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!
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!
$\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 $