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This lesson will work with real-life applications of exponential and logarithmic functions.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Logarithmic Function
Exponential Function
Exponential Growth
Exponential Decay

Explore

Graphing in the $(x, ln y)$ Coordinate Plane

Exponential functions have nonlinear graphs. In the diagram, the graph of $y = 2^{x}$ and some of its points are shown.

The graph of the exponential function and some points on it

Consider transforming these points into a different coordinate plane, where the horizontal axis represents the variable

x

and the vertical axis represents

ln y .

To do so, manipulate

y = 2^{x}

to find a rule for

ln y,

then make a table of values for values of

x

between

- 2

and

2 .

Finally, locate the points in the following

(x, ln y)

coordinate plane to draw the graph.

Identify the shape of the curve. When any exponential function is drawn on this coordinate plane, does a similar curve occur?

Example

Tables, Logarithmic Models, and Exponential Models

Heichi and Ignacio are good friends. However, when it comes to mathematics, they are pretty competitive. Neither Heichi nor Ignacio admits that the other person might be more knowledgeable in math.

Instead of letting this silly argument go, they decide to duel! They will challenge each other by asking one math question to the other.

a Ignacio presents a table that corresponds to the logarithmic function

y = 3 - 5 ln x .

$x$	$y$
$1$	$3$
$2$	$\approx - 0.47$
$3$	$\approx - 2.49$
$4$	$\approx - 3.93$
$5$	$\approx - 5.05$

Ignacio says it is impossible to find a logarithmic model of the form

x = a + b ln y

for this table. Heichi quickly counters, "It is totally possible." Ignacio replies, "Huh? That is bananas. No way, Heichi." Who is correct?

b Heichi, waiting for this moment to outsmart Ignacio, pulls out the following table from his notebook.

$x$	$y$
$1$	$1 k$
$2$	$2 k$
$3$	$4 k$
$4$	$8 k$
$5$	$16 k$

Assuming that

k

is a positive number, Heichi challenges Ignacio to write

y

as an exponential function of

x

in its simplest form. Help Ignacio solve Heichi's challenge!

Hint

a Isolate the variable

x .

What type of function is the resulting function?

b What is the constant multiplier between consecutive

y -

values?

Solution

a To find out who is correct, the

x -

variable will be isolated. Start by using inverse operations to isolate the natural logarithm.

y = 3 - 5 ln x

▼

Solve for

ln x

SubEqn

$LHS - 3 = RHS - 3$

y - 3 = - 5 ln x

DivEqn

$LHS / (- 5) = RHS / (- 5)$

\frac{y - 3}{- 5} = ln x

RearrangeEqn

Rearrange equation

ln x = \frac{y - 3}{- 5}

Now, use the definition of a natural logarithm to isolate

x .

ln x = \frac{y - 3}{- 5} \Leftrightarrow x = e^{\frac{y - 3}{- 5}}

When

y

is the independent variable and

x

the dependent variable, an exponential function is obtained.

Exponential Function x = e^{\frac{y - 3}{- 5}}

Therefore, it is not possible to find a logarithmic model when

y

is the independent variable and

x

the dependent variable. This means that a logarithmic model of the form

x = a + b ln y

is not possible for this table. Ignacio is correct.

b Paying close attention to the

y -

values of the table, a constant multiplier can be identified.

It is seen that the constant multiplier is

2 .

Also, the initial value is

1 k,

or just

k .

With this information, an exponential function can be written.

y = k (2)^{x - 1}

Ignacio pats himself on the back.

Example

Carbon $14$ in Vegetables

Heichi is upset that he lost to Ignacio in the last round. He decides to take a breather and turn to his favorite science book to practice some other problems. Here is one of his favorite topics: It is estimated that a preserved vegetable contains about $1$ microgram (a millionth of a gram) of Carbon $14 .$

The amount of Carbon

14

in the preserved vegetable can be modeled by the following exponential function.

