In the "Foundation Maths" book, chapter (20), "The Logarithm Function", I'm just having an issue of understanding how we got the result in the following (I have used "^" for power):

log (1/2)^-1/2 = log2^1/2

Thanks.

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- Jan 3rd 2011, 06:29 AMSWEngineerHow was this result found
In the "Foundation Maths" book, chapter (20), "The Logarithm Function", I'm just having an issue of understanding how we got the result in the following (I have used "^" for power):

log (1/2)^-1/2 = log2^1/2

Thanks. - Jan 3rd 2011, 06:38 AMe^(i*pi)
Standard rules of exponents - the log term is a red herring for this part

$\displaystyle a^{-b} = \dfrac{1}{a^b} = \left(\dfrac{1}{a}\right)^b$

In your case $\displaystyle a= \dfrac{1}{2}$ and $\displaystyle b = \dfrac{1}{2}$

$\displaystyle \left(\dfrac{1}{2}\right)^{-1/2} = \left(\dfrac{1}{\frac{1}{2}}\right)^{1/2} = 2^{1/2}$ - Jan 3rd 2011, 06:38 AMPlato
- Jan 3rd 2011, 06:56 AMSWEngineer
Thanks a lot for your replies.