1
$\begingroup$

The actual problem is:

$e^{2x} - 3e^x = 10$

I want to just natural log both sides, but I don't know if that's the right approach. I don't think that I can distribute an $\ln$, right?

  • 0
    I just edited it. It was suppose to be e^(2x)2012-11-25

1 Answers 1

3

Since $e^{2x} = (e^x)^2$, you can write this as a quadratic. Let $y = e^x$. Then it reduces to solving $ y^2 - 3y - 10 = 0. $

  • 0
    Thank you so much! I can't believe I didn't see that.2012-11-25