why is $2^{\log^{2}\frac{1}{2}} = 1/2$? I have no understanding of $\log^2n$, wouldn't $2^{\log^{2}\frac{1}{2}} = 1/4$ since $2^{\log\frac{1}{2}} = 1/2$?
$\log$ here has the base of $2$.
 A: $\textbf{Remark}$: Note that we have that $(a^b)^2 = a^{2b}$ and $a^{(b^2)} = a^{b^2}$, but in general we have that 
$$a^{b^2} \neq a^{2b}$$ (unless $b^2 = 2b$, which is only true for $b = 2$ and $b = 0$).
(Consider for example $(2^3)^2 = 2^(3 \cdot 2) = 2^6 = 64$, whereas $2^{(3^2)} = 2^9 = 512$). 
In your case, we have $a^{b^2}$ and with this in mind, we find
$$\log(\frac{1}{2}) = \log(1) - \log(2) = 0 -1 = -1,$$ 
since we work in base $2$ and therefore 
$$2 ^{\log^2(\frac{1}{2})} = 2^{(-1)^2} = 2^1 = 2.$$
A: Using a combination of exponent rules and the one log rule you mentioned,
$$2^{\log^2\frac{1}{2}} = 2^{\log\frac{1}{2} \cdot \log\frac{1}{2}} = (2^{\log\frac{1}{2}})^ {\log\frac{1}{2}} = \left(\frac{1}{2}\right)^{\log\frac{1}{2}}  = \frac{1}{2^{\log\frac{1}{2}}} = \frac{1}{\frac{1}{2}} = 2$$
A: You have if you use the calculus rules for logarithms:
$$\log_2(1/2)=-\log_2(2)$$
so
$$\log_2(1/2)^2=(-\log_2(2))^2=(-1)^2=1.$$
And since $\log_a(a)=1$, you get
$$\log_2(2)^2=1^2=1.$$
So
$$2^{\log_2(1/2)^2}=2^{1}=2.$$
