Algebra 2 · F.IF.C.8B · F.LE.A.1C

Doubling time and half life

A dose of medicine in the blood falls to half every six hours, whether there are 400 milligrams or 4. How can one length of time describe the whole decline?

You need

Growth and decay factorsAlgebra 1

Where you are 6 in Exponential growth and decay

Functions13

Exponential functions6

Exponential growth and decay6

Doubling time and half life

Growth and decay factors Building an exponential model from two points Linear versus exponential growth Logistic growth: when growth levels off Linear, polynomial and exponential growth compared

Why this exists

A growth or decay factor, from Growth and decay factors, says what an amount is multiplied by in one period. People rarely talk that way, though. A doctor says a medicine has a half-life of six hours. A biologist says a yeast culture doubles every twenty-five minutes. A single length of time is easier to picture than a factor such as 0.8909 per hour, and it answers the questions people actually ask: how long until the dose is mostly gone, how long until the culture is four times as big.

That only works because of a special property of exponential change. The time to double, or to halve, is the same from every starting point. A dose of 400 milligrams takes six hours to fall to 200; the remaining 200 takes six more hours to fall to 100, and so on forever. One number describes the whole process.

Definition

Let a quantity change exponentially, y = a \cdot b^t with a > 0.

If b > 1, its doubling time D is the length of time it takes to double. Over any stretch of time of length D, the amount is multiplied by 2.
If 0 < b < 1, its half-life H is the length of time it takes to fall to half. Over any stretch of time of length H, the amount is multiplied by \tfrac{1}{2}.

Written with the doubling time or the half-life in place of the factor, the two models are

y = a \cdot 2^{t/D} \qquad y = a\left(\tfrac{1}{2}\right)^{t/H}

where a is the amount at t = 0, and t, D and H are measured in the same unit of time. The fraction t/D counts how many doubling times have passed, and t/H counts how many half-lives have passed. It need not be a whole number.

Intuition

Figure 1 follows a 400 milligram dose of a medicine with a half-life of 6 hours.

Figure 1The dose M(t) = 400\left(\tfrac{1}{2}\right)^{t/6}. The dots are 6 hours apart and each sits at half the height of the one before.

Each dot is 6 hours to the right of the one before and half as high. The first six hours remove 200 milligrams, the next six remove 100, the next six 50. The amount removed keeps shrinking, but the time to lose half of whatever is left never changes. The curve flattens toward the time axis and never reaches it: half of a positive amount is still positive.

Figure 2 runs the other way: a yeast culture of 150 cells that doubles every 25 minutes.

Figure 2The culture Y(t) = 150 \cdot 2^{t/25}. Every step of the staircase is 25 minutes wide, and every rise is as tall as the height it starts from, so each step doubles the count.

The dashed staircase makes the pattern visible. Every step is the same width, 25 minutes, and every rise is exactly as tall as the height the step starts from, which is what doubling means. Equal steps across, ever-taller steps up: that is the shape of exponential growth. A straight line would rise by the same amount on each step instead, and would take longer and longer to double.

Results

The time to double or halve does not depend on where you start. For y = a \cdot b^t, if b^D = 2 then y doubles over every stretch of time of length D, whatever the starting amount a and whatever the starting time. In the same way, if b^H = \tfrac{1}{2} then y halves over every stretch of length H.

Whole numbers of periods. After n doubling times the amount is multiplied by 2^n. After n half-lives it is multiplied by \left(\tfrac{1}{2}\right)^n = \dfrac{1}{2^n}. Three half-lives leave \tfrac{1}{8}; ten leave \tfrac{1}{1024}, just under a thousandth.

From the doubling time or half-life to a factor per unit of time. A quantity that doubles every D units of time has the factor

b = 2^{1/D}

per unit, and one with half-life H has the factor b = \left(\tfrac{1}{2}\right)^{1/H} per unit.

From a factor to the doubling time or half-life. The doubling time is the number D with b^D = 2, and the half-life is the number H with b^H = \tfrac{1}{2}. These are rarely whole numbers. Listing the powers b, b^2, b^3, \dots shows which two whole numbers of periods the answer lies between; Logarithms find it exactly.

