Can anybody explain how these logarithms are transformed? I'm learning for my algorithms exam and I can't derive two logarithm transformations:


*

*$ 3^{log_{4}(n)}=n^{log_{4}(3)} $

*$ log_{3}(n)=log_{3}(e)*ln(n) $


I'm not very strong in logarithms, anybody can explain how I get in both cases from the left side of the equation to the right side?
Sorry if the terminology is wrong, I'm from Germany.
 A: Use $\log _a b = \frac{{\ln b}}{{\ln a}}$ and $e^{\ln x}  = x$:
$$
3^{\log _4 n}  = (e^{\ln 3} )^{^{\ln n/\ln 4} }  = e^{\ln 3\ln n/\ln 4}  = (e^{\ln n} )^{\ln 3/\ln 4}  = n^{\ln 3/\ln 4}  = n^{\log _4 3}, 
$$
$$
\log _3 n = \frac{{\ln n}}{{\ln 3}} = \frac{1}{{\ln 3}}\ln n = \frac{{\ln e}}{{\ln 3}}\ln n = (\log _3 e)\ln n.
$$
A: For the first one: Remember that for any number $a$, $4^{\log_4(a)}=a$. Also, $(a^b)^c=a^{bc}$. 
This gives us: $3^{\log_4(n)}=(4^{\log_4(3)})^{\log_4(n)}=4^{\log_4(3)\times\log_4(n)}$. 
Similarly, $n^{\log_4(3)}=(4^{\log_4(n)})^{\log_4(3)}=4^{\log_4(n)\times\log_4(3)}$.
Finally, since $ab=ba$ for any numbers $a,b$, this gives us the result.
For the second, it is enough to check that $3^{\log_3(e)\times\ln(n)}=n$, because (by definition) the only number $a$ such that $3^a=n$ is $a=\log_3(n)$.
Now: $3^{\log_3(e)\times\ln(n)}=(3^{\log_3(e)})^{\ln(n)}=e^{\ln(n)}=n$, remembering that $\ln$ is just an abbreviation for $\log_e$.
Usually this is presented in a more condensed way, as a change of basis formula, saying that $$\log_a(b)/\log_a(c)=\log_c(b)$$ for any positive numbers  $a,b,c$. This formula can be proved in exactly the same way as we checked the second of the transformations you asked about. 
A: Let me also give some strategy points with these problems that I think will help.
One rule that should help is the equality $log_b(c^p) = p \cdot \log_b(c)$.
Because the second question is asking for an equality with multiplication, you can try substituting $b = 3$, $c = e$, and $p = ln(n)$ to get:
$log_3(e^{ln(n)}) = ln(n) \cdot \log_3(e)$ and $e^{ln(n)}$ simplifies to $n$ so you're good.
In the first problem, you want to change that $3$ to a $4$ because you know that $4^{log_4(n)} = n$.  Thus you want $4^m = 3$ so $m = log_4(3)$.  Expressing $3^{log_4(n)} = (4^{log_4(3)})^{log_4(n)} = (4^{log_4(n)})^{log_4(3)} = n^{log_4(3)}$. 
A: *

*Use $\rm\ x = 4^{\:\ell\ x}\ \ $ in $\rm\ \ (4^{\:\ell\ 3})^{\:\ell\ n}\ =\ (4^{\:\ell\ n})^{\:\ell\ 3}\quad$ where $\rm\quad \ell\ x\ :=\ log_{\:4}\ x$

*Use $\rm\ n = e^{ln\ n}\:,\ \ $ take $\rm\ log_{\:3}\ $ of it  
See also the closely related prior questions here and here.
