Section 5.4 Doubling Time and Half-Life

Objectives

After completing this section, you should be able to do the following.

State the definitions of doubling time and half-life for exponential growth and decay.
Identify the doubling time or half-life of an exponential function written in terms of powers of $2$ or $\frac{1}{2}\text{.}$
Rewrite an exponential function in the form $A \cdot c^{t/T}$ to determine the factor by which a quantity changes over a given time period $T\text{.}$

In this section we introduce two important concepts associated with exponential processes: the doubling time of a process of exponential growth and the half-life of a process of exponential decay.

Subsection The Doubling Time

In Example 5.1.3 we looked at a population of E. coli bacteria that grows exponentially according to the formula

\begin{equation*} f(t) = 2000(1.0353)^t, \end{equation*}

where $t$ is measured in minutes and $f(t)$ is the number of bacteria. We noticed that the population doubles every $20$ minutes:

\begin{equation*} f(20) \approx 4000, \quad f(40) \approx 8000, \quad f(60) \approx 16000. \end{equation*}

For every exponentially increasing function the time needed for the current value to double is constant, and it is called the doubling time.

Definition 5.4.1. Doubling Time.

Let $f(t) = A \cdot b^t\text{,}$ $b \gt 1\text{,}$ be an increasing exponential function. The time needed for the value $f(t)$ to double is called the doubling time of the function $f\text{.}$

Example 5.4.2.

Show algebraically that every increasing exponential quantity $f(t) = A \cdot b^t\text{,}$ $b \gt 1\text{,}$ has a fixed doubling time; that is, it always takes the same amount of time for the quantity $f(t)$ to double.

Solution.

Let $t_d$ be the time after which the initial amount $f(0) = A$ doubles. In other words, after time $t_d\text{,}$ the value $f(t_d) = A \cdot b^{t_d}$ is twice the initial value $A\text{:}$

\begin{equation*} A \cdot b^{t_d} = A \cdot 2. \end{equation*}

Dividing both sides by $A\text{,}$ the time $t_d$ is such that

\begin{equation*} b^{t_d} = 2. \end{equation*}

After another $t_d$ units of time, the quantity doubles again and so

\begin{equation*} f(t_d + t_d) = A \cdot b^{t_d + t_d} = A \cdot b^{t_d} \cdot b^{t_d} = A \cdot 2 \cdot 2. \end{equation*}

In fact, no matter what time $t$ we start from, after $t_d$ units of time the quantity $f(t)$ will double.

It is important to realize that only exponentially increasing quantities have a constant doubling time. Processes modeled by different functions do not, as illustrated below.

Example 5.4.3.

Show that the function $g(t) = t^2 + 4$ does not have a fixed doubling time.

Solution.

The initial value is $g(0) = 4\text{.}$ How much time does it take for the value $4$ to double? We need to find a positive number $t$ such that $g(t) =2 \cdot 4 = 8\text{;}$ that is, to solve the equation

\begin{equation*} t^2 + 4 = 8. \end{equation*}

A solution to this equation is $t=2\text{;}$ that is, it takes $2$ units of time for the initial amount to double. How long does it take for the quantity $g(t)$ to double again from $8$ to $16\text{?}$ The solution to $g(t) = 2 \cdot 8 = 16$ is $t = \sqrt{12} \approx 3.46\text{.}$

So it took $2$ units of time for $g(t)$ to double from $4$ to $8\text{,}$ but $3.46 - 2 = 1.46$ units of time after that to double again from $8$ to $16\text{.}$ The time needed for $g(t)$ to double is not constant, and so $g(t)$ does not have a fixed doubling time.

Example 5.4.4.

You deposit $\$3000$ into a savings account that pays $4.43\%$ interest annually. Let $B = B(t)$ be your balance after $t$ years. As we saw in Section 5.1, the formula for $B(t)$ is

\begin{equation*} B(t) = 3000(1.0443)^t. \end{equation*}

How long will it take for your money to double?

Solution.

