The Central Idea
A quantity that changes by a constant percentage each period follows an exponential model.
If \(A_0\) is the initial amount and \(b\) is the multiplier applied each period, then after \(t\) periods,
\[ \boxed{A(t)=A_0b^t}. \]Growth occurs when
\[ b>1, \]while decay occurs when
\[ 0<b<1. \]From Percent Change to Multiplier
If a quantity increases by rate \(r\), expressed as a decimal, the growth multiplier is
\[ \boxed{1+r}. \]If a quantity decreases by rate \(r\), the decay multiplier is
\[ \boxed{1-r}. \]Example: Growth Multiplier
A population grows by \(6\%\) per year.
The annual multiplier is
\[ 1+0.06=1.06. \]Therefore,
\[ \boxed{b=1.06}. \]Example: Decay Multiplier
A vehicle depreciates by \(18\%\) per year.
The annual multiplier is
\[ 1-0.18=0.82. \]Therefore,
\[ \boxed{b=0.82}. \]The Basic Growth Model
If an initial quantity \(A_0\) grows by rate \(r\) each period, then
\[ \boxed{ A(t)=A_0(1+r)^t }. \]The rate \(r\) must be written as a decimal.
The Basic Decay Model
If an initial quantity \(A_0\) decreases by rate \(r\) each period, then
\[ \boxed{ A(t)=A_0(1-r)^t }. \]Example: Exponential Growth
A population begins at \(12{,}000\) and grows by \(4\%\) per year. Find the population after \(5\) years.
\[ A(t)=12000(1.04)^t. \] \[ A(5)=12000(1.04)^5. \] \[ A(5)\approx14599.84. \]Therefore, the model predicts approximately
\[ \boxed{14{,}600\text{ people}}. \]Example: Depreciation
A machine is worth \(\$40{,}000\) and depreciates by \(15\%\) per year. Find its modeled value after \(4\) years.
\[ V(t)=40000(0.85)^t. \] \[ V(4)=40000(0.85)^4. \] \[ V(4)\approx20880.25. \]Therefore,
\[ \boxed{\$20{,}880.25\text{ approximately}}. \]Why Exponential Decay Is Not Linear Subtraction
A \(15\%\) annual depreciation does not subtract the same dollar amount each year.
Instead, each year keeps
\[ 85\% \]of the previous year's value.
Therefore, the dollar decrease becomes smaller as the asset value falls.
Repeated Percentage Change
Suppose a quantity increases by \(10\%\) for three consecutive periods.
The total multiplier is
\[ (1.10)^3. \]It is not
\[ 1+0.30. \]Repeated percentage change compounds multiplicatively.
Example: Repeated Growth
A quantity begins at \(500\) and grows by \(10\%\) per period for \(3\) periods.
\[ A=500(1.10)^3. \] \[ A=500(1.331). \] \[ A=665.5. \]Therefore,
\[ \boxed{665.5}. \]Successive Percentage Changes Need Not Cancel
An increase of \(20\%\) followed by a decrease of \(20\%\) does not return a quantity to its starting value.
The combined multiplier is
\[ 1.20(0.80)=0.96. \]Therefore, the final value is
\[ \boxed{96\%} \]of the original.
Population Models
A simple population model with constant percentage growth has the form
\[ \boxed{ P(t)=P_0(1+r)^t }. \]Here \(P_0\) is the initial population and \(r\) is the growth rate per time period.
Example: Population Projection
A town has population \(25{,}000\) and grows by \(2.5\%\) annually. Estimate the population after \(8\) years.
\[ P(8)=25000(1.025)^8. \] \[ P(8)\approx30460.07. \]Therefore,
\[ \boxed{30{,}460\text{ people approximately}}. \]Population Decline
If a population decreases by a constant percentage each year,
\[ \boxed{ P(t)=P_0(1-r)^t }. \]Example: Population Decline
A population of \(80{,}000\) declines by \(3\%\) annually. Estimate the population after \(6\) years.
\[ P(6)=80000(0.97)^6. \] \[ P(6)\approx66637.76. \]Therefore,
\[ \boxed{66{,}638\text{ people approximately}}. \]Depreciation Models
If an asset loses a constant fraction \(r\) of its value each period, then
\[ \boxed{ V(t)=V_0(1-r)^t }. \]This model is often called declining-balance depreciation.
