The Central Idea
The standard exponential model is
\[ \boxed{f(t)=ab^t}, \]where \(a\) is the initial value and \(b\) is the repeated multiplication factor.
The value of \(b\) determines the basic behavior:
\[ \boxed{b>1\quad\Longrightarrow\quad\text{exponential growth}} \]and
\[ \boxed{0Exponential Growth
Exponential growth occurs when each new value is greater than the previous value by the same multiplicative factor.
For example,
\[ 5,\quad10,\quad20,\quad40,\quad80,\ldots \]grows exponentially because every value is obtained by multiplying the previous value by
\[ 2. \]The corresponding model is
\[ \boxed{f(t)=5(2)^t}. \]Exponential Decay
Exponential decay occurs when each new value is a positive fraction of the previous value.
For example,
\[ 160,\quad80,\quad40,\quad20,\quad10,\ldots \]decays exponentially because every value is obtained by multiplying the previous value by
\[ \frac12. \]The model is
\[ \boxed{f(t)=160\left(\frac12\right)^t}. \]Growth and Decay Are Both Repeated Multiplication
The difference between growth and decay is not the operation being used. Both use multiplication.
Growth multiplies by a factor larger than \(1\):
\[ 1.08,\quad1.5,\quad2,\quad3,\ldots \]Decay multiplies by a factor between \(0\) and \(1\):
\[ 0.95,\quad0.8,\quad0.5,\quad0.1,\ldots \]The magnitude of the factor determines how aggressively the quantity grows or decays.
Growth Rate and Growth Factor
If a quantity grows by rate \(r\) during each equal period, then the growth factor is
\[ \boxed{1+r}. \]Therefore, the standard growth model is
\[ \boxed{f(t)=a(1+r)^t}. \]Here \(r\) must be written as a decimal.
Example: \(6\%\) Growth
A \(6\%\) increase means
\[ r=0.06. \]The growth factor is
\[ 1+0.06=1.06. \]Thus a quantity beginning at \(500\) is modeled by
\[ \boxed{f(t)=500(1.06)^t}. \]Decay Rate and Decay Factor
If a quantity decreases by rate \(r\) during each equal period, then the remaining fraction is
\[ 1-r. \]Therefore, the standard decay model is
\[ \boxed{f(t)=a(1-r)^t}. \]Example: \(18\%\) Decay
An \(18\%\) decrease means
\[ r=0.18. \]The amount remaining after each period is
\[ 1-0.18=0.82. \]A quantity beginning at \(900\) is therefore modeled by
\[ \boxed{f(t)=900(0.82)^t}. \]Rate and Factor Should Not Be Confused
In a growth model with rate \(12\%\),
\[ r=0.12 \]but the repeated factor is
\[ 1.12. \]In a decay model with rate \(12\%\),
\[ r=0.12 \]but the repeated factor is
\[ 0.88. \]The rate describes the portion gained or lost. The factor describes the portion carried into the next period.
Why Percentage Growth Accelerates
Suppose \(100\) grows by \(10\%\) per period.
The values are
\[ 100,\quad110,\quad121,\quad133.1,\quad146.41,\ldots \]The increases are
\[ 10,\quad11,\quad12.1,\quad13.31,\ldots \]The percentage remains constant, but the absolute increase becomes larger because \(10\%\) is applied to a growing current value.
Why Percentage Decay Slows in Absolute Amount
Suppose \(100\) decays by \(20\%\) per period.
The values are
\[ 100,\quad80,\quad64,\quad51.2,\quad40.96,\ldots \]The decreases are
\[ 20,\quad16,\quad12.8,\quad10.24,\ldots \]The decay rate remains \(20\%\), but the absolute amount lost becomes smaller because the percentage is applied to a shrinking current value.
Constant Ratio Is the Signature
For equal input intervals, exponential data have a constant ratio.
Consider
\[ 3,\quad6,\quad12,\quad24,\quad48. \]Consecutive ratios are
\[ \frac63 = \frac{12}{6} = \frac{24}{12} = \frac{48}{24} = 2. \]Thus the data represent exponential growth with factor \(2\).
