- #1
TSN79
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Can one under any circumstances say that log x^(-y) = -y ?? I'm having some truble with this myself, but someone told me it is so... 
Is log( x^(-y) ) = -y, somebody told me this.
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Homework Help Overview
The discussion revolves around the properties of logarithms, specifically the expression log(x^(-y)) and its equivalence to -y. Participants are exploring the conditions under which this relationship holds true, particularly in the context of real and complex numbers.
Discussion Character
- Conceptual clarification, Assumption checking
Approaches and Questions Raised
- Participants are questioning the validity of the equation log(x^(-y)) = -y, with some noting that it may only hold under certain conditions, such as when y is negative or in the context of complex numbers. Others are discussing specific cases and examples to illustrate their points.
Discussion Status
The discussion is active, with various interpretations being explored. Some participants have offered insights into the conditions required for the equation to hold, while others are clarifying the relationship between logarithmic and exponential forms. There is no explicit consensus yet, but the dialogue is productive.
Contextual Notes
There are assumptions regarding the domain of x and y, particularly concerning the positivity of x and the sign of y. The discussion also touches on the implications of working within real versus complex number systems.
- #2
dextercioby
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In the complex numbers,your equality (or equation,but i don't know which the unknown is) makes perfect sense...
Daniel.
Daniel.
- #3
TSN79
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Well, I'm aware of the fact that log(x^(-y))=-y*log(x). But you're saying that what I wrote before makes sense only with complex numbers?
- #4
dextercioby
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Well,that "-y" has to be greater than 0,in order to make sense among the reals...So "y" should be negative.
Daniel.
Daniel.
- #5
VietDao29
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Hi,
I think that y can be whatever you like (positive, negative, or 0). eg: You can have:
[tex]\lg{10^{-3}} = -3[/tex]
So I think:
[itex]\log_{a}{x^{-y}} = -y[/itex] only if a = x
Hope I am right,
Viet Dao,
I think that y can be whatever you like (positive, negative, or 0). eg: You can have:
[tex]\lg{10^{-3}} = -3[/tex]
So I think:
[itex]\log_{a}{x^{-y}} = -y[/itex] only if a = x
Hope I am right,
Viet Dao,
- #6
dextercioby
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Yes,apparently my discussion skipped the logarithm part...
Disussed only the exponential.His initial equation is very valid within the reals for x>0 and has the solution,for [itex]y\neq 0[/itex],[itex]x=e[/itex],the Euler's number.
Daniel.
Disussed only the exponential.His initial equation is very valid within the reals for x>0 and has the solution,for [itex]y\neq 0[/itex],[itex]x=e[/itex],the Euler's number.Daniel.
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