Example: Vehicle Depreciation
A vehicle costs \(\$32{,}000\) and loses \(20\%\) of its value each year.
\[ V(t)=32000(0.80)^t. \]After \(3\) years,
\[ V(3)=32000(0.80)^3. \] \[ V(3)=16384. \]Therefore,
\[ \boxed{\$16{,}384}. \]Value Lost versus Value Remaining
If an asset depreciates by \(20\%\), then it retains
\[ 80\% \]of its value.
The model uses the retained multiplier
\[ \boxed{0.80}, \]not the lost fraction \(0.20\) as the exponential base.
Finding the Growth Rate from Two Values
Suppose
\[ A(t)=A_0(1+r)^t. \]If \(A_0\), \(A(t)\), and \(t\) are known, then
\[ \frac{A(t)}{A_0}=(1+r)^t. \]Therefore,
\[ \boxed{ r= \left( \frac{A(t)}{A_0} \right)^{1/t}-1 }. \]Example: Find a Population Growth Rate
A population grows from \(20{,}000\) to \(24{,}200\) in \(2\) years. Find the annual growth rate.
\[ r= \left( \frac{24200}{20000} \right)^{1/2}-1. \] \[ r=(1.21)^{1/2}-1. \] \[ r=1.10-1=0.10. \]Therefore,
\[ \boxed{10\%\text{ per year}}. \]Finding the Decay Rate
For
\[ A(t)=A_0(1-r)^t, \]divide by \(A_0\):
\[ \frac{A(t)}{A_0}=(1-r)^t. \]Therefore,
\[ \boxed{ r= 1- \left( \frac{A(t)}{A_0} \right)^{1/t} }. \]Example: Find a Depreciation Rate
A machine declines from \(\$50{,}000\) to \(\$32{,}000\) in \(2\) years under constant percentage depreciation.
\[ r= 1- \left( \frac{32000}{50000} \right)^{1/2}. \] \[ r=1-\sqrt{0.64}. \] \[ r=1-0.80=0.20. \]Therefore,
\[ \boxed{20\%\text{ per year}}. \]Finding Time in an Exponential Model
If
\[ A=A_0b^t, \]then
\[ \frac{A}{A_0}=b^t. \]Take logarithms:
\[ \ln\left(\frac{A}{A_0}\right)=t\ln b. \]Therefore,
\[ \boxed{ t= \frac{ \ln(A/A_0) }{ \ln b } }. \]Example: Time to Reach a Population
A population begins at \(10{,}000\) and grows by \(5\%\) annually. How long until it reaches \(15{,}000\)?
\[ 15000=10000(1.05)^t. \] \[ 1.5=(1.05)^t. \] \[ t= \frac{\ln1.5}{\ln1.05}. \] \[ t\approx8.31. \]Therefore,
\[ \boxed{8.31\text{ years approximately}}. \]Doubling Time
Doubling occurs when
\[ A=2A_0. \]For
\[ A=A_0(1+r)^t, \]we obtain
\[ 2=(1+r)^t. \]Therefore,
\[ \boxed{ t_d= \frac{\ln2}{\ln(1+r)} }. \]Example: Doubling Time
Find the doubling time for a population growing at \(6\%\) per year.
\[ t_d= \frac{\ln2}{\ln1.06}. \] \[ t_d\approx11.90. \]Therefore,
\[ \boxed{11.9\text{ years approximately}}. \]Half-Life Structure
Half-life is the time required for a decaying quantity to reach one-half of its original value.
If
\[ A=A_0b^t, \qquad 0<b<1, \]then half-life \(t_h\) satisfies
\[ \frac12=b^{t_h}. \]Therefore,
\[ \boxed{ t_h= \frac{\ln(1/2)}{\ln b} }. \]Example: Half-Life from a Decay Rate
A quantity decreases by \(12\%\) per year.