Why it is true

The same time from any start. Suppose y(t) = a \cdot b^t and b^D = 2. Starting at any time t and waiting D more,

\begin{aligned} y(t + D) &= a \cdot b^{t + D} \\ &= a \cdot b^t \cdot b^D \\ &= 2 \cdot a\,b^t = 2\,y(t) \end{aligned}

The middle step is Multiplying and dividing powers with the same base. Nothing in the argument depended on a or on t, so the amount doubles over every stretch of length D. Replacing 2 by \tfrac{1}{2} and D by H gives the same argument for the half-life.

Every exponential has one. When b > 1, the powers b^t grow without bound and rise through every height above 1 as t increases, each height exactly once, so there is exactly one D > 0 with b^D = 2. When 0 < b < 1, the powers fall toward 0 through every height below 1, so there is exactly one H > 0 with b^H = \tfrac{1}{2}.

The models. Call the factor per unit of time m. A doubling time D means D units multiply by 2, so m^D = 2, and the positive number with that property is m = 2^{1/D} (Rational exponents). Then, by Powers of powers, products and quotients,

a\,m^t = a\left(2^{1/D}\right)^t = a \cdot 2^{t/D}

The half-life model comes out the same way from m^H = \tfrac{1}{2}.

Worked examples

Example 1

A dose of medicine

A patient takes a 400 milligram dose of a medicine with a half-life of 6 hours. Write the amount M(t) in the blood after t hours, make a table every 6 hours for one day, and find the amount after 9 hours.

Solution

The start is a = 400 and the half-life is H = 6, so

M(t) = 400\left(\tfrac{1}{2}\right)^{t/6}

Each 6 hours halves the amount, so the table is found by halving:

t (hours)	M(t) (milligrams)
0	400
6	200
12	100
18	50
24	25

One day is 24 hours, which is 24/6 = 4 half-lives, and 400 \times \tfrac{1}{16} = 25 milligrams, matching the table.

After 9 hours, 9/6 = 1.5 half-lives have passed:

M(9) = 400\left(\tfrac{1}{2}\right)^{1.5} = \frac{400}{2\sqrt{2}} = 100\sqrt{2} \approx 141.42

About 141.42 milligrams remain. That is between the 200 at 6 hours and the 100 at 12 hours, as it must be, but it is not 150, the halfway value. The curve in Figure 1 bends, so the amount halfway through a half-life is not halfway between the two amounts.

Example 2

A growing yeast culture

A yeast culture starts with 150 cells and doubles every 25 minutes. Write the number of cells Y(t) after t minutes, and find it after 2 hours 30 minutes and after 1 hour.

Solution

The start is a = 150 and the doubling time is D = 25 minutes, so

Y(t) = 150 \cdot 2^{t/25}

Here t is in minutes, so the times must be changed to minutes before they go in. Two hours thirty minutes is 150 minutes, which is 150/25 = 6 doubling times:

Y(150) = 150 \cdot 2^6 = 150 \times 64 = 9{,}600

One hour is 60 minutes, which is 60/25 = 2.4 doubling times. A calculator gives 2^{2.4} \approx 5.27803, so

Y(60) = 150 \cdot 2^{2.4} \approx 791.70

A count of cells is a whole number, so the model predicts about 792 cells after one hour. As a check, 60 minutes lies between 50 minutes (600 cells) and 75 minutes (1{,}200 cells), and so does 792.

Example 3

How long until the dose is small

The 400 milligram dose from Example 1 has a half-life of 6 hours. When is exactly 12.5 milligrams left? Then find the first whole number of hours after which less than 10 milligrams is left.

Solution

The amount falls from 400 to 12.5, and 400 \div 12.5 = 32 = 2^5. So 12.5 milligrams is \tfrac{1}{32} of the dose, which takes 5 half-lives: 5 \times 6 = 30 hours. Halving step by step confirms it: 400, 200, 100, 50, 25, 12.5.

For 10 milligrams there is no whole number of half-lives that lands on it exactly. After 30 hours there are 12.5 milligrams, still above 10, and after 36 hours there are 6.25, below it. So the moment comes between 30 and 36 hours. Trying the whole hours in between with M(t) = 400\left(\tfrac{1}{2}\right)^{t/6}:

\begin{aligned} M(31) &= 400\left(\tfrac{1}{2}\right)^{31/6} \approx 11.14 \\ M(32) &= 400\left(\tfrac{1}{2}\right)^{32/6} \approx 9.92 \end{aligned}

The amount is still above 10 milligrams after 31 hours and is first below it after 32 hours. Finding the exact moment needs Logarithms, which turn the unknown exponent into something an equation can solve for.

