Finding domain with e functions help

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Discussion Overview

The discussion revolves around finding the domain of the function 1/[e^(x+y)-3]. Participants explore the conditions under which the expression is defined, particularly focusing on the inequality e^(x+y) >= 3.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant states the need to solve e^(x+y) >= 3 to find the domain.
  • Another participant expresses that e^(x+y) can be rewritten as e^y * e^x, suggesting a separation of variables.
  • A later post indicates that separating x and y could help in expressing y as a function of x, which may aid in determining the domain.
  • One participant mentions that e^(x+y) cannot equal 3, implying that x + y must not equal a certain value to avoid division by zero.
  • There is a suggestion to take the natural logarithm of both sides to solve the equation e^x = a, but it is unclear how this directly applies to the original problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to find the domain, and multiple perspectives on how to manipulate the equation are presented.

Contextual Notes

There are limitations regarding the assumptions made about the variables x and y, and the discussion does not resolve how to definitively express the domain based on the given conditions.

bart11
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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?
 
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e^(x+y) = e^y*e^x
 
pickslides said:
e^(x+y) = e^y*e^x

Thanks! But hmm, I'm still confused
 
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.
 
bart11 said:
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.
 
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