The Central Idea
An exponential process can often be summarized by two questions:
- What quantity do we start with?
- By what factor do we multiply at each equal step?
If the initial value is \(a\) and the repeated multiplication factor is \(b\), then after \(x\) steps the value is
\[ \boxed{f(x)=ab^x}. \]The coefficient \(a\) establishes the starting scale. The base \(b\) generates the repeated multiplicative change.
Starting with an Initial Value
Suppose a quantity begins at
\[ 5. \]If nothing has happened yet, the value is still \(5\).
In an exponential model, this starting quantity is the initial value.
When the independent variable counts elapsed steps from zero, the initial value is the output at
\[ x=0. \]Therefore,
\[ \boxed{f(0)=a}. \]Why the Initial Value Appears as a Coefficient
Consider the general exponential form
\[ f(x)=ab^x. \]At \(x=0\),
\[ f(0)=ab^0. \]Because
\[ b^0=1, \]we obtain
\[ f(0)=a(1)=a. \]Thus the coefficient \(a\) is not placed in front arbitrarily. It is the value that remains before any repeated multiplication has occurred.
One Step of Multiplicative Change
Suppose an initial quantity \(a\) is multiplied by a factor \(b\).
After one step, the value is
\[ ab. \]Using exponent notation,
\[ ab=ab^1. \]So the first step of an exponential process already has the form
\[ \boxed{ab^1}. \]Two Steps of Multiplicative Change
Apply the same factor \(b\) again:
\[ (ab)b. \]Associativity of multiplication gives
\[ a(b\cdot b). \]Therefore,
\[ \boxed{ab^2}. \]The exponent records how many times the repeated factor has been applied.
Three Steps of Multiplicative Change
Apply the factor once more:
\[ (ab^2)b. \]This becomes
\[ a(b\cdot b\cdot b) \]and therefore
\[ \boxed{ab^3}. \]The pattern is now visible.
From Repeated Multiplication to an Exponential Formula
The first several stages are
\[ \begin{array}{c|c} \text{Step} & \text{Value}\\ \hline 0 & a\\ 1 & ab\\ 2 & ab^2\\ 3 & ab^3\\ 4 & ab^4 \end{array} \]The natural generalization is
\[ \boxed{\text{value after }x\text{ steps}=ab^x}. \]Exponential notation compresses an entire chain of repeated multiplications into one expression.
Example: Repeated Doubling
Suppose a process begins with \(3\) and doubles during every step.
The initial value is
\[ a=3, \]and the repeated factor is
\[ b=2. \]The values are
\[ 3,\quad6,\quad12,\quad24,\quad48,\ldots \]Therefore, the exponential function is
\[ \boxed{f(x)=3(2)^x}. \]What Is a Growth Factor?
A growth factor is a repeated multiplication factor greater than \(1\).
If
\[ b>1, \]then multiplying a positive quantity by \(b\) makes the quantity larger.
For example,
\[ b=1.4 \]means that each new value is \(1.4\) times the preceding value.
A Growth Factor Does Not Mean Add the Factor
If the growth factor is \(1.5\), we do not add \(1.5\) at each step.
We multiply:
\[ \text{new value} = 1.5(\text{previous value}). \]Starting with \(20\),
\[ 20,\quad30,\quad45,\quad67.5,\ldots \]because
\[ 20(1.5)=30, \] \[ 30(1.5)=45, \]and
\[ 45(1.5)=67.5. \]What Is a Decay Factor?
A repeated multiplication factor between \(0\) and \(1\) produces exponential decay.
Thus
\[ 0For example,
\[ b=0.75 \]means that each stage retains \(75\%\) of the previous quantity.
Example: Repeated Halving
Suppose a quantity begins at \(160\) and is halved during every step.
The initial value is
\[ a=160, \]and the repeated factor is
\[ b=\frac12. \]The values are
\[ 160,\quad80,\quad40,\quad20,\quad10,\ldots \]Therefore,
\[ \boxed{f(x)=160\left(\frac12\right)^x}. \]The Constant-Ratio Test
Repeated multiplication creates a constant ratio between consecutive outputs.
