How do you solve this log equation

3xe^-x + x^2e^-x = 0

First of all, this is an exponential equation, not a logarithmic one.

3xe^(-x) + x^2 e^(-x) = 0
e^(-x) (3x + x^2) = 0
e^(-x) ≠ 0, so x(3 + x) = 0
x = 0 or x = -3

3xe^(-x) + x^2*e^(-x) = 0

e^(-x)*(x)(3 + x) = 0

x = 0 or x = -3

Note that e^(-x) = 0 is not possible.

3xe^-x + x^2e^-x = 0
3xe-^x =-x²e^-x
3x/e^x=-x²/e^x
3x=-x²
x=0
and x=-3