When you invest money, you usually earn interest. There are two main ways you can earn interest: simple and compound.
This means compound interest usually leads to greater returns than simple interest because the principal is updated each period after interest is earned.
Let's compare the two with an example of earning $5%$5% interest on an initial investment principal of $\$1000$$1000.
Year | Principal Used | Simple interest earned | New total |
---|---|---|---|
1 | $\$1000$$1000 | $\$50$$50 | $\$1050$$1050 |
2 | $\$1000$$1000 | $\$50$$50 | $\$1100$$1100 |
3 | $\$1000$$1000 | $\$50$$50 | $\$1150$$1150 |
Year | Principal Used | Compound interest earned | New total |
---|---|---|---|
1 | $\$1000$$1000 | $\$50$$50 | $\$1050$$1050 |
2 | $\$1050$$1050 | $\$52.50$$52.50 | $\$1102.50$$1102.50 |
3 | $\$1102.50$$1102.50 | $\$55.13$$55.13 | $\$1157.63$$1157.63 |
We can see that after just two years, there is already a financial benefit in earning compound interest over simple interest.
Of course, compound interest can also apply to loans, where the amount we need to repay increases in the same way.
In the previous example on compound interest, the new total after $1$1 year was $1000+1000\times0.05$1000+1000×0.05. If we factor out the principal amount, we can see that our new total after $1$1 year is equal to $1000\times1.05$1000×1.05.
Notice that this is the same way we calculate simple interest over one period.
However, since compound interest calculates interest based on the new total, the principal amount for the second year's interest will be equal to the total amount after the first year. This means that the new total after two years will be equal to $1000\times1.05\times1.05$1000×1.05×1.05.
Since the total after each year will be equal to $1.05$1.05 times the previous year's total, we can calculate the total after any number of years by multiplying by $1.05$1.05 the required number of times.
For example: if John takes out a loan of $\$2000$$2000 at an interest rate of $4%$4%, his total amount to repay at the end of each year will be equal to $1.04$1.04 times the total of the previous year. After $5$5 years, his total amount to repay will be equal to $\$2000\times1.04\times1.04\times1.04\times1.04\times1.04$$2000×1.04×1.04×1.04×1.04×1.04.
Since we are multiplying by the same value multiple times, we can condense the expression using indices. For the case above, we can rewrite John's total amount after $5$5 years as $\$2000\times1.04^5$$2000×1.045.
Writing the total in this form leads us to the compound interest formula.
If a principal amount $P$P is invested at an interest rate of $r$r for $n$n periods, the future value $A$A of the investment after $n$n periods will be:
$A=P\left(1+r\right)^n$A=P(1+r)n
Note that $r$r is a decimal value corresponding to some percentage interest rate. For example, if the interest rate is $4%$4% then $r=0.04$r=0.04.
If we only want to find the amount of interest earned, we can simply subtract the principal amount from the future value of the investment.
$\text{Interest }=\text{Future value}-\text{Principal amount}$Interest =Future value−Principal amount
Now that we have a formula for calculating compound interest, as long as we know the principal amount $P$P, the interest rate $r$r and the number of periods $n$n, we can calculate the future value $A$A.
In addition to this, since we have a formula relating the different variables of an investment, if we know any three of the four variables, we can use them to calculate the last unknown variable.
We can do this by substituting our known values into the formula and then solving the equation for the unknown variable.
Dave is planning to invest $\$3000$$3000 for $4$4 years. If he wants his investment to grow to $\$4000$$4000 at the end of $4$4 years, what annually compounding interest rate $r$r will he need?
Think: Since the period length is one year, we know that $P=3000$P=3000, $A=4000$A=4000 and $n=4$n=4. We can find the required interest rate $r$r by substituting our known values into the compound interest formula and solving for $r$r.
Do: Substituting our known values into the formula gives us:
$4000=3000\left(1+r\right)^4$4000=3000(1+r)4
We can then solve for r using the steps:
$4000$4000 | $=$= | $3000\left(1+r\right)^4$3000(1+r)4 | |
$\frac{4000}{3000}$40003000 | $=$= | $\left(1+r\right)^4$(1+r)4 |
Divide both sides by the principal amount |
$\left(\frac{4000}{3000}\right)^{\frac{1}{4}}$(40003000)14 | $=$= | $1+r$1+r |
Reverse the power of $4$4 by taking both sides to the power of $\frac{1}{4}$14 |
$\left(\frac{4000}{3000}\right)^{\frac{1}{4}}-1$(40003000)14−1 | $=$= | $r$r |
Subtract $1$1 from both sides |
$r$r | $=$= | $0.0746$0.0746 |
Evaluate for $r$r, rounding to four decimal places |
We can also convert our interest rate into a percentage value, so we can see that Dave will require an annual interest rate of:
$r=7.46%$r=7.46%
Reflect: Since we were given three of the four factors in the compound interest formula, we could substitute them into the formula to solve for the required missing factor.
We can perform similar, if easier, calculations to find the principal or future value if we know the other three factors.
However, finding the number of periods is a bit harder since we don't have a way to isolate the index of a term at this level of mathematics.
Instead, we can refer back to our repeated applications of simple interest, stopping after we have reached a desired value.
Jemma is investing $\$200$$200 with a interest rate of $5%$5% p.a., compounding annually. How many years will it take for her investment to earn at least $\$30$$30 in interest?