The balance function $B(t)$ is an increasing exponential function. To find its doubling time, set $B(t)$ equal to twice its initial value and solve for $t\text{.}$ That is,

\begin{equation*} B(t) = 3000(1.0443)^{t} = 2 \cdot 3000. \end{equation*}

Dividing both sides by $3000$ results in

\begin{equation*} 1.0443^t = 2. \end{equation*}

This equation cannot be solved algebraically as the unknown, $t\text{,}$ is in the exponent. Solving such equations requires logarithms—the topic of Chapter 6.

However, an approximate solution can be found by either graphing the function $y = 1.0443^x$ and then finding the point on the graph where $y = 2\text{,}$ or by evaluating $B(t) = 3000(1.0443)^{t}$ for a few suitable values of $t\text{:}$

$t$	$10$	$15$	$16$	$17$	$32$
$B(t)$	$4627.79$	$5747.78$	$6002.41$	$6268.32$	$12009.64$

This shows that it takes approximately $16$ years for your initial deposit $\$3000$ to double to $\$6000\text{.}$ That is, the doubling time is approximately $16$ years. After approximately another $16$ years, the balance $\$6000$ doubles again, and so on. After $48$ years, your balance will be roughly $\$24000\text{.}$

If the base of an increasing exponential function is written in terms of powers of $2\text{,}$ the doubling time can be found algebraically without logarithms.

Example 5.4.5.

The value $V(t)\text{,}$ in dollars, of an investment $t$ years after an initial amount of $\$2000$ was invested is given by

\begin{equation*} V(t) = 2000 \cdot 2^{t / 10}. \end{equation*}

How much will the investment be worth after $10\text{,}$ $20\text{,}$ and $30$ years? What pattern do you observe unfolding?
What is the doubling time of the investment?
Write the expression for $V(t)$ in the form $A \cdot b^t\text{.}$ Identify the growth factor and the annual percent growth rate.

Solution.

The following gives the value of the investment after $10\text{,}$ $20\text{,}$ and $30$ years:

\begin{align*} V(10) \amp= 2000 \cdot 2^{10 / 10} = 2000 \cdot 2^1 = 4000\\ V(20) \amp= 2000 \cdot 2^{20 / 10} = 2000 \cdot 2^{2} = 8000\\ V(30) \amp= 2000 \cdot 2^{30 / 10} = 2000 \cdot 2^{3} = 16000. \end{align*}

Therefore, the value of the investment doubles every $10$ years.
As we saw in (a), the doubling time of $V(t)$ is $10$ years.
To rewrite $V(t)$ in the form $V(t) = A \cdot b^t\text{,}$ observe that

\begin{equation*} V(t) = 2000 \cdot 2^{t / 10} = 2000 \cdot \left(2^{1 / 10}\right)^t. \end{equation*}

Using a calculator, $2^{1 / 10} \approx 1.072\text{,}$ rounded to three decimal places. Writing in the form $A\cdot b^t\text{,}$

\begin{equation*} V(t) = 2000 \cdot 1.072^t. \end{equation*}

The growth factor $b$ is $1.072\text{,}$ and the annual percent growth rate is $r = b - 1 = 1.072 - 1 = 0.072 = 7.2\%\text{.}$

As the example above illustrates, if an exponential function is written in terms of powers of $2\text{,}$ identifying the doubling time is straightforward.

Result 5.4.6.

For an exponential function $f(t)$ written in the form:

\begin{equation*} f(t) = A \cdot 2^{\frac{t}{T}} \end{equation*}

the doubling time is $T\text{.}$ The growth factor is

\begin{equation*} b = 2^{\frac{1}{T}}. \end{equation*}

Subsection The Half-Life

A decreasing exponential function $f(t) = A \cdot b^t\text{,}$ $b \lt 1\text{,}$ doesn’t have a doubling time—it never doubles as it is decreasing. Instead, for an exponentially decaying quantity, the time needed for the quantity to be reduced by half is constant, and called the half-life. If we denote the half-life by $t_h\text{,}$ then after time $t_h$ the initial value is reduced by half. After another $t_h$ units of time it is halved again, and so on. Half-life is exceptionally important in pharmacology. The leaflets that come with medications always give the half-life—the time needed for half of the medication to be eliminated from your body following a single dose.

