• January 3rd 2011, 06:29 AM
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

Thanks.
• January 3rd 2011, 06:38 AM
e^(i*pi)
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 $b = \dfrac{1}{2}$

$\left(\dfrac{1}{2}\right)^{-1/2} = \left(\dfrac{1}{\frac{1}{2}}\right)^{1/2} = 2^{1/2}$
• January 3rd 2011, 06:38 AM
Plato
Quote:

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

$\log \left( {2^{\frac{{ - 1}}{2}} } \right) = \frac{1}{2}\left[ {\log (2) - \log (1)} \right] = \log \left( {2^{\frac{1}{2}} } \right)$
• January 3rd 2011, 06:56 AM
SWEngineer
Thanks a lot for your replies.