The Central Idea

Linear change is additive:

\[ \boxed{\text{new value}=\text{old value}+\text{constant amount}}. \]

Exponential change is multiplicative:

\[ \boxed{\text{new value}=(\text{constant factor})(\text{old value})}. \]

For equal input intervals:

\[ \boxed{\text{linear}\longrightarrow\text{constant difference}} \]

while

\[ \boxed{\text{exponential}\longrightarrow\text{constant ratio}}. \]

The Basic Linear Form

A linear function can be written

\[ \boxed{L(x)=mx+b}. \]

Here \(b\) is the initial value at \(x=0\), and \(m\) is the constant rate of additive change.

Every one-unit increase in \(x\) changes the output by

\[ \boxed{m}. \]

The Basic Exponential Form

An exponential function can be written

\[ \boxed{E(x)=ab^x}. \]

Here \(a\) is the initial value at \(x=0\), and \(b\) is the constant multiplicative factor.

Every one-unit increase in \(x\) multiplies the output by

\[ \boxed{b}. \]

Initial Values Look Similar

Both

\[ L(x)=mx+b \]

and

\[ E(x)=ab^x \]

can be organized around a value at \(x=0\).

For the linear model,

\[ L(0)=b. \]

For the exponential model,

\[ E(0)=a. \]

The difference lies in how the function moves away from that initial value.

Linear Change Adds

Suppose a quantity begins at \(100\) and increases by \(20\) each period.

\[ 100,\quad120,\quad140,\quad160,\quad180,\ldots \]

Each difference is

\[ \boxed{20}. \]

Therefore,

\[ \boxed{L(t)=100+20t}. \]

Exponential Change Multiplies

Suppose a quantity begins at \(100\) and increases by \(20\%\) each period.

The growth factor is

\[ 1+0.20=1.20. \]

The values are

\[ 100,\quad120,\quad144,\quad172.8,\quad207.36,\ldots \]

Therefore,

\[ \boxed{E(t)=100(1.2)^t}. \]

Same First Step, Different Pattern

The two previous examples both move from

\[ 100 \]

to

\[ 120 \]

during the first period.

But the next values differ:

\[ \text{linear: }140 \]

versus

\[ \text{exponential: }144. \]

A single observed change is not enough to distinguish the models.

Constant Differences

Consider

\[ 12,\quad17,\quad22,\quad27,\quad32. \]

Consecutive differences are

\[ 5,\quad5,\quad5,\quad5. \]

Therefore, the pattern is linear.

Constant Ratios

Consider

\[ 12,\quad18,\quad27,\quad40.5,\quad60.75. \]

Consecutive ratios are

\[ \frac{18}{12} = \frac{27}{18} = \frac{40.5}{27} = 1.5. \]

Therefore, the pattern is exponential.

The Difference Test

For equally spaced inputs, compute

\[ f(x+1)-f(x). \]

If this difference is constant, the data are consistent with a linear relationship:

\[ \boxed{f(x+1)-f(x)=m}. \]

The Ratio Test

For equally spaced inputs and nonzero outputs, compute

\[ \frac{f(x+1)}{f(x)}. \]

If this ratio is constant, the data are consistent with an exponential relationship:

\[ \boxed{\frac{f(x+1)}{f(x)}=b}. \]

Equal Input Intervals Matter

Difference and ratio tests must compare equal changes in the independent variable.

If a table uses

\[ x=0,2,4,6, \]

then adjacent rows represent two-unit input changes.

A constant ratio between rows is a two-unit factor, not necessarily the one-unit factor.

Linear Change over a General Interval

If

\[ L(x)=mx+b, \]

then for any input increment \(h\),

\[ L(x+h)-L(x)=mh. \]

Therefore, equal input increments produce equal output differences.

Exponential Change over a General Interval

If

\[ E(x)=ab^x, \]

then

\[ E(x+h)=ab^{x+h}=ab^xb^h. \]

Therefore,

\[ \boxed{\frac{E(x+h)}{E(x)}=b^h}. \]

Equal input increments produce equal multiplicative factors.

Slope versus Growth Factor

The central parameter in a linear function is its slope:

\[ \boxed{m}. \]

The central change parameter in an exponential function is its growth or decay factor:

\[ \boxed{b}. \]

Slope measures additive change per input unit. The factor measures multiplicative change per input unit.

Linear Rate Has Output Units per Input Unit

Suppose

\[ L(t)=500+30t, \]

where \(t\) is measured in years and \(L\) is measured in dollars.

