Applications

Exponential models in the real world

In this chapter we use exponential functions to describe and solve real-life problems. The focus is not on the algebra of exponentials (covered elsewhere), but on recognizing when a situation is exponential, building the correct formula, and interpreting the results.

Throughout, an exponential model will have the general shape
$$
y = a \cdot b^t
$$
or, in continuous form,
$$
y = a \, e^{kt},
$$
where $t$ is usually time, $a$ is an initial value, and $b$ or $k$ describe the growth or decay rate.

We will look at several key application types:

• Percentage growth and decay
• Compound interest
• Population growth
• Radioactive decay and half‑life
• Depreciation
• Logistic-type constraints (very briefly, at an intuitive level)

You do not need to memorize every formula. What matters is understanding the structure of an exponential situation and choosing a function that matches it.

Recognizing exponential behavior

A situation is typically exponential if:

• The quantity changes by the same percentage in each equal time step, not by the same fixed amount.
• In tables, the ratio $\dfrac{\text{new value}}{\text{old value}}$ is constant from one step to the next.
• In graphs, the curve bends upward for growth, becoming steeper as time increases; for decay, the curve bends downward, flattening out but never reaching zero.

Linear vs. exponential (contrast):

• Linear: “adds 5 every year” → same absolute change.
• Exponential: “increases by 5% every year” → same relative (percentage) change.

Percentage growth and decay models

If a quantity starts at $a$ and changes by a fixed percentage $r$ per time step, the model is:

$$
y = a(1 + r)^t \quad \text{(for growth, } r>0\text{)}
$$

$$
y = a(1 - r)^t \quad \text{(for decay, } 0<r<1\text{)}
$$

Here:

• $a$ = initial amount (value at $t = 0$).
• $r$ = rate per time step, written as a decimal (e.g. $5\% \to r = 0.05$).
• $t$ = number of time steps (e.g. years, months, etc.).
• $1 + r$ or $1 - r$ is called the growth factor or decay factor.

Example structure (without full numeric work):

• “A town’s population is 20,000 and grows by 3% per year.”
• Initial value: $a = 20000$.
• Rate: $r = 0.03$.
• Model: $P(t) = 20000(1.03)^t$, with $t$ in years.
• “A car loses 15% of its value each year after purchase.”
• Initial value: $a$ = price when new.
• Rate: $r = 0.15$ (a decay rate).
• Model: $V(t) = a(0.85)^t$, with $t$ in years.

Compound interest

Compound interest is a key financial application of exponential functions.

Suppose:

• Principal (initial amount): $P$
• Annual interest rate: $r$ (as a decimal)
• Compounded $n$ times per year
• Time: $t$ years

Then the account balance is:

$$
A(t) = P\left(1 + \frac{r}{n}\right)^{nt}.
$$

Interpretation:

• $r/n$ is the interest rate per compounding period.
• $nt$ is the number of compounding periods that have occurred.
• The factor $\left(1 + \dfrac{r}{n}\right)$ is applied once for each period.

Common compounding frequencies:

• Annually: $n = 1$
• Quarterly: $n = 4$
• Monthly: $n = 12$
• Daily (bank convention): $n = 365$ or $360$

Effective annual rate

When compounding more than once per year, the effective annual rate is the actual percentage increase in one year, taking compounding into account.

Set $t = 1$ in the compound interest formula:

$$
A(1) = P\left(1 + \frac{r}{n}\right)^{n}.
$$

Then the effective annual rate $r_{\text{eff}}$ satisfies
$$
1 + r_{\text{eff}} = \left(1 + \frac{r}{n}\right)^{n},
$$
so
$$
r_{\text{eff}} = \left(1 + \frac{r}{n}\right)^{n} - 1.
$$

You can compare different financial offers by comparing their effective annual rates.

Continuous compounding

If interest is compounded continuously, the model is

$$
A(t) = P e^{rt}.
$$

Here:

• $r$ is still the annual interest rate (as a decimal).
• $e^{rt}$ arises as the limit of $\left(1+\dfrac{r}{n}\right)^{nt}$ as $n \to \infty$.

Continuous compounding is mathematically convenient and often used in theoretical finance and calculus contexts, even if real accounts often use monthly or daily compounding.

Population growth models

When a population grows in proportion to its size—each individual has a certain probability per unit time of producing new individuals—the idealized model is exponential.

A basic population model is:

$$
P(t) = P_0 (1 + r)^t \quad \text{(discrete time)}
$$

or

$$
P(t) = P_0 e^{kt} \quad \text{(continuous time)}.
$$

• $P_0$ = initial population.
• $r$ or $k$ = growth rate (positive for growth, negative for decline).

Exponential population growth assumes unlimited resources and no crowding effects. In reality, this is only a good model for limited time spans or early growth phases. Long-term models will often use logistic-type functions (which level off).

Doubling time

For exponential growth, the doubling time is how long it takes for the quantity to double.

Given $P(t) = P_0 (1 + r)^t$, doubling time $T$ satisfies:

$$
2P_0 = P_0 (1 + r)^T.
$$

Cancel $P_0$:

$$
2 = (1 + r)^T.
$$

Then

$$
T = \frac{\ln 2}{\ln (1 + r)}.
$$

For small $r$, a useful approximation is the “rule of 70”:

$$
T \approx \frac{70}{(\text{percentage rate})},
$$

where the percentage rate is $100r$ (for example, if $r = 0.05$, the percentage rate is $5$).

