MHB Finding domain with e functions help

[bart11]

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?

 

[pickslides]

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

 

[bart11]

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

Thanks! But hmm, I'm still confused

 

[pickslides]

This is supposed to help you separate x & y so you can find a function in the form of y = f(x) and find the domain from there.

 

[HallsofIvy]

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?

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?

(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.

 

[The Chaz]

View attachment 86

Then taking (natural) logs on both sides, we find that

View attachment 87