The annual multiplier is
\[ b=0.88. \]Therefore,
\[ t_h= \frac{\ln(0.5)}{\ln(0.88)}. \] \[ t_h\approx5.42. \]Thus,
\[ \boxed{5.42\text{ years approximately}}. \]Half-Life Model
If the half-life \(h\) is known directly, an especially convenient model is
\[ \boxed{ A(t)=A_0\left(\frac12\right)^{t/h} }. \]Every increase of \(h\) time units multiplies the amount by \(1/2\).
Example: Use a Half-Life Model
A quantity begins at \(800\) units and has half-life \(6\) years. Find the amount after \(18\) years.
\[ A(18) = 800 \left(\frac12\right)^{18/6}. \] \[ A(18)=800\left(\frac12\right)^3. \] \[ A(18)=100. \]Therefore,
\[ \boxed{100\text{ units}}. \]Continuous Growth and Decay
Some models use the continuous exponential form
\[ \boxed{ A(t)=A_0e^{kt} }. \]Growth occurs when
\[ k>0, \]while decay occurs when
\[ k<0. \]Example: Continuous Population Growth
A population begins at \(15{,}000\) and follows
\[ P(t)=15000e^{0.03t}. \]After \(10\) years,
\[ P(10)=15000e^{0.30}. \] \[ P(10)\approx20248.82. \]Therefore,
\[ \boxed{20{,}249\text{ people approximately}}. \]Continuous Doubling Time
If
\[ A=A_0e^{kt}, \]then doubling time satisfies
\[ 2=e^{kt}. \]Therefore,
\[ \boxed{ t_d=\frac{\ln2}{k} }. \]Continuous Half-Life
For continuous decay,
\[ A=A_0e^{kt}, \qquad k<0, \]half-life satisfies
\[ \frac12=e^{kt_h}. \]Therefore,
\[ \boxed{ t_h=\frac{\ln(1/2)}{k} }. \]Discrete versus Continuous Models
Discrete percentage growth uses
\[ A=A_0(1+r)^t. \]Continuous growth uses
\[ A=A_0e^{kt}. \]These are different parameterizations.
Do not substitute a discrete annual rate \(r\) directly for the continuous constant \(k\) without conversion.
Equivalent Discrete and Continuous Rates
If the two models produce the same one-period multiplier, then
\[ 1+r=e^k. \]Therefore,
\[ \boxed{k=\ln(1+r)} \]and
\[ \boxed{r=e^k-1}. \]Example: Convert a Discrete Rate to Continuous Form
A \(5\%\) annual growth multiplier is
\[ 1.05. \]The equivalent continuous parameter is
\[ k=\ln(1.05). \] \[ k\approx0.04879. \]Therefore, a discrete \(5\%\) annual increase corresponds to continuous growth constant
\[ \boxed{k\approx0.04879}. \]Building a Model from Two Data Points
Suppose an exponential model has form
\[ A(t)=A_0b^t. \]If \(A_0\) is known and a later value \(A(T)\) is known, then
\[ b^T=\frac{A(T)}{A_0}. \]Therefore,
\[ \boxed{ b= \left( \frac{A(T)}{A_0} \right)^{1/T} }. \]Example: Fit an Exponential Model
A population is \(8{,}000\) initially and \(10{,}648\) after \(3\) years.
Let
\[ P(t)=8000b^t. \]Then
\[ 10648=8000b^3. \] \[ b^3=1.331. \] \[ b=1.10. \]Therefore,
\[ \boxed{P(t)=8000(1.10)^t}. \]Interpreting the Initial Value
In
\[ A(t)=A_0b^t, \]setting \(t=0\) gives
\[ A(0)=A_0. \]Therefore, \(A_0\) is the model's starting amount.
Interpreting the Base
In
\[ A(t)=A_0b^t, \]the base \(b\) is the multiplier applied during each period.
If
\[ b=1.08, \]the quantity grows by \(8\%\) per period.
