Exponential and Logarithmic Applications

In Grade 11 you built growth and decay models and used them to predict amounts. The hardest question, “when will it reach this amount?”, needed guess and check. With logarithms you can now answer it exactly: how long until money doubles, how old a fossil is, or when a cup of tea is cool enough to drink.

Key ideas

The models

Situation	Model
Growth or decay at rate $r$ per period	$A = A_0(1 + r)^t$ or $A = A_0(1 - r)^t$
Doubling time $D$	$A = A_0(2)^{\frac{t}{D}}$
Half-life $H$	$A = A_0\left(\tfrac{1}{2}\right)^{\frac{t}{H}}$
Compound interest, rate $i$ per period	$A = P(1 + i)^n$
Cooling toward room temperature $T_s$	$T = T_s + (T_0 - T_s)\,b^t$, with $0 \lt b \lt 1$

In the cooling model, it’s the difference between the object’s temperature and the room’s that decays exponentially.

Solving for the time

1. Substitute the known values.
2. Isolate the power: divide by the coefficient (and, for cooling, subtract $T_s$ first).
3. Take the log of both sides and use the power law to bring the exponent down.
4. Solve, then round in a way that makes sense in context.

Finding a rate or a half-life

If you know two amounts and the time between them, you can solve for the unknown in the base or the exponent. To find a rate, take a root: $b^8 = 1.2$ gives $b = 1.2^{\frac{1}{8}}$. To find a half-life or doubling time, use logs.

Rounding in context

• Interest is added at the end of each compounding period, so round up to a whole number of periods.
• For continuous processes (decay, cooling, population), give a sensible decimal, and say what it means.
• Keep full calculator values until the end.

Reading from a graph

A graph (drawn by hand or with technology such as Desmos) lets you estimate the answer by finding where the curve meets a horizontal line. Use the graph to estimate, then use logarithms to get the exact value. Example 4 does both.

Worked examples

Example 1: Half-life

Iodine-131, used in medical treatments, has a half-life of about $8$ days. How long does it take a $50$ mg sample to decay to $5$ mg?

Solution. The model is $A = 50\left(\tfrac{1}{2}\right)^{\frac{t}{8}}$, with $t$ in days.

$$\begin{aligned} 50\left(\tfrac{1}{2}\right)^{\frac{t}{8}} &= 5 \\ \left(\tfrac{1}{2}\right)^{\frac{t}{8}} &= 0.1 && \text{divide by } 50 \\ \tfrac{t}{8}\log 0.5 &= \log 0.1 && \text{take logs} \\ t &= \frac{8\log 0.1}{\log 0.5} \approx 26.6 \end{aligned}$$

It takes about $26.6$ days.

Sense check: after $3$ half-lives ($24$ days) there are $6.25$ mg left, and after $4$ ($32$ days) there are $3.125$ mg. So $5$ mg should happen between $24$ and $32$ days. ✓

Example 2: Doubling time

A city’s population is growing at $2.4\%$ per year. How long will it take to double?

Solution. The starting population doesn’t matter. Call it $P_0$; we want $2P_0$:

$$\begin{aligned} P_0(1.024)^t &= 2P_0 \\ 1.024^t &= 2 && \text{divide by } P_0 \\ t &= \frac{\log 2}{\log 1.024} \approx 29.2 \end{aligned}$$

At this rate the population doubles in about $29.2$ years.

Example 3: Finding a half-life from data

A patient takes a $400$ mg dose of a medication. Three hours later, $290$ mg remains in the body. Find the half-life of the medication, to one decimal place.

Solution. Use $A = 400\left(\tfrac{1}{2}\right)^{\frac{t}{H}}$ and substitute $t = 3$, $A = 290$:

$$\begin{aligned} 400\left(\tfrac{1}{2}\right)^{\frac{3}{H}} &= 290 \\ \left(\tfrac{1}{2}\right)^{\frac{3}{H}} &= 0.725 \\ \tfrac{3}{H}\log 0.5 &= \log 0.725 \\ \frac{3}{H} &= \frac{\log 0.725}{\log 0.5} \approx 0.4640 \\ H &\approx \frac{3}{0.4640} \approx 6.5 \end{aligned}$$

The half-life is about $6.5$ hours.