Definition 5.4.7. Half-Life.

Let $f(t) = A \cdot b^t\text{,}$ $b \lt 1\text{,}$ be a decreasing exponential function. The time needed for the value $f(t)$ to be reduced by half is called the half-life of the function $f\text{.}$

Similarly, without knowledge of logarithms we cannot algebraically solve for the half-life for an exponential function unless the base of the function is expressed in terms of powers of $\frac{1}{2}\text{.}$

Example 5.4.8.

Let $Q(t) = 100\left(\frac{1}{2}\right)^{t / 5}$ be the amount of a medication in a patient’s bloodstream, in mg, $t$ hours after a single dose of $100$ mg.

How much of the medication is left in the patient’s bloodstream after $5$ hours? $10$ hours? $15$ hours? What pattern do you observe unfolding? What is the half-life of the medication?
Rewrite $Q(t)$ in the form $Q(t) = A \cdot b^t$ for constants $A$ and $b\text{.}$ Give the growth factor and the hourly percent growth rate.

Solution.

We calculate:

\begin{align*} Q(5) \amp= 100\left(\dfrac{1}{2}\right)^{5 / 5} = 100\left(\dfrac{1}{2}\right) = 50\\ Q(10) \amp= 100\left(\dfrac{1}{2}\right)^{10 / 5} = 100\left(\dfrac{1}{2}\right)^{2} = 25\\ Q(15) \amp= 100\left(\dfrac{1}{2}\right)^{15 / 5} = 100\left(\dfrac{1}{2}\right)^{3} = 12.5. \end{align*}

Every $5$ hours the amount of the medication left in the bloodstream is cut in half. Hence, the half-life of the medication is $t_h=5$ hours.
The formula can be alternatively written as $Q(t) = 100\left(\left(\frac{1}{2}\right)^{1 / 5}\right)^t\text{.}$ Using $\left(\frac{1}{2}\right)^{1 / 5} \approx 0.871\text{,}$ rounded to three decimal places, then

\begin{equation*} Q(t) = 100 \cdot 0.871^t. \end{equation*}

The growth factor is $b=0.871\text{.}$ The hourly percent growth rate is $r=-12.9\%$ as $0.871 - 1 = -0.129\text{.}$ We can also say that the hourly percent decay rate is $12.9\%\text{.}$

As the example above illustrates, if an exponential function is written in terms of powers of $\frac{1}{2}\text{,}$ identifying the half-life is straightforward.

Result 5.4.9.

For an exponential function $f(t)$ written in the form:

\begin{equation*} f(t) = A \cdot \left(\frac{1}{2}\right)^{t / T} \end{equation*}

the half-life is $T\text{.}$ The growth factor is

\begin{equation*} b = \left(\frac{1}{2}\right)^{1 / T}. \end{equation*}

Example 5.4.10.

A patient takes a $40$ mg tablet of a common anti-heartburn medication Famotidine. The amount of the medication in the bloodstream, $F(t)\text{,}$ in mg, $t$ hours after the dose, decays exponentially with half-life $3.8$ hours.

Find a formula for $F(t)\text{.}$
Find the growth factor and the percent decay rate.

Solution.

The initial amount is $A = 40$ and the half-life is $T = 3.8$ hours. The formula for $F(t)\text{,}$ in terms of powers of $\frac{1}{2}\text{,}$ is

\begin{equation*} F(t) = 40 \cdot \left(\frac{1}{2}\right)^{t / 3.8}. \end{equation*}
To rewrite $F(t)$ in the form $F(t) = 40 \cdot b^t\text{,}$ notice that:

\begin{equation*} F(t) = 40 \cdot \left( \left(\frac{1}{2}\right)^{1/3.8}\right)^t. \end{equation*}

Hence, the growth factor $b = \left(\frac{1}{2}\right)^{1/3.8} \approx 0.833$ (rounded to three decimal places), so $F(t) = 40(0.833)^t\text{.}$ The percent growth rate is $0.833 - 1 = -0.167 = -16.7\%\text{.}$

Subsection Additional Observations and Examples

Example 5.4.11.

