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

Printable View

- September 18th 2011, 03:59 PMjmanna98Help 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 - September 18th 2011, 04:05 PMTheChazRe: Help with natural logarithms
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. - September 19th 2011, 07:13 AMsbhatnagarRe: Help with natural logarithms
http://latex.codecogs.com/gif.latex?e^{2x}-3e^{x}+2=0

Let, http://latex.codecogs.com/gif.latex?v=e^{x}

http://latex.codecogs.com/gif.latex?v^{2}-3v+2=0

http://latex.codecogs.com/gif.latex?...rt{9-4(2)}}{2}

http://latex.codecogs.com/gif.latex?v= 2,1

http://latex.codecogs.com/gif.latex?e^{x}=2,1

http://latex.codecogs.com/gif.latex?x=\ln{2},0