}$. Thus, the equation becomes $2\left(2^{x}\right)=2^{2 x}+2^{3}$. We then have $2\left(2^{2 x}\right)-2^{2 x}=2^{6}$, which simplifies to $2^{2 x}(2-1)=2^{6}$, or $2^{2 x}=2^{6}$. Therefore, $2 x=6$ and it follows that $x=3$.

Solution 2: $4^{x}+4^{x}=2\left(4^{x}\right)=2\left(2^{2 x}\right)=4^{x}+4^{3}$, hence $4^{x}=4^{3}$ and it follows that $x=3$.