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.
 A: Firstly, as $4^2+1=17$, $x=2$ is an obvious solution.
Then as the LHS of Moo's solution has a term $4^x$ and the RHS $2^x$ the LHS grows faster than the RHS (or does this veer toward using using too much calculus?) so there can't be another equality as LHS will always be greater than RHS.
A: Assuming that we know that the solution is an integer, there are few possibilities, as $x2^x$ grows fast. $x$ is certainly positive. $3$ is certainly too large, and it is enough to test $1$ and $2$.
If we don't know, I don't think there is an analytical way to solve, even using Lambert's $W$ function, which can solve $x2^x=\frac{17}2$ only.
A: 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.
A: When $x\leq 0$,  there are no solutions because $x\left(2^x+2^{-x}\right)\leq 0$. Let $x>0$ and rewrite as
$$2^x+2^{-x}=\frac{17}{2x} $$
For $x>0$, the RHS is a strictly decreasing function ($a>b>0\Rightarrow \frac 1a<\frac 1b$). And we can show that the LHS is a strictly increasing function. Since $2^x+2^{-x}=2^x+\frac{1}{2^x}$ and $2^x>1$ when $x>0$, we need to show $t+\frac 1t$ is strictly increasing for $t>1$. Suppose $u>v>1$. Then
$$\left(u+\frac 1u\right)-\left(v+\frac 1v\right)=u-v-\frac{u-v}{uv}=(u-v)\left(1-\frac{1}{uv}\right)>0 $$ since $uv>1$. Thus, there can be no more than $1$ solution when $x>0$. We notice $x=2$ is a solution. That means $x=2$ is the only solution.
