Math Core

Now you have every tool you need: exponential functions to describe change, and logarithms to solve for the time it takes. This lesson puts them together on real questions. How long until an investment doubles? How old is a sample? What model fits two measurements, and what does it predict?

Two forms of the same model

An exponential model can be written in either of two forms:

y = a\,b^{t} \qquad\text{or}\qquad y = a\,e^{kt}.

They describe the same kinds of curves. Since $e^{kt} = \left(e^k\right)^t$ , the two are linked by $b = e^k$ , or equivalently $k = \ln b$ . Use whichever matches the information you're given: percent rates per period suggest $b$ ; continuous rates suggest $k$ .

When the unknown is the time, you'll almost always finish by taking a logarithm.

Solving an exponential model for time

Substitute the known values and isolate the exponential expression.
Take $\ln$ of both sides.
Solve for $t$ . Check that the answer is reasonable in context.

Doubling time

The doubling time is how long it takes a growing quantity to double. It doesn't depend on the starting amount: set $y = 2a$ and the $a$ 's cancel.

Worked example: How long to double?

You invest money at $5\%$ interest. How long does it take to double if interest is compounded (a) annually and (b) continuously?

Solution.

(a) Solve $2a = a(1.05)^t$ . Divide by $a$ : $1.05^t = 2$ . Then

t = \frac{\ln 2}{\ln 1.05} \approx \frac{0.693147}{0.048790} \approx 14.21 \text{ years}.

(b) Solve $2a = ae^{0.05t}$ , so $e^{0.05t} = 2$ and $0.05t = \ln 2$ :

t = \frac{\ln 2}{0.05} \approx 13.86 \text{ years}.

Continuous compounding doubles the money about four months sooner.

Tip

The rule of 70 estimates doubling time: divide $70$ by the percent rate. At $5\%$ , that's $70 \div 5 = 14$ years, close to both answers above. It works because $\ln 2 \approx 0.693$ , and $\dfrac{\ln 2}{r} \approx \dfrac{70}{100r}$ for small $r$ .

Half-life

The half-life $h$ of a decaying substance is the time for half of it to disappear. The model is $A = A_0\left(\tfrac{1}{2}\right)^{t/h}$ , where $A_0$ is the starting amount.

Worked example: Radioactive decay

A radioactive isotope used in medicine has a half-life of $8$ days. How long until a $100$ mg sample decays to $15$ mg?

Solution. Set up the model and isolate the power:

15 = 100\left(\tfrac{1}{2}\right)^{t/8} \;\Longrightarrow\; \left(\tfrac{1}{2}\right)^{t/8} = 0.15.

Take $\ln$ of both sides:

\frac{t}{8}\ln 0.5 = \ln 0.15 \;\Longrightarrow\; t = \frac{8\ln 0.15}{\ln 0.5} \approx \frac{8(-1.897120)}{-0.693147} \approx 21.90 \text{ days}.

Reasonableness check: after $2$ half-lives ( $16$ days) there are $25$ mg; after $3$ ( $24$ days), $12.5$ mg. $15$ mg falls between, and so does $21.9$ days.

Building a model from data

Given two data points, you can find the model and then use it to predict. If the first point is at $t = 0$ , you have $a$ immediately; the second point gives the growth factor.

Worked example: Fit and predict

A city had $12{,}000$ residents in 2015 and $15{,}000$ in 2020. Assume continuous exponential growth.

Find a model $P = ae^{kt}$ , with $t$ in years since 2015.
Predict the population in 2030.
In what year does the model predict the population will reach $30{,}000$ ?

Solution.

At $t = 0$ , $P = 12{,}000$ , so $a = 12{,}000$ . Then $15{,}000 = 12{,}000e^{5k}$ gives $e^{5k} = 1.25$ and

k = \frac{\ln 1.25}{5} \approx 0.04463.

The model is $P \approx 12{,}000e^{0.04463t}$ : continuous growth of about $4.46\%$ per year.

In 2030, $t = 15$ . Since $e^{5k} = 1.25$ exactly, $e^{15k} = 1.25^3$ , so

P = 12{,}000(1.25)^3 = 12{,}000(1.953125) = 23{,}437.5.

The model predicts about $23{,}438$ residents.

Solve $30{,}000 = 12{,}000e^{kt}$ : $e^{kt} = 2.5$ , so $t = \dfrac{\ln 2.5}{k} = \dfrac{5\ln 2.5}{\ln 1.25} \approx 20.53$ years after 2015. The population reaches $30{,}000$ during 2035.

Common mistake

Don't round $k$ too early. Using $k = 0.04$ instead of $0.04463$ in part 3 would give $\dfrac{\ln 2.5}{0.04} \approx 22.9$ years, off by more than two years. Keep $k$ exact (as $\tfrac{\ln 1.25}{5}$ ) or store it in your calculator.

Newton's law of cooling

A hot object cools quickly at first and then more slowly as its temperature approaches the room's. Newton's law of cooling models this as exponential decay of the difference between the object and the room:

T = T_{\text{room}} + (T_0 - T_{\text{room}})e^{-kt}.

Coffee cooling from 90 °C toward a room temperature of 20 °C (dashed asymptote).Open in grapher →

Worked example: Cooling coffee

A cup of coffee at $90^\circ\text{C}$ is placed in a $20^\circ\text{C}$ room. After $10$ minutes it is $60^\circ\text{C}$ . How long until it cools to $40^\circ\text{C}$ ?

Solution. The model is $T = 20 + 70e^{-kt}$ . Use the data point to find $k$ :

60 = 20 + 70e^{-10k} \;\Longrightarrow\; e^{-10k} = \frac{40}{70} = \frac{4}{7} \;\Longrightarrow\; k = \frac{\ln(7/4)}{10} \approx 0.05596.

Now solve for the time when $T = 40$ :

40 = 20 + 70e^{-kt} \;\Longrightarrow\; e^{-kt} = \frac{2}{7} \;\Longrightarrow\; t = \frac{\ln 3.5}{k} = \frac{10\ln 3.5}{\ln 1.75} \approx 22.39 \text{ minutes}.

Practice

For rounded answers, round to $2$ decimal places unless the problem says otherwise.

Practice 1

A quantity starts at $a$ and is cut in half every $3$ years. Which model gives the amount after $t$ years?

Practice 2

How many years does it take an investment to double at $7\%$ interest compounded annually?