Isolating Exponent Variables for Deriving Equations Our Math class right now is all about Geometric Sequences
and we all know the equation for them which is
$$A_n = A_1 \cdot r^{n-1}$$
We were only given this particular equation so to make things faster for myself
, I derived some equations for specific variables
$$A_1 = \frac{A_n}{r^{n-1}} $$ $$ r = \sqrt[n-1]{\frac {A_n}{A_1}} $$
However, after a couple of days of thinking I cannot seem to get a proper derived equation for the variable n in that particular equation. In the end I resorted to this method where I manually think up of an exponent and base... for Example:
$A_n = 2916$
$A_1 = 4$
$r = 3$
$2916 = 4 \cdot3^{n-1} $
isolating $r^{n-1}$ so we divide 4 to both sides cancelling 4 and 2916 will become 729
$729 = 3^{n-1} $
using 3 as a base, 729 can be shortened to $3^6$
$3^6 = 3^{n-1} $
using simple logic we know that n = 7

But enough context, on the formula above, I want to know how exactly do I isolate the n variable like how I can divide both sides to 'transpose' a variable being multiplied to another term
 A: After researching a bit I found out the existence of logarithms which are apparently the inverse of exponents, the same as how divide is the inverse of multiply.
After studying logarithm laws for a good hour I found out another way to finish the same question:
$2916 = 4 \cdot3^{n-1}$
divide 2916 by 4 to isolate $3^{n-1}$
$729 = 3^{n-1}$
Apply log() to both sides
$log(729) = log(3^{n-1})$
one law of log states that $log(x^y)$ becomes $y \cdot log(x)$ so..
$log(729) = (n-1)\cdot log(3)$
then divide log(3) on both sides, that means $\frac{log(729)}{log(3)}$ which is equal to 6
$ 6 = n -1 $ or simply $n = 7$

Correct me if im wrong but this is my attempt at deriving the original equation to a one-line equation
$A_n = A_1 \cdot r^{n-1}$
divide $A_1$ on both sides
$\frac{A_n}{A_1} = r^{n-1}$
Apply log() to both sides
$log(\frac{A_n}{A_1}) = log(r^{n-1})$
Apply that law of log about exponents
$log(\frac{A_n}{A_1}) = (n-1)\cdot log(r)$
Divide both sides by $log(r)$
$\frac{log(\frac{A_n}{A_1})}{log(r)} = n-1$
Add one to both sides
$$n = \frac{log(\frac{A_n}{A_1})}{log(r)}+1$$
Hopefully I answered my own question, if I did, well...
might as well keep this here for future reference