Think: Since Jemma wants to earn at least $\$30$$30 on her investment, she wants the future value $A$A to be at least $\$230$$230. We are given that the $P=200$P=200 and $r=0.05$r=0.05. We can substitute these values into the compound interest formula with some unknown value for $n$n. By calculating the new totals one year at a time, we can find the number of years required.
Do: Substituting our values into the formula gives us:
$A=200\times1.05^n$A=200×1.05n
Calculating the new total of the investment one year at a time gives us:
As we can see, the future value $A$A will reach (and surpass) the target of $\$230$$230 after $3$3 years.
So far, the examples we have looked at have all had interest calculated annually. However, interest can be calculated over any period and the interest rate per period will need to match it.
To match the interest rate to the period duration, we can use interest rate conversions.
It is also important to match the total number of periods to the duration of the investment. To do this, we can simply divide the duration of the investment by the duration of the period.
Cam is investing $\$1000$$1000 into an account that accumulates interest at a rate of $2.4%$2.4% p.a., compounding monthly. What will the account's total be after $3$3 years?
Think: Since interest is compounding monthly, interest will be calculated $12$12 times a year.
Since the interest rate of $2.4%$2.4% is an annual interest rate, we need to divide it by $12$12. This will give us a monthly interest rate of $\frac{2.4%}{12}$2.4%12.
Since there are $12$12 months in a year, after $3$3 years Cam's investment will have experienced $3\times12$3×12 periods.
Do: Substituting our values into the compound interest formula gives us:
$A=1000\left(1+\frac{2.4%}{12}\right)^{3\times12}$A=1000(1+2.4%12)3×12
Simplifying our values gives us:
$A=1000\times1.005^{36}$A=1000×1.00536
Evaluating this and rounding to the nearest cent tells us that, after $3$3 years, Cam's investment will reach a total of:
$A=\$1196.68$A=$1196.68
As we saw when calculating compound interest rate using a table, the interest earned each period was greater than the interest earned last period. This tells us that an investment earning compound interest will be increasing at an increasing rate.
To see what this looks like on a graph, we can compare an investment with simple interest to an investment with compound interest.
The graph below shows both an investment with simple interest, and one with compound interest, labelled Investment A and Investment B respectively.
Which investment has a higher principal amount?
Investment A
Investment B
Investment A
Investment B
Which investment has a higher final amount after $10$10 years?
Investment A
Investment B
Investment A
Investment B
After how many years are the investments equal in value?
$9$9 years
$6$6 years
$10$10 years
$12$12 years
$9$9 years
$6$6 years
$10$10 years
$12$12 years
As we can see, the curve representing compound interest is increasing at an increasing rate, while the simple interest curve is linear.
These relationships are reflected in the formulas for simple and compound interest. Simple interest is calculated using the formula $A=P+Prn$A=P+Prn which is a linear equation in terms of $n$n, the number of periods. Meanwhile, compound interest uses the formula $A=P\left(1+r\right)^n$A=P(1+r)n which is non-linear equation in terms of $n$n.
While the compound interest formula is used for calculating the result of compound interest, it can also be used for situations that follow the same non-linear pattern.
For example: appreciation or depreciation over one year is a single percentage change, but we can model them over multiple years using the compound interest formula.
In fact, any percentage change that happens multiple times can be modelled using the compound interest formula.
Another way to think of this is that compound interest is simply the percentage increase of money happening multiple times.
Victoria borrows $\$35000$$35000 at a rate of $4.8%$4.8% p.a, compounding monthly.
After $4$4 months, Victoria repays the loan all at once. How much money does she pay back in total?
Round your answer to the nearest cent.
How much interest was generated on the loan over the four months?
Valerie borrows $\$1250$$1250 at a rate of $7.6%$7.6% p.a, compounding annually.
Which of the following expressions represents the amount Valerie must repay after $15$15 years, assuming that she hasn't paid anything back?
$1250\times\left(1+0.076\right)\times15$1250×(1+0.076)×15
$1250\times\left(1+0.076\right)\times\frac{15}{12}$1250×(1+0.076)×1512
$1250\times\left(1+0.076\right)^{\frac{15}{12}}$1250×(1+0.076)1512
$1250\times\left(1+0.076\right)^{15}$1250×(1+0.076)15
$1250\times\left(1+0.076\right)\times15$1250×(1+0.076)×15
$1250\times\left(1+0.076\right)\times\frac{15}{12}$1250×(1+0.076)×1512
$1250\times\left(1+0.076\right)^{\frac{15}{12}}$1250×(1+0.076)1512
$1250\times\left(1+0.076\right)^{15}$1250×(1+0.076)15
How much interest was generated on the loan over the fifteen years?
Round your answer to two decimal places.
Bart is planning to invest $\$110000$$110000 into a saving scheme which accumulates interest weekly. If he aims to have his investment reach $\$120000$$120000 after $2$2 years, what interest rate per annum, $r$r, will he need?
Enter each line of working as an equation.
Round your answer to four decimal places.
A swimming pool is losing water at the rate of $2%$2% per week due to evaporation.
The pool currently holds $850$850 kilolitres of water.
How much water will be lost in the next $27$27 days?
Give your answer in kilolitres and correct to two decimal places .
How many weeks will it take for the pool to lose at least half its water?
Connect the compound interest formula to repeated applications of simple interest using appropriate digital technologies
Describe, interpret and sketch parabolas, hyperbolas, circles and exponential functions and their transformations.