The slope

\[ 30 \]

has units of dollars per year.

Exponential Factor Is a Ratio

Suppose

\[ E(t)=500(1.06)^t. \]

The factor

\[ 1.06 \]

means each year's value is \(1.06\) times the previous year's value.

It corresponds to \(6\%\) growth per year.

Fixed Amount versus Fixed Percentage

A fixed amount added each period produces linear change.

\[ \boxed{+25\text{ each period}\longrightarrow\text{linear}} \]

A fixed percentage applied to the current amount produces exponential change.

\[ \boxed{+25\%\text{ each period}\longrightarrow\text{exponential}} \]

Example: Salary Increase by Fixed Dollars

A salary begins at \(\$40{,}000\) and rises by \(\$1500\) each year.

The change is a constant amount, so a linear model is appropriate:

\[ \boxed{S(t)=40000+1500t}. \]

Example: Salary Increase by Fixed Percentage

A salary begins at \(\$40{,}000\) and rises by \(3\%\) each year.

The growth factor is

\[ 1.03. \]

Therefore,

\[ \boxed{S(t)=40000(1.03)^t}. \]

The Percentage Increase in a Linear Model Is Not Constant

Suppose a quantity gains \(20\) every period:

\[ 100,\quad120,\quad140,\quad160,\ldots \]

The first increase is

\[ \frac{20}{100}=20\%. \]

The next is

\[ \frac{20}{120}\approx16.67\%. \]

The absolute increase is constant, but the percentage increase changes.

The Absolute Increase in an Exponential Model Is Not Constant

Suppose a quantity grows by \(20\%\):

\[ 100,\quad120,\quad144,\quad172.8,\ldots \]

The increases are

\[ 20,\quad24,\quad28.8,\ldots \]

The percentage is constant, but the absolute increase changes.

Recognizing Linear Change from a Table

\[ \begin{array}{c|c} x & y\\ \hline 0 & 7\\ 1 & 11\\ 2 & 15\\ 3 & 19 \end{array} \]

Consecutive differences are

\[ 4,\quad4,\quad4. \]

Therefore,

\[ \boxed{y=4x+7}. \]

Recognizing Exponential Change from a Table

\[ \begin{array}{c|c} x & y\\ \hline 0 & 7\\ 1 & 14\\ 2 & 28\\ 3 & 56 \end{array} \]

Consecutive ratios are

\[ 2,\quad2,\quad2. \]

Therefore,

\[ \boxed{y=7(2)^x}. \]

Building a Linear Model from a Table

If the table begins at

\[ y=12 \]

when \(x=0\), and each one-unit input step adds \(5\), then

\[ b=12 \]

and

\[ m=5. \]

Therefore,

\[ \boxed{L(x)=12+5x}. \]

Building an Exponential Model from a Table

If the table begins at

\[ y=12 \]

when \(x=0\), and each output is \(1.5\) times the previous output, then

\[ a=12 \]

and

\[ b=1.5. \]

Therefore,

\[ \boxed{E(x)=12(1.5)^x}. \]

Graph Shape: Linear

The graph of

\[ y=mx+b \]

is a straight line.

Equal horizontal movements produce equal vertical changes.

Its slope is constant everywhere.

Graph Shape: Exponential Growth

For

\[ y=ab^x,\qquad a>0,\quad b>1, \]

the graph curves upward and becomes increasingly steep as \(x\) increases.

Equal horizontal movements multiply the output by the same factor rather than adding the same amount.

Graph Shape: Exponential Decay

For

\[ y=ab^x,\qquad a>0,\quad0 the graph falls quickly at first and then flattens toward its horizontal asymptote.

The decrease remains proportional rather than constant in absolute size.

A Linear Function Has Constant First Differences

For equally spaced inputs, a linear function produces a sequence whose first differences are constant.

This makes first differences a practical diagnostic for tables of data.

An Exponential Function Has Constant First Ratios

For equally spaced inputs, an exponential function produces a sequence whose consecutive ratios are constant.

This makes first ratios the corresponding multiplicative diagnostic.

Long-Run Linear Growth

If \(m>0\), then

\[ L(x)=mx+b \]

grows without bound as \(x\to\infty\).

But each additional input unit contributes only the same fixed amount \(m\).

Long-Run Exponential Growth

If

\[ a>0,\qquad b>1, \]

then

\[ E(x)=ab^x \]

also grows without bound.

But its absolute increase grows with the current value, so sustained exponential growth eventually outpaces sustained positive linear growth.