C (t) = I (\frac{1}{2})^{\frac{t}{5 5 0 0}}

Here,

t

is the time in years that has passed since the vegetable's death, and

I

is the amount of Carbon

14

present in the vegetable at death, measured in micrograms. Heichi, feeling good about seeing a familiar topic, dives into solving the following problems.

a If the vegetable died

2000

years ago, how much Carbon

14

was present in the living vegetable? Round the answer to two decimal places.

b If the vegetable died

10000

years ago, how much Carbon

14

was present in the living vegetable? Round the answer to two decimal places.

c The amount it takes for half of the Carbon

14

to decay is called its half-life. What half-life does the expression for the given function

C (t)

imply for Carbon

14 ?

Hint

a Substitute

2000

for

t

in the given function.

b Substitute

10000

for

t

in the given function.

c The initial amount of Carbon

14

I .

Therefore, half the initial amount is

\frac{I}{2} .

Solution

a It is known that there is

1

microgram of Carbon

14

present in a vegetable that died

2000

years ago. To calculate how much Carbon

14

was present in the living vegetable,

2000

will be substituted for

t

and

1

will be substituted for

C (2000) .

Then, the equation can be solved for

I .

C (t) = I (\frac{1}{2})^{\frac{t}{5 5 0 0}}

Substitute

$t = 2000$

C (2000) = I (\frac{1}{2})^{\frac{2 0 0 0}{5 5 0 0}}

Substitute

$C (2000) = 1$

1 = I (\frac{1}{2})^{\frac{2 0 0 0}{5 5 0 0}}

▼

Solve for

I

ReduceFrac

$\frac{a}{b} = \frac{a / 5 0 0}{b / 5 0 0}$

1 = I (\frac{1}{2})^{\frac{4}{1 1}}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

1 = I (\frac{1}{2 ^{\frac{4}{1 1}}})

MultEqn

$LHS \cdot 2^{\frac{4}{1 1}} = RHS \cdot 2^{\frac{4}{1 1}}$

2^{\frac{4}{1 1}} = I (1)

IdPropMult

Identity Property of Multiplication

2^{\frac{4}{1 1}} = I

RearrangeEqn

Rearrange equation

I = 2^{\frac{4}{1 1}}

UseCalc

Use a calculator

I = 1.286664 \dots

RoundDec

Round to $2$ decimal place(s)

I \approx 1.29

Therefore, about

1.29

micrograms of Carbon

14

were present in the living vegetable.

b Similar to Part A, it is known that there is

1

microgram of Carbon

14

present in a vegetable that died

10000

years ago. This sentence can also be expressed mathematically.

C (10000) = 1

With this information, the amount of Carbon

14

that was present in the living vegetable

I

can be found by substituting

10000

for

t

and

1

for

C (10000) .

C (t) = I (\frac{1}{2})^{\frac{t}{5 5 0 0}}

Substitute

$t = 10000$

C (10000) = I (\frac{1}{2})^{\frac{1 0 0 0 0}{5 5 0 0}}

Substitute

$C (10000) = 1$

1 = I (\frac{1}{2})^{\frac{1 0 0 0 0}{5 5 0 0}}

▼

Solve for

I

ReduceFrac

$\frac{a}{b} = \frac{a / 5 0 0}{b / 5 0 0}$

1 = I (\frac{1}{2})^{\frac{2 0}{1 1}}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

1 = I (\frac{1}{2 ^{\frac{2 0}{1 1}}})

MultEqn

$LHS \cdot 2^{\frac{2 0}{1 1}} = RHS \cdot 2^{\frac{2 0}{1 1}}$

2^{\frac{2 0}{1 1}} = I (1)

IdPropMult

Identity Property of Multiplication

2^{\frac{2 0}{1 1}} = I

RearrangeEqn

Rearrange equation

I = 2^{\frac{2 0}{1 1}}

UseCalc

Use a calculator

I = 3.526365 \dots

RoundDec

Round to $2$ decimal place(s)

I \approx 3.53

Therefore, about

3.53

micrograms of Carbon

14

were present in the living vegetable.

c Half the initial amount of Carbon

14

present in the vegetable is

\frac{I}{2} .

Therefore, the half-life of Carbon

14

present in the vegetable can be found by substituting

\frac{I}{2}

for

C (t),

and isolating the variable

t .

