Compound Interest and Depreciation 💡

Welcome, students! In this lesson, you will learn how money and value change over time using compound interest and depreciation. These ideas are important in everyday life: savings accounts grow, loans can increase, and items like cars and phones lose value. In IB Mathematics Analysis and Approaches SL, this topic connects strongly to exponentials, powers, and algebraic reasoning.

Lesson Objectives

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

• explain the meaning of compound interest and depreciation,
• use formulas for growth and decay correctly,
• solve practical problems involving percentage change over time,
• recognize how exponential models fit into Number and Algebra,
• interpret results using real-world examples and mathematical evidence.

Think about this: if you put money in a bank account, the interest may be added again and again, not just once. That repeated growth is what makes compound interest so powerful. On the other hand, a car, laptop, or phone often loses value each year, which is a form of depreciation. 📈📉

1. What Is Compound Interest?

Compound interest means that interest is added to the original amount, and then future interest is calculated on the new total. This creates growth on growth.

The amount after $n$ periods is often written as:

$$A = P\left(1 + r\right)^n$$

where:

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

If interest is compounded more than once per year, the formula becomes:

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

where:

• $m$ is the number of compounding periods per year,
• $t$ is the number of years.

Example 1

Suppose students deposits $1000$ into a savings account with annual interest of $5\%$. After $3$ years, the amount is:

$$A = 1000\left(1 + 0.05\right)^3$$

$$A = 1000(1.157625)$$

$$A = 1157.625$$

So the final amount is about $\$1157.63.

Notice that the interest was added each year to the growing balance. After the first year, the amount is $1050$. In the second year, the interest is earned on $1050$, not just on the original $1000$. That is the key idea of compounding.

Why It Matters

Compound interest appears in:

• savings accounts,
• investments,
• loans and credit cards,
• population growth models,
• science and technology models.

In IB mathematics, this is an example of an exponential function because the variable $n$ appears in the exponent. The output changes by a constant factor each period rather than by a constant amount.

2. Understanding Depreciation

Depreciation is the decrease in value of an item over time. It is the opposite of compound growth. Many objects lose value because they are used, become outdated, or wear out. A common example is a car. A brand-new car may lose a large amount of value in the first year.

The depreciation formula is:

$$V = P\left(1 - r\right)^n$$

where:

• $V$ is the value after $n$ periods,
• $P$ is the original value,
• $r$ is the depreciation rate per period as a decimal,
• $n$ is the number of periods.

Example 2

A phone costs $800$ and depreciates by $20\%$ per year. After $2$ years, its value is:

$$V = 800\left(1 - 0.20\right)^2$$

$$V = 800(0.8)^2$$

$$V = 800(0.64)$$

$$V = 512$$

So the phone is worth $\$512$ after $2 years.

This model shows exponential decay because each year the value is multiplied by the same factor, $1-r$. The value does not drop by the same dollar amount every year; instead, it drops by the same percentage.

3. Compound Interest vs. Simple Interest

It is important not to mix up compound interest and simple interest.

Simple interest is calculated only on the original principal. The formula is:

$$I = Prt$$

where:

• $I$ is the interest earned,
• $P$ is the principal,
• $r$ is the rate,
• $t$ is time.

The total amount is:

$$A = P + I = P\left(1 + rt\right)$$

Comparison Example

Suppose $1000$ is invested at $5\%$ for $3$ years.

For simple interest:

$$A = 1000\left(1 + 0.05 \cdot 3\right) = 1150$$

For compound interest:

$$A = 1000(1.05)^3 = 1157.625$$

The compound interest amount is larger because the interest earns interest too. Over longer periods, the difference becomes much bigger. This is why compounding is such a powerful mathematical idea. 🚀

4. Exponential Patterns and Algebraic Reasoning

Compound growth and depreciation are examples of exponential patterns. In Number and Algebra, you study how expressions behave when repeated multiplication is involved.

