Find $f(x)$ for $f'(x) = f(x) \ln(f(x))$ and $f(0) = 1$ $$f'(x) = f(x) \ln\big(f(x)\big)$$
$$f(0) = 1, \qquad  f(x) > 0$$
I am studying for finals on my own and this exercise is really bothering me because I can't seem to solve it. If I divide by $f(x)$ I get $\ln'f(x) = \ln f(x)$ but then I don't know what to do. Any help is appreciated! 

With $f(x)>0$, let $g(x)=\ln(f(x))$, i.e. $f(x)=e^{g(x)}$.
The equation turns to 
$$e^{g(x)}g'(x)=e^{g(x)}g(x),$$
which can be simplified as
$$g'(x)-g(x)=0,$$ with $g(0)=0$.
Multiplying by $e^{-x}$,
$$g'(x)e^{-x}-g(x)e^{-x}=\left(g(x)e^{-x}\right)'=0,$$
and 
$$g(x)e^{-x}=C^{st}=g(0)e^{-0}=0,$$ hence
$$g(x)=0,f(x)=1.$$