Let $P(t)$ be the population of a village, in the number of people, $t$ years after the village was founded. The population that was $250$ people initially triples every $15$ years.

Find a formula for $P(t)\text{.}$
Write $P(t)$ in the form $P(t) = A \cdot b^t\text{.}$

Solution.

By letting $P(t) = 250 \cdot 3^t\text{,}$ the population is multiplied by the factor of $3$ every year. This is not correct! To have the population multiplied by $3$ every $15$ years, we should instead use

\begin{equation*} P(t) = 250 \cdot 3^{\frac{t}{15}}. \end{equation*}

Observe that $P(15) = 250 \cdot 3^{15/15} = 250 \cdot 3 = 750\text{,}$ $P(30) = 250 \cdot 3^{30/15} = 250 \cdot 3^2 = 750 \cdot 3 = 2250\text{,}$ and so on. This indicates that the population triples every $15$ years, as desired.
We can rewrite $P(t)$ as

\begin{equation*} P(t) = 250 \cdot 3^{t/15} = 250 \cdot (3^{1/15})^t. \end{equation*}

Or, by using $3^{1/15} \approx 1.076\text{,}$ then

\begin{equation*} P(t) = 250 \cdot 1.076^t. \end{equation*}

The previous example gives insight into the following result.

Result 5.4.12.

For an exponential function of the form

\begin{equation*} f(t) = A \cdot c^{t / T}, \end{equation*}

then $c$ is the factor by which the quantity increases or decreases every $T$ units of time.

Example 5.4.13.

Find a formula $V(t) = A \cdot b^t$ for the value of an investment initially worth $1500$ that grows $12\%$ every $6$ years.

Solution.

A quantity grows by $12\%$ when it is multiplied by the factor of $1.12\text{.}$ Hence, $V(t)$ is multiplied by $1.12$ every $6$ years. By the previous result,

\begin{equation*} V(t) = 1500 \cdot 1.12^{t / 6}. \end{equation*}

To get $V(t)$ in the form $V(t) = A \cdot b^t\text{,}$ note that:

\begin{equation*} V(t) = 1500 \cdot 1.12^{t / 6} = 1500 \cdot (1.12^{1/6})^t = 1500(1.0191)^t. \end{equation*}

Exercises Exercises

Finding the Doubling Time.

For each exponential function, find the initial amount (in units of $y$) and the doubling time (in units of $t$).

1.

$\displaystyle y=47\cdot 2^{\frac{t}{5}}$

Solution.

initial amount: $47\text{;}$ doubling time: $5$

2.

$\displaystyle y=800\cdot 2^{\frac{t}{9}}$

Solution.

initial amount: $800\text{;}$ doubling time: $9$

3.

$\displaystyle y=89\cdot 4^{\frac{t}{6}}$

Solution.

initial amount: $89\text{;}$ doubling time: $3$

4.

$\displaystyle y=120\cdot 8^{\frac{t}{12}}$

Solution.

initial amount: $120\text{;}$ doubling time: $4$

Finding the Half-Life.

For each exponential function, find the initial amount (in units of $y$) and the half-life (in units of $t$).

5.

$\displaystyle y=75\cdot \left(\frac{1}{2}\right)^{\frac{t}{7}}$

Solution.

initial amount: $75\text{;}$ half-life: $7$

6.

$\displaystyle y=920\cdot \left(\frac{1}{2}\right)^{\frac{t}{10}}$

Solution.

initial amount: $920\text{;}$ half-life: $10$

7.

$\displaystyle y=875\cdot \left(\frac{1}{16}\right)^{\frac{t}{12}}$

Solution.

initial amount: $875\text{;}$ half-life: $3$

8.

