How was this result found

**SWEngineer** · January 3rd 2011, 06:29 AM

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.

**e^(i*pi)** · January 3rd 2011, 06:38 AM

Standard rules of exponents - the log term is a red herring for this part

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

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

$\left(\dfrac{1}{2}\right)^{-1/2} = \left(\dfrac{1}{\frac{1}{2}}\right)^{1/2} = 2^{1/2}$

**Plato** · January 3rd 2011, 06:38 AM

Originally Posted by SWEngineer

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

$\log \left( {2^{\frac{{ - 1}}{2}} } \right) = \frac{1}{2}\left[ {\log (2) - \log (1)} \right] = \log \left( {2^{\frac{1}{2}} } \right)$

**SWEngineer** · January 3rd 2011, 06:56 AM

Thanks a lot for your replies.

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