How to calculate the exponent of a given number. Here's my problem : I can't figure how to get back my exponent
Here the formula I use to get a given number :
$$a(\text{const}) = 200,\quad  b(\text{const}) = 1.1,\quad  c(\text{var}) = 50$$
So $c$ is the parameter I pass to my formula and that I want to get back.
$\left(a\left(b^c\right)\right)-a = \text{value}$
Using my values this formula give $23278.170575939063301333299198072$
I need to get the exponent I used to achieve this number.
So by logic I tried to do my formula backward.
$(\text{value}+a)/a$ But after this point I can't figure how to get back to my $c$ variable.
 A: 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 $b>0$.
A: The reason you're struggling with this is because there is no way of doing it with conventional arithmetical operations (addition, multiplication, exponentiation, and taking roots). Or if there is, I have no idea how it could be done.
What you need is an operation that is the inverse of
$$1.1^x$$
Inverting an exponentiation is a special operation called the logarithm. In this case, the base $1.1$ logarithm, written $\log_{1.1}$:
$$\log_{1.1}(1.1^x)=x$$
There is of course a logarithm associated with any particular base:
$$log_b(b^x)=x$$
If you need this for programming, then whatever language you're using almost certainly has a built-in logarithm function. However, it may only provide logarithms for a few select bases, like $2$ and $10$. In that case, you'll need the change of base formula.
A: The situation is that one knows the values of $b$ and $d$ where $d=b^c$ for some unknown $c$ and that one wants to compute the value of $c$. As explained by others there is no (exact) formula for that, which would avoid the function logarithm since 
$$
c=\frac{\log d}{\log b}=\log_bd.
$$
However, to compute approximate values of $c$, one can use the fact that, for every positive $x$,
$$
\lim_{\varepsilon\to0}\frac{x^\varepsilon-1}\varepsilon=\log x.
$$
Thus, for $\varepsilon$ small,
$$
c\approx\frac{d^\varepsilon-1}{b^\varepsilon-1}.
$$
This requires to be able to compute $x^\varepsilon$ accurately for small values of $\varepsilon$.
