Solve $x(2^x+2^{-x})=\frac{17}{2}$ analytically Is it possible to solve $x(2^x+2^{-x})=\frac{17}{2}$ analytically? I was able to rearrange to get $x\cosh(x\ln(2))=\frac{17}{4}$ but can't get any further. WA gives 2 as the solution, but no steps provided.

First we simplify the question
$$2^x + \frac{1}{2^x} = \frac{17}{2x}$$
$$\frac{4^x + 1}{2^x} = \frac{17}{2x}$$
Assigning $2^x$ as $t$
$$\frac{t^2 + 1}{t} = \frac{17}{2x}$$
Comparing numerators and denominators
$$t = 2x$$
$$t^2 + 1 = 17$$
$$t^2 = 16$$
$$x^2 = 4$$
$$x = +2, -2$$
But, on substituting the obtained values in the main equation, we get only $x= 2$ as solution.
