Is log( x^(-y) ) = -y, somebody told me this.

[TSN79]

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...

 

[dextercioby]

In the complex numbers,your equality (or equation,but i don't know which the unknown is) makes perfect sense...

Daniel.

 

[TSN79]

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?

 

[dextercioby]

Well,that "-y" has to be greater than 0,in order to make sense among the reals...So "y" should be negative.

Daniel.

 

[VietDao29]

Hi,
I think that y can be whatever you like (positive, negative, or 0). eg: You can have:
\lg{10^{-3}} = -3
So I think:
\log_{a}{x^{-y}} = -y only if a = x
Hope I am right,
Viet Dao,

 

[dextercioby]

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 y\neq 0,x=e,the Euler's number.

Daniel.