If
\[ f(x)=ab^x, \]then
\[ f(x+1)=ab^{x+1}. \]Divide:
\[ \frac{f(x+1)}{f(x)} = \frac{ab^{x+1}}{ab^x} = b. \]Therefore,
\[ \boxed{\frac{f(x+1)}{f(x)}=b}. \]Finding a Growth Factor from Two Consecutive Values
Suppose a quantity changes from \(40\) to \(50\) in one equal time step.
The multiplicative factor is
\[ \frac{50}{40}. \]Simplify:
\[ \frac{50}{40}=1.25. \]Therefore, the growth factor is
\[ \boxed{1.25}. \]Finding a Decay Factor from Two Consecutive Values
Suppose a quantity changes from \(200\) to \(150\) in one equal time step.
The repeated factor is
\[ \frac{150}{200}=0.75. \]Therefore, the decay factor is
\[ \boxed{0.75}. \]Each new value is \(75\%\) of the previous value.
Finding the Factor from a Table
Consider
\[ \begin{array}{c|c} x & f(x)\\ \hline 0 & 8\\ 1 & 24\\ 2 & 72\\ 3 & 216 \end{array} \]Compute consecutive ratios:
\[ \frac{24}{8}=3, \] \[ \frac{72}{24}=3, \]and
\[ \frac{216}{72}=3. \]Therefore, the repeated factor is
\[ \boxed{b=3}. \]Since \(f(0)=8\), the function is
\[ \boxed{f(x)=8(3)^x}. \]Equal Input Steps Matter
The constant-ratio idea applies when the input advances by equal amounts.
For a table with
\[ x=0,1,2,3,\ldots, \]compare consecutive outputs.
If the inputs instead advance by \(2\) each time, such as
\[ x=0,2,4,6,\ldots, \]then the corresponding output ratio represents two units of exponential change rather than one.
Always interpret the ratio together with the size of the input interval.
Example: A Two-Unit Input Step
Suppose
\[ f(x)=5(2)^x. \]Compare \(x=0\) and \(x=2\):
\[ f(0)=5 \]and
\[ f(2)=20. \]The ratio is
\[ \frac{20}{5}=4. \]That does not mean the one-unit factor is \(4\). Two one-unit doublings occurred:
\[ 2^2=4. \]From Percentage Increase to Growth Factor
A percentage increase can be converted into a multiplication factor.
Suppose a quantity increases by \(r\), where \(r\) is written as a decimal.
The new value is
\[ \text{old value}+r(\text{old value}). \]Factor the old value:
\[ (1+r)(\text{old value}). \]Therefore, the growth factor is
\[ \boxed{1+r}. \]Example: A \(7\%\) Increase
Convert \(7\%\) to decimal form:
\[ 7\%=0.07. \]Add the increase to \(100\%\):
\[ 1+0.07=1.07. \]Therefore, a \(7\%\) increase corresponds to the growth factor
\[ \boxed{1.07}. \]Example: A \(35\%\) Increase
A \(35\%\) increase means that the new value is
\[ 100\%+35\%=135\% \]of the previous value.
As a decimal multiplier,
\[ 135\%=1.35. \]Therefore,
\[ \boxed{b=1.35}. \]From Percentage Decrease to Decay Factor
Suppose a quantity decreases by rate \(r\), written as a decimal.
The new value is
\[ \text{old value}-r(\text{old value}). \]Factor the old value:
\[ (1-r)(\text{old value}). \]Therefore, the decay factor is
\[ \boxed{1-r}. \]Example: A \(12\%\) Decrease
Convert \(12\%\) to decimal form:
\[ 12\%=0.12. \]Subtract from \(1\):
\[ 1-0.12=0.88. \]Therefore, a \(12\%\) decrease corresponds to the decay factor
\[ \boxed{0.88}. \]Rate and Factor Are Different Quantities
A growth or decay rate describes how much is gained or lost.
