Exponential Modeling

In earlier courses, you learned the shape and behavior of exponential functions. This page focuses on the practical skill of building exponential models from real data — determining the equation, extracting meaningful quantities like doubling time and half-life, and converting between different forms of the exponential function.

Two Standard Forms

Exponential models come in two common forms:

These are interchangeable. If $b = e^k$, then $ab^x = ae^{kx}$. The conversion is:

$k = \ln b \qquad \text{and} \qquad b = e^k$

Building a Model from Two Points

Given two data points $(x_1, y_1)$ and $(x_2, y_2)$, you can determine $a$ and $b$ (or $a$ and $k$).

Method for $y = ab^x$

1. Write two equations: $y_1 = ab^{x_1}$ and $y_2 = ab^{x_2}$
2. Divide to eliminate $a$: $\frac{y_2}{y_1} = b^{x_2 - x_1}$
3. Solve for $b$: $b = \left(\frac{y_2}{y_1}\right)^{1/(x_2 - x_1)}$
4. Substitute back to find $a$: $a = \frac{y_1}{b^{x_1}}$

Example 1: A bacterial culture has 500 cells at $t = 0$ hours and 4000 cells at $t = 3$ hours. Find the exponential model.

Given: $(0, 500)$ and $(3, 4000)$.

Step 1: $500 = ab^0 = a$, so $a = 500$.

Step 2: $4000 = 500 \cdot b^3$

Step 3: $b^3 = 8$, so $b = 2$.

Model: $y = 500 \cdot 2^t$ (the population doubles every hour).

Continuous form: $k = \ln 2 \approx 0.693$, so $y = 500e^{0.693t}$.

Example 2: A sample of a radioactive isotope has 120 grams at year 0 and 45 grams at year 10. Find the exponential decay model.

Given: $(0, 120)$ and $(10, 45)$.

From $(0, 120)$: $a = 120$.

$45 = 120 \cdot b^{10} \implies b^{10} = \frac{45}{120} = 0.375$

$b = (0.375)^{1/10} = 0.375^{0.1} \approx 0.9066$

Model: $y = 120(0.9066)^t$

Continuous form: $k = \ln(0.9066) \approx -0.0981$, so $y = 120e^{-0.0981t}$.

The negative value of $k$ confirms decay (the quantity is decreasing).

When Neither Point Has $x = 0$

If neither data point has $x = 0$, the division method still works.

Example 3: A population is 2000 at $t = 5$ and 7000 at $t = 12$. Find the model.

$b^{12-5} = \frac{7000}{2000} = 3.5$

$b = 3.5^{1/7} \approx 1.1960$

Find $a$: $2000 = a(1.1960)^5 \implies a = \frac{2000}{1.1960^5} = \frac{2000}{2.448} \approx 817.0$

Model: $y \approx 817.0(1.1960)^t$

Doubling Time

The doubling time $T_d$ is the time it takes for an exponential quantity to double. For $y = ae^{kx}$:

$ae^{kT_d} = 2a \implies e^{kT_d} = 2 \implies T_d = \frac{\ln 2}{k}$

For $y = ab^x$:

$T_d = \frac{\ln 2}{\ln b}$

Example 4: The bacterial culture from Example 1 has $k = 0.693$. Its doubling time is:

$T_d = \frac{\ln 2}{0.693} = \frac{0.693}{0.693} = 1 \text{ hour}$

This confirms what we saw: $b = 2$ means the population doubles every 1 unit of time.

Example 5: A city’s population grows at a continuous rate of $k = 0.032$ per year. How long until the population doubles?

$T_d = \frac{\ln 2}{0.032} = \frac{0.693}{0.032} \approx 21.7 \text{ years}$

Half-Life

The half-life $T_{1/2}$ is the time for an exponential quantity to decrease to half its value. For $y = ae^{kx}$ (with $k < 0$):

$ae^{kT_{1/2}} = \frac{a}{2} \implies e^{kT_{1/2}} = \frac{1}{2} \implies T_{1/2} = \frac{\ln(1/2)}{k} = \frac{-\ln 2}{k}$

Since $k$ is negative for decay, $T_{1/2}$ is positive.

Example 6: From Example 2, $k \approx -0.0981$. The half-life is:

$T_{1/2} = \frac{-\ln 2}{-0.0981} = \frac{0.693}{0.0981} \approx 7.07 \text{ years}$

Verification: Starting with 120 g, after 7.07 years: $120e^{-0.0981 \times 7.07} = 120e^{-0.694} = 120 \times 0.500 = 60$ g. Exactly half.

Converting Between Growth Rate and Continuous Rate

The growth rate $r$ (as a percentage) and the continuous rate $k$ are related but not identical:

• If the annual growth rate is $r$ (as a decimal), then $b = 1 + r$ and $k = \ln(1 + r)$
• If the continuous rate is $k$, then $r = e^k - 1$

For small $r$, $k \approx r$. But for larger rates, the difference matters.

Example 7: An investment grows at 8 percent per year. What is the continuous growth rate?

$k = \ln(1.08) = 0.07696$

The continuous rate is about 7.7 percent, slightly less than the 8 percent annual rate. This is because continuous compounding gets a “head start” on reinvesting growth throughout the year.

Example 8: A radioactive substance has a continuous decay rate of $k = -0.05$. What fraction remains after one year?

$b = e^{-0.05} = 0.9512$

So about 95.12 percent remains after one year, meaning the annual decay rate is about 4.88 percent.

Real-World Application: Carbon-14 Dating

Carbon-14 has a half-life of 5730 years. Archaeologists use this to date organic material.

The continuous decay rate is:

$k = \frac{-\ln 2}{5730} = \frac{-0.6931}{5730} \approx -0.0001210 \text{ per year}$

Model: $N(t) = N_0 e^{-0.0001210 t}$, where $N_0$ is the original amount.

If a bone fragment has 35 percent of its original carbon-14, how old is it?

$0.35 N_0 = N_0 e^{-0.0001210 t}$

$0.35 = e^{-0.0001210 t}$

$\ln(0.35) = -0.0001210 t$

$t = \frac{\ln(0.35)}{-0.0001210} = \frac{-1.0498}{-0.0001210} \approx 8676 \text{ years}$

The bone is approximately 8,676 years old.

Recognizing Exponential vs. Non-Exponential Data

Not all growing data is exponential. To test whether data is approximately exponential, check for a constant ratio between consecutive $y$-values (assuming equally spaced $x$-values):

Constant ratio of 1.50 confirms exponential growth with $b = 1.5$.

If the ratios are not approximately constant, the data is not exponential — it might be linear, quadratic, logistic, or something else entirely.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The two standard forms $y = ab^x$ and $y = ae^{kx}$ are interchangeable via $k = \ln b$
• To build a model from two points, divide the equations to eliminate $a$, solve for $b$, then back-substitute for $a$
• Doubling time $= \frac{\ln 2}{k}$ and half-life $= \frac{-\ln 2}{k}$ (where $k < 0$ for decay)
• Growth rate $r$ and continuous rate $k$ are close for small values but diverge for larger ones: $k = \ln(1 + r)$
• Check for constant ratios in equally spaced data to confirm exponential behavior

Return to College Algebra for more topics in this section.