How to get the logarithm of both sides of this equation?

Question

anonymous · Answer

$ a_{n}=(\frac{a_{n-2}}{a_{n−1}})^{\frac{1}{2}} $

loser66 · Answer

why do you have to do that? I mean why do you have to take log?

anonymous · Answer

Is it like $\log_{\frac{a_{n-2}}{a_n-1}}x = \frac{1}{2}$ where x = a_n?

loser66 · Answer

If you HAVE to, then
$$ln (a_n)= \dfrac{1}{2} ln (a_{n-2}) -\dfrac{1}{2} ln(a_{n-1})$$

anonymous · Answer

Why ln not log?

loser66 · Answer

the same, just because ln is simpler than log

solomonzelman · Answer

ln is a log base e, log, is a log base 10, (when a base in log is unspecified).

doesn't make much difference, but when it comes to calculus, ln is much simpler, as looser said...

just do ln... once again, it doesn't matter.