How does my professor go from this logarithm to the following one of a different base?

Question

I don't understand on the second last line how my professor goes from $2^{log_3(n)} = n^{log_3(2)}$ how is that relation formed?

Answer 1

Note that $\log_b a = \frac{\log b}{\log a}$, so

$$2^{\log_3 n} = 2^{\frac{\log n}{\log 3}} = \exp\left(\log 2 \cdot \frac{\log n}{\log 3}\right) = \exp \left(\log n \cdot\frac{\log 2}{\log 3}\right) = n^{\frac{\log 2}{\log 3}} = n^{\log_3 2}.$$

Answer 2

Hint: Take the base-three logarithm of them both. Apply known logarithm properties.

Answer 3

Change of basis property of logarithms:

$$\log_3n=\frac{\log_2n}{\log_23}\implies 2^{\log_3n}=\left(2^{\log_2n}\right)^{1/\log_23}=n^{1/\log_23}=n^{\log_32}$$