4.5 Modeling with Exponential and Logarithmic Functions
4.5.1 Learning Objectives
By the end of this section, you should be able to:
- construct exponential growth and decay models;
- use doubling time and half-life;
- apply Newton’s Law of Cooling;
- interpret logistic growth; and
- use logarithmic scales for acidity, sound, and earthquake intensity.
4.5.2 Exponential Growth and Decay
A quantity changing at a rate proportional to its current amount can be modelled by
\[ P(t)=P_0e^{kt}, \]
where \(P_0\) is the initial amount, \(t\) is time, and \(k\) is the continuous growth rate. Growth has \(k>0\), while decay has \(k<0\).
Example 4.20 Building a Growth Model
A culture begins with \(800\) cells and doubles every \(6\) hours. Find a model and predict the number after \(15\) hours.
Solution
Using the doubling-time form,
\[ P(t)=800(2)^{t/6}. \]
Then
\[ P(15)=800(2)^{2.5}\approx4525. \]
\[ \boxed{P(t)=800(2)^{t/6};\quad P(15)\approx4525\text{ cells}} \]
Exercise 4.20 A population begins at \(1200\) and grows continuously at \(3\%\) per year. Write the model.
Show answer
Answer: \(\boxed{P(t)=1200e^{0.03t}}\), where \(t\) is measured in years.
4.5.3 Doubling Time and Half-Life
For continuous growth \(P=P_0e^{kt}\), the doubling time is
\[ t_d=\frac{\ln2}{k}. \]
If a decaying quantity has half-life \(h\), then
\[ A(t)=A_0\left(\frac12\right)^{t/h}. \]
Example 4.21 Using Half-Life
A sample begins with \(240\text{ mg}\) and has a half-life of \(12\) days. How much remains after \(30\) days?
Solution
\[ A(t)=240\left(\frac12\right)^{t/12}. \]
Therefore,
\[ A(30)=240\left(\frac12\right)^{2.5}\approx42.43. \]
\[ \boxed{42.43\text{ mg}} \]
Exercise 4.21 A quantity grows continuously at \(5\%\) per year. Find its doubling time.
Show answer
Answer: \(\boxed{t_d=\ln2/0.05\approx13.86\text{ years}}\)
4.5.4 Newton’s Law of Cooling
When an object is placed in surroundings with constant temperature \(T_s\), its temperature can be modelled by
\[ T(t)=T_s+(T_0-T_s)e^{-kt}, \]
where \(T_0\) is its initial temperature and \(k>0\) is a cooling constant.
Example 4.22 Predicting a Temperature
A drink at \(90^\circ\text{C}\) is placed in a \(20^\circ\text{C}\) room. Its cooling constant is \(k=0.04\) per minute. Find its temperature after \(25\) minutes.
Solution
\[ \begin{aligned} T(25)&=20+(90-20)e^{-0.04(25)}\\ &=20+70e^{-1}\\ &\approx45.75. \end{aligned} \]
\[ \boxed{T(25)\approx45.75^\circ\text{C}} \]
Exercise 4.22 An object at \(10^\circ\text{C}\) warms in a \(22^\circ\text{C}\) room according to \(T(t)=22-12e^{-0.08t}\). Find \(T(10)\).
Show answer
Answer: \(\boxed{T(10)=22-12e^{-0.8}\approx16.61^\circ\text{C}}\)
4.5.5 Logistic Growth
Unlimited exponential growth is often unrealistic. A logistic model
\[ P(t)=\frac{L}{1+Ae^{-kt}} \]
begins with approximately exponential growth but levels off near the carrying capacity \(L\). The graph is S-shaped.
Example 4.23 Interpreting a Logistic Model
For
\[ P(t)=\frac{5000}{1+9e^{-0.6t}}, \]
identify the carrying capacity and initial population.
Solution
The carrying capacity is the numerator \(L=5000\). At \(t=0\),
\[ P(0)=\frac{5000}{1+9}=500. \]
\[ \boxed{\text{Carrying capacity }5000;\quad P(0)=500} \]
Exercise 4.23 For \(P(t)=\dfrac{12000}{1+5e^{-0.2t}}\), identify the carrying capacity.
