Thread: natural log question

e^x - 6e^(-x) = 1
solve for x

I think you can make 1=lne or e^0 and that helps but i'm still having trouble at the next step.

If you multiply through by e^x you should end up with an equation you can solve

Multiply both sides by e^x to get:

$\displaystyle e^{2x} - 6 = e^x$

Rearrange it into a quadratic:

$\displaystyle e^{2x} - e^x - 6 = 0$

Notice that this factors into:

$\displaystyle (e^x-3)(e^x+2)=0$

Can you continue?

got it, thanks alot guys!