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2A. Exponential and Logarithmic Functions
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AP Precalculus

2.2B Change in linear and exponential functions

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Lesson

Introduction

Learning objectives

  • 2.2.A Describe how the input and output values of a function vary together by comparing function values.
  • 2.2.B Describe similarities and differences between linear and exponential functions.

Exponential functions and geometric sequences

Geometric sequences have an exponential relationship because the terms share a common ratio.

Recall an exponential function can be written in the form:

\displaystyle f\left(x\right)=y_i r^{\left(x-x_i\right)}
\bm{r}
common ratio
\bm{\left(x_i,y_i\right)}
a known point

A geometric sequence is represented by the formula:

\displaystyle g_n=g_k r^{\left(n-k\right)}
\bm{g_n}
nth term
\bm{g_k}
a known (kth) term
\bm{r}
common ratio
\bm{n}
term number
\bm{k}
term number of the known (kth) term

Geometric sequences with a common ratio greater than 1 can model exponential growth.

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a(n)=\left(2\right)^{n-1}
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f(x)=\dfrac{1}{2}\left(2\right)^x

Geometric sequences with a common ratio between 0 and 1, non-inclusive, can model exponential decay.

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a(n)=32\left(\dfrac{1}{2}\right)^{n-1}
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f(x)=64\left(\dfrac{1}{2}\right)^x

Geometric sequences with a negative common ratio cannot be used to model exponential growth or decay, as we can see in this graph:

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This is because the base of an exponential function is restricted. The base, or constant factor, must be greater than zero but not equal to one.

Geometric sequences can have a negative common ratio, but exponential functions cannot have a negative constant factor.

Explicit rule for this geometric sequence:

a_n=(-2)^{n-1}

Just like arithmetic sequences, the domain of any geometric sequence is a subset of the integers, usually starting from 0 or 1.

Examples

Example 1

A group of students is working on a project that is due in 6 weeks. To determine how long it would take them to complete the project, they created little tasks that they could do along the way and came up with a total of 324 tasks. They worked on the project frequently at the beginning of the month but had to work on other projects later in the month.

The table below shows the number of tasks left to complete each week which can be represented by a geometric sequence.

n1234
a_n3241083612
a

Identify the common ratio.

Worked Solution
Create a strategy

We can find the common ratio by dividing a term by the previous term.

Apply the idea

Since we already know this is a geometric sequence, we can use any pair of values whose n-values are 1 unit apart. For example: \dfrac{12}{36}=\dfrac{1}{3}

The common ratio is \dfrac{1}{3}.

Reflect and check

We could have chosen other values and arrived at the same result. For example, \dfrac{108}{324}=\dfrac{1}{3} and \dfrac{36}{108}=\dfrac{1}{3}.

b

Determine the value of a_5.

Worked Solution
Create a strategy

Using the constant factor found in part (a), we know that as n increases by 1, a_n decreases by a factor of \dfrac{1}{3}. We can substitute r=\dfrac{1}{3} into the recursive formula and solve for the next term.

Apply the idea
\displaystyle a_n\displaystyle =\displaystyle \dfrac{1}{3} \cdot a_{n-1}Recursive formula
\displaystyle a_5\displaystyle =\displaystyle \dfrac{1}{3}\cdot a_4Substitute n=5
\displaystyle a_5\displaystyle =\displaystyle \dfrac{1}{3}\cdot 12Substitute a_4=12
\displaystyle a_5\displaystyle =\displaystyle 4Evaluate the multiplication

The 5th term is 4, so there are 4 tasks left to complete in the 5th week.

Reflect and check

You may have noticed that we are dividing each term by 3, but this is the same as multiplying each term by \dfrac{1}{3}.

324\div3=108

324\cdot \dfrac{1}{3}=108

Idea summary

A geometric sequence is an exponential function because it has a common ratio.