C (t) = I (\frac{1}{2})^{\frac{t}{5 5 0 0}}

Substitute

$C (t) = \frac{I}{2}$

\frac{I}{2} = I (\frac{1}{2})^{\frac{t}{5 5 0 0}}

DivEqn

$LHS / I = RHS / I$

\frac{1}{2} = (\frac{1}{2})^{\frac{t}{5 5 0 0}}

$a = a^{1}$

(\frac{1}{2})^{1} = (\frac{1}{2})^{\frac{t}{5 5 0 0}}

Next, the Property of Equality for Exponential Equations can be used.

(\frac{1}{2})^{1} = (\frac{1}{2})^{\frac{t}{5 5 0 0}} \Leftrightarrow 1 = \frac{t}{5 5 0 0}

Finally, the value of

t

can be found by multiplying both sides of the obtained equation by

5500 .

1 = \frac{t}{5 5 0 0} \Leftrightarrow t = 5500

Therefore, the half-life of Carbon

14

implied by the given function is

5500

years. Heichi is feeling re-energized after solving for these solutions.

Example

Uranium $238$

Ignacio, feeling good about showing his prowess in math, goes to explore the chemistry lab. He sees none other than Heichi studying. Heichi invites Ignacio to see that Uranium $238$ is the most common isotope of uranium found in nature. It also has many applications in nuclear technology. Uranium $238$ decays exponentially and its half-life is about $4.5$ billion years.

Ignacio feels a sense of joy that Heichi has shown him this.

a Heichi, unable to control his competitive urge, metaphorically flips the tables and once again challenges Ignacio. "Write an exponential function expressing how much Uranium

238

remains after

t

billion years if the initial sample size is

40

grams. If you can!"

b Ignacio will not be taken down so easily. He asks Heichi which of the following graphs, in the context of this situation, corresponds to the function found in Part A.

Help Heichi determine the graph!

Hint

a Knowing that the uranium decays exponentially, the general equation for the remaining amount of uranium

238

after

t

billion years is

U (t) = a e^{- k t},

where

a

and

k

are positive numbers.

b Graph the function and compare it with the given choices. What is the

y -

intercept of the graph if the initial sample size is

40 ?

Solution

a Let

U (t)

be the amount of uranium

238

left from the sample, in grams, after

t

billions years. Knowing that the uranium decays exponentially, a general equation can be written.

U (t) = a e^{- k t}

Here,

a

and

c

are positive real numbers, where

a

is the initial amount. Since the initial sample size is

40

grams,

a = 40 .

U (t) = 40 e^{- k t}

Furthermore, the half-life of uranium

238

4.5

billion years. Therefore,

U (4.5)

is equal to

\frac{4 0}{2} = 20 .

With this information, starting with substitution, the value of

k

can be found.

U (t) = 40 e^{- k t}

Substitute

$t = 4.5$

U (4.5) = 40 e^{- k (4.5)}

Substitute

$U (4.5) = 20$

20 = 40 e^{- k (4.5)}

DivEqn

$LHS / 40 = RHS / 40$

\frac{2 0}{4 0} = e^{- k (4.5)}

ReduceFrac

$\frac{a}{b} = \frac{a / 2 0}{b / 2 0}$

\frac{1}{2} = e^{- k (4.5)}

MultNegPos

$(- a) b = - a b$

\frac{1}{2} = e^{- 4.5 k}

Now the definition of a natural logarithm can be applied to remove the variable from the exponent.

\frac{1}{2} = e^{- 4.5 k} \Leftrightarrow ln \frac{1}{2} = - 4.5 k

Finally, the value of

k

can be found by dividing both sides of the obtained equation by

- 4.5 .

ln \frac{1}{2} = - 4.5 k \Leftrightarrow k \approx 0.154

Now that the value of

k

was found, the desired function can be written.

U (t) = 40 e^{- 0.145 t}

Ignacio, once again, feels he has defeated Heichi's challenge. But wait! Heichi points out there is a mistake in Ignacio's calculation. The value of

k

is approximately

0.154,

as found before, and not

0.145,

as Ignacio wrote in the function rule. This means that the following is the desired function.

