14^log14 of 21
yes
it is not hard to prove though. i will show you how to prove it. that way, if you are not allowed to use this rule, this is how you would attempt the problem in general.
Let $\displaystyle a^{\log_a x} = y$, we will show that $\displaystyle y = x$
$\displaystyle a^{\log_a x} = y$ ...........................................take $\displaystyle \log$ to the base $\displaystyle a$ of both sides
$\displaystyle \Rightarrow \log_a a^{\log_a x} = \log_a y$
$\displaystyle \Rightarrow \log_a x \cdot \log_a a = \log_a y$ .................since $\displaystyle \log_a (x^n) = n \log_a x$
$\displaystyle \Rightarrow \log_a x = \log_a y$ ...............................since $\displaystyle \log_a a = 1$
$\displaystyle \Rightarrow x = y$ ..............................................by taking the antilog of both sides.