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

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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?

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  • $\begingroup$ Will you like to support my question as well? $\endgroup$ – Shadow Apr 14 at 0:01
  • $\begingroup$ @Shadow, what is meant by support here? $\endgroup$ – lab bhattacharjee Apr 14 at 2:31
  • $\begingroup$ Upvote @Iab Bhattacharjee $\endgroup$ – Shadow Apr 16 at 23:59
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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 $x=10000.$

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  • $\begingroup$ Will you like to support my question as well? $\endgroup$ – Shadow Apr 14 at 0:01
  • $\begingroup$ Upvote @Michael Rozenberg $\endgroup$ – Shadow Apr 17 at 0:00

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