Decay Also Has a Constant Ratio
Consider
\[ 625,\quad500,\quad400,\quad320,\quad256. \]Consecutive ratios are
\[ \frac{500}{625} = \frac{400}{500} = \frac{320}{400} = \frac{256}{320} = 0.8. \]Therefore, the process has decay factor
\[ \boxed{0.8} \]and decay rate
\[ \boxed{20\%}. \]Finding a Growth Rate from a Factor
Suppose the repeated factor is
\[ 1.27. \]Subtract \(1\):
\[ 1.27-1=0.27. \]Therefore, the growth rate is
\[ \boxed{27\%}. \]Finding a Decay Rate from a Factor
Suppose the repeated factor is
\[ 0.73. \]The amount lost each period is
\[ 1-0.73=0.27. \]Therefore, the decay rate is
\[ \boxed{27\%}. \]Finding a Factor from a Table
Consider
\[ \begin{array}{c|c} t & P(t)\\ \hline 0 & 200\\ 1 & 230\\ 2 & 264.5\\ 3 & 304.175 \end{array} \]Compute the ratios:
\[ \frac{230}{200}=1.15, \] \[ \frac{264.5}{230}=1.15. \]Therefore, the growth factor is
\[ 1.15, \]and the growth rate is
\[ 15\%. \]Writing a Model from a Table
Using the previous table, the initial value is
\[ P(0)=200, \]and the growth factor is
\[ 1.15. \]Therefore,
\[ \boxed{P(t)=200(1.15)^t}. \]Evaluating a Growth Model
Suppose
\[ P(t)=800(1.04)^t. \]To find the value after \(5\) periods,
\[ P(5)=800(1.04)^5. \]Since
\[ (1.04)^5\approx1.2166529, \]we obtain
\[ P(5)\approx973.32. \]Therefore,
\[ \boxed{P(5)\approx973.32}. \]Evaluating a Decay Model
Suppose
\[ Q(t)=1200(0.9)^t. \]After \(4\) periods,
\[ Q(4)=1200(0.9)^4. \]Since
\[ (0.9)^4=0.6561, \]we obtain
\[ \boxed{Q(4)=787.32}. \]Interpreting a Growth Model
Consider
\[ P(t)=1500(1.08)^t. \]The model tells us:
- the initial value is \(1500\);
- the growth factor is \(1.08\);
- the growth rate is \(8\%\) per input period;
- each new value is \(108\%\) of the previous value.
Interpreting a Decay Model
Consider
\[ A(t)=2400(0.94)^t. \]The model tells us:
- the initial value is \(2400\);
- the decay factor is \(0.94\);
- the quantity retains \(94\%\) per period;
- the decay rate is \(6\%\) per period.
The Graph of Exponential Growth
For a basic positive growth function
\[ f(x)=ab^x,\qquad a>0,\quad b>1, \]the graph rises from left to right.
It passes through
\[ (0,a) \]and grows increasingly rapidly as \(x\) increases.
When there is no vertical shift, the horizontal asymptote is
\[ \boxed{y=0}. \]The Graph of Exponential Decay
For
\[ f(x)=ab^x,\qquad a>0,\quad 0It also passes through
\[ (0,a) \]and approaches
\[ y=0 \]as \(x\) increases.
Growth and Decay Are Reflections through Reciprocal Bases
If \(b>1\), then
\[ \left(\frac1b\right)^x=b^{-x}. \]Therefore, replacing \(x\) with \(-x\) converts a growth base \(b\) into its reciprocal decay base \(1/b\).
For example,
\[ \left(\frac12\right)^x=2^{-x}. \]The graphs of \(2^x\) and \(2^{-x}\) are reflections of one another across the \(y\)-axis.
Long-Run Behavior of Growth
If
\[ a>0 \]and
\[ b>1, \]then
\[ ab^t \]becomes arbitrarily large as \(t\) increases.
Symbolically,
\[ \boxed{f(t)\to\infty\text{ as }t\to\infty}. \]Long-Run Behavior of Decay
If
\[ a>0 \]and
\[ 0approaches zero as \(t\) increases.
Symbolically,
\[ \boxed{f(t)\to0\text{ as }t\to\infty}. \]Approaching Zero Does Not Mean Reaching Zero
For a standard positive decay model,
\[ f(t)=ab^t, \]both \(a\) and \(b^t\) remain positive.
