Maths Advanced: Growth, Statistics and Reasoning
Exponential Growth and Logarithms
In this lesson you will
- Model growth and decay with exponential formulas
- Use logarithms to solve for time
- Explain why exponential change feels sudden
Introduction
When something changes by a fixed percentage each period, instead of a fixed amount, the pattern is exponential. Savings with compound interest, populations, the spread of some infections, radioactive decay and depreciating cars all follow it. This is different from linear change, and it behaves very differently over time.
The model
amount = start x (1 + r) ^ t- r is the growth rate per period (as a decimal), and t is the number of periods.
- For decay, r is negative: 1 - 0.1 = 0.9 for a 10 percent drop per period.
Example: 1,000 growing 5 percent a year for 10 years:
1,000 x 1.05^10 = 1,628.89.
Linear growth of 50 per year would give 1,500. The extra comes from growth on growth.
Decay example: a car loses 15 percent of its value per year. Starting at 20,000, after 3 years: 20,000 x 0.85^3 = 20,000 x 0.614125 = 12,282.50.
Quick quiz
A value of 200 grows 10 percent per period. What is it after 2 periods?
Exponential versus linear
A linear process adds the same amount each step. An exponential process multiplies by the same factor each step. Early on they can look similar. Later, the exponential one is far ahead. A classic example: the amount from doubling a grain of rice on each square of a chessboard. Square 1 is 1 grain, square 10 is 512, and square 64 is 2^63, about 9.2 x 10^18 grains.
This is why exponential processes seem to come from nowhere: the numbers look small for a long time, then change very quickly.
Logarithms: solving for time
A logarithm is the inverse of exponentiation. log base 10 of 1,000 is 3, because 10^3 = 1,000. The natural logarithm, written ln, uses the base e, about 2.718.
How long for 1,000 to double at 5 percent a year? Solve 1.05^t = 2:
t = ln(2) / ln(1.05) = 0.6931 / 0.04879 = 14.2 yearsThe rule of 72 estimates this as 72 / 5 = 14.4 years, which is close.
A general formula for time to reach a target multiple M: t = ln(M) / ln(1 + r).
Doubling and half-life
- Doubling time: time to double under growth.
- Half-life: time to halve under decay. For decay of 10 percent per period, 0.9^t = 0.5, so t = ln(0.5) / ln(0.9) = about 6.6 periods.
Where logs appear
- Orders of magnitude and scales such as the pH scale and decibels, where equal steps mean multiplying by a factor.
- Log graphs that turn exponential curves into straight lines, which make growth rates easier to compare.
- Information and computing: the number of bits for n items is about log base 2 of n.
Cautions
- Real-world growth usually cannot continue forever: limits such as resources slow it. Models like logistic curves describe this.
- Rates need care: 5 percent per year compounded monthly is not exactly the same as 5 percent per year compounded annually.
- Predictions from fitted curves can be very sensitive to small errors in the rate.
Try it yourself
Calculate 2,000 growing at 3 percent for 20 years. Find how long it takes for a value to triple at 8 percent a year using the log formula. Compare your result with a spreadsheet that multiplies year by year.
Summary
- Exponential change multiplies by a fixed factor each period: start x (1 + r)^t.
- Logarithms solve for the time needed, for instance t = ln(M) / ln(1 + r).
- Exponential processes seem slow, then sudden, and real growth is usually limited.
Lesson complete
+20 XP
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Sources and further reading