$x^{4\log x} = \frac{x^{12}}{{10}^8} $ all sum of x?

Question

${10}^8 = \frac{x^{12}}{x^{4\log x}}$

${10}^8 = x^{12 - 4\log x}$

$x > 0$

$x \neq 1$

by guessing, x = 10 is true. but how sum of all x?

Answer 1

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 $100$.

Answer 2

As you wrote: $x\ne1\;\;\&\;\;x>0$

Another way:

$\log_xx^{4\log{x}}=\log_x{\frac{x^{12}}{10^8}}$

$4\log x\cdot\log_xx=\log_x{x^{12}}-\log_x{10^8}$

$4\log x=12-8\log_x10\Bigg/:4$

$\log{x}=3-2\log_x10$

$\log{x}=\frac{\log_xx}{\log_x10}=\frac{1}{\log_x10}$

$\frac{1}{\log_x10}=3-2\log_x10\Bigg/\cdot\log_x10$

$2\log^2_x10-3\log_x10+1=0$

$2\log^2_x10-2\log_x10-\log_x10+1=0$

$2\log_x10(\log_x10-1)-(\log_x10-1)=0$

$(\log_x10-1)(2\log_x10-1)=0$

$\log_x10=1\implies x_1=10$

$\log_x10^2=1\implies x_2=100$

$x_1+x_2=110$