Solve this logarithmic equation: $2^{2-\ln x}+2^{2+\ln x}=8$

Question

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?

Answer 1

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 $\ln x = 0$.

Answer 2

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 $$\dfrac4a + 4a = 8 \implies a^2 + 1 =2a \implies a =1 \implies 2^{\ln(x)} = 1 \implies x = 1$$

Answer 3

Let $\displaystyle 2^{\ln x} =t$, then your equation is quadratic, right?

Answer 4

Hint:

Dividing by $4$ we get the equivalent equation $$2^{-\ln x}+2^{\ln x}=2$$ then let $y=2^{\ln x}$ so $$y^{-1}+y=2\iff1+y^2=2y$$