How to prove $3^{\log_4n} = n^{\log_43}$? I got this from "4.4 The recursion-tree method for solving recurrences" in book "Introduction to Algorithms"
The recurrence that try to use recursion-tree to solve is:
$T(n) = 3T(n/4) + cn^2$

 A: $$\Large{3^{\log_4n}=e^{\ln3\frac{\ln n}{\ln4}}=e^{\ln n\frac{\ln 3}{\ln4}}=n^{\log_43}}$$
A: Let $\log_4n=y\implies4^y=n$
Now $n^{\log_43}=4^{y\log_43}=(4^{\log_43})^y=3^y$
as for if $\log_43=z,3=4^z=4^{\log_43}$
Alternatively,
if $3^{\log_4n}=y,\log_4y=\log_43\log_4n$
and if $n^{\log_43}=z,\log_4z=\log_4n\log_43$
$$\implies\log_4y=\log_4z\implies y=z$$
A: Take the log base 3 of both sides. You get $log_4(n) \stackrel{?}{=} log_4(3) * log_3(n)$
Then, note that $ log_3(n) *log_4(3) = log_4(3^{log_3(n)}) = log_4(n)$
A: Let, $\log_4{n}=p$, then $n=4^p$ from definition of $\log$. Now, $n^{\log_4{3}}=4^{p\log_4{3}}=(4^{log_4{3}})^p=3^p=3^{\log_4{n}}$.
Also you can do like this: 
Use , $3=4^{\log_4{3}}$, then $3^{\log_4{n}}=(4^{\log_4{3}})^{\log_4{n}}=(4^{\log_4{n}})^{\log_4{3}}=n^{\log_4{3}}$
A: You have $\log_4$ here, so write both numbers as powers of $4$, and we see immediately that
$$
(4^{\log_43})^{\log_4n}=(4^{\log_4n})^{\log_43}
$$
Alternatively (or rather equivalently), take $\log_4$ of both numbers, and the numbers are clearly seen as equal.