\displaystyle g_n=g_k r^{\left(n-k\right)}
\bm{g_n}
nth term
\bm{g_k}
a known (kth) term
\bm{r}
common ratio
\bm{n}
term number
\bm{k}
term number of the known (kth) term

Comparing linear and exponential functions

Key features of a function are useful in helping to sketch the function, as well as to interpret information about the function in a given context.

The characteristics, or key features, of a function include its:

  • domain and range

  • x- and y-intercepts

  • maximum or minimum value(s)

  • rate of change over specific intervals

  • end behavior

  • positive and negative intervals

  • increasing and decreasing intervals

Exploration

Leilani and Koda each open a bank account with \$100. Leilani's account will earn 3\% interest every month. Koda's account will earn \$9 every month.

  1. Who will have more money in the short term?
  2. Who will have more money in the long term?
  3. How do the rates of change differ?

We can use key features to compare linear and exponential functions. Many of their features are similar, but their rates of change are different. A linear function has a constant rate of change while an exponential function has a constant percent rate of change.

This means that we are adding the same number to each output of a linear function, but we are multiplying the same number to each output of an exponential function. Multiplication grows faster than addition, so a quantity increasing exponentially will always exceed a quantity increasing linearly over time.

Examples

Example 2

Consider the two functions shown in the graphs below.

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a

State the intercepts of each function.

Worked Solution
Create a strategy

We need to look for both the x-intercepts and the y-intercepts. We also need to state them as ordered pairs.

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Apply the idea

Both functions have a y-intercept at \left(0, -1\right).

Also, both functions have an x-intercept at \left(1, 0\right).

b

Compare the end behavior of the two functions.

Worked Solution
Apply the idea

On the right side, both functions take larger and larger positive values as x gets further from zero. That is, as x \to \infty, y \to \infty for both functions.

A line passing through points (1,0) and (0, -1) plotted on a four quadrant coordinate plane. On the first quadrant, there is a vertically upward arrow labeled y approaches infinity and a right arrow labeled x approaches infinity. On the third quadrant, there is a vertically downward arrow labeled y approaches negative infinity and a left arrow labeled x approaches negative infinity.

On the left side, f\left(x\right) takes larger and larger negative values as x gets further from zero. That is, as x \to -\infty, y \to -\infty for f\left(x\right).

An exponential curve passing through points (1,0) and (0, -1) and has an asymptote at y = negative 2 plotted on a four quadrant coordinate plane. On the first quadrant, there is a vertically upward arrow labeled y approaches infinity and a right arrow labeled x approaches infinity. Just above the negative x axis, there is a left arrow labeled x approaches negative infinity. On the third quadrant, there is a left arrow labeled y approaches negative 2.

On the other hand, the values of g\left(x\right) get closer and closer to y=-2 as x gets further from zero on the left side. That is, as x \to -\infty, y \to -2 for g\left(x\right).

Reflect and check

Note that although both functions tend towards infinity to the right, the way they do so is different. The first function increases at a constant rate, while the second function increases at an increasing rate.

c

Using the graph of each function, find where f\left(x\right)=g\left(x\right).

Worked Solution
Create a strategy

We can look for the points where f\left(x\right) and g\left(x\right) would intersect if drawn on the same plane.

Apply the idea

If we draw f\left(x\right) on the same plane as g\left(x\right), we get:

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We can see that our two functions intersect at \left(0,-1\right) and \left(1,0\right). This means that at x=0 and x=1 the two functions are equal to eachother.

Reflect and check

If we tried to solve this equation algebraically, we would have to solve something of the form ab^{x}=mx+b. Since this is hard to solve, we can plot the graphs of each side of the equation and use the point of intersection to solve the equations instead.

d

Compare the average rate of change of each function over the following intervals:

  • -1<x<0
  • 0<x<1
  • 1<x<2
Worked Solution
Create a strategy

We calculate the average rate of change with the formula \dfrac{f\left(b\right)-f\left(a\right)}{b-a}.

