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

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

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?

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

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 $$x=10000.$$

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