Simple log equality

Question

I just can't remember how to justify the following:

The question would be how would we derive to the equality above starting from 3^log(4)n.

Thanks.

Answer 1

Hint:

$$a^{\log b}=b^{\log a}$$

Let $y=a^{\log b}$

$\log y = \log a^{\log b}$

$\log y = (\log b) ( \log a)$

$\log y = \log b^{\log a}$

$y=b^{\log a}$

Answer 2

Take log of both sides: $$\log_4(3^{\log_4 n}) = \log_4(n^{\log_4 3})$$ $$\log_4 n\cdot \log_4 3 = \log_4 3\cdot\log_4 n$$

Starting from $3^{\log_4 n}$ you can get:
$$3^{\log_4 n} = 4^{\log_4(3^{\log_4 n})} = 4^{\log_4 3\cdot\log_4 n} = 4^{\log_4(n^{\log_4 3})} = n^{\log_4 3}$$

Answer 3

I have always found it difficult to work with $log_a(n)$ functions, as well as $a^n$ ones, so I prefer coming back to definition everytime and work from there, even if there are shortcuts or tricks involving these.

definitions $\begin{cases} x^y=e^{y\ln(x)}\\ \log_x(y)=\frac{\ln(y)}{\ln(x)} \end{cases}$

$3^{\log_4(n)}=\exp(\frac{\ln(n)}{\ln(4)}\times\ln(3))$

$n^{\log_4(3)}=\exp(\frac{\ln(3)}{\ln(4)}\times\ln(n))$

And from there I see they are actually equal, it is only a re-arrangement of things.