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- Thread starter bart11
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- Feb 1, 2012

- 57

e^(x+y) = e^y*e^x

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Thanks! But hmm, I'm still confusede^(x+y) = e^y*e^x

- Feb 1, 2012

- 57

- Jan 29, 2012

- 1,151

The only possible "difficulty" with that is that you cannot divide by 0. That, in turn, means that e^{x+y} cannot be equal to 3. So, what can x+ y not be equal to?I'm asked to find the domain of 1/[e^(x+y)-3]

Which means I'm solving for e^(x+y)>=3

But how would I go about solving this?

(To solve e^x= a, take the natural logarithm of both sides.)

Pickslide's observation that e^{x+y}= e^xe^y is true but I don't believe using that is particularly useful here.

Last edited:

- Jan 26, 2012

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