- #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.
- Thread starter TSN79
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In summary, the conversation discusses the validity of the equation log x^(-y) = -y in both real and complex numbers. It is agreed that the equation makes sense in complex numbers, but in real numbers, the value of y must be negative for the equation to hold. It is also mentioned that the equation only holds for x=e when y is not equal to 0.
- #2
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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
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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
Homework Helper
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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
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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.
FAQ: Is log( x^(-y) ) = -y, somebody told me this.
Is log( x^(-y) ) equal to -y?
Yes, this statement is true. According to the logarithm rules, when the base (x) and the argument (-y) are both raised to the power of a logarithm, the result is the exponent (-y) itself. Therefore, log( x^(-y) ) is equal to -y.
Can you explain why log( x^(-y) ) = -y?
Logarithms are the inverse operations of exponents. This means that the logarithm of a number is the power to which the base must be raised to get that number. In this case, the base (x) must be raised to the power of -y to get the argument (x^(-y)). Therefore, the result of the logarithm is -y.
Is there a specific condition for log( x^(-y) ) = -y to be true?
Yes, for this statement to be true, the base (x) must be a positive number and the argument (-y) must be a non-zero real number. Logarithms with negative bases or zero arguments are undefined.
Are there any other equivalent forms of log( x^(-y) ) = -y?
Yes, there are two other equivalent forms: x^(-y) = 10^(-y) and x^(1/y) = 10. These forms can be derived from the logarithmic rule log(a^b) = b*log(a).
How can I use log( x^(-y) ) = -y in scientific calculations?
Logarithms are useful for simplifying complex equations and for converting between different units of measurement. In scientific calculations, the statement log( x^(-y) ) = -y can be used to solve for the unknown variable when given the other two variables. It can also be used to convert between logarithmic and exponential expressions.