Example 4

Doubling time from a percent

The number of electric cars registered in a town grows 14\% a year. Between which two whole numbers of years does the number double? Does it double in the same time from any starting year?

Solution

A 14\% growth is the factor b = 1.14 a year. Doubling means the factor over the whole time is 2, so the doubling time is the D with 1.14^D = 2. List the powers:

years n	factor 1.14^n
1	1.14
2	1.2996
3	1.4815
4	1.6890
5	1.9254
6	2.1950

After 5 years the factor is 1.9254, not yet 2; after 6 years it is 2.1950, past 2. So the doubling time is between 5 and 6 years, nearer 5. Logarithms later give it as about 5.29 years.

The factor for any 5 years is 1.14^5, wherever those 5 years start, by Multiplying and dividing powers with the same base. So from every starting year, 5 years fall a little short of doubling and 6 years more than double. Notice the doubling time is not 100 \div 14 \approx 7.1 years, the time it would take if the cars were added at the same rate every year: growth on growth gets there sooner.

Example 5

The factor per minute

The culture from Example 2 doubles every 25 minutes. Find its growth factor per minute and its percent growth per minute, and write the model with that factor.

Solution

The factor per minute m must multiply to 2 over 25 minutes, so m^{25} = 2 and

m = 2^{1/25} \approx 1.0281

A calculator gives 2^{1/25} \approx 1.028114. Since m = 1 + 0.0281, the culture grows about 2.81\% a minute. By Powers of powers, products and quotients, 150 \cdot 2^{t/25} = 150\left(2^{1/25}\right)^t \approx 150(1.0281)^t.

Again, 25 \times 2.81\% \approx 70\% is far short of the 100\% that doubling means. Each minute’s growth is taken of an amount that has already grown, so a small rate repeated many times compounds into a doubling.

Example 6

Reading a half-life from a table

A tank holds a pollutant that breaks down exponentially. Its mass is measured every 4 hours.

t (hours)	mass (grams)
0	3{,}000
4	2{,}121.32
8	1{,}500
12	1{,}060.66
16	750
20	530.33
24	375

Find the half-life, and the mass after two days.

Solution

From 3{,}000 grams at 0 hours, half is 1{,}500 grams, reached at 8 hours. The half-life must be the same from any start, so check another pair: from 2{,}121.32 at 4 hours, half is 1{,}060.66, reached at 12 hours, again 8 hours later. The half-life is 8 hours. (Each 4-hour step multiplies by about 0.7071, and two such steps make 0.7071^2 \approx 0.5.)

Two days is 48 hours, which is 48/8 = 6 half-lives, so the mass is

3000 \times \left(\tfrac{1}{2}\right)^6 = \frac{3000}{64} = 46.875

grams, about 46.9 grams.

Example 7

When two samples hold the same amount

A laboratory has two radioactive samples. Sample A holds 800 milligrams and has a half-life of 3 days. Sample B holds 200 milligrams and has a half-life of 6 days. After how many days do they hold the same amount, and what is that amount?

Solution

The amounts after t days are

\begin{aligned} A(t) &= 800\left(\tfrac{1}{2}\right)^{t/3} \\ B(t) &= 200\left(\tfrac{1}{2}\right)^{t/6} \end{aligned}

A starts larger but halves twice as often, so B should catch up. Setting the two equal and dividing both sides by 200 and by \left(\tfrac{1}{2}\right)^{t/3}, which is never zero,

\begin{aligned} 4 &= \left(\tfrac{1}{2}\right)^{t/6 - t/3} \\ &= \left(\tfrac{1}{2}\right)^{-t/6} = 2^{t/6} \end{aligned}

using the quotient of powers and then \left(\tfrac{1}{2}\right)^{-k} = 2^k. Since 4 = 2^2, and two powers of 2 are equal only when their exponents are equal, t/6 = 2 and t = 12.

After 12 days, Sample A has gone through 12/3 = 4 half-lives and holds 800 \times \tfrac{1}{16} = 50 milligrams, and Sample B has gone through 12/6 = 2 half-lives and holds 200 \times \tfrac{1}{4} = 50 milligrams. They agree at 50 milligrams, and after that B holds more.

