f(x) = (x^3)(lnx)
f'(x) = (3x^2)(lnx) + (x^3)(1/x)
how do solve for x when
(3x^2)(lnx) + (x^3)(1/x) = 0
You have:
$\displaystyle f'(x)=3x^2\cdot \ln(x)+x^2=0$
It can be very useful to wright:
$\displaystyle x^2[3\ln(x)+1]=0$
EDIT:
Also remember: $\displaystyle \ln(0)$ is undefined!
Almost, $\displaystyle x=e^{\frac{-1}{3}}$ is a solution, but see my EDIT, $\displaystyle \ln(0)$ is undefined, so $\displaystyle x=0$ is not a solution.
What you always have to do if you solve logarithmic equations is first determine where the function is defined and undefined (the 'existence conditions'), so at the end you can compare with your 'existence conditions' if the solution(s) satisfies.
(Sorry, I had to say this before )