We have a function such that $\displaystyle f(t) = 10 \cdot 2^{kt}$ and $\displaystyle f(\tfrac{1}{2}) = 3$. Find $\displaystyle k$.

$\displaystyle f(\tfrac{1}{2}) = 3$, therefore:

$\displaystyle 3 = 10 \cdot 2^{\tfrac{k}{2}}$

Taking the log to the base 2 on both sides:

$\displaystyle log_2 3 = log_2 10 + \frac{k}{2}$

Now, I know that $\displaystyle log_a xy = log_a x + log_a y$, but this must mean that we had:

$\displaystyle log_2 3 = log_2 10 + log_2 2^{\tfrac{k}{2}}$

Can someone show me why (proof) that $\displaystyle log_2 2^{\tfrac{k}{2}} = \frac{k}{2}$?

And is it just me, or do those log renders look a bit squashed? The logs and the fractional exponent. Is there a better way to render these?