Exponential Decay

Halving at every step

Start with 80 and halve it at every step: 80, 40, 20, 10, 5. Halving is multiplying by $\frac{1}{2}$1/2, so this is the same rule as exponential growth, multiplying by the same factor at every step, with a factor between 0 and 1.

After n steps the 80 has been halved n times, so the amount is $80\times 0.5^n$80 × 0.5ⁿ. After 3 steps it is $80\times 0.5^3=80\times \frac{1}{8}=10$80 × 0.5³ = 80 × 1/8 = 10. Halving n times divides by $2^n$2ⁿ, so the fraction left after n halvings is $\frac{1}{2^n}$1/2ⁿ. A factor b between 0 and 1 gives exponential decay, and a factor above 1 gives growth.

The amount after 0, 1, 2, 3 and 4 halvings: 80, 40, 20, 10, 5. Each bar is half the one before it.

Falling by less each time

Look at the drops: from 80 to 40 is a drop of 40, then 20, then 10, then 5. Each drop is half the one before it, because each step takes away half of what is left, and less is left every time.

Compare taking away the same amount at every step. Subtracting 20 each time gives 80, 60, 40, 20, 0: a straight line down, which reaches 0 after four steps and would go below it after that. The halving curve falls steeply at first and then flattens out.

The gold curve is $y=80\times 0.5^x$y = 80 × 0.5ˣ, through 80, 40, 20, 10 and 5. The white line is y = 80 − 20x, which takes 20 away at every step and reaches 0 at x = 4.

Never reaching zero

Half of a positive amount is still positive, so however many times 80 is halved, something is left. After 10 halvings it is $80/2^{10}=80/1024=0.078$80 / 2¹⁰ = 80 / 1024 = 0.078, to 3 decimal places. After 20 halvings it is $80/2^{20}$80 / 2²⁰, less than a ten-thousandth, and still more than 0.

So the curve closes in on the x-axis forever without touching it. A line that a curve gets closer and closer to without reaching is called an asymptote, and the x-axis is an asymptote of every exponential decay curve.

$y=80\times 0.5^x$y = 80 × 0.5ˣ over ten steps. By x = 10 the height is 80 ÷ 1024, about 0.08: too close to the x-axis to see the gap, and still above it.

Half-life

The time an amount takes to halve is called its half-life. It is the decay twin of doubling time, and in the same way it does not depend on where you start: 80 takes one half-life to reach 40, and 40 takes exactly as long again to reach 20.

A radioactive substance has a half-life of 3 days, and a sample of it weighs 80 g. After 3 days 40 g is left, after 6 days 20 g, and after 9 days 10 g. After t days there have been $\frac{t}{3}$t/3 half-lives, so the mass left is $80\times 0.5^{\frac{t}{3}}$80 × 0.5^(t/3) grams. At t = 9 that is $80\times 0.5^3=10\text{g}$80 × 0.5³ = 10 g, as counted.

The formula also works between the half-lives. After 1 day the mass is $80\times 0.5^{\frac{1}{3}}=63.5\text{g}$80 × 0.5^(1/3) = 63.5 g, to 1 decimal place. It is not $80-\frac{40}{3}=66.7\text{g}$80 − 40/3 = 66.7 g: the substance does not lose a third of the first 40 g in each of the first three days, because every day it loses the same fraction of what is left, not the same number of grams.

A percentage fall is a factor

A fall of 15% keeps 85% of the amount, and 85% = 0.85. So losing 15% a year means multiplying by 0.85 every year. A car bought for $20,000 that loses 15% of its value each year is worth $20000\times 0.85^t$20000 × 0.85ᵗ dollars after t years.

After one year it is worth $17,000, after two $14,450, and after three $12,282.50. The yearly loss was $3000, then $2550, then $2167.50: each year it loses 15% of a smaller value, so the loss shrinks, just as the drops did when halving.

The usual mistakes

Subtracting a fixed amount. Halving 80 twice is not 80 − 40 − 40 = 0. The second halving takes half of 40, which is 20, and leaves 20.

Counting one halving too many. With a half-life of 3 days, 9 days is 9 ÷ 3 = 3 halvings, which leaves $80/2^3=10\text{g}$80 / 2³ = 10 g. A fourth halving gives 5 g, the mass after 12 days.

Using the fall as the factor. A 15% fall multiplies by 0.85, the part that is kept. Multiplying by 0.15 keeps only the part that was lost.

When the time is the unknown

The next problem asks when an amount that halves every 4 hours, from 320 mg, falls to 100 mg. After x hours it is $320\times 0.5^{\frac{x}{4}}$320 × 0.5^(x/4). Setting this equal to 100 and dividing by 320 gives $0.5^{\frac{x}{4}}=0.3125$0.5^(x/4) = 0.3125. A power of $\frac{1}{2}$1/2 is 1 over the same power of 2, so turning both sides over gives $2^{\frac{x}{4}}=\frac{1}{0.3125}=3.2$2^(x/4) = 1/0.3125 = 3.2. Now the unknown $\frac{x}{4}$x/4 is the power of 2 that gives 3.2. Since $2^1=2$2¹ = 2 and $2^2=4$2² = 4, it lies between 1 and 2.

To find it, write both sides as powers of 10. A calculator gives the common logarithm of 2, the power of 10 that makes 2, as log 2 = 0.3010 to 4 decimal places. So $2=10^{0.3010}$2 = 10^0.3010. Raising a power to a power multiplies the indices, so $2^k=10^{0.3010k}$2ᵏ = 10^(0.3010k), and the logarithm of $2^k$2ᵏ is k × 0.3010. This is the power law: the logarithm of a power is the index times the logarithm of the base. Also $3.2=32/10=2^5/10$3.2 = 32 / 10 = 2⁵ / 10, so its logarithm is 5 × 0.3010 − 1 = 0.5050, because dividing by 10 takes 1 off the index.

Taking the logarithm of both sides of $2^{\frac{x}{4}}=3.2$2^(x/4) = 3.2 therefore gives $\frac{x}{4}\times 0.3010=0.5050$x/4 × 0.3010 = 0.5050, so $\frac{x}{4}=0.5050/0.3010=1.678$x/4 = 0.5050 / 0.3010 = 1.678 half-lives, and x is 4 times that.