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?
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?
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. $