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.

2. 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 $b = \dfrac{1}{2}$

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

3. 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
$\log \left( {2^{\frac{{ - 1}}{2}} } \right) = \frac{1}{2}\left[ {\log (2) - \log (1)} \right] = \log \left( {2^{\frac{1}{2}} } \right)$

4. Thanks a lot for your replies.