Apply the idea
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For the linear function:

\dfrac{f(0)-f(1)}{0-(-1)}=\dfrac{-1-(-2)}{1}=1

\dfrac{f(1)-f(0)}{1-0}=\dfrac{0-(-1)}{1}=1

\dfrac{f(2)-f(1)}{2-1}=\dfrac{1-0}{1}=1

The average rate of change is constant. The values are increasing by 1 each time.

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For the exponential function:

\dfrac{f(0)-f(-1)}{0-(-1)}=\dfrac{-1-(-1.5)}{1}=0.5

\dfrac{f(1)-f(0)}{1-0}=\dfrac{0-(-1)}{1}=1

\dfrac{f(2)-f(1)}{2-1}=\dfrac{2-0}{1}=2

The average rate of change is not constant. The values are changing by a greater amount each time.

Although both functions are increasing, the exponential function is increasing at a faster rate.

Reflect and check

Over the interval before the first point of intersection, -1<x<0, the exponential function actually has an average rate of change that is less than the linear function. However, as x increased we saw a rapid increase in the average rate of change. So, the exponential function is still increasing at a faster rate.

Example 3

Consider the functions shown in the graph and table below.

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g\left(x\right)0.40960.5120.640.811.251.56251.95312.4414
a

State whether each function is linear or exponential.

Worked Solution
Create a strategy

We can see from the graph of f\left(x\right) that it forms a straight line.

For the table of g\left(x\right), we need to look at the differences or ratios of the outputs to determine if it in linear or exponential.

Apply the idea

We will start by finding the differences in the outputs of g\left(x\right).

0.512-0.4096=0.1024

0.64-0.512=0.128

0.8-0.64=0.16

Since there isn't a common difference, we know the function is not linear. Next, we will find the ratios of the outputs.

0.512\div 0.4096=1.25

0.64\div 0.512=1.25

0.8\div 0.64=1.25

Since the outputs have a common ratio, g\left(x\right) is exponential. Since f\left(x\right) increases by a constant amount, it is linear.

b

Compare the intervals where the function is increasing and decreasing for each function.

Worked Solution
Create a strategy

Since f\left(x\right) is linear, we know it will only increase or decrease over its domain. We can determine whether f\left(x\right) is increasing or decreasing by looking at its graph.

Similarly, exponential functions only increase or decrease over their domain. We can determine whether g\left(x\right) is increasing or decreasing by looking at the outputs in the table.

Apply the idea

Looking at the end behavior of the graph of f\left(x\right), we see that the y-values are growing as x increases.

Looking at the function values (outputs) of the table of g\left(x\right), the values are getting larger as x increases.

Both functions are increasing over their domain.

c

Determine which function will have a higher value as x increases.

Worked Solution
Create a strategy

We already know that f\left(x\right) increases linearly while g\left(x\right) increases exponentially. We can use what we know about the rates of change of linear and exponential functions to determine which one will exceed the other.

Apply the idea

Since f\left(x\right) will grow by a constant amount but g\left(x\right) will grow by a constant percentage, g\left(x\right) will have a higher value as x increases.

Reflect and check

Using technology to graph both functions and extending the x-axis, we can confirm that the exponential function will exceed the linear function for higher values of x.

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Example 4

Two objects are depreciating in value as shown in the table below:

Number of years01234
Object A\$7\,500\$6\,000\$4\,800\$3\,840\$3\,072
Object B\$12\,000\$9\,000\$6\,750\$5\,062.50\$3\,796.88
a

Determine whether each object is decreasing linearly or exponentially.

Worked Solution
Create a strategy

A linear function has a constant rate of change. An exponential function has a constant percent rate of change. We need to determine if the values of each object have a common difference or a common ratio.

Apply the idea

By observing the first 3 values of each object, we can see that the objects are decreasing by different amounts.

