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