<s>[INST] 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?
 [/INST] We have $2^{2-\ln(x)} = \dfrac4{2^{\ln(x)}}$ and $2^{2+\ln(x)} = 4\cdot2^{\ln(x)}$. Hence, setting $2^{\ln(x)} = a$, we obtain
$$\dfrac4a + 4a = 8 \implies a^2 + 1 =2a \implies a =1 \implies 2^{\ln(x)} = 1 \implies x = 1$$
</s>