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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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    WOW. This is a mouth full. I did not think it would lead to so much math.2011-09-04
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    @J. M.: Do you think the [special-functions] tag should be added to the question? Or even create one for the Lambert function.2011-09-04
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    @Américo: I'm not quite sure if there should be a [tag:special-functions] tag, much less a more specialized [tag:lambert-w] tag. On the other hand, we now have quite a pile of questions where the Lambert function shows up in the solution... tell you what, let's create a tag for Lambert questions if at least five people upvote your comment. How's that sound?2011-09-04
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    It sounds good.2011-09-04
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    @Doug: Math only leads to more math! And J.M., I was thinking the same thing (not the part about 5 upvotes) as soon as I read the question. There sure are a few L-W function questions/answers out there.2011-09-04
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    @Americo, J.M.: I am actually against the Lambert function, mostly since I am not sure whether or not it is only a tool or a topic for questions as well. One could argue the same about [forcing], but in that case I can answer. How about we open this for a meta discussion?2011-09-04
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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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    Why is x the exponent for y? I don't find that really intuitive, could you please explain?2011-09-04
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    You forgot some braces there, I hope you did not mind.2011-09-04
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    @Doug: it is a basic property of logarithms: see the [power formula here](http://en.wikipedia.org/wiki/Logarithm#Product.2C_quotient.2C_power.2C_and_root)2011-09-04
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    Henry and @Asaf, don't forget to upvote Américo's comment on J.M.'s answer :)2011-09-04
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    @The Chaz: What if I disagree? Do I still have to upvote or can I vote against?2011-09-04
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    @Asaf: 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.