Mortgage payment calculation without annuty.

Question

I have been asked the following problem by a student of mine and there is a specific method that he requested.

A mortgage of $\$450,000$ is loaned for a monthly payment for $30$ years with nominal annual interest rate of $6\%$. This loaned is planned to be paid off by a monthly payment of $\$2697.98$. If one decides to pay $\$3500$ each month, how many years earlier will he/she finish paying the debt?

Fortunately I have been studying financial math these days, so I understand that

$$450,000a_{\overline{360}\rceil.005}=2697.98$$

However, in order to solve the number of years that it will take to pay off the loan with monthly payment of $\$3500$, I am thinking that the number of months $n$ needed can be calculated by

$$450,000(1.005)^{n}=3500s_{\overline{n}\rceil.005}$$

which implies

$$n \approx 206.44$$

or about 17 years (therefore the answer is about 13 years earlier), which is a fairly complex calculation for a student who just started precalc.

The student knows the straight forward compound interest, and he claims that he has never seen or heard the word annuity. Is there a way to solve this problem without using the concept of annuity but compound interest?

Answer 1

Your equation is

$450000\cdot 1.005^n=3500\cdot \frac{1.005^n-1}{1.005-1}$. This is the right equation. There is no other way as to solve the equation for n. I Show how I did it. I use parameters. It make it easier to stay on top of things.

\begin{align} & S_0\cdot q^n=r\cdot \frac{q^n-1}{i} \quad \text{| muliplying by i} \\ & S_0\cdot q^n \cdot i=r\cdot (q^n-1) \quad \text{| muliplying out the brackets} \\ & S_0\cdot q^n \cdot i=r\cdot q^n-r \quad \text{| r to the LHS} \\ & r+S_0\cdot q^n \cdot i=r\cdot q^n \quad \text{| $S_0\cdot q^n \cdot i$ to the RHS}\\ & r=r\cdot q^n -S_0\cdot q^n \cdot i\quad \text{| factoring out $q^n$}\\ &r=q^n(r -S_0\cdot i)\quad \text{} \\ & \frac{r}{r -S_0\cdot i}=q^n\quad \text{| logarithmizing both sides} \\ & \ln \left( \frac{r}{r -S_0\cdot i} \right) =\ln \left( q^n \right) \quad \text{} \\ &\ln \left( \frac{r}{r -S_0\cdot i} \right) =n\cdot \ln \left( q \right) \quad \text{} \end{align}

$$\huge{\text{ $\frac{\ln \left( \frac{r}{r -S_0\cdot i} \right)}{ \ln \left( q \right)}=n$}}$$

Now you can insert the values for the parameters and calculate $n$. You can memorize this formula. But i think it is a better way, to do this kind of calculations several times. After some calculations you won´t have difficulties to do such a calculation in about 3 minutes or less.