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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?

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right. Instead, $e^x-3e^x = -2e^x$ so $-2e^x=10$. Now why can't you take natural log yet? – James S. Cook Nov 25 '12 at 3:34
Did you intend $3e^x$ or $(3e)^x$? – Austin Mohr Nov 25 '12 at 3:35
I just edited it. It was suppose to be e^(2x) – Gabby Nov 25 '12 at 3:56
up vote 3 down vote accepted

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

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Thank you so much! I can't believe I didn't see that. – Gabby Nov 25 '12 at 4:03

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