Boundaries

Only exponential change has a doubling time. The time to double is the same from every start only when the amount is multiplied by a constant factor. A quantity growing by a fixed amount, such as 100 + 20t, takes 5 units to go from 100 to 200 but 10 more to go from 200 to 400. See Linear, polynomial and exponential growth compared.

Growth has no half-life, and decay has no doubling time. When b > 1 the amount never falls, and when 0 < b < 1 it never rises.

A half-life is not a time to empty. After two half-lives a quarter remains, not nothing. In the model the amount stays positive for every t, approaching 0 without reaching it, as the graph’s horizontal asymptote shows.

Units must match. In 2^{t/D}, t and D are in the same unit. With a doubling time in minutes, one hour enters as t = 60, not t = 1.

Exact times usually need logarithms. Without them, a time can be found exactly only when the target is a whole power of 2 times the start (or a simple fraction of a doubling time); otherwise powers and tables give two whole numbers it lies between.

Real quantities follow the model only for a while. A yeast culture runs out of food and space, and its growth levels off, as Logistic growth: when growth levels off describes. A body clears a medicine at a rate that may change with the dose. The doubling time and half-life describe the model, and the model is a good description only while the factor really stays constant.

Where this goes next

Logarithms answer the question this lesson keeps approaching from both sides: for what exponent does b^D = 2? With them, a doubling time comes out as one number, and so does the time a dose takes to fall to any level. In Building an exponential model from two points, two measurements give the start and the factor, and the doubling time or half-life follows. Graphs of exponential functions shows how the shape in Figures 1 and 2 changes with the base, and interest added several times a year, in Compound interest and compounding periods, has a doubling time of its own. Later, equations involving logarithms bring their own care over which answers are allowed, in solving logarithmic equations.

Exercises

Whole numbers of periods

A 720 milligram sample has a half-life of 4 days. How much is left after 12 days?
A colony of 80 algae cells doubles every 3 hours. How many cells are there after 15 hours?
A quantity halves every 6 hours. How long does it take to fall to \tfrac{1}{8} of its starting amount?
A quantity doubles every 7 years. By what factor does it grow in 35 years?
After how many half-lives is \tfrac{1}{64} of a sample left?

Writing and using a model

Write the amount A(t) left after t days of the sample in Exercise 1, and find the amount after 10 days, to two decimal places.
Write the number B(t) of algae cells after t hours for the colony in Exercise 2, and find the number after 4 hours.
For P(t) = 45 \cdot 2^{t/12}, give the starting amount and the doubling time, and find P(30) to two decimal places.
For Q(t) = 1200\left(\tfrac{1}{2}\right)^{t/30}, give the starting amount and the half-life, and find Q(90).

The factor per unit of time

A quantity doubles every 10 years. Find its yearly growth factor to four decimal places, and its percent growth a year.
A quantity has a half-life of 5 hours. Find its hourly decay factor to four decimal places, and its percent decay an hour.

From a factor or a table

A quantity grows by the factor 1.09 a year. Between which two whole numbers of years is its doubling time? Show the powers you compare.
A quantity decays by the factor 0.9 a year. Between which two whole numbers of years is its half-life? Show the powers you compare.
A quantity decays exponentially. After 0, 3, 6, 9 and 12 days it measures 500, 353.55, 250, 176.78 and 125. Find its half-life, then find the amount after 30 days.
A quantity grows exponentially. After 0, 10, 20, 30 and 40 minutes it measures 48, 72, 108, 162 and 243. Between which two of these times does it first double? Starting instead from t = 10, does it double within the next 20 minutes?

How long until

The sample in Exercise 1 starts at 720 milligrams and has a half-life of 4 days. After how many whole half-lives is it first below 20 milligrams? Give the time in days.
The colony in Exercise 2 starts at 80 and doubles every 3 hours. After how many hours does it reach 10{,}240?
Population A starts at 90 and doubles every 4 hours. Population B starts at 360 and doubles every 6 hours. After how many hours are they equal, and what is the common size?

Reasoning

A quantity starts at 60 and grows by 30 each hour, so it is 60 + 30t after t hours. Find how long it takes to go from 60 to 120, and from 120 to 240. Explain why it has no doubling time in the sense of this lesson.
Imke says that after two half-lives a sample is all gone, because two halves make a whole. Explain what is wrong, and give the fraction left after 2 half-lives and after 10.
Show that if y = a \cdot b^t and b^H = \tfrac{1}{2}, then y(t + H) = \tfrac{1}{2}\,y(t) for every t. Which law of exponents does the argument use?
A quantity has doubling time D. Explain why it takes exactly 2D to become four times as big, and why it takes more than D but less than 2D to become three times as big.