Therefore,
\[ f(t)>0 \]for every finite real \(t\).
The values can become extremely small, but the basic model does not equal zero at a finite input.
Growth Eventually Outpaces Linear Change
Compare
\[ L(t)=100+50t \]with
\[ E(t)=100(1.5)^t. \]The linear model adds \(50\) each period.
The exponential model multiplies by \(1.5\) each period.
Although their values may be similar at first, sustained exponential growth eventually dominates sustained linear growth.
Growth Depends on the Current Amount
A \(5\%\) growth rate does not represent a fixed absolute increase.
On \(100\),
\[ 5\%=5. \]On \(1000\),
\[ 5\%=50. \]The same percentage produces different absolute changes because the current quantity is different.
Decay Depends on the Current Amount
A \(10\%\) decay rate also acts on the current value.
Ten percent of \(500\) is
\[ 50, \]while ten percent of \(200\) is
\[ 20. \]As a decaying quantity shrinks, the absolute amount lost each period also tends to shrink.
Example: Population Growth
Suppose a population begins at \(20{,}000\) and grows by \(2.5\%\) per year.
The growth factor is
\[ 1+0.025=1.025. \]A model is
\[ \boxed{P(t)=20000(1.025)^t}. \]Here \(t\) represents years after the beginning of the model.
Example: Depreciation
Suppose an asset is valued at \(\$30{,}000\) and loses \(12\%\) of its value each year.
The remaining factor is
\[ 1-0.12=0.88. \]The value after \(t\) years is modeled by
\[ \boxed{V(t)=30000(0.88)^t}. \]Example: Medication Remaining
Suppose \(80\%\) of a substance remains after each hour.
If the initial amount is \(250\) milligrams, then
\[ \boxed{A(t)=250(0.80)^t}. \]The factor \(0.80\) is a decay factor.
The corresponding decay rate is
\[ 1-0.80=0.20=20\%. \]Example: Bacterial Growth
Suppose a culture begins with \(600\) cells and increases by \(40\%\) during each hour.
The growth factor is
\[ 1.40. \]Therefore,
\[ \boxed{N(t)=600(1.40)^t}. \]Example: Repeated Loss
Suppose a process retains \(65\%\) of its current amount each cycle.
Then
\[ b=0.65. \]The decay rate is
\[ 1-0.65=0.35. \]Thus the process loses
\[ \boxed{35\%} \]per cycle.
Finding an Initial Value from a Model
Consider
\[ P(t)=480(1.07)^t. \]At \(t=0\),
\[ P(0)=480(1.07)^0=480. \]Therefore, the initial value is
\[ \boxed{480}. \]Finding an Initial Value from a Later Value
Suppose
\[ P(t)=a(1.2)^t \]and
\[ P(3)=864. \]Then
\[ 864=a(1.2)^3. \]Since
\[ (1.2)^3=1.728, \]we have
\[ a=\frac{864}{1.728}=500. \]Therefore,
\[ \boxed{a=500}. \]Finding a Factor across Multiple Periods
Suppose a quantity grows from \(100\) to \(172.8\) in three equal periods.
If the one-period factor is \(b\), then
\[ 172.8=100b^3. \]Divide:
\[ b^3=1.728. \]Therefore,
\[ b=1.2. \]The growth rate is
\[ \boxed{20\%}. \]Successive Percentage Changes Compound
Suppose a quantity grows by \(10\%\) during each of two periods.
The two-period multiplier is
\[ 1.10^2=1.21. \]Therefore, the total two-period increase is
\[ 21\%. \]Repeated percentage changes compound multiplicatively rather than adding arithmetically.
Growth Followed by Decay
Suppose a quantity increases by \(25\%\) and then decreases by \(25\%\).
The combined factor is
\[ 1.25(0.75)=0.9375. \]Thus the final value is
\[ 93.75\% \]of the original.
Equal percentage growth and decay rates do not cancel because they act on different current values.
Changing the Time Unit
A growth or decay factor belongs to a particular input interval.
