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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.

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    $\begingroup$ Do you mean $x^{\log x} = 100x$? $\endgroup$ – Antonio Vargas Jan 30 '14 at 2:02
  • $\begingroup$ 1.-Substitute $x=10^y$ 2.- ??? 3.-Profit $\endgroup$ – chubakueno Jan 30 '14 at 2:04
  • $\begingroup$ yes. that is what i mean $\endgroup$ – user3175999 Jan 30 '14 at 2:04
  • $\begingroup$ 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). $\endgroup$ – chubakueno Jan 30 '14 at 2:06
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    $\begingroup$ @user3175999 - engineers use it to mean base $10$, mathematicians base $e$. $\endgroup$ – nbubis Jan 30 '14 at 2:10
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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.

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  • $\begingroup$ can you explain in the second line where you went from log(x^logx) = log(x)log(x) $\endgroup$ – user3175999 Jan 30 '14 at 2:17
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    $\begingroup$ @user3175999 - Remember that $\log (a^b) = b\log(a)$. $\endgroup$ – nbubis Jan 30 '14 at 2:18
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Hint: Take the logarithm of both sides. You will get a quadratic equation in $y=\log x$.

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  • $\begingroup$ thanks, this helped, but the other person was more clear $\endgroup$ – user3175999 Jan 30 '14 at 2:34

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