For a sequence of values modeled by compound interest, each term is found by multiplying the previous term by the same ratio.

For example:

• Year 0: $1000$
• Year 1: $1000(1.05)$
• Year 2: $1000(1.05)^2$
• Year 3: $1000(1.05)^3$

This is a geometric sequence, where each term is multiplied by a common ratio. The common ratio here is $1.05$.

For depreciation at $20\%$ each year, the common ratio is $0.8$.

Real-World Connection

If a business buys equipment, it may record a lower value each year because of depreciation. If a student saves money for university, compound interest helps estimate future savings. In both cases, algebra lets us model what happens over time without checking every single year by hand.

5. Solving for Time, Rate, or Initial Value

IB Mathematics often asks you to solve for an unknown in the compound interest or depreciation formula. This may involve rearranging expressions and using logarithms.

If

$$A = P\left(1 + r\right)^n$$

then to solve for $n$, you can use logarithms:

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

Taking logarithms of both sides gives:

$$\log\left(\frac{A}{P}\right) = n\log\left(1 + r\right)$$

So:

$$n = \frac{\log\left(\frac{A}{P}\right)}{\log\left(1 + r\right)}$$

Example 3

How long will it take for $2000$ to grow to $3000$ at an annual rate of $4\%$?

$$3000 = 2000(1.04)^n$$

Divide by $2000$:

$$1.5 = (1.04)^n$$

Use logarithms:

$$n = \frac{\log(1.5)}{\log(1.04)}$$

This gives approximately:

$$n \approx 10.43$$

So it takes a little more than $10$ years.

This shows an important IB skill: using algebraic manipulation to solve real problems. Sometimes the answer is not a whole number of years, and that is completely acceptable in modeling.

6. Interpreting Growth and Decay Graphs

Graphs make compound interest and depreciation easier to understand. A growth graph rises slowly at first, then faster and faster. A decay graph falls quickly at first, then levels off.

For a function like

$$f(x) = 1000(1.08)^x$$

the graph increases because $1.08 > 1$.

For a function like

$$g(x) = 1000(0.9)^x$$

the graph decreases because $0 < 0.9 < 1$.

Key graph features include:

• the y-intercept, which is the initial value when $x=0$,
• the curve shape, which shows exponential behavior,
• the rate of change, which is proportional to the current value.

A common mistake is thinking exponential growth adds the same amount each time. It does not. It multiplies by the same factor each time. That is why the graph curves upward or downward instead of forming a straight line.

Conclusion

Compound interest and depreciation are two sides of exponential change. Compound interest describes money or value that grows by a fixed percentage each period, while depreciation describes value that decreases by a fixed percentage each period. Both are modeled using exponential expressions, which are a major part of Number and Algebra in IB Mathematics Analysis and Approaches SL.

students, the key mathematical ideas are repeated multiplication, percentage change, and algebraic manipulation. These ideas help you solve practical problems in finance, technology, and everyday life. By understanding compound interest and depreciation, you are also strengthening your understanding of exponential functions, geometric sequences, and logarithms.

Study Notes

• Compound interest means interest is earned on both the original principal and previous interest.
• The compound interest formula is $A = P\left(1 + r\right)^n$.
• If interest is compounded multiple times per year, use $A = P\left(1 + \frac{r}{m}\right)^{mt}$.
• Depreciation means an asset loses value over time.
• The depreciation formula is $V = P\left(1 - r\right)^n$.
• Compound interest is exponential growth; depreciation is exponential decay.
• These situations form geometric sequences with a common ratio.
• Simple interest uses $I = Prt$ and does not include interest on interest.
• Logarithms are useful for solving for $n$ when the exponent is unknown.
• Real-world examples include savings accounts, loans, cars, phones, and equipment.
• In IB Mathematics Analysis and Approaches SL, this topic connects number systems, exponentials, logarithms, and algebraic manipulation.