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Set $t=2^x>0$. The exponential function identity $4^x=(2^x{)}^2$ reduces the problem to the quadratic equationPositivity excludes $t=-1$, so $2^x=2$. Since the exponential function $2^x$ is strictly increasing, the unique answer is $\boxed{x=1}$.

Expanding the quadratic form givesMatching coefficients of $xy$ and $y^2$ yields