Check (with the unrounded value $H \approx 6.466$): $400\left(\tfrac{1}{2}\right)^{\frac{3}{6.466}} \approx 290$. ✓ Also, $290$ mg is more than half of $400$ mg after $3$ hours, so the half-life should be more than $3$ hours. ✓

Example 4: Cooling, from a graph and with logs

A cup of tea is poured at $90\,^\circ\text{C}$ in a $21\,^\circ\text{C}$ room. Its temperature after $t$ minutes is

$$T = 21 + 69(0.94)^t$$

Use the graph to estimate when the tea reaches $60\,^\circ\text{C}$, then find the time algebraically.

The curve crosses $T = 60$ a little after $t = 9$ minutes.

Solution. From the graph, the curve crosses $T = 60$ at about $t = 9$ minutes.

Algebraically, subtract the room temperature first, then divide:

$$\begin{aligned} 21 + 69(0.94)^t &= 60 \\ 69(0.94)^t &= 39 && \text{subtract } 21 \\ 0.94^t &= \tfrac{39}{69} && \text{divide by } 69 \\ t &= \frac{\log\left(\frac{39}{69}\right)}{\log 0.94} \approx 9.2 \end{aligned}$$

The tea reaches $60\,^\circ\text{C}$ after about $9.2$ minutes, which agrees with the graph. Notice that the tea never gets below $21\,^\circ\text{C}$ in this model: the asymptote is the room temperature.

Common mistakes

Taking logs before isolating the power. In $21 + 69(0.94)^t = 60$, you can’t take the log of each term. Subtract $21$ and divide by $69$ first, so the equation has the form $b^t = \text{number}$.

Using the wrong growth or decay factor. A $2.4\%$ increase is a factor of $1.024$, not $1.24$ or $2.4$. A $15\%$ decrease is a factor of $0.85$, not $0.15$.

Flipping the half-life exponent. In $A_0\left(\tfrac{1}{2}\right)^{\frac{t}{H}}$, the time $t$ goes on top. After one half-life, $t = H$ and the exponent is $1$, which is what you want.

Rounding the number of compounding periods down. If the equation gives $n \approx 39.8$ quarters, the money isn’t there until the end of the $40$th quarter. Round up, and convert to years if asked.

Rounding in the middle. Rounding $\log 0.725$ or $\tfrac{3}{H}$ too early can change the last digit of the answer. Keep the calculator values and round once, at the end.

Practice

1. (Warm-up) How many years does it take an investment to double at $5\%$ per year, compounded annually?

Solution

$$1.05^n = 2 \quad\Rightarrow\quad n = \frac{\log 2}{\log 1.05} \approx 14.2$$

Interest is added once a year, so it takes $15$ years.

2. (Warm-up) A substance has a half-life of $12$ hours. How long does it take $80$ g to decay to $10$ g? (Hint: you don’t need logs for this one.)

Solution

$80 \to 40 \to 20 \to 10$ is $3$ half-lives, so it takes $3 \times 12 = 36$ hours. (With logs: $\left(\tfrac{1}{2}\right)^{\frac{t}{12}} = \tfrac{1}{8}$ gives $\tfrac{t}{12} = 3$.)

3. (Core) A bacteria culture starts with $500$ cells and doubles every $25$ minutes. When will there be $20\,000$ cells?

Solution

$$500(2)^{\frac{t}{25}} = 20\,000 \quad\Rightarrow\quad 2^{\frac{t}{25}} = 40 \quad\Rightarrow\quad t = \frac{25\log 40}{\log 2} \approx 133$$

After about $133$ minutes (about $2$ hours and $13$ minutes).

4. (Core) Mateo invests $3500 at $3.6\%$ per year, compounded quarterly. How long until it’s worth at least $5000?

