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.

  • 1
    $\begingroup$ Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. $\endgroup$ – Zev Chonoles Jul 13 '13 at 16:58
  • $\begingroup$ Why was this put on hold? The question is perfectly clear: the OP wants to know how to get back $x$ from $b^x$. $\endgroup$ – Jack M Jul 13 '13 at 18:50

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 $b>0$.

  • $\begingroup$ If the OP knew there was such a thing as $\log_b$, they wouldn't have asked this question. $\endgroup$ – Jack M Jul 13 '13 at 18:51

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


Inverting an exponentiation is a special operation called the logarithm. In this case, the base $1.1$ logarithm, written $\log_{1.1}$:


There is of course a logarithm associated with any particular base:


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.


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 $\varepsilon$.

  • $\begingroup$ The rumor has it that this trick is at the basis of Giorgio Parisi's solution of Sherrington-Kirpatrick model using the replica method ("take a positive integer $n$, write down the solution for $n$, make $n\to0$, the limiting object is what you are after"...) and of other accomplishments in theoretical physics. $\endgroup$ – Did Jul 14 '13 at 8:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.