$x^{4\log x} = \frac{x^{12}}{{10}^8} $ all sum of x? ${10}^8 = \frac{x^{12}}{x^{4\log x}}$
${10}^8 = x^{12 - 4\log x}$
$x > 0$
$x \neq 1$
by guessing, x = 10 is true. but how sum of all x?
 A: Assuming you are wondering how to systematically get all $x$ for which the original expression is true, you can take logs of both sides (base 10 log) to obtain
\begin{align*}
8 = \log x^{12-4\log x} = (12-4\log x) (\log x) = 12\log x - 4(\log x)^2.
\end{align*}
Organizing a bit gives the equivalent statement,
\begin{align*}
(\log x)^2 - 3\log x + 2 = 0.
\end{align*}
The quadratic formula now gives
\begin{align*}
\log x = \frac{3\pm \sqrt{9-8}}{2} = \{2,1\}.
\end{align*}
The zeros of the function $f(x) = (\log x)^2-3\log x + 2$ are then $x = 10$ or $100$.
A: As you wrote: $x\ne1\;\;\&\;\;x>0$
Another way:
$\log_xx^{4\log{x}}=\log_x{\frac{x^{12}}{10^8}}$
$4\log x\cdot\log_xx=\log_x{x^{12}}-\log_x{10^8}$
$4\log x=12-8\log_x10\Bigg/:4$
$\log{x}=3-2\log_x10$
$\log{x}=\frac{\log_xx}{\log_x10}=\frac{1}{\log_x10}$
$\frac{1}{\log_x10}=3-2\log_x10\Bigg/\cdot\log_x10$
$2\log^2_x10-3\log_x10+1=0$
$2\log^2_x10-2\log_x10-\log_x10+1=0$
$2\log_x10(\log_x10-1)-(\log_x10-1)=0$
$(\log_x10-1)(2\log_x10-1)=0$
$\log_x10=1\implies x_1=10$
$\log_x10^2=1\implies x_2=100$
$x_1+x_2=110$