If
\[ b=0.92, \]it decreases by \(8\%\) per period.
Recovering the Percent Rate from the Base
If
\[ b>1, \]then the growth rate is
\[ \boxed{r=b-1}. \]If
\[ 0<b<1, \]then the decay rate is
\[ \boxed{r=1-b}. \]Linear versus Exponential Change
Linear change adds a constant amount:
\[ A(t)=A_0+dt. \]Exponential change multiplies by a constant factor:
\[ A(t)=A_0b^t. \]Constant difference suggests a linear model.
Constant ratio suggests an exponential model.
Example: Recognize Exponential Data
Consider the values
\[ 100,\quad120,\quad144,\quad172.8. \]Successive ratios are
\[ \frac{120}{100} = \frac{144}{120} = \frac{172.8}{144} = 1.2. \]Therefore, the data show \(20\%\) exponential growth per period.
Example: Recognize Linear Data
Consider
\[ 100,\quad120,\quad140,\quad160. \]Successive differences are all
\[ 20. \]Therefore, the data show linear rather than exponential growth.
Threshold Problems
Many applications ask when a quantity exceeds or falls below a target.
Write an exponential inequality such as
\[ A_0b^t\ge T \]or
\[ A_0b^t\le T. \]Solve with logarithms, then interpret any discrete-time restrictions.
Example: Minimum Whole Number of Years
A population of \(5{,}000\) grows \(8\%\) annually. When will it first exceed \(8{,}000\)?
\[ 5000(1.08)^t>8000. \] \[ (1.08)^t>1.6. \] \[ t> \frac{\ln1.6}{\ln1.08}. \] \[ t>6.11. \]If the model is checked only at whole-year intervals, the first whole year exceeding the target is
\[ \boxed{7\text{ years}}. \]Depreciation to a Target Value
Suppose an asset follows
\[ V(t)=V_0b^t, \qquad 0<b<1. \]To find when its value falls to \(V_t\), use
\[ \boxed{ t= \frac{\ln(V_t/V_0)}{\ln b} }. \]Example: Time to a Target Value
A \(\$30{,}000\) asset retains \(85\%\) of its value each year. When will it fall below \(\$10{,}000\)?
\[ 30000(0.85)^t<10000. \] \[ (0.85)^t<\frac13. \]Solving,
\[ t> \frac{\ln(1/3)}{\ln(0.85)} \approx6.76. \]Therefore, after whole annual depreciation steps, it first falls below the target after
\[ \boxed{7\text{ years}}. \]Population Models Have Limits
Constant percentage growth cannot continue indefinitely in many real populations.
Resources, migration, policy, disease, environmental conditions, and carrying capacity can change the rate.
Exponential population models are most useful over intervals where a roughly constant percentage rate is a reasonable approximation.
Depreciation Models Have Limits
Real asset values may not follow one constant percentage indefinitely.
Salvage value, market conditions, maintenance, age, and obsolescence can change the pattern.
The model should be used within the range where its assumptions are plausible.
Checking Growth Models
For positive growth rate,
\[ A(t+1)>A(t). \]The ratio
\[ \frac{A(t+1)}{A(t)} \]should equal the growth multiplier.
Checking Decay Models
For a positive decay rate less than \(100\%\),
\[ 0<A(t+1)<A(t). \]A model that produces negative values from an ordinary positive exponential-decay setup has been constructed incorrectly.
Checking the Initial Condition
Substitute
\[ t=0. \]A valid model should return
\[ \boxed{A(0)=A_0}. \]Common Error: Adding the Percent Each Period
A \(5\%\) annual increase does not mean adding \(0.05\) units each year.
It means multiplying by
\[ \boxed{1.05} \]each year.
