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

Question

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.

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 02:06
@user3175999 - engineers use it to mean base $10$, mathematicians base $e$. — Nathaniel Bubis, Jan 30 '14 at 02:10

score 9 · Accepted Answer · answered Jan 30 '14 at 02:13

9

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.

answered Jan 30 '14 at 02:13

Nathaniel Bubis

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can you explain in the second line where you went from log(x^logx) = log(x)log(x) – user3175999 Jan 30 '14 at 02:17
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@user3175999 - Remember that $\log (a^b) = b\log(a)$. – Nathaniel Bubis Jan 30 '14 at 02:18

score 7 · Answer 2 · answered Jan 30 '14 at 02:11

7

Hint: Take the logarithm of both sides. You will get a quadratic equation in $y=\log x$.

answered Jan 30 '14 at 02:11

Lucian

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thanks, this helped, but the other person was more clear – user3175999 Jan 30 '14 at 02:34

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

2 Answers2