U (t) = 40 e^{- 0.154 t}

Ignacio gasps in despair, as Heichi gloats.

b It is now Heichi's turn to take on Ignacio's challenge to draw the graph of the function

U (t) = 40 e^{- 0.154 t} .

To do this, Heichi will make a table of values. Since time cannot be negative, only non-negative values of

t

will be considered.

$x$	$40 e^{- 0.154 t}$	$y$
$0$	$40 e^{- 0.154 (0)}$	$40$
$2$	$40 e^{- 0.154 (2)}$	$\approx 29.4$
$4$	$40 e^{- 0.154 (4)}$	$\approx 21.6$
$6$	$40 e^{- 0.154 (6)}$	$\approx 15.9$
$8$	$40 e^{- 0.154 (8)}$	$\approx 11.7$
$10$	$40 e^{- 0.154 (10)}$	$\approx 8.6$

Now, the points can be plotted and connected with a smooth curve. Recall that

t

cannot take negative values. Therefore, the negative part of the horizontal axis will not be considered.

This graph corresponds to choice A. Heichi answered correctly. He shrugs his shoulders and is ready to crown himself as the champion of Math.

Example

Modeling a Village's Population

Ignacio and Heichi see that this competition has turned ugly. They make a truce and decide to team up in studying for a geography test — together, as buddies.

However, the formulas did not allow the rivalry between Ignacio and Heichi to fade. While learning the concepts related to population, they find out that the population

P

of a small village in the north of Argentina is modeled by an exponential function.

P (t) = 500 (1.04)^{2 t}

Here,

t

is the time in years that has passed since the foundation of the village in the year

2020 .

Heichi and Ignacio cannot help themselves and start challenging each other.

a Ignacio says that by the year

2050

the population, rounded to the nearest integer, will be about

5260

people. Is this true or false?

b Heichi wants to find the year in which the population will be

3037

people, but does not know how to do it. Help Heichi find this year before Ignacio realizes that he cannot find the answer! Round the answer to the nearest integer.

Hint

a How many years are there between

2020

and

2050 ?

b Substitute

3037

for

P (t)

and solve for

t .

Then, add this value to

2020 .

Solution

a To determine whether Ignacio is correct, the population in the village for the year

2050

will be calculated. To do so, the difference between

2050

and

2020

needs to be found.

2050 - 2020 = 30 years

Next,

t = 30

will be substituted into the given equation.

P (t) = 500 (1.04)^{2 t}

Substitute

$t = 30$

P (30) = 500 (1.04)^{2 (30)}

▼

Evaluate right-hand side

Multiply

P (30) = 500 (1.04)^{60}

UseCalc

Use a calculator

P (30) = 5259.813704

RoundInt

Round to nearest integer

P (30) \approx 5260

The population of the village by the year

2050

will be about

5260

people. Therefore, Ignacio is correct. His eyes are gleaming with pride.

b To determine the year in which the population of the village will be about

3037

people, this number will be substituted for

P (t) .

Then, the variable

t

will be isolated.

P (t) = 500 (1.04)^{2 t}

Substitute

$P (t) = 3037$

3037 = 500 (1.04)^{2 t}

DivEqn

$LHS / 500 = RHS / 500$

\frac{3 0 3 7}{5 0 0} = (1.04)^{2 t}

UseCalc

Use a calculator

6.074 = (1.04)^{2 t}

Now, to eliminate the

t

from the exponent, the definition of a logarithm can be used.

6.074 = (1.04)^{2 t} ⇕ lo g_{1.04} 6.074 = 2 t

Next, the Change of Base Formula will be used to find the value of

t .

lo g_{1.04} 6.074 = 2 t

$lo g_{c} a = \frac{lo g _{b} a}{lo g _{b} c}$

\frac{lo g 6 . 0 7 4}{lo g 1 . 0 4} = 2 t

▼

Solve for

t

UseCalc

Use a calculator

45.996546 \dots = 2 t

DivEqn

$LHS / 2 = RHS / 2$

22.998273 \dots = t

RearrangeEqn

Rearrange equation

t = 22.998273 \dots

RoundInt

Round to nearest integer

t \approx 23

The population of the village will be

3037

people about

23

years after its foundation. This is in the year

2020 + 23 = 2043 .

