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