# How do I solve this equation involving a logarithm?

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?

-
Wait, suddenly I'm not so sure it is something for [algebra-precalculus] anymore. –  Asaf Karagila Sep 4 '11 at 9:24

## 3 Answers

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

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

-
WOW. This is a mouth full. I did not think it would lead to so much math. –  Doug Sep 4 '11 at 9:48
@J. M.: Do you think the [special-functions] tag should be added to the question? Or even create one for the Lambert function. –  Américo Tavares Sep 4 '11 at 12:42
@Américo: I'm not quite sure if there should be a special-functions tag, much less a more specialized 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? –  J. M. is back. Sep 4 '11 at 12:47
It sounds good. –  Américo Tavares Sep 4 '11 at 12:49
Here is the link for a meta thread about adding a tag for the Lambert W function: meta.math.stackexchange.com/q/2908/622 –  Asaf Karagila Sep 4 '11 at 15:52

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.

-

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.

-
Why is x the exponent for y? I don't find that really intuitive, could you please explain? –  Doug Sep 4 '11 at 9:16
You forgot some braces there, I hope you did not mind. –  Asaf Karagila Sep 4 '11 at 9:16
@Asaf: thank you –  Henry Sep 4 '11 at 9:42
@Doug: it is a basic property of logarithms: see the power formula here –  Henry Sep 4 '11 at 9:44
@Asaf: To the meta-cave! –  The Chaz 2.0 Sep 4 '11 at 15:54