The easiest way to solve equations like this is to use the substitution z = ex . With this, the equation becomes
z2 - 4 e3 z = 0 . The solutions for z are
z = 0 and z = 4e3 . The z =0 solution is erroneous because there is no value of x which makes ex = 0
The other solution for x is ln(4 e3) = 3 + ln(4).
If the problem had been e2x - 4 ex + 3 = 0 , method of solution would have been similar. After the substitution
z2 - 4 z + 3 = 0. The solutions are z = 1 and z = 3 giving the two values of x as
x = ln(1) = 0 and x = ln(3)