Logarithmic and Exponential Equations

Lessons

Senior 5 · Mathematics

Logarithmic and Exponential Equations

4 lessons

Lesson 3 of 4

Compound interest and mortgages

25 min

Money that earns money

A bank pays interest on savings and charges interest on loans, and in both cases the interest is compounded: added to the balance, so that it earns (or costs) interest itself. In S3 you met $A = P (1 + r)^{n}$ for yearly compounding. This lesson adds the two tools that real finance needs: logarithms, to find the time or the rate, and the geometric sum from Unit 2, to handle regular payments such as savings plans and mortgages. All rates and amounts in this lesson are illustrative.

Definition

Compound interest: interest that is added to the balance at the end of each period, so that later interest is calculated on the larger balance. A principal $P$ at a nominal annual rate $r$ , compounded $k$ times a year for $t$ years, grows to

A = P\left(1 + \frac{r}{k}\right)^{kt}.

Effective annual rate: the yearly rate that gives the same growth with one compounding a year, $\left(1 + \frac{r}{k}\right)^k - 1$ .

Example: 9% a year compounded monthly is $\frac{9\%}{12} = 0.75\%$ per month, and $1.0075^{12} - 1 = 9.38\%$ effective.

Worked example 1: amount, time and rate

Gisele saves 500,000 Frw at 9% a year compounded monthly. (a) How much does she have after 3 years? (b) How long until her money doubles? (c) What yearly rate, compounded once a year, would double money in 8 years?

Solution

$A = 500\,000 \times 1.0075^{36} = 654\,323$ Frw (with yearly compounding it would be $500\,000 \times 1.09^3 = 647\,515$ Frw).
${1.0075}^{m} = 2$ , so $m = \frac{\log 2}{\log 1.0075} = 92.8$ . Interest is added only at the end of a month, so she needs 93 months: 7 years and 9 months.
$(1 + r)^{8} = 2$ , so $1 + r = 2^{\frac{1}{8}} = 1.0905$ : a rate of 9.05% a year.

Warning

Match the rate to the period. With monthly compounding, use the monthly rate $\frac{r}{12}$ and count the periods in months. Writing $1.09^{36}$ for 3 years of monthly compounding at 9% a year multiplies the money by about 22 instead of 1.31.

Regular payments: an annuity

Suppose you deposit the same amount $D$ at the end of every month, at a monthly rate $i$ . After $n$ months the first deposit has grown for $n - 1$ months, the next for $n - 2$ , and the last not at all. The total is a geometric sum with ratio $1 + i$ (Unit 2):

F = D + D(1 + i) + \cdots + D(1 + i)^{n-1} = D\,\frac{(1 + i)^n - 1}{i}.

Worked example 2: a savings plan

A teacher deposits 50,000 Frw at the end of every month for 5 years in an account paying 0.5% a month. How much is in the account at the end?

Solution $n = 60$ , $i = 0.005$ : $F = 50\,000 \times \frac{1.005^{60} - 1}{0.005} = 3\,488\,502$ Frw. She paid in $60 \times 50\,000 = 3\,000\,000$ Frw; the rest, 488,502 Frw, is interest.

Definition

Mortgage (amortised loan): a loan, often to buy a house or land, that is repaid by equal payments at regular intervals; each payment covers that period's interest and repays part of the debt (the principal).

Amortisation schedule: a table showing, for each payment, the interest, the principal repaid and the balance still owed.

Example: on a balance of 30,000,000 Frw at 1.25% a month, the first month's interest is 375,000 Frw.

The mortgage formula

Borrow $L$ at a rate $i$ per month and repay $P$ at the end of each month. If $B_{m}$ is the balance after $m$ payments, each month the balance grows by interest and then falls by the payment:

B_0 = L, \qquad B_{m+1} = (1 + i)B_m - P.

This is the recurrence $u_{n+1} = ru_n + c$ of Unit 2. Its fixed point is $\frac{P}{i}$ , and $B_m - \frac{P}{i} = (1 + i)^m\left(L - \frac{P}{i}\right)$ , which rearranges to

B_m = L(1 + i)^m - P\,\frac{(1 + i)^m - 1}{i}.

The loan is repaid after $n$ payments when $B_{n} = 0$ . Solving for $P$ , or for $n$ with logarithms:

Important

Loan repayment formulae (monthly rate $i$ , $n$ monthly payments)

P = \frac{L\,i}{1 - (1 + i)^{-n}}, \qquad n = -\frac{\log\left(1 - \frac{L\,i}{P}\right)}{\log(1 + i)}.

The second formula needs $P \gt L\,i$ : the payment must be larger than the first month's interest, or the debt never falls.

