Find the zeros of $f(x) = x^2 - 150 x\log(x)$ I have to find the zeros of $f(x) = x^2 - 150 x\log(x)$
It is is pretty straightforward to find $x = 0$ because $x(x-150 \log(x)) = 0$, but how can I find the zeros of $x - 150 \log(x)$?
 A: As Joshua said, this is not solvable by elementary algebra.  But it can be solved using the Lambert W function.  Recall the definition: $$xe^x = y \;\Longleftrightarrow\; x = W(y).$$
Start with $x^2 - 150 x \log x = 0$.  Factor, $x(x-150 \log x) = 0$.  So either $x=0$ or $x-150 \log x = 0$.  For the first case, $x=0$, we should note that $\log 0$ is not defined, but $\lim_{x \to 0^+}(x^2 - 150 x \log x) = 0$.  So $x=0$ is a solution in that sense.
Here is the second case.
$$
x = 150\log x\\
e^x = x^{150}\\
e^{x/150} = x\\
1=xe^{-x/150}\\
\frac{-1}{150}=\frac{-x}{150}e^{-x/150}\\
W\left(\frac{-1}{150}\right) = \frac{-x}{150}\\
-150W\left(\frac{-1}{150}\right)= x
$$
Actually, there are two real branches of the $W$ function, so we get two real solutions
$$
-150W_0\left(\frac{-1}{150}\right) \approx 1.006734134,\qquad
-150W_{-1}\left(\frac{-1}{150}\right) \approx 1042.390835.
$$
Here are two graphs of $x^2 - 150 x\log(x)$.
 
Here is the classic paper on Lambert W. Perhaps few papers can be labeled "classic" after only 25 years, but this is one!
Corless, R. M.; Gonnet, G. H.; Hare, D. E. G.; Jeffrey, D. J.; Knuth, D. E., On the Lambert (w) function, Adv. Comput. Math. 5, No. 4, 329-359 (1996). ZBL0863.65008.
