Lesson 24: Solving Exponential Equations
Transcript of Lesson 24: Solving Exponential Equations
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
386
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Lesson 24: Solving Exponential Equations
Student Outcomes
Students apply properties of logarithms to solve exponential equations.
Students relate solutions to π(π₯) = π(π₯) to the intersection point(s) on the graphs of π¦ = π(π₯) and π¦ = π(π₯)
in the case where π and π are constant or exponential functions.
Lesson Notes
Much of our previous work with logarithms in Topic B provided students with the particular skills needed to manipulate
logarithmic expressions and solve exponential equations. Although students have solved exponential equations in
earlier lessons in Topic B, this is the first time that they solve such equations in the context of exponential functions. In
this lesson, students solve exponential equations of the form ππππ‘ = π using properties of logarithms developed in
Lessons 12 and 13 (F-LE.A.4). For an exponential function π, students solve equations of the form π(π₯) = π and write a
logarithmic expression for the inverse (F-BF.B.4a). Additionally, students solve equations of the form π(π₯) = π(π₯)
where π and π are either constant or exponential functions (A-REI.D.11). Examples of exponential functions in this
lesson draw from Lesson 7, in which the growth of a bacteria population was modeled by the function π(π‘) = 2π‘ , and
Lesson 23, in which students modeled the growth of an increasing number of beans with a function π(π‘) = π(ππ‘), where
π β 1 and π β 1.5.
Students use technology to calculate logarithmic values and to graph linear and exponential functions.
Classwork
Opening Exercise (4 minutes)
The Opening Exercise is a simple example of solving an exponential equation of the form ππππ‘ = π. Allow students to
work independently or in pairs to solve this problem. Circulate around the room to check that all students know how to
apply a logarithm to solve this problem. Students may choose to use either a base-2 or base-10 logarithm.
Opening Exercise
In Lesson 7, we modeled a population of bacteria that doubled every day by the function π·(π) = ππ, where π was the time
in days. We wanted to know the value of π when there were ππ bacteria. Since we did not know about logarithms at the
time, we approximated the value of π numerically, and we found that π·(π) = ππ when π β π. ππ.
Use your knowledge of logarithms to find an exact value for π when π·(π) = ππ, and then use your calculator to
approximate that value to π decimal places.
Since π·(π) = ππ, we need to solve ππ = ππ.
ππ = ππ
π π₯π¨π (π) = π₯π¨π (ππ)
π =π
π₯π¨π (π)
π β π. ππππ
Thus, the population will reach ππ bacteria in approximately π. ππππ days.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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Discussion (2 minutes)
Ask students to describe their solution method for the Opening Exercise. Make sure that solutions are discussed using
both base-10 and base-2 logarithms. If all students used the common logarithm to solve this problem, then present the
following solution using the base-2 logarithm:
2π‘ = 10
log2(2π‘) = log2(10)
π‘ = log2(10)
π‘ =log(10)
log(2)
π‘ =1
log(2)
π‘ β 3.3219
The remaining exercises ask students to solve equations of the form π(π₯) = π or π(π₯) = π(π₯), where π and π are
exponential functions (F-LE.A.4, F-BF.B.4a, A-REI.D.11). For the remainder of the lesson, allow students to work either
independently or in pairs or small groups on the exercises. Circulate to ensure students are on task and solving the
equations correctly. After completing Exercises 1β4, debrief students to check for understanding, and ensure they are
using appropriate strategies to complete problems accurately before moving on to Exercises 5β10.
Exercises 1β4 (25 minutes)
Exercises
1. Fiona modeled her data from the bean-flipping experiment in Lesson 23 by the function π(π) = π. πππ(π. πππ)π, and
Gregor modeled his data with the function π(π) = π. πππ(π. πππ)π.
a. Without doing any calculating, determine which student, Fiona or Gregor, accumulated πππ beans first.
Explain how you know.
Since the base of the exponential function for Gregorβs model, π. πππ, is larger than the base of the
exponential function for Fionaβs model, π. πππ, Gregorβs model will grow more quickly than Fionaβs, and he
will accumulate πππ beans before Fiona does.
b. Using Fionaβs model β¦
i. How many trials would be needed for her to accumulate πππ beans?
We need to solve the equation π(π) = πππ for π.
π. πππ(π. πππ)π = πππ
π. ππππ =πππ
π. πππ
π π₯π¨π (π. πππ) = π₯π¨π (πππ
π. πππ)
π π₯π¨π (π. πππ) = π₯π¨π (πππ) β π₯π¨π (π. πππ)
π =π β π₯π¨π (π. πππ)
π₯π¨π (π. πππ)
π β ππ. ππ
So, it takes ππ trials for Fiona to accumulate πππ beans.
