Compound Interest and Depreciation 💡

Introduction

students, money does not just stay the same over time. In real life, bank savings can grow, while cars, phones, and machines often lose value. This lesson explains how that works using compound interest and depreciation. These ideas are important in finance, science, business, and everyday decision-making 📈📉

Learning objectives

By the end of this lesson, students, you should be able to:

explain the key ideas and terms behind compound interest and depreciation;
use formulas to calculate growth and decay over time;
solve exam-style problems involving repeated percentage change;
connect these ideas to sequences, algebra, and exponential functions in Number and Algebra;
interpret real-world examples using clear mathematical reasoning.

A key idea in this topic is that change is often multiplicative, not additive. That means the amount changes by the same percentage each time, not by the same number. This is why savings accounts and falling asset values are modeled with powers of a factor.

Compound Interest: growth over time

Compound interest means interest is earned on both the original amount and the interest already added. Suppose you deposit $P$ dollars in a bank account at an annual interest rate of $r$. If the interest is compounded once per year, after one year the balance is $P(1+r)$. After two years, the interest is earned on that new balance, giving $P(1+r)^2$. After $n$ years, the amount is

$$A=P(1+r)^n$$

Here:

$A$ is the final amount,
$P$ is the principal, or starting amount,
$r$ is the annual interest rate written as a decimal,
$n$ is the number of compounding periods.

This formula appears constantly in IB Mathematics because it is an example of an exponential relationship. The same pattern connects to sequences: $P$, $P(1+r)$, $P(1+r)^2$, $P(1+r)^3$, and so on form a geometric sequence.

Example 1: yearly compounding

students, imagine you invest $2000$ at $5\%$ interest per year. What is the value after $3$ years?

Use $A=P(1+r)^n$ with $P=2000$, $r=0.05$, and $n=3$:

$$A=2000(1.05)^3$$

$$A=2315.25$$

So the account grows to $\$2315.25. Notice that the interest in year 3 is calculated on a larger amount than in year 1. That is the meaning of compounding.

Compound interest with more frequent compounding

Sometimes interest is compounded more than once per year, such as monthly. If the annual nominal rate is $r$ and interest is compounded $m$ times per year, then after $t$ years:

$$A=P\left(1+\frac{r}{m}\right)^{mt}$$

This is still exponential growth, but the growth is divided into smaller steps. For example, monthly compounding means $m=12$. The more frequently interest is compounded, the slightly larger the final amount will usually be, assuming the same nominal annual rate.

Example 2: monthly compounding

Suppose $P=1000$, $r=0.06$, and the money is compounded monthly for $2$ years. Then

$$A=1000\left(1+\frac{0.06}{12}\right)^{24}$$

$$A\approx 1127.16$$

This shows how compound interest can be used to predict future account balances. In exams, students, you should be careful with the number of periods and make sure the rate matches the compounding frequency.

Depreciation: loss in value over time

Depreciation is the opposite idea: an item loses value by a fixed percentage each period. A car, phone, or machine often decreases in value because of wear, age, or new technology. If something is worth $P$ now and depreciates at a rate of $r$ per period, then after $n$ periods its value is

$$V=P(1-r)^n$$

Here:

$V$ is the final value,
$P$ is the original value,
$r$ is the depreciation rate as a decimal,
$n$ is the number of periods.

This is also exponential behavior, but now the factor $1-r$ is less than $1$, so the value decreases over time.

Example 3: depreciation of a car

A car is bought for $\$18{,}000$ and depreciates by $15\% each year. What is its value after $4$ years?

Use $V=P(1-r)^n$ with $P=18000$, $r=0.15$, and $n=4$:

$$V=18000(0.85)^4$$

$$V\approx 9392.48$$

So the car is worth about $\$9392.48$ after $4 years.

Depreciation problems are often easier if you think of the multiplying factor first. A $15\%$ decrease means multiply by $0.85$. A $7\%$ increase means multiply by $1.07$. This conversion between percentage and factor is essential.

Solving problems and finding missing values

In IB Mathematics: Analysis and Approaches HL, you are not only expected to calculate final values, but also to rearrange formulas and interpret results. Compound interest and depreciation questions often ask for the time, the rate, or the original value.

Finding the rate

If you know the starting value $P$, final value $A$, and time $n$, you can solve

$$A=P(1+r)^n$$

for $r$.

For example, if $1000$ grows to $1331$ in $3$ years, then

$$1331=1000(1+r)^3$$

$$1.331=(1+r)^3$$

$$1+r=1.1$$

$$r=0.1$$

So the annual interest rate is $10\%$.

Finding the time

If the amount doubles, you may need to use logarithms. Suppose $500$ is invested at $4\%$ compound interest and grows to $1000$. Then

$$1000=500(1.04)^n$$

$$2=(1.04)^n$$

Taking logarithms gives

$$n=\frac{\log 2}{\log 1.04}$$

$$n\approx 17.67$$

So it takes about $17.67$ years. This links compound interest to logarithms, another important part of Number and Algebra.

Real-world reasoning

These calculations are useful in practical decisions. A student might compare savings accounts, a business might estimate equipment replacement costs, and a family might predict how much a car will be worth later. The mathematics helps answer questions like: Is the growth enough to reach a goal? How long until an item loses half its value? Should an investment be left for more time?

How this topic fits into Number and Algebra

Compound interest and depreciation fit naturally into Number and Algebra because they use several core ideas from the topic:

Sequences and series: repeated multiplication creates a geometric sequence;
Symbolic manipulation: formulas such as $A=P(1+r)^n$ must be rearranged and interpreted;
Exponents and logarithms: powers model repeated change, and logarithms help solve for time;
Reasoning and proof: you can justify formulas by showing the pattern after $1$, $2$, and $3$ periods, then generalizing to $n$ periods.

For example, after one year of compound growth, the amount is $P(1+r)$. After two years, multiply again by $1+r$ to get $P(1+r)^2$. After three years, the pattern becomes $P(1+r)^3$. This is a simple example of mathematical induction-style thinking: identify a pattern and extend it logically.

You should also understand the difference between linear and exponential change. If something increases by a fixed amount each year, that is linear. If it increases by a fixed percentage each year, that is exponential. A $5\%$ increase on $1000$ is $50$, but a $5\%$ increase on $1050$ is $52.50$. The amount of change gets larger because the base is growing.

Conclusion

students, compound interest and depreciation describe how values change repeatedly over time. Compound interest models growth with a factor greater than $1$, while depreciation models decay with a factor between $0$ and $1$. Both are based on exponentials and geometric sequences, which makes them central to Number and Algebra. Mastering these ideas helps you solve financial problems, interpret graphs, and use algebra confidently in real situations 💰

Study Notes

Compound interest uses the formula $A=P(1+r)^n$ when interest is compounded once per period.
If compounding happens $m$ times per year, use $A=P\left(1+\frac{r}{m}\right)^{mt}$.
Depreciation uses the formula $V=P(1-r)^n$.
Convert percentages to decimal form before substituting into a formula.
A growth factor is $1+r$; a decay factor is $1-r$.
These situations create geometric sequences because each term is found by multiplying by the same factor.
Exponential models are different from linear models because the percentage change stays constant, not the amount.
Logarithms are useful for solving equations where the unknown is in the exponent.
Real-life examples include savings accounts, loans, cars, phones, and business equipment.
In IB Mathematics: Analysis and Approaches HL, this topic connects algebra, sequences, logarithms, and problem solving.