# Thread: Finding x in a simple algebra problem

1. ## Finding x in a simple algebra problem

OK, I'm old and this is not so simple for me, and this is not for a test. It's for my own use and knowledge. And I'll greatly appreciate any help.

I'm not sure how to set this up properly in a formula, so I'll do this in a story problem. Here goes:

Joe has a loan of $4,000. His interest rate is 15% (1.25% monthly) and he will pay$100 each month until the loan is paid off. How many months (x) will it take for Joe to pay off this loan?

How would this look set up in the proper formula to find x, and how many months?

Thanks again for any help.

2. This is called "amortization". The formula is too complicated for me to type in LaTex (because I'm bad at LaTex...), but a web search should give you what you're looking for.

At any age, this can be a baffling problem.

Joe has a loan of $4,000. His interest rate is 15% (1.25% monthly) and he will pay$100 each month until the loan is paid off.
How many months $\,n$ will it take for Joe to pay off this loan?

This is an Amortization, with this formula: . $A \;=\;P\,\dfrac{i(1+i)^n}{(1+i)^n - 1}$

. . where: . $\begin{Bmatrix} A &=& \text{periodic payment} \\ P &=& \text{principal borrowed} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}$

First, we'll solve for $\,n.$

We have: . $A[(1+i)^n-1] \;=\;Pi(1+i)^n$

. . . . . . . . $A(1+i)^n - A \;=\;Pi(1+i)^n$

. . $A(1+i)^n - Pi(1+i)^n \;=\;A$

. . . . . . . $(1+i)^n(A - Pi) \;=\;A$

. . . . . . . . . . . . . $(1+i)^n \;=\;\dfrac{A}{A-Pi}$

Take logs:. . . . $\ln(1+i)^n \;=\;\ln\left(\frac{A}{A-Pi}\right)$

. . . . . . . . . . $n\!\cdot\!\ln(1+i) \;=\;\ln\left(\frac{A}{A-Pi}\right)$

Therefore:. . . . . . . . . $n \;=\;\dfrac{\ln\left(\frac{A}{A-Pi}\right)}{\ln(1+i)}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We are given: . $\begin{Bmatrix}A &=& 100 \\ P &=& 4000 \\ i &=& 0.0125 \end{Bmatrix}$

. . Hence: . $A - Pi \:=\:100 - (4000)(0.0125) \:=\:100-50 \:=\:50$

$\text{Substitute: }\:n \;=\;\dfrac{\ln(\frac{100}{50})}{\ln(1.0125)} \;=\;55.79\hdots$

It will take 56 months for Joe to pay off his loan.

4. Soroban's solution is somewhat complex for ordinary folks.In this problem the capital recovery factor is 100/4000 = .025. If you go to an interest table for 1.25 rate per month you find that this factor occurs at 55 months. Interest tables are available on line.They do require study on how used.

bjh

5. Wow Soroban, that looked like a lot of work. Thank you so much!! And yes BJH, Soroban's answer is complex for me, but if I put the numbers in, it makes sense to me. I also think I grasped your answer with the tables...I checked it out with a search and there is lots of info for me to peruse.Thank you too. Thanks also to TheChaz. I did find a loan amortization site, but they always want the "years of the loan," which is what I'm trying to solve, thanks.

Great Forum!! Thanks everyone.