Here's a better quality link to the picture:
[link removed]
And this is what I'm referring to when I say EMI
[link removed]
Sorry for any confusions
No apology needed for my poor eyesight.
The purpose of these problems is to teach you how to manipulate a formula.
\(\displaystyle a = p * \dfrac{r(1 + r)^n}{(1 + r)^n - 1}, \ r > 0 < p, \ n \in \mathbb Z^+, \text { and } n \ge 1.\)
Technical point 1.
\(\displaystyle a = p * \dfrac{r(1 + r)^n}{(1 + r)^n - 1} \implies a = pr * \dfrac{(1 + r)^n}{(1 + r)^n - 1}.\)
\(\displaystyle r > 0 \text { and } n \ge 1 \implies (1 + r) > 1 \implies (1 + r)^n > 1 \implies\)
\(\displaystyle (1 + r)^n - 1 > 0 \text { and obviously } (1 + r)^n > (1 + r)^n - 1 \implies \dfrac{(1 + r)^n}{(1 + r)^n - 1} > 1 \implies\)
\(\displaystyle pr * \dfrac{(1 + r)^n}{(1 + r)^n - 1} > pr \implies a > pr.\)
This is pure common sense: the amount to pay off a loan must exceed the interest, but it is important mathematically.
That formula has four variables. It is set up to give you a if p, r, and n are known. But what if you know a different set of three variables and need to find the fourth? You re-arrange the formula mathematically.
Suppose you know a, r, and n but need to know p. You isolate p on one side of the equation.
\(\displaystyle a = p * \dfrac{r(1 + r)^n}{(1 + r)^n - 1} \implies a * \{(1 + r)^n - 1\} = p * r(1 + r)^n \implies p = a * \dfrac{(1 + r)^n - 1}{r(1 + r)^n}.\)
Basic algebra.
Suppose you know a, r, and p but need to know n. You isolate n on one side of the equation.
\(\displaystyle a = p * \dfrac{r(1 + r)^n}{(1 + r)^n - 1} \implies a * \{(1 + r)^n - 1\} = pr(1 + r)^n \implies\)
\(\displaystyle a(1 + r)^n - a = pr(1 + r)^n \implies a = a(1 + r)^n - pr(1 + r)^n \implies\)
\(\displaystyle a = (a - pr)(1 + r)^n \implies (1 + r)^n = \dfrac{a}{a - pr}.\)
Now what? You have n on one side but as an exponent. Remember logarithms: they let you deal with exponents.
\(\displaystyle (1 + r)^n = \dfrac{a}{a - pr} \implies log\{(1 + r)^n\} = log \left ( \dfrac{a}{a - pr} \right ) \implies \)
\(\displaystyle n * log(1 + r) = log(a) - log(a - pr) \implies n = \dfrac{log(a) - log(pr)}{log(1+ r)}.\)
What if you do not know r? Sadly, that gets into more advanced math than you know yet.
