Can anybody explain how these logarithms are transformed?

Question

I'm learning for my algorithms exam and I can't derive two logarithm transformations:

1. $ 3^{log_{4}(n)}=n^{log_{4}(3)} $
2. $ 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.

Answer 1

Use $\log _a b = \frac{{\ln b}}{{\ln a}}$ and $e^{\ln x} = x$: $$ 3^{\log _4 n} = (e^{\ln 3} )^{^{\ln n/\ln 4} } = e^{\ln 3\ln n/\ln 4} = (e^{\ln n} )^{\ln 3/\ln 4} = n^{\ln 3/\ln 4} = n^{\log _4 3}, $$ $$ \log _3 n = \frac{{\ln n}}{{\ln 3}} = \frac{1}{{\ln 3}}\ln n = \frac{{\ln e}}{{\ln 3}}\ln n = (\log _3 e)\ln n. $$

Answer 2

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.

Answer 3

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 $3^{log_4(n)} = (4^{log_4(3)})^{log_4(n)} = (4^{log_4(n)})^{log_4(3)} = n^{log_4(3)}$.

Answer 4

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

2. Use $\rm\ n = e^{ln\ n}\:,\ \ $ take $\rm\ log_{\:3}\ $ of it

See also the closely related prior questions here and here.