Common Error: Using the Decay Rate as the Base
A \(20\%\) annual depreciation uses multiplier
\[ \boxed{0.80}, \]not
\[ 0.20. \]Common Error: Confusing Percentage Points and Percent Change
A quantity falling from \(80\) to \(60\) decreases by \(20\) units, but the percent decrease is
\[ \frac{20}{80}=0.25. \]Therefore, the decrease is
\[ \boxed{25\%}. \]Common Error: Assuming Equal Percentage Changes Cancel
An increase of \(10\%\) followed by a decrease of \(10\%\) gives multiplier
\[ 1.10(0.90)=0.99. \]Therefore, the final amount is \(99\%\) of the original, not \(100\%\).
Common Error: Using Linear Interpolation on Exponential Change
Exponential models change by ratios, not equal differences.
A midpoint in time does not generally correspond to the arithmetic mean of the endpoint values.
Common Error: Ignoring the Time Unit
If a growth rate is annual, the exponent counts years.
A monthly rate requires months or an equivalent conversion.
Always match the exponent's unit to the period associated with the multiplier.
Common Error: Rounding Too Early
Exponential calculations repeatedly magnify rounding error.
Keep extra digits during intermediate steps and round only the final reported result.
Common Error: Extending a Model Too Far
A model that fits several years of data may become unrealistic over much longer periods.
Algebra can extend a formula indefinitely, but interpretation must respect the assumptions of the application.
A Reliable Procedure for Growth Models
- Identify the initial value \(A_0\).
- Convert the growth percent to decimal \(r\).
- Form the multiplier \(1+r\).
- Match the time variable to the rate period.
- Write \(A(t)=A_0(1+r)^t\).
- Evaluate or solve for the requested quantity.
- Check that the result grows in the expected direction.
A Reliable Procedure for Depreciation and Decay
- Identify the initial value \(A_0\).
- Convert the decay percent to decimal \(r\).
- Form the retained multiplier \(1-r\).
- Write \(A(t)=A_0(1-r)^t\).
- Use logarithms if time is unknown.
- Check that the modeled value stays positive and decreases appropriately.
A Reliable Procedure for Fitting an Exponential Model
- Choose a form \(A(t)=A_0b^t\).
- Use the initial data value for \(A_0\).
- Substitute a second data point.
- Solve for \(b\) using a root.
- Convert \(b\) to a percent growth or decay rate.
- Substitute the data points back into the model.
Practice 1: Growth Model
A population begins at \(18{,}000\) and grows by \(3\%\) per year. Write the model.
Practice 2: Evaluate Growth
Using Practice 1, estimate the population after \(6\) years.
Practice 3: Depreciation Model
A vehicle is worth \(\$28{,}000\) and depreciates by \(18\%\) annually. Write the model.
Practice 4: Evaluate Depreciation
Using Practice 3, find the modeled value after \(4\) years.
Practice 5: Find a Growth Rate
A population grows from \(10{,}000\) to \(12{,}100\) in \(2\) years. Find the annual growth rate.
Practice 6: Find a Decay Rate
An asset falls from \(\$40{,}000\) to \(\$25{,}600\) in \(2\) years. Find the annual depreciation rate.
Practice 7: Doubling Time
Find the doubling time for a quantity growing by \(8\%\) per year.
Practice 8: Half-Life from Decay Rate
Find the half-life of a quantity that decreases by \(10\%\) per year.
Practice 9: Half-Life Model
A substance begins with \(640\) units and has half-life \(4\) years. How much remains after \(12\) years?
Practice 10: Time to Target
A population begins at \(20{,}000\) and grows by \(4\%\) per year. How long until it reaches \(30{,}000\)?
Practice 11: Fit a Growth Model
A population is \(5{,}000\) initially and \(6{,}655\) after \(3\) years. Find an exponential model.
Practice 12: Continuous Growth
Evaluate
\[ P(t)=12000e^{0.025t} \]at \(t=8\).
Practice 13: Successive Percent Changes
A quantity increases by \(25\%\) and then decreases by \(20\%\). What fraction of the original value remains?
Practice 14: Recognize the Model
A sequence of values is
\[ 200,\quad230,\quad264.5,\quad304.175. \]Determine whether the data are linear or exponential and find the percent change per period.