A growth or decay factor describes what multiplies the old value to produce the new value.
For a \(20\%\) increase,
\[ r=0.20 \]but
\[ b=1.20. \]For a \(20\%\) decrease,
\[ r=0.20 \]but
\[ b=0.80. \]A Useful Conversion Summary
For a decimal rate \(r\):
\[ \boxed{\text{growth factor}=1+r} \]and
\[ \boxed{\text{decay factor}=1-r}. \]Conversely, if the factor \(b\) is already known:
\[ \boxed{\text{growth rate}=b-1\qquad(b>1)} \]and
\[ \boxed{\text{decay rate}=1-b\qquad(0Finding a Percentage Rate from a Growth Factor
Suppose the growth factor is
\[ 1.18. \]Subtract \(1\):
\[ 1.18-1=0.18. \]Convert to a percentage:
\[ 0.18=18\%. \]Therefore, a factor of \(1.18\) represents
\[ \boxed{18\%\text{ growth per step}}. \]Finding a Percentage Rate from a Decay Factor
Suppose the decay factor is
\[ 0.84. \]The amount lost each step is
\[ 1-0.84=0.16. \]Therefore, the decay rate is
\[ \boxed{16\%}. \]The quantity retains \(84\%\) and loses \(16\%\) during each step.
Repeated Percentage Growth
Suppose an initial quantity \(P_0\) grows by rate \(r\) during each equal time period.
After one period,
\[ P_1=P_0(1+r). \]After two periods,
\[ P_2=P_0(1+r)(1+r) = P_0(1+r)^2. \]After three periods,
\[ P_3=P_0(1+r)^3. \]After \(t\) periods,
\[ \boxed{P(t)=P_0(1+r)^t}. \]Repeated Percentage Decay
If the initial quantity \(Q_0\) decreases by rate \(r\) during each equal time period, then the repeated factor is
\[ 1-r. \]Therefore,
\[ \boxed{Q(t)=Q_0(1-r)^t}. \]This is the standard repeated-decay structure.
Example: Repeated \(10\%\) Growth
Suppose \(500\) grows by \(10\%\) each year.
The growth factor is
\[ 1+0.10=1.10. \]Therefore,
\[ P(t)=500(1.10)^t. \]After three years,
\[ P(3)=500(1.10)^3 \] \[ =500(1.331) \] \[ \boxed{=665.5}. \]Why Repeated \(10\%\) Growth Is Not Repeated Addition of \(50\)
Ten percent of the initial \(500\) is \(50\), but the same dollar or unit increase does not repeat.
The values are
\[ 500,\quad550,\quad605,\quad665.5,\ldots \]The increases are
\[ 50,\quad55,\quad60.5,\ldots \]The percentage stays constant while the absolute increase changes because the percentage is applied to the current value.
Example: Repeated \(15\%\) Decay
Suppose a quantity begins at \(800\) and loses \(15\%\) per period.
The amount retained is
\[ 100\%-15\%=85\%. \]Therefore, the decay factor is
\[ 0.85. \]The model is
\[ \boxed{Q(t)=800(0.85)^t}. \]Recursive and Explicit Descriptions
Repeated multiplication can be described recursively.
If a sequence begins with \(a\) and repeatedly multiplies by \(b\), then
\[ u_0=a \]and
\[ \boxed{u_{n+1}=bu_n}. \]The corresponding explicit formula is
\[ \boxed{u_n=ab^n}. \]The recursive rule tells us how to get the next value. The explicit rule lets us jump directly to any indexed value.
Example: Recursive to Explicit Form
Suppose
\[ u_0=12 \]and
\[ u_{n+1}=1.5u_n. \]The initial value is \(12\), and the repeated factor is \(1.5\).