If a quantity grows by factor \(1.1\) each month, then after two months the factor is
\[ (1.1)^2=1.21. \]The two-month growth factor is \(1.21\), even though the one-month factor is \(1.1\).
Always interpret the factor together with the units of the input variable.
Discrete and Continuous Interpretations
Exponential formulas can be evaluated at integer inputs such as
\[ t=0,1,2,3,\ldots \]to describe repeated discrete stages.
They can also be evaluated at noninteger inputs when the model treats time as continuous.
For example,
\[ P(2.5) \]can represent the modeled quantity halfway between the second and third time units.
Exponential Models Are Models, Not Automatic Laws
An exponential formula is appropriate only when the underlying situation reasonably supports proportional or multiplicative change.
Real populations encounter resource limits. Asset values may change irregularly. Biological processes can respond to changing conditions.
A model can be mathematically correct while still having a limited range of practical usefulness.
Model Assumptions Matter
When using
\[ P(t)=a(1+r)^t, \]we are typically assuming that:
- the same proportional rate applies during each modeled interval;
- the time intervals are measured consistently;
- the process is not interrupted by unmodeled structural changes;
- the initial value is measured at the model's chosen starting point.
Common Error: Adding the Rate Instead of Multiplying by a Factor
A \(5\%\) growth rate does not mean
\[ P(t)=P_0+0.05t. \]Instead, the current amount is repeatedly multiplied by
\[ 1.05. \]The exponential model is
\[ \boxed{P(t)=P_0(1.05)^t}. \]Common Error: Using \(r\) Instead of \(1+r\)
If a quantity grows by \(9\%\), then
\[ r=0.09, \]but the growth factor is
\[ 1.09. \]Therefore,
\[ \boxed{P(t)=P_0(1.09)^t}. \]Common Error: Using \(r\) Instead of \(1-r\)
If a quantity decays by \(9\%\), then the amount remaining is
\[ 91\%=0.91. \]Therefore,
\[ \boxed{P(t)=P_0(0.91)^t}. \]Using \(0.09\) would incorrectly say that only \(9\%\) remains after each period.
Common Error: Assuming a Decay Factor Is Negative
Exponential decay normally uses
\[ 0produces decay because each new value is smaller than the previous value.
The base does not need to be negative.
Common Error: Ignoring Units
A factor of \(1.02\) per month is not the same statement as a factor of \(1.02\) per year.
The numerical factor and the time interval form one modeling statement.
Always identify both.
Common Error: Assuming Exponential Growth Continues Forever in Reality
Mathematically,
\[ a(1+r)^t \]can grow without bound.
Real systems often face limitations that eventually make a different model more appropriate.
The lesson is not that every real process grows forever, but that exponential functions describe sustained proportional growth under the model's assumptions.
A Reliable Procedure for Exponential Growth Problems
- Identify the initial value \(a\).
- Identify the growth rate.
- Convert the rate to decimal form.
- Compute the growth factor \(1+r\).
- Identify the time unit represented by one exponent step.
- Write \(P(t)=a(1+r)^t\).
- Substitute the desired input.
- Interpret the result in context and with units.
A Reliable Procedure for Exponential Decay Problems
- Identify the initial value \(a\).
- Identify the decay rate.
- Convert the rate to decimal form.
- Compute the remaining factor \(1-r\).
- Identify the time unit represented by one exponent step.
- Write \(P(t)=a(1-r)^t\).
- Evaluate the function at the desired input.
- Interpret the result with the correct units.
Example: Full Growth Workflow
A quantity begins at \(2{,}400\) and grows by \(7\%\) per year.
Initial value:
\[ a=2400. \]Growth rate:
\[ r=0.07. \]Growth factor:
\[ 1+r=1.07. \]Model:
\[ \boxed{P(t)=2400(1.07)^t}. \]After four years,
\[ P(4)=2400(1.07)^4\approx3145.91. \]So the model predicts approximately
\[ \boxed{3146} \]units after four years.
Example: Full Decay Workflow
A quantity begins at \(5{,}000\) and decreases by \(11\%\) per year.