Answers

90 milligrams. 12 days is 3 half-lives, and 720 \times \tfrac{1}{8} = 90.
2{,}560 cells. 15 hours is 5 doublings, and 80 \times 2^5 = 80 \times 32 = 2{,}560.
18 hours. \tfrac{1}{8} = \left(\tfrac{1}{2}\right)^3 is 3 half-lives, and 3 \times 6 = 18.
A factor of 32. 35 years is 5 doubling times, and 2^5 = 32.
6 half-lives, since 2^6 = 64, so \left(\tfrac{1}{2}\right)^6 = \tfrac{1}{64}.
A(t) = 720\left(\tfrac{1}{2}\right)^{t/4}, and A(10) = 720\left(\tfrac{1}{2}\right)^{2.5} \approx 127.28 milligrams.
B(t) = 80 \cdot 2^{t/3}, and B(4) = 80 \cdot 2^{4/3} \approx 201.59, so about 202 cells.
Start 45, doubling time 12. P(30) = 45 \cdot 2^{2.5} \approx 254.56.
Start 1{,}200, half-life 30. Q(90) = 1200 \times \tfrac{1}{8} = 150, since 90 is 3 half-lives.
2^{1/10} \approx 1.0718, a growth of about 7.18\% a year.
\left(\tfrac{1}{2}\right)^{1/5} \approx 0.8706, a decay of about 12.94\% an hour, since 1 - 0.8706 = 0.1294.
Between 8 and 9 years. 1.09^8 \approx 1.9926, just under 2, and 1.09^9 \approx 2.1719, over 2.
Between 6 and 7 years. 0.9^6 = 0.531441, still above \tfrac{1}{2}, and 0.9^7 = 0.4782969, below it.
Half-life 6 days: 500 halves to 250 from t = 0 to t = 6, and 353.55 halves to about 176.78 from t = 3 to t = 9. After 30 days, 5 half-lives, the amount is 500 \div 32 = 15.625.
Between t = 10 and t = 20: doubling 48 means reaching 96, and the amount is 72 at t = 10 and 108 at t = 20. Yes: from 72 at t = 10, double is 144, and the amount is 108 at t = 20 and 162 at t = 30, so it doubles within those 20 minutes, as it must, since every 10 minutes multiplies by 1.5 and 1.5^2 = 2.25 > 2.
After 6 half-lives, which is 24 days. Halving gives 360, 180, 90, 45, 22.5 (after 20 days, still above 20) and then 11.25 (after 24 days).
After 21 hours. 10{,}240 \div 80 = 128 = 2^7, which is 7 doublings, and 7 \times 3 = 21.
After 24 hours, at 5{,}760. Setting 90 \cdot 2^{t/4} = 360 \cdot 2^{t/6} and dividing gives 2^{t/4 - t/6} = 4, so 2^{t/12} = 2^2 and t = 24. Then 360 \cdot 2^4 = 90 \cdot 2^6 = 5{,}760.
From 60 to 120 takes 2 hours, since 60 + 30 \times 2 = 120; from 120 to 240 takes 4 more hours, since 60 + 30 \times 6 = 240. Adding the same 30 each hour is a smaller and smaller fraction of a growing amount, so each doubling takes longer. A doubling time exists only when each hour multiplies by the same factor.
Each half-life removes half of what is left at that moment, not half of the original amount, so the second half-life removes only a quarter of the original. After 2 half-lives \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4} is left, and after 10 half-lives \left(\tfrac{1}{2}\right)^{10} = \tfrac{1}{1024}. In the model some always remains.
y(t + H) = a \cdot b^{t + H}, which is a \cdot b^t \cdot b^H, which is \tfrac{1}{2}\,a\,b^t = \tfrac{1}{2}\,y(t). The step b^{t+H} = b^t \cdot b^H is the product of powers law; nothing depended on t, so the halving happens over every stretch of length H.
Over 2D the amount doubles twice, from any start, so it is multiplied by 2 \times 2 = 4. The amount is always increasing, so it passes 3 times the start after it passes 2 times the start (at D) and before it reaches 4 times the start (at 2D).