Prove Identity involving logarithms and exponents

Question

Can you give me some hints on how to solve the following identity? $a^{\ln(n)} = n^{\ln(a)}$

Answer 1

$\ln a \ln n=\ln n \ln a \implies \ln n^{\ln a}=\ln a^{\ln n}$ (since $\ln a^b=b\ln a$)

Now by the injectivity of $\ln$ (i.e. because it is strictly increasing) so $n^{\ln a}=a^{\ln n}$

Answer 2

HINT

Use that

$$\log n=\frac{\log_a n}{\log_a e}$$

therefore

$$a^{\log n}=a^{\frac{\log_a n}{\log_a e}}=(a^{\log_a n})^{\frac1{\log_a e}}=\dots$$

Answer 3

If $a,b,c >0$, then the general result is $$a^{\log_{~b} c}= c^{\log_{~b} a}.~~~(1)$$ You may prove (1) by taking $\log$ on both w.r.t. and unknown base we get. $$ \log_{b} c ~\log a = \log_{~b} a ~ \log c$$ this is nothing but $$\frac{\log c}{\log b} \log a= \frac{\log a}{\log b} \log c,$$ Hence (1) is proved.

Note: it is known that $$\log_{~q}~ p= \frac{\log_{~r} p}{\log_{~r}~ q}$$