<s>[INST] $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?
 [/INST] As you wrote: $x\ne1\;\;\&\;\;x>0$
Another way:
$\log_xx^{4\log{x}}=\log_x{\frac{x^{12}}{10^8}}$
$4\log x\cdot\log_xx=\log_x{x^{12}}-\log_x{10^8}$
$4\log x=12-8\log_x10\Bigg/:4$
$\log{x}=3-2\log_x10$
$\log{x}=\frac{\log_xx}{\log_x10}=\frac{1}{\log_x10}$
$\frac{1}{\log_x10}=3-2\log_x10\Bigg/\cdot\log_x10$
$2\log^2_x10-3\log_x10+1=0$
$2\log^2_x10-2\log_x10-\log_x10+1=0$
$2\log_x10(\log_x10-1)-(\log_x10-1)=0$
$(\log_x10-1)(2\log_x10-1)=0$
$\log_x10=1\implies x_1=10$
$\log_x10^2=1\implies x_2=100$
$x_1+x_2=110$
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