Sorry for the possibly stupid question. But I can't solve this equation. I'm need to find X from:

$$N = \frac{\log(\frac{Bd-X}{Ad-X})}{\log(D)} - \frac{\log(1-\frac{Bi}{P+X})}{\log(I)}$$

I don't know how to kill log in this equation.

This is a credit card loan analysis problem.

$A$ - the loan Amount (initial loan balance, $B_0$)
$N$ - the total Number of payments/periods/month for the entire loan (from $B_0$ till $B_N$)
$B_n$ - the Balance after n payments have been made
$i$ - the Interest rate per period, not per year, $0 < i < 1$
$p$ - the Payment rate, $0 < p < 1$, $i <= p$
$P$ - the min. Payment amount
Actual payment for loan (of this type) $Payment = MIN(A*p ; P)$
$X$ - additional eXtra payment
$B$ - loan balance at some "threshold point" $B_t = P/p$

When loan balance $B_n > B_t$ then payment $=B_n*p$ and number of periods: $$N_{B_n>B_t} = \frac{\log(\frac{Bd-X}{Ad-X})}{\log(D)}$$ When loan balance $B_n <= B_t$ then payment $=P$ and number of periods: $$N_{B_n<=B_t}=-\frac{\log(1-\frac{Bi}{P+X})}{\log(I)}$$ $$N = N_{B_n>B_t} + N_{B_n<=B_t}$$

Variable Replacements:
$I = 1 + i$, $1 < I < 2$
$d = i - p$, $-1 < d < 0$
$D = 1 + i - p$, $0 < D < 1$

$log$ base is $10$ but I can replace it with $e$ or anything else.
Please send me in the right direction.
Thanks in advance.

  • 2
    $\begingroup$ Can you share the source of the problem? Also, I'm assuming you are defining $\log x = \log_{10} x$. $\endgroup$ – Toby Mak Aug 23 at 7:11
  • 2
    $\begingroup$ This screams for numerical solution ... $\endgroup$ – Hagen von Eitzen Aug 23 at 7:27
  • 1
    $\begingroup$ How are the several symbols defined? For example, is $i$ the imaginary unit or not? Are we dealing with real numbers alone or not? What other things do you know about the quantities involved. This may be important. $\endgroup$ – Allawonder Aug 23 at 8:11
  • $\begingroup$ @Cesareo, how to do it? This is my whole problem. Please, explain :) $\endgroup$ – Dmitriy Stennikov Aug 23 at 8:31
  • $\begingroup$ Chuck it into C my good lad. $\endgroup$ – rodger_kicks Aug 23 at 8:53

By a combination of homography and antilogarithms, you can turn this equation to the form

$$t^a(rt+1)=pt+q$$ where the unknown is $t$.

The details are unimportant, this is just to show that except for a few very specific cases (such as $a=1,2,3$), there is no analytical solution. You need to use a numerical solver.


$$ N = \log\left(\left(\frac{Bd-X}{Ad-X}\right)^{\frac{1}{\log D}}/\left(1-\frac{Bi}{P+X}\right)^{\frac{1}{\log I}}\right) $$


$$ 10^{N\log D} = \left(\frac{Bd-X}{Ad-X}\right)/\left(1-\frac{Bi}{P+X}\right)^{\log_D I} $$


$$ 10^N D = \frac{\frac{Bd-X}{Ad-X}}{\left(1-\frac{Bi}{P+X}\right)^{\log_D I}} $$


$$ f(X)=\alpha\left(1-\frac{Bi}{P+X}\right)^{\beta} -\frac{Bd-X}{Ad-X} = 0 $$

This equation can be solved numerically by using an iterative process like Newton's in which

$$ X_{k+1} = X_k - \frac{F(X_k)}{F'(X_k)} $$

  • 1
    $\begingroup$ How does this help ? $\endgroup$ – Yves Daoust Aug 23 at 8:57
  • 1
    $\begingroup$ What is the benefit of transforming the equation before using Newton ? $\endgroup$ – Yves Daoust Aug 23 at 9:08

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.