That was super close — Ignacio almost noticed Heichi's struggle but not before Heichi found the solution.

Discussion

Relation Between Exponential and Linear Models

Recall the exploration seen at the beginning. In general, a set of more than two points

(x, y)

fits an exponential model if and only if the set of transformed points

(x, ln y)

fits a linear model. For example, consider the following function.

y = 2^{x}

This function can be graphed on an

(x, y)

coordinate plane.

Now, considering the given function, natural logarithms can be taken on both sides of the equation. Then, to simplify, the Power Property of Logarithms can be used.

y = 2^{x}

$ln (LHS) = ln (RHS)$

ln y = ln 2^{x}

$ln (a^{b}) = b \cdot ln (a)$

ln y = x ln 2

Next,

ln y = x ln 2

will be graphed on an

(x, ln y)

coordinate plane. To do this, a table of values will be constructed.

$x$	$x ln 2$	$ln y = x ln 2$
$- 2$	$- 2 ln 2$	$\approx - 1.39$
$- 1$	$- 1 ln 2$	$\approx - 0.69$
$0$	$0 ln 2$	$0$
$1$	$1 ln 2$	$\approx 0.69$
$2$	$2 ln 2$	$\approx 1.39$

Now, the points found in the table can be plotted and connected on an

(x, ln y)

coordinate plane.

While the graph of

y = 2^{x}

is an exponential curve in the

(x, y)

plane, the graph of

ln y = x ln 2

is a straight line in the

(x, ln y)

plane. It is worth noting that, in the

(x, ln y)

plane, the equation of a line is

ln y = m x + b,

where

m

is the slope and

b

the

y -

intercept.

Example

Visual Near Point

Heichi realizes that he might not be as great at math as Ignacio. To regain some confidence, he tells Ignacio that Ignacio's eyesight will worsen over time. "How can you prove that, Heichi?" asks Ignacio. "The visual near point of a person is the closest point at which an object can be placed and distinctly be observed by the person, and it varies with age," proudly states Heichi.

Ignacio admits his interest in the topic. He decides to learn more and wants to create a scatter plot for the data pairs

(x, ln y),

where

x

is the age in years and

y

the distance of the visual near point in centimeters. Then, he needs to write an exponential model for the original data pairs

(x, y) .

Give Ignacio a hand.

Answer

Scatter Plot:

Example Exponential Model: $y = 4.137 (1.054)^{x}$

Hint

To create the scatter plot make a table of values and plot the obtained points on an $(x, ln y)$ coordinate plane. To write the exponential model use any two of the obtained points.

Solution

To make the scatter plot, a table of data pairs $(x, ln y)$ will be created using the given information.

$x$	$y$	$ln y$
$20$	$12$	$ln 12 \approx 2.48$
$30$	$15$	$ln 15 \approx 2.71$
$40$	$25$	$ln 25 \approx 3.22$
$50$	$40$	$ln 40 \approx 3.69$
$60$	$100$	$ln 100 \approx 4.61$

Next, the obtained points will be plotted on an $(x, ln y)$ coordinate plane. Since all the values are positive, only the first quadrant will be considered.

The points are not collinear. However, they lie close to a straight line. Consider the line passing through the first and last points.

Since the set of points

(x, ln y)

fits a linear model, an exponential model should be a good fit for the original data. Now recall the slope-intercept form of a line to write the equation for the line of fit. Remember that, in this case, the dependent variable is

ln y,

therefore the form of the equation should be as follows.

ln y = m x + b

Two points can be used to find the slope

m .