Exponential Growth Can Start Below Linear Growth

A common misconception is that an exponential function must immediately be larger than a linear function.

It may begin below the linear function and remain below it for some time.

The key long-run fact is that a positive exponential growth function with base greater than \(1\) eventually outgrows any fixed linear function.

Example: Same Initial Value

Compare

\[ L(t)=100+20t \]

and

\[ E(t)=100(1.2)^t. \]

Both begin at

\[ 100. \]

Both equal \(120\) after one period.

Their later behavior diverges because one adds \(20\) while the other multiplies by \(1.2\).

Compare the Models after Six Periods

Linear:

\[ L(6)=100+20(6)=220.00. \]

Exponential:

\[ E(6)=100(1.2)^6\approx298.60. \]

By this point, the multiplicative process is already growing faster in absolute amount.

Comparing Absolute Changes

For

\[ L(t)=100+20t, \]

every one-period change is

\[ 20. \]

For

\[ E(t)=100(1.2)^t, \]

the one-period change at time \(t\) is

\[ E(t+1)-E(t) = 0.2E(t). \]

The exponential increase is \(20\%\) of the current value.

When Does the Exponential Increase Exceed the Linear Increase?

In the preceding example, the linear model always gains \(20\) per period.

The exponential model gains

\[ 0.2E(t). \]

This increase exceeds \(20\) whenever

\[ 0.2E(t)>20, \]

or

\[ E(t)>100. \]

After the initial period, the exponential model's absolute increases become larger.

Crossover Points

A linear and exponential model can intersect one or more times.

A crossover point satisfies

\[ L(x)=E(x). \]

Such equations often require numerical methods, graphing, or logarithms, depending on the specific forms.

A Simple Crossover Equation

Compare

\[ L(t)=600 \]

and

\[ E(t)=100(1.2)^t. \]

The crossover occurs when

\[ 100(1.2)^t=600. \]

Therefore,

\[ (1.2)^t=6 \]

and

\[ \boxed{ t=\frac{\ln6}{\ln1.2}\approx9.83 }. \]

Linear Decay

Linear change can also decrease.

For example,

\[ L(t)=100-8t \]

subtracts \(8\) during each period.

The constant difference is

\[ \boxed{-8}. \]

Exponential Decay

Exponential decay repeatedly multiplies by a factor between \(0\) and \(1\).

For example,

\[ E(t)=100(0.92)^t \]

loses \(8\%\) of the current amount each period.

The constant ratio is

\[ \boxed{0.92}. \]

Linear Decay Can Reach Zero in Finite Time

For

\[ L(t)=100-8t, \]

set \(L(t)=0\):

\[ 100-8t=0. \]

Therefore,

\[ t=12.5. \]

The mathematical line continues below zero unless the model's context stops it.

Positive Exponential Decay Approaches Zero

For

\[ E(t)=100(0.92)^t, \]

every finite-time value is positive.

The function approaches zero:

\[ \boxed{E(t)\to0\text{ as }t\to\infty}, \]

but does not reach zero at a finite real input.

Words That Often Suggest Linear Change

Phrases such as these often indicate additive structure:

The important clue is a fixed amount per equal interval.

Words That Often Suggest Exponential Change

Phrases such as these often indicate multiplicative structure:

The important clue is a fixed factor or percentage applied repeatedly.

Do Not Classify by the Word “Growth” Alone

Both linear and exponential relationships can describe growth.

The phrase

“grows by 500 people each year”

is linear.

The phrase

“grows by 5% each year”

is exponential.

Do Not Classify by Curving or Increasing Alone

Many nonlinear functions curve, and many nonexponential functions increase.

A graph that rises is not automatically exponential.

Look for multiplicative structure, constant ratios, or an exponential formula rather than relying only on visual appearance.

Comparing Two Models after Ten Periods

Compare

\[ L(t)=500+40t \]

with

\[ E(t)=500(1.08)^t. \]

After \(10\) periods,

\[ L(10)=900.00 \]

while

\[ E(10)\approx1079.46. \]

A comparison at one time is useful, but model type is still determined by the additive or multiplicative rule.

A Larger Linear Value Can Still Belong to the Slower Long-Run Model

At a particular time, a linear function may be larger than an exponential growth function.

That does not make the linear function exponential, nor does it overturn the eventual-growth comparison.

Model type describes structure, not which output happens to be larger at one selected input.

