# How do you solve $x^{\log x}=100x$

How do you solve $x^{\log x}=100x$?

Can you please thoroughly explain the left side of the equation.

Please explain very clearly because I have only been learning logarithms for about a week.

• Do you mean $x^{\log x} = 100x$? – Antonio Vargas Jan 30 '14 at 2:02
• 1.-Substitute $x=10^y$ 2.- ??? 3.-Profit – chubakueno Jan 30 '14 at 2:04
• yes. that is what i mean – user3175999 Jan 30 '14 at 2:04
• You should specify that you are using the logarithm base $10$ (It is not wrong as stated, but I suspect by the $100=10^2$ that you intended that). – chubakueno Jan 30 '14 at 2:06
• @user3175999 - engineers use it to mean base $10$, mathematicians base $e$. – nbubis Jan 30 '14 at 2:10

Take $\log$ from both sides: $$\log \left( x^{\log x}\right)=\log(100x)$$ $$\log (x) \log (x)=\log(100x)=\log(100)+\log(x)$$ Or: $$(\log x)^2-(\log x)=2$$ Now you have a quadratic equation which you should be able to solve.
• @user3175999 - Remember that $\log (a^b) = b\log(a)$. – nbubis Jan 30 '14 at 2:18
Hint: Take the logarithm of both sides. You will get a quadratic equation in $y=\log x$.