Therefore, the explicit rule is
\[ \boxed{u_n=12(1.5)^n}. \]Example: Explicit to Recursive Form
Suppose
\[ f(n)=200(0.9)^n. \]The initial value is
\[ f(0)=200, \]and each new value is \(0.9\) times the preceding value.
A recursive description is therefore
\[ \boxed{f(0)=200,\qquad f(n+1)=0.9f(n)}. \]The Initial Value Need Not Occur at Clock Time Zero
In applications, the word initial means the beginning of the model, not necessarily the beginning of historical time.
If an account contains \(\$2{,}000\) on the day we begin measuring, then \(2{,}000\) is the initial value for that model.
If \(t\) counts years since that day, then
\[ t=0 \]refers to the selected starting date.
Units Belong to the Model
An exponential formula should be interpreted together with its units.
In
\[ P(t)=1200(1.03)^t, \]the number \(1200\) might represent people, dollars, cells, kilograms, or another quantity.
The input \(t\) might represent years, months, hours, or another interval.
A factor of \(1.03\) means multiplication by \(1.03\) per one unit of the chosen input scale.
Changing the Time Unit Changes the Factor
Suppose a quantity doubles every year.
If \(t\) is measured in years, the one-year factor is
\[ 2. \]Over two years, the factor is
\[ 2^2=4. \]The underlying process has not changed, but the factor associated with a single input step depends on how that input step is defined.
Recovering the Initial Value from a Known Point
Sometimes the factor \(b\) is known, but the initial value is not.
Suppose
\[ f(x)=a(2)^x \]and
\[ f(3)=40. \]Substitute:
\[ 40=a(2)^3. \]Since \(2^3=8\),
\[ 40=8a. \]Therefore,
\[ \boxed{a=5}. \]Recovering a One-Step Factor from Two Values
Suppose an exponential process has
\[ f(2)=18 \]and
\[ f(3)=27. \]Because these inputs differ by one unit,
\[ b=\frac{f(3)}{f(2)}. \]Therefore,
\[ b=\frac{27}{18}=\frac32. \]The one-step growth factor is
\[ \boxed{\frac32=1.5}. \]Recovering a Factor across Several Steps
Suppose
\[ f(1)=10 \]and
\[ f(4)=80. \]Three equal one-unit steps separate the inputs.
Therefore,
\[ 80=10b^3. \]Divide by \(10\):
\[ b^3=8. \]Thus
\[ \boxed{b=2}. \]The process doubles during each one-unit step.
Repeated Multiplication and Powers
Powers arise naturally because the same factor is multiplied repeatedly.
If a factor \(b\) is applied \(m\) times and then another \(n\) times, the total effect is
\[ b^m\cdot b^n. \]The repeated factors combine into
\[ \boxed{b^{m+n}}. \]This connection is one reason exponent laws are structurally important for exponential functions.
Multiplicative Change Can Compound
Suppose a quantity grows by factor \(1.2\) during one period and then by factor \(1.2\) again during the next.
The combined two-period factor is
\[ 1.2\cdot1.2=1.44. \]Therefore, two successive \(20\%\) increases produce a total multiplier of
\[ \boxed{1.44}. \]The final value is \(44\%\) above the original value, not merely \(40\%\) above it.
Why Successive Percentage Changes Do Not Simply Add
Percentage changes act on the current value.
Start with \(100\) and increase by \(10\%\):
\[ 100(1.10)=110. \]Increase the new value by another \(10\%\):
\[ 110(1.10)=121. \]The total increase is
\[ 21\%. \]The correct combined factor is
\[ 1.10^2=1.21. \]Growth Followed by Equal Percentage Decay Does Not Cancel
Suppose \(100\) increases by \(20\%\):
\[ 100(1.20)=120. \]Now decrease \(120\) by \(20\%\):
\[ 120(0.80)=96. \]The final value is not \(100\).
The combined multiplier is
\[ 1.20(0.80)=0.96. \]Therefore,
\[ \boxed{100\longrightarrow96}. \]Why Equal Percentage Changes Do Not Cancel
The increase and decrease are applied to different base quantities.
