03 Exponential Functions

A structured guide to modeling growth and decay, compound interest, logarithmic solutions, and the assumptions behind exponential models.

An exponential relationship describes repeated multiplication rather than repeated addition. The general form is

f(x)=ab^x,

where $a$ is the initial value and $b$ is the base. The base must satisfy $b>0$ and $b\ne 1$ .

If $b>1$ , the function increases and represents growth.
If $0<b<1$ , the function decreases and represents decay.
The domain is all real numbers, and for $a>0$ the range is the set of positive real numbers.
The $y$ -intercept is $(0,a)$ , because $b^0=1$ .
The is $y=0$ when there is no vertical translation.

For example, $f(x)=3(2)^x$ doubles whenever $x$ increases by $1$ . In contrast, $g(x)=80(0.75)^x$ retains $75\%$ of its previous value during each interval, so it decreases by $25\%$ .

The key distinction from a linear model is that a linear function changes by a constant amount, while an changes by a constant percentage or multiplicative factor.

Takeaway: Inspect the base to identify the behavior, then interpret the coefficient as the starting value.

Discrete Growth and Decay

For changes that occur at separate, equal intervals, use a discrete model. A growth rate $r$ is converted from a percentage to a decimal and added to $1$ :

A(t)=A_0(1+r)^t.

For decay, subtract the decimal rate from $1$ :

A(t)=A_0(1-r)^t.

For example, a population of $12{,}000$ growing by $2.5\%$ per year is modeled by

P(t)=12{,}000(1.025)^t.

After $10$ years,

P(10)=12{,}000(1.025)^{10}\approx 15{,}361.

A car purchased for $\$28{,}000$ that loses $15\%$ of its value each year is modeled by

V(t)=28{,}000(0.85)^t.

After $4$ years,

V(4)=28{,}000(0.85)^4\approx \$14{,}573.

The model assumes that the same percentage rate continues throughout the interval being considered.

Takeaway: Growth uses a factor of $1+r$ , while decay uses a factor of $1-r$ .

Continuous Growth and Decay

Some quantities change continuously rather than at separate time intervals. The continuous model is

A(t)=A_0e^{kt},

where $k>0$ indicates growth and $k<0$ indicates decay. The rate is proportional to the amount currently present, so a larger quantity changes more rapidly while the relative rate remains constant.

For a substance with initial mass $500$ grams that decays continuously at $8\%$ per hour,

M(t)=500e^{-0.08t}.

After $6$ hours,

M(6)=500e^{-0.48}\approx 309.2,

so approximately $309.2$ grams remain.

Two useful time measures follow from the continuous model. For growth, the doubling time is

t_d=\frac{\ln 2}{k},\qquad k>0.

For decay, the is

t_{1/2}=\frac{\ln 2}{\lvert k\rvert},\qquad k<0.

Takeaway: Use $e^{kt}$ when the process is continuous and the rate is proportional to the amount present.

is an exponential process because each compounding period multiplies the current balance by the same factor. When interest is compounded $n$ times per year, use

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

where $P$ is the principal, $r$ is the annual rate as a decimal, $n$ is the number of compounding periods per year, and $t$ is measured in years.

Common compounding frequencies include $n=1$ for annually, $n=2$ for semiannually, $n=4$ for quarterly, $n=12$ for monthly, and approximately $n=365$ for daily compounding. Continuous compounding uses

A(t)=Pe^{rt}.

Example: for a deposit of $\$2{,}500$ at an annual rate of $4.8\%$ , compounded monthly for $3$ years,

A(3)=2{,}500\left(1+\frac{0.048}{12}\right)^{12(3)}

A(3)=2{,}500(1.004)^{36}\approx \$2{,}887.37.

The interest earned is approximately

\$2{,}887.37-\$2{,}500=\$387.37.

To find the time required to reach a target balance, solve the compound-interest equation with logarithms:

t=\frac{\ln(A/P)}{n\ln\left(1+r/n\right)}.

Takeaway: Match the rate, compounding frequency, and time units carefully before substituting.

Solving Exponential Equations

There are two main strategies for solving an equation with an unknown exponent.

If both sides can be expressed with the same base, use the one-to-one property:

b^u=b^v\Longrightarrow u=v,

provided $b>0$ and $b\ne 1$ . For example,

2^{x+1}=16=2^4

leads to

x+1=4,

so $x=3$ .

When a common base is not convenient, take logarithms. For

5^x=42,

taking natural logarithms gives

\ln(5^x)=\ln(42),

then the power rule gives

x\ln 5=\ln 42,

x=\frac{\ln 42}{\ln 5}\approx 2.321.

In general,

b^x=c\Longrightarrow x=\log_b(c)=\frac{\ln c}{\ln b}.

Some equations require substitution first. For

4^x-5(2^x)+4=0,

use $4^x=(2^x)^2$ and let $u=2^x$ . Then

u^2-5u+4=0=(u-1)(u-4),

so $u=1$ or $u=4$ . Substituting back gives $x=0$ or $x=2$ . Check candidates in the original equation, especially after substitutions or other transformations.

Takeaway: Match bases when possible; otherwise use logarithms to isolate the exponent.

Modeling and Interpreting Results

To build and interpret an exponential model, follow a consistent sequence:

Identify the initial amount $A_0$ .
Decide whether the quantity grows or decays.
Convert the percentage rate to a decimal.
Choose a discrete model, $A(t)=A_0(1+r)^t$ or $A(t)=A_0(1-r)^t$ , or a continuous model, $A(t)=A_0e^{kt}$ .
Substitute the given values and solve for the requested quantity.
Report units and round appropriately.
Check whether the constant-rate assumption is reasonable for the time interval.

Example: a bacterial culture begins with $800$ bacteria and grows by $12\%$ per hour. To find when it reaches $2{,}000$ bacteria, write

2{,}000=800(1.12)^t.

After dividing by $800$ and taking natural logarithms,

t=\frac{\ln(2.5)}{\ln(1.12)}\approx 7.72.

The culture reaches the target after approximately $7.72$ hours under the stated model.

Exponential models may be reliable over a limited interval but unrealistic indefinitely. A population can encounter limits such as food or space, an investment rate can change, and a substance can be affected by multiple removal processes. Compare predictions with observations and state the assumptions behind the model.

Takeaway: A numerical result is meaningful only when its units, rounding, and modeling assumptions fit the situation.

Exponential Function

Discrete Growth and Decay

Continuous Growth and Decay

Compound Interest

Solving Exponential Equations

Modeling and Interpreting Results