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

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In summary, the conversation discusses the domain for the function ln(x^2 + y^2) and whether the answer should be x ≠ 0 and y ≠ 0 or (x^2 + y^2) > 0. It is concluded that the correct answer is either (x^2 + y^2) > 0 or (x,y) ≠ (0,0), meaning that x and y cannot both be 0 at the same time.
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chetzread
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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 ?
 

1. What is the domain of z = ln (x^2 + y^2)?

The domain of z = ln (x^2 + y^2) is all real numbers except for x = 0 and y = 0. This is because the natural logarithm function is undefined for zero and negative numbers.

2. Can negative numbers be used in the domain of z = ln (x^2 + y^2)?

No, negative numbers cannot be used in the domain of z = ln (x^2 + y^2). The expression x^2 + y^2 must be positive for the natural logarithm to be defined.

3. How is the domain of z = ln (x^2 + y^2) affected by the value of x and y?

The domain of z = ln (x^2 + y^2) is affected by the value of x and y in the sense that the expression x^2 + y^2 must be positive for the natural logarithm to be defined. This means that the values of x and y must result in a positive number when squared.

4. Can the domain of z = ln (x^2 + y^2) include complex numbers?

No, the domain of z = ln (x^2 + y^2) cannot include complex numbers. The natural logarithm function is only defined for real numbers.

5. What happens to the domain of z = ln (x^2 + y^2) when x or y approaches infinity?

The domain of z = ln (x^2 + y^2) does not change when x or y approaches infinity. The natural logarithm function is still defined as long as the expression x^2 + y^2 remains positive.

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