A \(20\%\) increase on \(100\) adds
\[ 20. \]But a \(20\%\) decrease on \(120\) subtracts
\[ 24. \]Equal percentage rates do not automatically represent equal absolute changes.
Linear Change versus Exponential Change
Suppose one quantity begins at \(100\) and increases by \(20\) each period.
Its values are
\[ 100,\quad120,\quad140,\quad160,\quad180,\ldots \]This is additive change.
Now suppose another quantity begins at \(100\) and grows by \(20\%\) per period:
\[ 100,\quad120,\quad144,\quad172.8,\quad207.36,\ldots \]This is multiplicative change.
Constant Difference versus Constant Ratio
A useful diagnostic is:
\[ \boxed{\text{constant difference}\longrightarrow\text{linear}} \]while
\[ \boxed{\text{constant ratio}\longrightarrow\text{exponential}}. \]These tests assume equal increments of the independent variable.
Common Error: Using the Rate as the Factor
A quantity that grows by \(8\%\) per period does not use
\[ 0.08 \]as its growth factor.
Instead,
\[ b=1+0.08=1.08. \]Therefore, a model beginning at \(500\) is
\[ \boxed{500(1.08)^t}, \]not
\[ 500(0.08)^t. \]Common Error: Using the Decay Rate as the Remaining Factor
A quantity that loses \(30\%\) per period retains
\[ 100\%-30\%=70\%. \]Therefore, its decay factor is
\[ 0.70, \]not \(0.30\).
The number \(0.30\) represents the proportion lost. The number \(0.70\) represents the proportion remaining.
Common Error: Treating the Initial Value as the First Multiplied Value
Suppose a process begins at \(10\) and doubles each step.
At step \(0\), the value is
\[ 10. \]At step \(1\), it is
\[ 20. \]Therefore, the correct model is
\[ \boxed{10(2)^x}, \]because
\[ 10(2)^0=10. \]Common Error: Looking for Constant Differences
Consider
\[ 4,\quad12,\quad36,\quad108. \]The differences are
\[ 8,\quad24,\quad72, \]which are not constant.
But the ratios are
\[ 3,\quad3,\quad3. \]Therefore, the pattern is exponential with repeated factor \(3\).
Common Error: Ignoring the Size of the Input Step
If a table lists values every two years, the ratio between adjacent rows is a two-year factor.
It should not automatically be interpreted as the one-year factor.
The input scale is part of the model, and multiplicative factors must be tied to a specific interval.
A Reliable Procedure for Building an Exponential Model
- Identify the starting point of the model.
- Record the initial value \(a\).
- Determine the size and units of one input step.
- Identify the repeated multiplication factor \(b\).
- If given a percentage increase \(r\), use \(b=1+r\).
- If given a percentage decrease \(r\), use \(b=1-r\).
- Write the model \(f(x)=ab^x\).
- Check that \(f(0)=a\).
- Check that one additional step multiplies the output by \(b\).
- Interpret the result with the correct units.
Example: Full Growth-Factor Workflow
A quantity begins at \(1{,}200\) and increases by \(6\%\) each year.
Initial value:
\[ a=1200. \]Growth rate:
\[ r=0.06. \]Growth factor:
\[ b=1+0.06=1.06. \]Therefore,
\[ \boxed{f(t)=1200(1.06)^t}. \]Check:
\[ f(0)=1200(1.06)^0=1200. \]Example: Full Decay-Factor Workflow
A quantity begins at \(900\) and decreases by \(4\%\) each month.
Initial value:
\[ a=900. \]Decay rate:
\[ r=0.04. \]Decay factor:
\[ b=1-0.04=0.96. \]Therefore,
\[ \boxed{Q(t)=900(0.96)^t}. \]Practice 1: Identify the Initial Value and Factor
For
\[ f(x)=70(1.4)^x, \]identify the initial value and repeated multiplication factor.