For object A:

6000-7500=-1500

4800-6000=-1200

For object B:

9000-12000=-3000

6750-9000=-2250

This means that neither function is linear, so let's see if they are exponential by finding the ratios of the amounts.

For object A:

6\,000\div 7\,500=0.8

4\,800\div 6\,000=0.8

3\,840\div 4\,800=0.8

The values do have a common ratio, so object A depreciates exponentially. For object B:

9\,000\div 12,000=0.75

6\,750\div 9\,000=0.75

5\,062.5\div 6\,750 =0.75

The outputs share a common ratio, so both functions are exponential.

Reflect and check

Remember that exponential functions are in the form y=ab^x. When 0<b<1,the function decays exponentially.

Exponential decay function for object A: y=7500\left(0.8\right)^x

Exponential decay function for object B: y=12000\left(0.75\right)^x

b

Describe the rate of change of each object.

Worked Solution
Create a strategy

We already know the constant factors of the exponential functions that model the decay, but let's describe the rate of change as a constant percentage. Remember: {\text{Decay rate}=1-\text{Decay factor}}.

Apply the idea

The decay factor of object A is 0.8, so the decay rate is r=1-0.8=0.2. Multiplying that by 100 gives us 20 \%.

The decay factor of object B is 0.75, so the decay rate is r=1-0.75=0.25. Multiplying that by 100 gives us 25 \%.

Object A depreciates by 20\% while object B depreciates by 25\%.

c

Determine which object will have a higher value after 10 years.

Worked Solution
Create a strategy

Both functions are decreasing at a decreasing rate. However, the function with the lower percentage will decrease slower than the other function over time.

Apply the idea

Since object A only loses 20\% of its value each year while object B loses 25\% of its value each year, Object A will decrease at a slower rate. This means Object A have a higher value after 10 years.

Reflect and check

We can verify our answer by using the functions we found in the reflection from part (a):

Object A: y=7500\left(0.8\right)^{10}\approx \$805.31

Object B: y=12000\left(0.75\right)^{10}\approx \$675.76

Example 5

Ethan is playing a new game on his phone. After successfully playing his first game on day 1, he was awarded 25 diamonds. The game then rewards him with 3 diamonds for each consecutive day he plays after day 1.

a

Determine if the number of diamonds he has after n days of consecutive play is linear or exponential.

Worked Solution
Create a strategy

To identify whether the given problem is linear or exponential, we need to check if the situation represents an arithmetic or geometric sequence. To do this, we should determine if there is a common difference or a common ratio.

Apply the idea

The number of diamonds increases by 3 each day, so there is a common difference of d=3. Therefore, the problem involves linear growth.

b

Determine the number of diamonds Ethan will have after playing 6 consecutive days.

Worked Solution
Create a strategy

We can use the formula for the nth term of an arithmetic sequence, {a\left(n\right)=a\left(1\right) + d\left(n-1\right)} to write an explicit rule in function notation that models this scenario.

Apply the idea

First, we must find an explicit rule for the number of diamonds Ethan gets each day on his game:

\displaystyle a\left(n\right)\displaystyle =\displaystyle a\left(1\right) + d\left(n-1\right)General rule for an arithmetic sequence
\displaystyle a\left(n\right)\displaystyle =\displaystyle 25 + 3\left(n-1\right)Substitute a\left(1\right) = 25 and d=3
\displaystyle a\left(n\right)\displaystyle =\displaystyle 25+3n-3Distributive property
\displaystyle a\left(n\right)\displaystyle =\displaystyle 3n+22Combine like terms

Now we can use our explicit rule to find a\left(6\right).

\displaystyle a\left(n\right)\displaystyle =\displaystyle 3n+22Explicit rule
\displaystyle a\left(6\right)\displaystyle =\displaystyle 3\left(6\right)+22Substitute n=6
\displaystyle a\left(6\right)\displaystyle =\displaystyle 40Evaluate the multiplication and addition

Ethan will have 40 diamonds on the 6th day.

