Exponential and logarithmic models are used when a quantity changes by a percentage, approaches a limiting
value, decays by a fixed fraction, or is measured on a logarithmic scale.
1. General modeling idea
Many real-world problems follow an exponential pattern:
\[
\text{new amount}=\text{starting amount}\times \text{growth factor}
\]
When the unknown is in the exponent, logarithms are used to solve the equation.
\[
b^x=M
\]
\[
x=\log_b(M)
\]
2. Newton’s Law of Cooling
Newton’s Law of Cooling models the temperature of an object moving toward the surrounding temperature.
\[
T(t)=T_s+(T_0-T_s)e^{-kt}
\]
Here:
- \(T_0\) is the initial temperature.
- \(T_s\) is the surrounding temperature.
- \(k\) is a positive cooling constant.
- \(t\) is time.
- \(T(t)\) is the temperature after time \(t\).
To solve for time, isolate the exponential part:
\[
\frac{T-T_s}{T_0-T_s}=e^{-kt}
\]
Then take the natural logarithm:
\[
t=-\frac{1}{k}\ln\left(\frac{T-T_s}{T_0-T_s}\right)
\]
3. Cooling example
Suppose an object cools from \(90^\circ\) to \(40^\circ\) in a room at \(20^\circ\), with \(k=0.08\).
\[
T(t)=20+(90-20)e^{-0.08t}
\]
Set \(T(t)=40\):
\[
40=20+70e^{-0.08t}
\]
\[
\frac{20}{70}=e^{-0.08t}
\]
\[
t=-\frac{1}{0.08}\ln\left(\frac{20}{70}\right)
\]
\[
t\approx15.66
\]
4. Population growth and decay
Continuous population growth or decay is modeled by:
\[
P(t)=P_0e^{rt}
\]
Here, \(P_0\) is the initial population, \(r\) is the continuous growth rate, and \(t\) is time.
If \(r>0\), the model shows growth.
\[
r>0
\]
If \(r<0\), the model shows decay.
\[
r<0
\]
To solve for time:
\[
t=\frac{\ln(P/P_0)}{r}
\]
To solve for rate:
\[
r=\frac{\ln(P/P_0)}{t}
\]
5. Radioactive decay
Radioactive decay is often modeled with half-life:
\[
N(t)=N_0\left(\frac{1}{2}\right)^{t/h}
\]
Here, \(N_0\) is the initial amount, \(h\) is the half-life, and \(N(t)\) is the remaining amount.
To solve for time:
\[
t=h\frac{\ln(N/N_0)}{\ln(1/2)}
\]
To solve for half-life:
\[
h=t\frac{\ln(1/2)}{\ln(N/N_0)}
\]
6. Radioactive example
Suppose \(100\) grams of a substance has half-life \(5\) days. Find the time for \(12.5\) grams to remain.
\[
12.5=100\left(\frac{1}{2}\right)^{t/5}
\]
\[
\frac{12.5}{100}=\left(\frac{1}{2}\right)^{t/5}
\]
\[
0.125=\left(\frac{1}{2}\right)^{t/5}
\]
Since \(0.125=1/8=(1/2)^3\), we get:
\[
\frac{t}{5}=3
\]
\[
t=15
\]
7. Sound intensity and decibels
Sound intensity level is logarithmic:
\[
L=10\log_{10}\left(\frac{I}{I_0}\right)
\]
Here, \(I/I_0\) is the intensity ratio. To reverse the formula:
\[
\frac{I}{I_0}=10^{L/10}
\]
If \(I/I_0=10000\), then:
\[
L=10\log_{10}(10000)
\]
\[
L=10\cdot4
\]
\[
L=40\text{ dB}
\]
8. Choosing the correct model
| Situation |
Model |
Main idea |
| Cooling or warming toward surroundings |
\(T(t)=T_s+(T_0-T_s)e^{-kt}\) |
The difference \(T-T_s\) decays exponentially. |
| Population growth or decay |
\(P(t)=P_0e^{rt}\) |
The rate is proportional to the current population. |
| Radioactive decay |
\(N(t)=N_0(1/2)^{t/h}\) |
Every half-life leaves half the amount. |
| Sound intensity |
\(L=10\log_{10}(I/I_0)\) |
A multiplicative intensity ratio becomes an additive dB value. |
9. Common mistakes
- Do not use a linear model when the change is percentage-based.
- When the unknown is in an exponent, use logarithms.
- For cooling, the target temperature must be between the starting temperature and the surrounding temperature.
- For radioactive decay, the remaining amount must be positive and cannot exceed the initial amount.
- For decibels, the intensity ratio must be positive.
- Enter percent rates carefully: \(3.2\%\) means \(r=0.032\).
10. Modeling workflow
A good modeling solution usually follows these steps:
- Identify the scenario.
- Choose the formula.
- Assign the known values to the correct symbols.
- Substitute carefully.
- Use logarithms if the unknown is in the exponent.
- Check that the result makes sense in the real-world context.
