Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer

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 $$ y^2 - 3y - 10 = 0. $$

share|improve this answer
    
Thank you so much! I can't believe I didn't see that. –  Gabby Nov 25 '12 at 4:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.