Linear and Exponential Models Can Approximate Each Other Briefly

Over a sufficiently short interval, a slowly growing exponential curve may look nearly straight.

This is one reason short data sets can make model selection difficult.

Additional observations, context, and residual analysis can help determine whether a linear or exponential model is more appropriate.

Model Selection Uses Context as Well as Data

A constant-ratio table strongly suggests exponential structure, but real measurements often include noise.

In applications, ask what mechanism makes sense:

Linear Models Have Constant Average Rate over Every Interval

For a linear function,

\[ L(x)=mx+b, \]

the average rate of change over any interval is

\[ \boxed{m}. \]

This is why the graph is a straight line.

Exponential Average Rate of Change Depends on the Interval

For

\[ E(x)=ab^x, \]

the average additive rate of change is not constant.

As a positive exponential growth function becomes larger, its absolute changes generally become larger as well.

Relative Change Is Natural for Exponentials

For one-unit steps,

\[ \frac{E(x+1)-E(x)}{E(x)} \]

measures relative change.

Since

\[ E(x+1)=bE(x), \]

we obtain

\[ \frac{E(x+1)-E(x)}{E(x)} = b-1. \]

Therefore,

\[ \boxed{\text{relative change per step}=b-1}. \]

Relative Change Is Not Constant for Ordinary Linear Growth

For

\[ L(x)=mx+b, \]

the one-step relative change is

\[ \frac{m}{L(x)}. \]

Unless \(m=0\), this quantity changes as \(L(x)\) changes.

Linear models naturally preserve absolute change, not relative change.

Constant Functions as a Boundary Case

A constant function

\[ f(x)=c \]

can be viewed as linear with slope

\[ m=0. \]

The exponential expression

\[ c(1)^x \]

is also constant, but standard exponential-function definitions exclude base \(1\) because it does not exhibit exponential growth or decay and is not one-to-one.

Common Error: Confusing \(+10\) with \(+10\%\)

Increasing by \(10\) each period is linear:

\[ f(t)=a+10t. \]

Increasing by \(10\%\) each period is exponential:

\[ f(t)=a(1.10)^t. \]

The percent sign changes the mathematical structure completely.

Common Error: Looking Only at the First Two Values

Any two distinct points determine a line, and many exponential models can also pass through those two points.

Two observations alone generally do not reveal whether the underlying process has a constant difference or constant ratio.

Additional data or contextual information are needed.

Common Error: Using Differences on Unequally Spaced Inputs

If input intervals differ, raw output differences should not be compared as though the time steps were equal.

First account for the different input spacing or compute rates over the actual intervals.

Common Error: Calling Any Percentage Problem Exponential

A percentage can describe a one-time calculation without defining an exponential process.

Exponential change requires the percentage or multiplicative factor to be applied repeatedly to the current quantity over equal intervals.

Common Error: Assuming Exponential Growth Is Always Realistic Forever

A mathematical exponential model can grow without bound.

Real systems may face resource limits, capacity constraints, policy changes, saturation, or other effects that eventually make a different model more appropriate.

Common Error: Assuming Linear Models Continue Forever

Linear models also have limits of usefulness.

A line describing a declining quantity may eventually predict impossible negative values.

Mathematical continuation and practical validity are different questions.

A Reliable Procedure for Classifying a Table

  1. Check whether the input values are equally spaced.
  2. Compute consecutive output differences.
  3. If the differences are constant, test a linear model.
  4. If the outputs are nonzero, compute consecutive output ratios.
  5. If the ratios are constant, test an exponential model.
  6. Identify the initial value if \(x=0\) is present.
  7. Construct the model and verify it against every available row.

A Reliable Procedure for Classifying a Verbal Model

  1. Identify the starting quantity.
  2. Identify what happens during each equal interval.
  3. If a fixed amount is added or subtracted, use a linear model.
  4. If a fixed factor or percentage acts on the current value, use an exponential model.
  5. Match the rate or factor to the correct units of the input variable.
  6. Check whether the model assumptions make sense for the intended interval.

Example: Full Classification Workflow

Consider

\[ \begin{array}{c|c} t & Q(t)\\ \hline 0 & 200\\ 1 & 230\\ 2 & 264.5\\ 3 & 304.175 \end{array} \]

Differences are

\[ 30,\quad34.5,\quad39.675, \]

so they are not constant.