Show answer
Answer: \(\boxed{12000}\)
4.5.6 Logarithmic Scales
Logarithmic scales compress wide ranges of positive values.
The pH of a solution with hydrogen-ion concentration \(H\), in moles per litre, is
\[ \text{pH}=-\log H. \]
Sound intensity level is
\[ \beta=10\log\left(\frac I{I_0}\right), \]
where \(I\) is intensity and \(I_0\) is a reference intensity. A difference of \(d\) units on a base-\(10\) logarithmic scale corresponds to an intensity ratio of \(10^d\).
Example 4.24 Calculating pH
Find the pH of a solution with \(H=3.2\times10^{-5}\) moles per litre.
Solution
\[ \text{pH}=-\log(3.2\times10^{-5})\approx4.49. \]
\[ \boxed{\text{pH}\approx4.49} \]
Exercise 4.24 Two earthquakes differ in magnitude by \(1.7\). Find the ratio of their intensities.
Show answer
Answer: The ratio is \(10^{1.7}\approx\boxed{50.1}\).
4.5.7 Conceptual Takeaways
- Exponential change depends on the current amount.
- Doubling time and half-life describe repeated multiplicative change.
- Newton’s Law models movement toward the surrounding temperature.
- Logistic growth includes a limiting carrying capacity.
- Equal differences on a logarithmic scale represent equal ratios.
4.5.8 Skills You Should Be Able to Do
- Build and evaluate exponential growth and decay models.
- Calculate doubling time and half-life quantities.
- Apply and interpret Newton’s Law of Cooling.
- Identify features of logistic models.
- Calculate and interpret values on logarithmic scales.
4.5.9 Practice Problems with Solutions
A population starts at \(2500\) and grows continuously at \(2.4\%\) per year. Write a model.
Show Solution
With \(P_0=2500\) and \(k=0.024\),
\[ \boxed{P(t)=2500e^{0.024t}}. \]
Use the model in Problem 1 to predict the population after \(12\) years.
Show Solution
\[ P(12)=2500e^{0.024(12)}\approx3334. \]
\[ \boxed{3334\text{ people, approximately}} \]
A sample has initial mass \(500\text{ mg}\) and half-life \(18\) days. Find the mass after \(45\) days.
Show Solution
\[ A(45)=500\left(\frac12\right)^{45/18} =500\left(\frac12\right)^{2.5} \approx88.39. \]
\[ \boxed{88.39\text{ mg}} \]
A population doubles every \(9\) years. How long will it take to become five times its initial size?
Show Solution
Using \(P/P_0=2^{t/9}\),
\[ 5=2^{t/9}. \]
Taking logarithms gives
\[ t=9\frac{\ln5}{\ln2}\approx20.90. \]
\[ \boxed{20.90\text{ years, approximately}} \]
A liquid at \(80^\circ\text{C}\) cools in a \(20^\circ\text{C}\) room according to \(T(t)=20+60e^{-0.05t}\). Find its temperature after \(30\) minutes.
Show Solution
\[ T(30)=20+60e^{-1.5}\approx33.39. \]
\[ \boxed{33.39^\circ\text{C}} \]
For \(P(t)=\dfrac{8000}{1+15e^{-0.4t}}\), find the initial population and carrying capacity.
Show Solution
\[ P(0)=\frac{8000}{1+15}=500. \]
The limiting value is \(8000\). Therefore,
\[ \boxed{P(0)=500;\quad \text{carrying capacity }8000}. \]
A sound has intensity \(10^6\) times the reference intensity. Find its sound level.
Show Solution
\[ \beta=10\log(10^6)=10(6)=60. \]
\[ \boxed{60\text{ dB}} \]
A solution has pH \(6.2\). Find its hydrogen-ion concentration.
Show Solution
From \(6.2=-\log H\),
\[ \log H=-6.2, \]
so
\[ H=10^{-6.2}\approx6.31\times10^{-7}. \]
\[ \boxed{H\approx6.31\times10^{-7}\text{ mol/L}} \]