How to solve $x$ for which $\frac{x-5}{x} = \left(\frac{5}{x}\right)^\sqrt{2x+2}$? $${\dfrac{x-5}{x} = \left(\dfrac{5}{x}\right)^\sqrt{2x+2}}$$
So at the first glance, I thought maybe this was a normal equation and that can be easily solved using logarithm. I was wrong... (after hours of trying).
There are two (possible) ideas which I think possible to solve this problem:


*

*Try to solve $x$ for which, $$0 = \frac {d}{dx} | lhs - rhs |$$I was
thinking that the functional graph must be at the minimum if $x$ is the right
answer for the equation.

*Brute-force all possible real numbers (using computer
programming to do this job).


However, I'm also thinking that it must have other ways to solve for $x$ which I have no skills/knowledge whatsoever in order to solve the equation.
This is not my homework, it's just random challenge that popped out while I'm surfing the net!
 A: I guess that only numerical methods could solve the problem and Newton method would probably be the simplest to use.
Starting from a "reasonable" guess $x_0$, Newton mpethod will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, consider that we are looking for the zero of function
$$f(x)=\frac{x-5}{x}- \left(\frac{5}{x}\right)^{\sqrt{2 x+2}}$$ I let you the pleasure of finding the derivative. Looking at the plot of $f(x)$, you can see that the $x$ intercept is close to $7$ which would be a perfect guess.
But, let us be lazy and use $x_0=5$ for which $f(5)=-1$. The method will generate the following iterates 
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 5 \\
 1 & 6.120046189 \\
 2 & 6.755330011 \\
 3 & 6.902818440 \\
 4 & 6.909062337 \\
 5 & 6.909072846 
\end{array}
\right)$$ You could do this with Excel computing the derivative numerically.
A: If you set $5/x=t$, then
$$
2x+2=\frac{10}{t}+2=\frac{10+2t}{t}
$$
so the equation can be written
$$
(1-t)^{\sqrt{t}}=t^{\sqrt{10+2t}}
$$
The limitation $0\le t\le 1$ is implicit in the original equation.

The left-hand side describes a decreasing function, the right-hand side an increasing function, so the solution is unique. It exists by comparing the values at $0$ and $1$.
