$\frac{\left(10^4\right)}{x^2}=\frac{\left(x^{\left(8-2\log x\right)}\right)}{10^4}$ Solve for x.

Question

$\frac{\left(10^4\right)}{x^2}=\frac{\left(x^{\left(8-2\log x\right)}\right)}{10^4}$ Solve for x.

$\frac{\left(10^4\right)}{x^2}=\frac{\left(x^{\left(8-2\log x\right)}\right)}{10^4}\Rightarrow 10^8=\frac{x^10}{x^{2\log x}}=\frac{x^{10}}{{(x^{\log_x x^2})}^{\frac {1}{\log_x10}}}=\frac{x^{10}}{{(x^2)}^{\frac 1{\log _x 10}}}$ [The last line comes from the fact that $\log_{10}x^2=\frac{\log_xx^2}{\log _x 10}$]

Now I am getting that one of the solutions is $x=10$. Now how to solve it after that?

Answer 1

Hint:

Take logarithm base $10$ assuming the given logarithm in the same $10$

$$4-2\log_{10}x=(8-2\log_{10}x)\log_{10}x-4\iff(\log_{10}x)^2-5\log_{10}x+4=0$$

which is a quadratic equation in $\log_{10}x$

Can you solve it?

Answer 2

It's $$10^8=x^{10-2\log{x}}$$ or $$\log10^8=\log{x^{10-2\log{x}}}$$ or $$8=(10-2\log{x})\log{x}$$ Can you end it now?

I got $x=10$ or $x=10000.$