$\displaystyle y=63\cdot \left(\frac{1}{4}\right)^{\frac{t}{10}}$

Solution.

initial amount: $63\text{;}$ half-life: $5$

Doubling Time & Half-Life Formulas.

For each of the following, write a formula for an exponential function in the form $\displaystyle f(t)=A\cdot c^{\frac{t}{T}}$ that describes the given scenario. Then convert the function to the form $f(t)=A\cdot b^t\text{.}$

9.

The population of a town begins with $1500$ people at $t=0$ and doubles every $8$ years.

Solution.

$f(t)=1500(2)^{\frac{t}{8}}=1500(1.0905)^t$

10.

The population of a town begins with $750$ people at $t=0$ and triples every $9$ years.

Solution.

$f(t)=750(3)^{\frac{t}{9}}=750(1.1298)^t$

11.

The initial amount of $50$ mg of a radioactive element is halved every $11$ days.

Solution.

$f(t)=50\left(\frac{1}{2}\right)^{\frac{t}{11}}=50(0.9389)^t$

12.

The initial amount of $60$ mg of a radioactive element is cut by one-third every $5$ months.

Solution.

$f(t)=60\left(\frac{1}{3}\right)^{\frac{t}{5}}=60(0.8027)^t$

Estimating Doubling Time & Half-Life.

For each of the following, estimate the half-life or the doubling time whichever applies.

13.

$t$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$f(t)$	$100.52$	$126.65$	$159.57$	$201.04$	$253.29$	$319.13$	$402.08$	$506.59$

Solution.

doubling time: $3$

14.

$t$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$g(t)$	$20.15$	$14.25$	$10.08$	$7.12$	$5.04$	$3.56$	$2.52$	$1.78$

Solution.

half-life: approximately $2$

15.

The graph of an exponential function below shows the population of bacteria, $P(t)\text{,}$ in a laboratory experiment $t$ minutes after the experiment began.

(a)

What is the initial amount, $P_0\text{,}$ of bacteria?

Solution.

$2000$ bacteria

(b)

Estimate the doubling time of the population.

Solution.

$20$ minutes

(c)

Write a formula for $P(t)$ in the form $\displaystyle P(t)=P_0\cdot 2^{\frac{t}{T}}\text{.}$ Then rewrite the formula in the form $\displaystyle P(t)=P_0\cdot b^t\text{.}$ Round off $b$ to four decimal places. What is the percent growth rate?

Solution.

$P(t) = 2000(2)^{\frac{t}{20}} \approx 2000(1.0353)^t\text{;}$ growth rate: $3.53\%$

16.

The amount of caffeine remaining in the body, $C=C(t)\text{,}$ in milligrams, $t$ hours after drinking a cup of coffee, is an exponential function and its graph is given below:

(a)

Estimate the amount of caffeine, $C_0\text{,}$ absorbed by the body from a cup of coffee.

Solution.

$96$ mg

(b)

Estimate the half-life of caffeine.

Solution.

$5$ hours

(c)

Write a formula for $C(t)$ in the form $\displaystyle C(t)=C_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{T}}\text{.}$ Rewrite the formula in the form $\displaystyle C(t)=C_0\cdot b^t\text{.}$ Round $b$ off to three decimal places. What is the percent growth rate?

Solution.

$C(t) = 96\left(\frac{1}{2}\right)^{\frac{t}{5}} \approx 96(0.8706)^t\text{;}$ growth rate: $-12.94\%$

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\(t\)	\(10\)	\(15\)	\(16\)	\(17\)	\(32\)
\(B(t)\)	\(4627.79\)	\(5747.78\)	\(6002.41\)	\(6268.32\)	\(12009.64\)

\(t\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)
\(f(t)\)	\(100.52\)	\(126.65\)	\(159.57\)	\(201.04\)	\(253.29\)	\(319.13\)	\(402.08\)	\(506.59\)

\(t\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)
\(g(t)\)	\(20.15\)	\(14.25\)	\(10.08\)	\(7.12\)	\(5.04\)	\(3.56\)	\(2.52\)	\(1.78\)