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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.

BlackAdder
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alyssaw20
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3 Answers3

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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 $b>0$.

nullUser
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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.

Jack M
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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 $\varepsilon$.

Did
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  • 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. – Did Jul 14 '13 at 08:59