This same idea of doubling time applies broadly to any exponential growth: bacteria, investments, social media followers, etc.

Radioactive decay and half‑life

Radioactive substances decay over time. Each nucleus has some probability per unit time of decaying, independent of others. This leads naturally to exponential decay.

A basic decay model is:

$$
N(t) = N_0 e^{kt}, \quad k < 0,
$$

where:

• $N(t)$ is the amount (or number of nuclei) at time $t$,
• $N_0$ is the initial amount ($t=0$),
• $k$ is a negative constant describing the decay rate.

Half‑life definition and formula

The half‑life $T_{1/2}$ of a substance is the time it takes for the amount to decrease to half of its current value.

By definition:

$$
N(T_{1/2}) = \frac{1}{2} N_0.
$$

Substitute into $N(t) = N_0 e^{kt}$:

$$
\frac{1}{2} N_0 = N_0 e^{k T_{1/2}}.
$$

Cancel $N_0$:

$$
\frac{1}{2} = e^{k T_{1/2}}.
$$

Take natural logarithms:

$$
\ln \frac{1}{2} = k T_{1/2}.
$$

So

$$
k = \frac{\ln \frac{1}{2}}{T_{1/2}} = -\frac{\ln 2}{T_{1/2}}.
$$

Thus the decay model can also be written as:

$$
N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}},
$$

which is often more intuitive: each half‑life multiplies the amount by $\tfrac12$.

Example structure:

• “A substance has a half‑life of 10 years and starts with 80 grams.”
• Model: $N(t) = 80\left(\frac{1}{2}\right)^{t/10}$.

You can then evaluate $N(t)$ for various $t$ to see how it declines.

Depreciation of assets

Depreciation is the loss of value of an asset (such as a car, machine, or computer) over time. Some depreciation is modeled linearly, but many real situations use exponential depreciation, which matches a constant percentage loss each year.

If an asset worth $V_0$ depreciates at a rate of $r$ per year (for example, loses $12\%$ of its value each year), the model is:

$$
V(t) = V_0 (1 - r)^t.
$$

Here:

• $r$ is a decimal: $12\% \to r = 0.12$.
• $(1-r)$ is the annual remaining portion after one year.

You can also analyze:

• How long until the asset’s value reaches a certain threshold.
• The percentage of value lost over a given time period, using the model.

Comparing different exponential scenarios

Different applications often lead to similar-looking functions. You should be able to identify:

• What quantity is being modeled (population, money, mass, value, etc.).
• What time unit $t$ represents.
• What the parameters (such as $a$, $P$, $P_0$, $k$, $r$, $T_{1/2}$) mean in the specific context.

Common tasks:

1. Building a model from words
• Identify the initial amount.
• Decide whether it is growth or decay.
• Identify the rate (percentage per time unit, half-life, doubling time, etc.).
• Choose a form:
• Percentage change per period → $y = a(1 \pm r)^t$.
• Continuous growth/decay → $y = a e^{kt}$.
• Half-life → $y = a \left(\tfrac{1}{2}\right)^{t/T_{1/2}}$.
2. Interpreting parameters

Given $y = 500(1.08)^t$:

• Initial amount: $500$ (value at $t=0$).
• Annual rate: $8\%$ increase.
• $t$: number of years.

Given $N(t) = 120 e^{-0.3 t}$:

• Initial amount: $120$.
• Decay constant: $k = -0.3$ (per unit of $t$).
• You can find a “half-life” from $T_{1/2} = \dfrac{\ln 2}{0.3}$ if desired.

3. Solving for time

Often you know the initial amount, the rate, and a target value, and you want to know how long it takes to reach that target. The pattern is:

• Start with the exponential equation:
  $$
  y = a b^t \quad \text{or} \quad y = a e^{kt}.
  $$
• Divide both sides by $a$.
• Take logarithms to solve for $t$:
  $$
  t = \frac{\ln(y/a)}{\ln b} \quad \text{or} \quad t = \frac{1}{k} \ln\left(\frac{y}{a}\right).
  $$

This “take logs to solve for the exponent” idea is fundamental in exponential applications.

Logistic-type limitations (qualitative)

Pure exponential models assume:

• Unlimited resources,
• No crowding effects,
• Constant growth or decay rates over all time.

In many real-world systems (especially populations, epidemics, and some economic phenomena), these assumptions fail for large times:

• Growth may start exponential, then slow down as resources become limited.
• Decay may be influenced by external factors.

Such systems are often better modeled by logistic functions, which start out behaving like an exponential, then level off toward a maximum value. The detailed algebra of logistic functions belongs to more advanced courses, but you should recognize:

• Exponential growth is often a short- to medium-term approximation.
• Long-term behavior may be bounded or may change rate.

Summary of key exponential application forms

For quick reference:

• Constant percentage growth:
  $$
  y = a(1 + r)^t
  $$
• Constant percentage decay:
  $$
  y = a(1 - r)^t
  $$
• Compound interest ($n$ times per year):
  $$
  A(t) = P\left(1 + \frac{r}{n}\right)^{nt}
  $$
• Continuous compounding / continuous growth or decay:
  $$
  y = a e^{kt}
  $$
• Radioactive decay with half-life $T_{1/2}$:
  $$
  N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}
  $$

In all of these, the exponential function captures the idea that the rate of change is proportional to the current amount, leading to doubling and halving behavior rather than simple addition or subtraction.