Ratios are

\[ \frac{230}{200} = \frac{264.5}{230} = \frac{304.175}{264.5} = 1.15. \]

Therefore, the model is exponential:

\[ \boxed{Q(t)=200(1.15)^t}. \]

Example: Full Linear Workflow

Consider

\[ \begin{array}{c|c} t & Q(t)\\ \hline 0 & 200\\ 1 & 230\\ 2 & 260\\ 3 & 290 \end{array} \]

Every difference is

\[ 30. \]

Therefore, the model is linear:

\[ \boxed{Q(t)=200+30t}. \]

Practice 1: Linear or Exponential?

Classify the sequence:

\[ 10,\quad15,\quad20,\quad25,\quad30. \]

Practice 2: Linear or Exponential?

Classify the sequence:

\[ 10,\quad15,\quad22.5,\quad33.75,\quad50.625. \]

Practice 3: Build a Linear Model

A quantity begins at \(75\) and increases by \(12\) each month. Write a model.

Practice 4: Build an Exponential Model

A quantity begins at \(75\) and increases by \(12\%\) each month. Write a model.

Practice 5: Classify a Table

\[ \begin{array}{c|c} x & y\\ \hline 0 & 4\\ 1 & 12\\ 2 & 36\\ 3 & 108 \end{array} \]

Classify the relationship and write a model.

Practice 6: Classify Another Table

\[ \begin{array}{c|c} x & y\\ \hline 0 & 4\\ 1 & 12\\ 2 & 20\\ 3 & 28 \end{array} \]

Classify the relationship and write a model.

Practice 7: Fixed Amount or Fixed Percentage?

A town gains \(900\) residents each year. Is a linear or exponential model more appropriate under that assumption?

Practice 8: Fixed Amount or Fixed Percentage?

A town grows by \(2.4\%\) each year. Is a linear or exponential model more appropriate under that assumption?

Practice 9: Compare Two Models

Let

\[ L(t)=100+25t \]

and

\[ E(t)=100(1.25)^t. \]

Find both values at \(t=4\).

Practice 10: Explain the Difference

Explain why

\[ 500+0.05t \]

does not represent \(5\%\) growth per period, while

\[ 500(1.05)^t \]

does.

Practice Answers

Practice 1

Consecutive differences are

\[ 5,\quad5,\quad5,\quad5. \]

Therefore,

\[ \boxed{\text{linear}}. \]

Practice 2

Consecutive ratios are

\[ 1.5,\quad1.5,\quad1.5,\quad1.5. \]

Therefore,

\[ \boxed{\text{exponential}}. \]

Practice 3

The initial value is \(75\), and the constant additive change is \(12\):

\[ \boxed{L(t)=75+12t}. \]

Practice 4

The growth factor is

\[ 1+0.12=1.12. \]

Therefore,

\[ \boxed{E(t)=75(1.12)^t}. \]

Practice 5

Ratios are constant:

\[ 3,\quad3,\quad3. \]

Therefore,

\[ \boxed{y=4(3)^x}. \]

Practice 6

Differences are constant:

\[ 8,\quad8,\quad8. \]

Therefore,

\[ \boxed{y=4+8x}. \]

Practice 7

A fixed number of residents is added each year, so

\[ \boxed{\text{linear}} \]

is the appropriate structure under the stated assumption.

Practice 8

A fixed percentage of the current population is applied each year, so

\[ \boxed{\text{exponential}} \]

is the appropriate structure.

Practice 9

\[ L(4)=100+25(4)=200. \]

For the exponential model,

\[ E(4)=100(1.25)^4. \] \[ E(4)=244.140625. \]

Therefore,

\[ \boxed{L(4)=200,\qquad E(4)\approx244.14}. \]

Practice 10

The expression

\[ 500+0.05t \]

adds a fixed amount of \(0.05\) per period. It does not repeatedly apply \(5\%\) to the current value.

The expression

\[ 500(1.05)^t \]

repeatedly multiplies by \(1.05\), so it models

\[ \boxed{5\%\text{ growth per period}}. \]

What to Remember

Linear and exponential relationships differ in the operation that repeats.

\[ \boxed{L(x)=mx+b} \]

preserves constant additive change, while

\[ \boxed{E(x)=ab^x} \]

preserves constant multiplicative change.

The central diagnostic is: ask whether each equal step adds the same amount or multiplies by the same factor.

Where We Go Next

Linear and exponential functions are two major models of change, but they are not the only ones useful in applications.

The next lesson, Exponential, Logarithmic, and Logistic Models, compares several related model families, including unrestricted exponential growth, inverse logarithmic relationships, and logistic growth that slows as a system approaches a limiting capacity.