Practice 2: Build a Function from Repeated Multiplication
A quantity begins at \(9\) and triples during every step.
Write an exponential function for the quantity after \(x\) steps.
Practice 3: Find the Constant Ratio
Find the repeated factor:
\[ 6,\quad15,\quad37.5,\quad93.75,\ldots \]Practice 4: Convert a Growth Rate to a Factor
Find the growth factor corresponding to a \(14\%\) increase per period.
Practice 5: Convert a Decay Rate to a Factor
Find the decay factor corresponding to a \(23\%\) decrease per period.
Practice 6: Convert a Factor to a Growth Rate
A process has repeated factor
\[ 1.125. \]Find the percentage growth rate per step.
Practice 7: Convert a Factor to a Decay Rate
A process has repeated factor
\[ 0.68. \]Find the percentage decay rate per step.
Practice 8: Write a Percentage-Growth Model
A quantity begins at \(2{,}500\) and grows by \(3.5\%\) per year.
Write an exponential model.
Practice 9: Write a Percentage-Decay Model
A quantity begins at \(640\) and loses \(18\%\) per month.
Write an exponential model.
Practice 10: Linear or Exponential?
Consider
\[ 25,\quad30,\quad36,\quad43.2,\quad51.84. \]Determine whether the pattern is linear or exponential, and identify the relevant difference or ratio.
Practice Answers
Practice 1
The initial value is
\[ \boxed{70}, \]and the repeated factor is
\[ \boxed{1.4}. \]Practice 2
The initial value is \(9\), and tripling gives factor \(3\):
\[ \boxed{f(x)=9(3)^x}. \]Practice 3
Divide consecutive values:
\[ \frac{15}{6}=2.5 \]and
\[ \frac{37.5}{15}=2.5. \]Therefore, the repeated factor is
\[ \boxed{2.5}. \]Practice 4
\[ 1+0.14=\boxed{1.14}. \]Practice 5
\[ 1-0.23=\boxed{0.77}. \]Practice 6
\[ 1.125-1=0.125. \]Therefore, the growth rate is
\[ \boxed{12.5\%}. \]Practice 7
\[ 1-0.68=0.32. \]Therefore, the decay rate is
\[ \boxed{32\%}. \]Practice 8
The growth factor is
\[ 1+0.035=1.035. \]Therefore,
\[ \boxed{P(t)=2500(1.035)^t}. \]Practice 9
The remaining factor is
\[ 1-0.18=0.82. \]Therefore,
\[ \boxed{Q(t)=640(0.82)^t}. \]Practice 10
The consecutive ratios are
\[ \frac{30}{25}=1.2, \] \[ \frac{36}{30}=1.2, \] \[ \frac{43.2}{36}=1.2. \]Therefore, the pattern is
\[ \boxed{\text{exponential with factor }1.2}. \]This corresponds to \(20\%\) growth per step.
What to Remember
An exponential relationship begins with an initial value and repeatedly multiplies by a constant factor:
\[ \boxed{f(x)=ab^x}. \]- \(a\) is the initial value.
- \(b\) is the repeated multiplication factor.
- \(f(0)=a\).
- For equal one-unit steps, consecutive outputs have ratio \(b\).
- If \(b>1\), the process grows.
- If \(0
- A growth rate \(r\) gives factor \(1+r\).
- A decay rate \(r\) gives factor \(1-r\).
- Repeated percentage changes compound because each change acts on the current value.
- Constant ratios indicate exponential structure; constant differences indicate linear structure.
The central operational habit is: identify what you start with, identify what multiplies it each step, and then let exponentiation record how many times that multiplication occurs.
Where We Go Next
Initial values and repeated factors tell us how exponential relationships are built. We can now analyze more directly what happens when those factors are greater than \(1\) or between \(0\) and \(1\).
The next lesson, Exponential Growth and Decay, develops these two major behaviors in greater depth, including growth and decay rates, model interpretation, long-run behavior, and applications in which quantities increase or decrease proportionally over repeated intervals.