Practice 15: Identify the Error
A machine depreciates by \(30\%\) annually. A student writes
\[ V(t)=V_0(0.30)^t. \]Explain the error and write the correct model.
Practice Answers
Practice 1
\[ \boxed{ P(t)=18000(1.03)^t }. \]Practice 2
\[ P(6)=18000(1.03)^6. \] \[ P(6)\approx21492.94. \]Therefore,
\[ \boxed{21{,}493\text{ people approximately}}. \]Practice 3
The retained multiplier is
\[ 1-0.18=0.82. \]Therefore,
\[ \boxed{ V(t)=28000(0.82)^t }. \]Practice 4
\[ V(4)=28000(0.82)^4. \] \[ V(4)\approx12659.41. \]Therefore,
\[ \boxed{\$12{,}659.41\text{ approximately}}. \]Practice 5
\[ r= \left( \frac{12100}{10000} \right)^{1/2}-1. \] \[ r=\sqrt{1.21}-1. \] \[ r=1.10-1. \] \[ \boxed{10\%}. \]Practice 6
\[ r= 1- \left( \frac{25600}{40000} \right)^{1/2}. \] \[ r=1-\sqrt{0.64}. \] \[ r=1-0.80. \] \[ \boxed{20\%}. \]Practice 7
\[ t_d= \frac{\ln2}{\ln1.08}. \] \[ t_d\approx9.01. \]Therefore,
\[ \boxed{9.01\text{ years approximately}}. \]Practice 8
Decay multiplier:
\[ b=0.90. \]Then
\[ t_h= \frac{\ln0.5}{\ln0.9}. \] \[ t_h\approx6.58. \]Therefore,
\[ \boxed{6.58\text{ years approximately}}. \]Practice 9
\[ A(12)=640\left(\frac12\right)^{12/4}. \] \[ A(12)=640\left(\frac12\right)^3. \] \[ A(12)=80. \]Therefore,
\[ \boxed{80\text{ units}}. \]Practice 10
\[ 30000=20000(1.04)^t. \] \[ 1.5=(1.04)^t. \] \[ t= \frac{\ln1.5}{\ln1.04}. \] \[ t\approx10.34. \]Therefore,
\[ \boxed{10.34\text{ years approximately}}. \]Practice 11
\[ 6655=5000b^3. \] \[ b^3=1.331. \] \[ b=1.10. \]Therefore,
\[ \boxed{ P(t)=5000(1.10)^t }. \]Practice 12
\[ P(8)=12000e^{0.025(8)}. \] \[ P(8)=12000e^{0.2}. \] \[ P(8)\approx14656.83. \]Therefore,
\[ \boxed{14{,}657\text{ approximately}}. \]Practice 13
Combined multiplier:
\[ 1.25(0.80)=1. \]Therefore,
\[ \boxed{100\%\text{ of the original value remains}}. \]Practice 14
Successive ratios are
\[ \frac{230}{200} = \frac{264.5}{230} = \frac{304.175}{264.5} = 1.15. \]Therefore, the data are exponential with
\[ \boxed{15\%\text{ growth per period}}. \]Practice 15
A \(30\%\) depreciation means the asset retains
\[ 70\% \]each year.
Therefore, the correct model is
\[ \boxed{ V(t)=V_0(0.70)^t }. \]What to Remember
Constant percentage growth uses
\[ \boxed{ A(t)=A_0(1+r)^t }. \]Constant percentage decay uses
\[ \boxed{ A(t)=A_0(1-r)^t }. \]The general exponential form is
\[ \boxed{A(t)=A_0b^t}. \]Doubling and half-life problems are solved by setting the ratio \(A/A_0\) equal to \(2\) or \(1/2\) and solving for time.
The most important habit is to convert a percent change into the correct multiplier before applying it repeatedly.
Where We Go Next
Growth and depreciation models describe how quantities change through time. Markets create another kind of applied relationship: quantities supplied and demanded depend on price, and equilibrium occurs where those relationships agree.
The next lesson, Supply, Demand, and Market Equilibrium, develops that algebraic framework.