Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Why ln not log?

Answer 6

the same, just because ln is simpler than log

Answer 7

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.