Solution

The quarterly rate is $\dfrac{0.036}{4} = 0.009$:

$$3500(1.009)^n = 5000 \quad\Rightarrow\quad 1.009^n = \tfrac{10}{7} \quad\Rightarrow\quad n = \frac{\log\left(\frac{10}{7}\right)}{\log 1.009} \approx 39.8$$

Round up to $40$ quarters, which is $10$ years. Check: $3500(1.009)^{40} \approx$ $5008.58, and $3500(1.009)^{39} \approx$ $4963.91. ✓

5. (Core) A new car costs $32 000 and loses $15\%$ of its value each year. When will it be worth $10 000?

Solution

$$32\,000(0.85)^t = 10\,000 \quad\Rightarrow\quad 0.85^t = 0.3125 \quad\Rightarrow\quad t = \frac{\log 0.3125}{\log 0.85} \approx 7.2$$

After about $7.2$ years. Check: after $7$ years it’s worth about $10 258, and after $8$ years about $8720. ✓

6. (Core) Carbon-14 has a half-life of about $5730$ years. A bone fragment found at a dig in Ontario has $35\%$ of the carbon-14 it had when the animal died. Estimate the age of the bone.

Solution

$$\left(\tfrac{1}{2}\right)^{\frac{t}{5730}} = 0.35 \quad\Rightarrow\quad t = \frac{5730\log 0.35}{\log 0.5} \approx 8679$$

The bone is about $8700$ years old. Sense check: $35\%$ is between $50\%$ ($1$ half-life) and $25\%$ ($2$ half-lives), so the age should be between $5730$ and $11\,460$ years. ✓

7. (Core) A town had $18\,000$ people in 2015 and $21\,500$ in 2023. Assume exponential growth.

• (a) Find the annual growth rate, as a percent to two decimal places.
• (b) In what year will the population reach $30\,000$?

Solution

(a) Let $b$ be the growth factor: $18\,000\,b^8 = 21\,500$, so

$$b = \left(\frac{21\,500}{18\,000}\right)^{\frac{1}{8}} \approx 1.02246$$

The growth rate is about $2.25\%$ per year.

(b) Using the unrounded $b$, with $t$ in years after 2015:

$$18\,000\,b^t = 30\,000 \quad\Rightarrow\quad b^t = \tfrac{5}{3} \quad\Rightarrow\quad t = \frac{\log\left(\frac{5}{3}\right)}{\log b} \approx 23.0$$

The population reaches $30\,000$ in about $2015 + 23 = 2038$.

8. (Challenge) A bowl of soup at $85\,^\circ\text{C}$ is left in a $20\,^\circ\text{C}$ kitchen, so $T = 20 + 65\,b^t$, with $t$ in minutes. After $10$ minutes it has cooled to $50\,^\circ\text{C}$.

• (a) Find $b$, to four decimal places.
• (b) When will the soup reach $30\,^\circ\text{C}$?

Solution

(a) $20 + 65\,b^{10} = 50$, so $b^{10} = \tfrac{30}{65}$ and

$$b = \left(\tfrac{30}{65}\right)^{\frac{1}{10}} \approx 0.9256$$

(b) $20 + 65\,b^t = 30$, so $b^t = \tfrac{10}{65}$. Take logs, using $\log b = \tfrac{1}{10}\log\left(\tfrac{30}{65}\right)$:

$$t = \frac{\log\left(\frac{10}{65}\right)}{\log b} = \frac{10\log\left(\frac{10}{65}\right)}{\log\left(\frac{30}{65}\right)} \approx 24.2$$

After about $24.2$ minutes.

9. (Challenge) Aisha invests $4000 at $5\%$ per year and Ben invests $5000 at $3\%$ per year, both compounded annually. After how many years will Aisha’s investment be worth more than Ben’s?

Solution

$$\begin{aligned} 4000(1.05)^t &= 5000(1.03)^t \\ \left(\frac{1.05}{1.03}\right)^t &= 1.25 && \text{divide both sides by } 4000(1.03)^t \\ t &= \frac{\log 1.25}{\log\left(\frac{1.05}{1.03}\right)} \approx 11.6 \end{aligned}$$

Interest is added at the end of each year, so Aisha’s investment is worth more after $12$ years. Check: after $12$ years, Aisha has about $7183.43 and Ben about $7128.80. After $11$ years, Aisha has about $6841.36 and Ben about $6921.17. ✓