Initial value:
\[ a=5000. \]Decay rate:
\[ r=0.11. \]Remaining factor:
\[ 1-r=0.89. \]Model:
\[ \boxed{Q(t)=5000(0.89)^t}. \]After three years,
\[ Q(3)=5000(0.89)^3\approx3524.85. \]Therefore,
\[ \boxed{Q(3)\approx3524.85}. \]Practice 1: Growth or Decay?
Classify
\[ f(x)=700(1.12)^x \]as exponential growth or exponential decay.
Practice 2: Growth or Decay?
Classify
\[ g(x)=900(0.74)^x. \]Practice 3: Find the Growth Rate
Find the percentage growth rate corresponding to the factor
\[ 1.16. \]Practice 4: Find the Decay Rate
Find the percentage decay rate corresponding to the factor
\[ 0.91. \]Practice 5: Write a Growth Model
A quantity begins at \(350\) and grows by \(8\%\) per month.
Write an exponential model.
Practice 6: Write a Decay Model
A quantity begins at \(1{,}200\) and loses \(14\%\) per year.
Write an exponential model.
Practice 7: Evaluate a Growth Model
Given
\[ P(t)=500(1.1)^t, \]find \(P(3)\).
Practice 8: Evaluate a Decay Model
Given
\[ Q(t)=800(0.75)^t, \]find \(Q(2)\).
Practice 9: Identify the Model from a Table
Find an exponential model for
\[ \begin{array}{c|c} t & y\\ \hline 0 & 50\\ 1 & 60\\ 2 & 72\\ 3 & 86.4 \end{array} \]Practice 10: Interpret a Decay Model
Consider
\[ A(t)=2500(0.83)^t. \]State the initial value, decay factor, percentage retained, and percentage lost during each period.
Practice Answers
Practice 1
Since
\[ 1.12>1, \]the function represents
\[ \boxed{\text{exponential growth}}. \]Practice 2
Since
\[ 0<0.74<1, \]the function represents
\[ \boxed{\text{exponential decay}}. \]Practice 3
\[ 1.16-1=0.16. \]Therefore, the growth rate is
\[ \boxed{16\%}. \]Practice 4
\[ 1-0.91=0.09. \]Therefore, the decay rate is
\[ \boxed{9\%}. \]Practice 5
The growth factor is
\[ 1+0.08=1.08. \]Therefore,
\[ \boxed{P(t)=350(1.08)^t}. \]Practice 6
The remaining factor is
\[ 1-0.14=0.86. \]Therefore,
\[ \boxed{Q(t)=1200(0.86)^t}. \]Practice 7
\[ P(3)=500(1.1)^3 \] \[ =500(1.331) \] \[ \boxed{=665.5}. \]Practice 8
\[ Q(2)=800(0.75)^2 \] \[ =800(0.5625) \] \[ \boxed{=450}. \]Practice 9
The initial value is \(50\).
The constant ratio is
\[ \frac{60}{50}=1.2. \]Therefore,
\[ \boxed{y=50(1.2)^t}. \]This represents \(20\%\) growth per period.
Practice 10
The initial value is
\[ \boxed{2500}. \]The decay factor is
\[ \boxed{0.83}. \]The quantity retains
\[ \boxed{83\%} \]and loses
\[ \boxed{17\%} \]during each period.
What to Remember
Exponential growth and decay share the same basic structure:
\[ \boxed{f(t)=ab^t}. \]- \(a\) is the initial value.
- \(b>1\) produces exponential growth.
- \(0
- A growth rate \(r\) gives factor \(1+r\).
- A decay rate \(r\) gives factor \(1-r\).
- Equal input intervals produce a constant output ratio.
- Growth accelerates in absolute amount because percentages act on a growing base.
- Decay slows in absolute amount because percentages act on a shrinking base.
- Positive exponential growth tends toward infinity in the long run.
- Positive exponential decay approaches zero without reaching it at a finite input.
The central modeling question is: what fraction or multiple of the current amount becomes the next amount?
Where We Go Next
Exponential growth becomes especially important when money earns interest. Repeated compounding applies a growth factor again and again, and changing the compounding interval leads naturally toward continuous exponential growth.
The next lesson, Compound Interest and Continuous Compounding, develops the standard financial growth formulas, compares compounding frequencies, and introduces the constant \(e\) as the natural base for continuous growth.