Solve this logarithmic equation: $2^{2-\ln x}+2^{2+\ln x}=8$ Solve this logarithmic equation: $2^{2-\ln x}+2^{2+\ln x}=8$. I thought to write
$$\dfrac{2^2}{2^{\ln(x)}} + 2^2 \cdot 2^{\ln(x)} = 2^3 \implies \dfrac{2^2 + 2^2 \cdot \left(2^{\ln(x)}\right)^2}{2^{\ln(x)}} = 2^3$$
What should I do else?

Since $2^2+2^2 = 4+4 = 8$, can you just let $\ln x = 0$ and find $x = 1$?
ETA: This can be shown to be unique by finding the derivative with respect to $x$:
$$
\frac{d}{dx} 2^{2-\ln x} + 2^{2+\ln x}
    = \frac{\ln 2}{x} \left(2^{2+\ln x}-2^{2-\ln x}\right)
$$
and provided that $x > 0$, this only equals $0$ at $\ln x = 0$.