Worked example 3: a house loan

A family borrows 30,000,000 Frw to build a house, at 15% a year compounded monthly, repaid over 15 years. Find the monthly payment, the total interest, and the first three lines of the amortisation schedule.

Solution $i = \frac{0.15}{12} = 0.0125$ and $n = 180$ :

P = \frac{30\,000\,000 \times 0.0125}{1 - 1.0125^{-180}} = 419\,876 \text{ Frw per month}.

The family repays $180 \times 419\,876 = 75\,577\,704$ Frw, so the interest is about 45,577,704 Frw, more than the loan itself.

Month	Interest (Frw)	Principal repaid (Frw)	Balance (Frw)
1	375,000	44,876	29,955,124
2	374,439	45,437	29,909,687
3	373,871	46,005	29,863,682

Each month the interest falls a little and the principal repaid rises a little; Figure 1 shows the whole story.

Figure 1. The balance of the mortgage in Worked example 3, computed from

B_m = L(1 + i)^m - P\,\frac{(1 + i)^m - 1}{i}

. Early payments are mostly interest, so the debt falls slowly at first.

Worked example 4: how many payments?

A cooperative borrows 5,000,000 Frw at 1% a month and can repay 120,000 Frw a month. How many payments are needed?

Solution $L\,i = 50\,000 \lt 120\,000$ , so the loan can be repaid.

n = -\frac{\log\left(1 - \frac{50\,000}{120\,000}\right)}{\log 1.01} = -\frac{\log 0.58333}{\log 1.01} = 54.2.

So 54 full payments and a smaller 55th payment: 55 payments in all.

Tip

Two quick checks. The monthly payment must be more than $L i$ (the first month's interest) and more than $\frac{L}{n}$ (the payment with no interest at all). For Worked example 3: $375\,000 \lt 419\,876$ and $\frac{30\,000\,000}{180} = 166\,667 \lt 419\,876$ . ✓

Mathispower4u: The Installment Loan Formula. Watch how the payment formula is used to find a monthly payment from the amount borrowed, the rate per period and the number of payments, and how the total interest follows.

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Practice

All figures are illustrative. Give amounts to the nearest franc. The last two are harder.

Find the value of 200,000 Frw invested for 3 years at 8% a year compounded quarterly.
How long does it take money to double (a) at 12% a year compounded yearly (b) at 1% a month compounded monthly?
1,000,000 Frw grows to 1,500,000 Frw in 5 years with yearly compounding. Find the yearly rate.
A car loan of 8,000,000 Frw is repaid monthly over 4 years at 1% a month. Find the payment and the total interest.
A loan of 2,000,000 Frw at 1.5% a month is repaid at 60,000 Frw a month. How many payments are needed?
Find the effective annual rate of 12% a year compounded monthly.
Find the value after 3 years of 20,000 Frw saved at the end of each month at 0.4% a month.
A borrower takes 1,500,000 Frw at 2% a month and offers to repay 30,000 Frw a month. Show that the loan is never repaid, and find the smallest whole-franc monthly payment that clears it in 5 years.

Show answersHide answers

$200\,000 \times 1.02^{12} = 253\,648$ Frw.
(a) $\frac{\log 2}{\log 1.12} = 6.1$ , so 7 years (b) $\frac{\log 2}{\log 1.01} = 69.7$ , so 70 months.
$(1 + r)^{5} = 1.5$ , $r = 1.5^{\frac{1}{5}} - 1 = 8.45\%$ .
$P = \frac{8\,000\,000 \times 0.01}{1 - 1.01^{-48}} = 210\,671$ Frw; interest $48 \times 210\,671 - 8\,000\,000 \approx 2\,112\,193$ Frw.
$n = -\frac{\log(1 - 0.5)}{\log 1.015} = 46.6$ : 47 payments (the last one smaller).
$1.01^{12} - 1 = 12.68\%$ .
$20\,000 \times \frac{1.004^{36} - 1}{0.004} = 772\,762$ Frw.
The first month's interest is $1\,500\,000 \times 0.02 = 30\,000$ Frw, the whole payment, so the balance never falls. For 60 months: $P = \frac{1\,500\,000 \times 0.02}{1 - 1.02^{-60}} = 43\,151.95$ , so the smallest whole-franc payment is 43,152 Frw.

Sources and credits

Written by Ganzaa: the explanations, the derivation, the graph, the schedule, the worked examples (all rates and amounts are illustrative), the practice items and the quiz questions.
Video: The Installment Loan Formula, Mathispower4u, embedded from YouTube.

1Money that earns money

2Regular payments: an annuity

3The mortgage formula

Money that earns money

Regular payments: an annuity

The mortgage formula