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I'm running in circles and I don't understand how to do this.

$x\log(x) = 100$

Where the $\log$ is in base $10$, I understand that $\log(y)=x$ is $10^x = y$. So is it the same for $x\log(x) = 100$? Would it be $10^{100}=x\cdot x$? It doesn't come out right when I do it, and it's clear that I have holes in my knowledge on logarithms, could someone please tell me my flaws and explain this to me?

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    Wait, suddenly I'm not so sure it is something for [algebra-precalculus] anymore.2011-09-04

3 Answers 3

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You will need the services of the Lambert function $W(x)$ to solve this equation. Briefly, the Lambert function is the inverse of the function $xe^x$: if $x=ye^y$, then $y=W(x)$.

To turn your equation into a form where the Lambert function's appearance becomes transparent, let's first turn everything into natural logarithms:

$x\ln\,x=100\ln\,10$

and then we make the left side a "little" complicated:

$(\ln\,x)e^{\ln\,x}=100\ln\,10$

We now recognize the Lambert form, and thus perform the inversion:

$\ln\,x=W(100\ln\,10)$

from which

$x=e^{W(100\ln\,10)}\approx 56.961248432\dots$

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    Here is the link for a meta thread about adding a tag for the Lambert W function: http://meta.math.stackexchange.com/q/2908/6222011-09-04
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If $x \log_b (y) = z$ then taking anti-logarithms you get $y^x = b^z$.

So in this case with $y=x$ and $b=10$ you get $x^x = 10^{100}$.

You will not find it easy to solve this explicitly for $x$; try reading about the Lambert W function or use numerical methods to get something just over 56.96.

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    @$A$saf: To the meta-cave!2011-09-04
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There is no way to solve $x\log_{10}x=100$ exactly using the methods of school algebra. There is a way, called Newton's Method, to get a solution to as many decimals as you want, using Calculus. Newton's Method is in a thousand intro Calculus textbooks, also a thousand websites. If you haven't done Calculus yet, you have something to look forward to.