What is the Domain of ln(x^2 + y^2)?

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Homework Help Overview

The discussion revolves around determining the domain of the function ln(x^2 + y^2). Participants are examining the conditions under which the logarithm is defined, particularly focusing on the implications of the expression being greater than zero.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the validity of the claim that x ≠ 0 and y ≠ 0 as the domain, suggesting instead that the condition should be expressed as (x^2 + y^2) > 0. There is a discussion about the implications of the point (0,0) and whether x and y can be zero simultaneously.

Discussion Status

Some participants appear to agree on the need to clarify the domain, with one suggesting that the domain excludes only the origin (0,0). Others are exploring the implications of the inequality and the definitions involved, indicating a productive exploration of the topic without reaching a consensus.

Contextual Notes

There is a focus on the interpretation of the domain in relation to the function's definition and the conditions under which the logarithm is valid. Participants are navigating through the nuances of the mathematical definitions involved.

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Homework Statement



for the domain of ln (x^2 + y^2 ) , it it given in my notes that the ans is x ≠ 0 and y ≠ 0

IMO , it's wrong to give x ≠ 0 and y ≠ 0 , because the meaning of x ≠ 0 and y ≠ 0 is that x and y can't be 0 all the times so just leave the ans (x^2 + y^2 ) > 0 , will do ?

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The Attempt at a Solution


when x = 0 , y not = 0 , z is defined , when y = 0 , z not = 0 , z is also undefined , So , x and y can be 0
 
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You are correct. The domain is the number plane excluding only the point (0,0) (the 'origin').
 
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The equation ##x^2 + y^2 = 0## can be thought of as a degenerate circle whose center is at (0, 0) and whose radius is 0. In other words, the point (0, 0). The graph of the solution set of the inequality ##x^2 + y^2 > 0## is all of the points in the plane outside that degenerate circle.
 
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Mark44 said:
The equation ##x^2 + y^2 = 0## can be thought of as a degenerate circle whose center is at (0, 0) and whose radius is 0. In other words, the point (0, 0). The graph of the solution set of the inequality ##x^2 + y^2 > 0## is all of the points in the plane outside that degenerate circle.
so the ans can't be x ≠0 , y≠0 ?
the correct ans is either (x^2) + (y^2) >0 or (x,y) ≠(0,0) which means x any y can't be 0 at the same time ?
 

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