Solve e^(2x) - e^x - 6 = 0 for x. Can't seem to set this up correctly!

May 2010
4
0
I need help setting up this properly... Solve e^(2x) - e^x - 6 = 0 for x.

I keep running into dead ends! Do I need to make some sort of substitution? Please help, I'd really appreciate it!
 
Last edited by a moderator:
Apr 2010
384
153
Canada
I need help setting up this properly... I keep running into dead ends! Do I need to make some sort of substitution? Please help, I'd really appreciate it!
\(\displaystyle e^{2x} - e^x - 6 = 0 \)

\(\displaystyle e^{x} e^x - e^x - 6 = 0 \)

This looks like a quadratic equation to me

Let \(\displaystyle e^x = u \)

\(\displaystyle u^2 - u - 6 = 0 \)

\(\displaystyle (u-3)(u+2) = 0 \)

It follows that,

\(\displaystyle e^x = 3 \) and \(\displaystyle e^x = -2 \)

But since the negative root produces imaginary values, we will leave it

Take,

\(\displaystyle e^x = 3 \)

\(\displaystyle x = ln3 \)