The points

(20, ln 12)

and

(60, ln 100)

can be substituted in the Slope Formula.

m = \frac{ln y _{2} - ln y _{1}}{x _{2} - x _{1}}

SubstitutePoints

Substitute $(20, l n 12)$ & $(60, l n 100)$

m = \frac{l n 1 0 0 - l n 1 2}{6 0 - 2 0}

▼

Evaluate right-hand side

$ln (a) - ln (b) = ln (\frac{a}{b})$

m = \frac{ln \frac{1 0 0}{1 2}}{6 0 - 2 0}

SubTerm

Subtract term

m = \frac{ln \frac{1 0 0}{1 2}}{4 0}

UseCalc

Use a calculator

m = 0.053006 \dots

RoundDec

Round to $3$ decimal place(s)

m \approx 0.053

Knowing that the slope is about

0.053,

a partial equation of the line can be written.

ln y = 0.053 x + b

To find the value of

b,

the intersection with the vertical axis, a point must be used. For simplicity, choose the first point of the table.

ln y = 0.053 x + b

Substitute

$x = 20$

ln y = 0.053 (20) + b

$ln y \approx 2.48$

2.48 \approx 0.053 (20) + b

▼

Solve for

b

Multiply

2.48 \approx 1.06 + b

SubEqn

$LHS - 1.06 = RHS - 1.06$

1.42 \approx b

RearrangeEqn

Rearrange equation

b \approx 1.42

The equation of the line can now be written.

ln y = 0.053 x + 1.42

Finally, to find the exponential model that represents the original data, the variable

y

must be isolated. To do so, the definition of a natural logarithm will be used.

ln y = 0.053 x + 1.42 ⇕ y = e^{0.053 x + 1.42}

To simplify the obtained equation, the Product of Powers Property and the Power of a Power Property will be used.

y = e^{0.053 x + 1.42}

▼

Simplify right-hand side

SumInExponent

$a^{m + n} = a^{m} \cdot a^{n}$

y = e^{0.053 x} e^{1.42}

CommutativePropMult

Commutative Property of Multiplication

y = e^{1.42} e^{0.053 x}

ProdInExponent

$a^{m \cdot n} = (a^{m})^{n}$

y = e^{1.42} (e^{0.053})^{x}

UseCalc

Use a calculator

y = 4.137120 \dots (1.054429 \dots)^{x}

RoundDec

Round to $3$ decimal place(s)

y = 4.137 (1.054)^{x}

Note that this is only an example exponential model. If any two other points are used, a different model will be obtained. Ignacio is happy to have solved this and has regained some trust in Heichi's math knowledge. Heichi, too, feels good about sharing a topic of interest with his friend.

Closure

Using Technology

While it is extremely impressive to do math by hand, an exponential model for the last example can also be found by using a graphing calculator. Recall the given data points.

$x$	$y$
$20$	$12$
$30$	$15$
$40$	$25$
$50$	$40$
$60$	$100$

It was previously discussed that if graphing the data points in the form $(x, ln y)$ results in a linear pattern, then an exponential function is a good model for the data points in the form $(x, y) .$ To find the equation of the exponential model using a graphing calculator, press $STAT,$ choose EDIT, and enter the values in the first two columns, $L 1$ and $L 2 .$

Now, to find the values of $ln y,$ place the cursor in the heading of $L 3$ and press $LN .$ After that, press $2ND,$ and then $2$ before finally closing the parentheses. By pressing $ENTER,$ $L 3$ will be filled automatically with the corresponding values for $ln y .$

Having entered the values, they can be plotted in a scatter plot by pressing $2nd$ and $Y = .$ Then, choose one of the plots in the list. Make sure to turn the desired plot ON. Choose the type to be a scatterplot, and assign $L 1$ and $L 3$ as XList and YList, respectively. Any mark can be picked.

After this, press $GRAPH .$ If needed, the scale and the minimum and maximum $x -$ and $y -$ values can be adjusted by pressing $WINDOW .$

The points approximate a line. Therefore, an exponential function is an appropriate model. The best fitting exponential function for the given data can be found by performing an exponential regression. To do this press $STAT,$ and from the CALC menu choose the exponential regression option.

Here,

a \approx 3.5

and

b \approx 1.054,

rounded to three decimal places. With this information, the exponential function can be written.

y = 3.5 (1.054)^{x}

Note that the value of

a

is not the same as in the last example, but the value of

b

is exactly the same.

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