in Analysis you define the exponential function mostly over the exponential series:
$$\exp(z) := \sum_{k=0}^{\infty}{\frac{z^k}{k!}}$$ you can proove that this is equivalent to $$\exp(x)=\lim_{k\rightarrow\infty}{\left(1+\frac{x}{k}\right)^k}$$
(and yes this is if you set x = 1 you get one definition for e)
$exp(z)*exp(w) = exp(z+w)$ can then be 'simply' proved with the Cauchy Product for series (because they are absolute convergent):
therefor we define: $a_k := \frac{z^k}{k!}$ and $b_k:=\frac{w^k}{k!}$
Now it's just a lot of calculation:
$$\exp(z)*\exp(w)=\left(\sum_{k=0}^{\infty}{\frac{z}{k!}}\right)*\left(\sum_{k=0}^{\infty}{\frac{w}{k!}}\right) = \left(\sum_{k=0}^{\infty}{a_k}\right)* \left(\sum_{k=0}^{\infty}{b_k}\right) \overset{Cauchy}{=} \sum_{k=0}^{\infty}{\sum_{j=0}^{k}{a_{k-j}b_j}} = \sum_{k=0}^{\infty}{\sum_{j=0}^{k}{\frac{z^{k-j}}{(k-j)!}*\frac{w^j}{j!}}} = \sum_{k=0}^{\infty}{\frac{1}{k!}\sum_{j=0}^{k}{\begin{pmatrix}k\\j\end{pmatrix}z^{k-j}w^j}} \underset{formula}{\overset{binomial}{=}}\sum_{k=0}^{\infty}{\frac{(z+w)}{k!}} = \exp(z+w)$$
.... there you can see, why you haven't found a proof yet.
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
Best Answer
I'll write $v$ for value. We have $$ v = a(b^c)-a $$ as you wrote in your question. Then $$ v + a = a(b^c) $$ $$ \iff \frac{v+a}{a} = b^c $$ $$ \iff \log_b\left( \frac{v+a}{a} \right) = c $$ assuming, of course that $b>0$.