Thread: Help with natural logarithms

Can someone please tell me if I am doing this problem properly?

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

ln2=ln3e^x-lne^2x

ln2=(ln3e^x)/(lne^2x)

ln2=ln3e^-x

ln2=-xln3e (lne cancels)

ln2=-3x and then solving for x to get -.23

Your second line is wrong.
This is how I would approach the problem:
Let A = e^x. Then you have
A^2 - 3A + 2 = 0
Factor to solve for A, then convert this to a solution involving x.

Let,