Reflect and check

The domain for this sequence, which represents the number of days Ethan plays the game, is the set of natural numbers.

c

Find the number of consecutive days Ethan will need to play to earn 160 diamonds.

Worked Solution
Create a strategy

We can use the rule from part (b), a\left(n\right)=3n+22, to find the number of days, n, it will take to get a\left(n\right)=160 diamonds.

Apply the idea
\displaystyle a(n)\displaystyle =\displaystyle 3n+22Explicit rule
\displaystyle 160\displaystyle =\displaystyle 3n+22Substitute a(n)=160
\displaystyle 138\displaystyle =\displaystyle 3nSubtract 22 from both sides
\displaystyle 46\displaystyle =\displaystyle nDivide both sides by 3
\displaystyle n\displaystyle =\displaystyle 46Symmetric property of equality

It will take Ethan 46 days to earn 160 diamonds.

Reflect and check

Alternatively, we can find the number of additional days after day 1 in a more conceptual way without using the equation:

\displaystyle 160-25\displaystyle =\displaystyle 135Number of diamonds Ethan needs to earn after day 1
\displaystyle 135 \div 3\displaystyle =\displaystyle 45Number of days Ethan needs to play after day 1
\displaystyle 45+1\displaystyle =\displaystyle 46Total number of days Ethan needs to play

Example 6

A ball is dropped onto the ground from a height of 8 \text{ m}. On each bounce, the ball reaches a maximum height of 60\% of its previous maximum height.

a

Determine if the heights of each bounce can be represented linearly or exponentially.

Worked Solution
Create a strategy

To identify whether the given problem is linear or exponential, we should determine if the situation represents an arithmetic or geometric sequence. To do this, we should determine if there is a common difference or a common ratio.

Apply the idea

In the problem, the height of the previous bounce is multiplied by 60\% or 0.6 to get the next height. So, the common ratio is r=0.6, and the sequence is geometric.

This situation can be modeled exponentially.

b

Determine the height that the ball reaches after the 4th bounce.

Worked Solution
Create a strategy

In part (a), we determined that the problem involves a geometric sequence with a common ratio of r=0.6. We also know that the ball had an initial height of 8\text{ m}, so a\left(0\right)=8. We are using a(0) since the initial height occurred before the first bounce, and n represents the number of the bounce.

To find a\left(1\right), we can multiply a\left(0\right) by the common ratio.

Apply the idea
\displaystyle a\left(1\right)\displaystyle =\displaystyle a\left(0\right)\cdot rRecursive rule
\displaystyle a\left(1\right)\displaystyle =\displaystyle 8 \cdot 0.8Substitute a\left(0\right)=8 and r=0.6
\displaystyle =\displaystyle 4.8Evaluate the multiplication

Now that we know the first term, we can use the explicit rule to find the height of the 4th bounce, where n=4:

\displaystyle a\left(n\right)\displaystyle =\displaystyle 4.8 \cdot 0.6^{n-1}Explicit rule
\displaystyle a\left(4\right)\displaystyle =\displaystyle 4.8 \cdot 0.6^{4-1}Substitute n=4
\displaystyle a\left(4\right)\displaystyle =\displaystyle 1.0368Evaluate the exponent and multiplication

The height of the 4th bounce will be 1.0368\text{ m}.

Reflect and check

To use the initial term instead of finding the first term, we can adjust the explicit rule to be:

a\left(n\right)=a\left(0\right)r^n

a\left(4\right)=8\left(0.6\right)^4=1.0368

Idea summary

A linear function has a constant rate of change while an exponential function has a constant percent rate of change. A quantity increasing exponentially will always exceed a quantity increasing linearly over time.

Outcomes

2.2.A

Construct functions of the real numbers that are comparable to arithmetic and geometric sequences.

2.2.B

Describe similarities and differences between linear and exponential functions.

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