Scaffolding:
Have struggling students begin
this exercise with functions
π(π‘) = 7(2π‘) and π(π‘) = 4(3π‘).
MP.2
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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ii. How many trials would be needed for her to accumulate π, πππ beans?
We need to solve the equation π(π) = ππππ for π.
π. πππ(π. πππ)π = ππππ
π. ππππ =ππππ
π. πππ
π π₯π¨π (π. πππ) = π₯π¨π (ππππ
π. πππ)
π π₯π¨π (π. πππ) = π₯π¨π (ππππ) β π₯π¨π (π. πππ)
π =π β π₯π¨π (π. πππ)
π₯π¨π (π. πππ)
π β ππ. ππ
So, it takes ππ trials for Fiona to accumulate ππππ beans.
c. Using Gregorβs model β¦
i. How many trials would be needed for him to accumulate πππ beans?
We need to solve the equation π(π) = πππ for π.
π. πππ(π. πππ)π = πππ
π. ππππ =πππ
π. πππ
π π₯π¨π (π. πππ) = π₯π¨π (πππ
π. πππ)
π π₯π¨π (π. πππ) = π₯π¨π (πππ) β π₯π¨π (π. πππ)
π =π β π₯π¨π (π. πππ)
π₯π¨π (π. πππ)
π β π. ππ
So, it takes ππ trials for Gregor to accumulate πππ beans.
ii. How many trials would be needed for him to accumulate π, πππ beans?
We need to solve the equation π(π) = ππππ for π.
π. πππ(π. πππ)π = ππππ
π. ππππ =ππππ
π. πππ
π π₯π¨π (π. πππ) = π₯π¨π (ππππ
π. πππ)
π π₯π¨π (π. πππ) = π₯π¨π (ππππ) β π₯π¨π (π. πππ)
π =π β π₯π¨π (π. πππ)
π₯π¨π (π. πππ)
π β ππ. ππ
So, it takes ππ trials for Gregor to accumulate ππππ beans.
d. Was your prediction in part (a) correct? If not, what was the error in your reasoning?
Responses will vary. Either students made the correct prediction, or they did not recognize that the base π
determines the growth rate of the exponential function so the larger base π. πππ causes Gregorβs function to
grow much more quickly than Fionaβs.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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2. Fiona wants to know when her model π(π) = π. πππ(π. πππ)π predicts accumulations of πππ, π, πππ, and ππ, πππ
beans, but she wants to find a way to figure it out without doing the same calculation three times.
a. Let the positive number π represent the number of beans that Fiona wants to have. Then solve the equation
π. πππ(π. πππ)π = π for π.
π. πππ(π. πππ)π = π
π. ππππ =π
π. πππ
π π₯π¨π (π. πππ) = π₯π¨π (π
π. πππ)
π π₯π¨π (π. πππ) = π₯π¨π (π) β π₯π¨π (π. πππ)
π =π₯π¨π (π) β π₯π¨π (π. πππ)
π₯π¨π (π. πππ)
b. Your answer to part (a) can be written as a function π΄ of the number of beans π, where π > π. Explain what
this function represents.
The function π΄(π) =π₯π¨π (π)βπ₯π¨π (π.πππ)
π₯π¨π (π.πππ) calculates the number of trials it will take for Fiona to accumulate π
beans.
c. When does Fionaβs model predict that she will accumulate β¦
i. πππ beans?
π΄(πππ) =π₯π¨π (πππ) β π₯π¨π (π. πππ)
π₯π¨π (π. πππ)β ππ. ππ
According to her model, it will take Fiona ππ trials to accumulate πππ beans.
ii. π, πππ beans?
π΄(ππππ) =π₯π¨π (ππππ) β π₯π¨π (π. πππ)
π₯π¨π (π. πππ)β ππ. ππ
According to her model, it will take Fiona ππ trials to accumulate ππππ beans.
iii. ππ, πππ beans?
π΄(πππππ) =π₯π¨π (πππππ) β π₯π¨π (π. πππ)
π₯π¨π (π. πππ)β ππ. ππ
According to her model, it will take Fiona ππ trials to accumulate πππππ beans.
3. Gregor states that the function π that he found to model his bean-flipping data can be written in the form
π(π) = π. πππ(πππ₯π¨π (π.πππ)π). Since π₯π¨π (π. πππ) β π. ππππ, he is using π(π) = π. πππ(πππ.πππππ) as his new model.
a. Is Gregor correct? Is π(π) = π. πππ(πππ₯π¨π (π.πππ)π) an equivalent form of his original function? Use properties
of exponents and logarithms to explain how you know.
Yes, Gregor is correct. Since πππ₯π¨π (π.πππ) = π. πππ, and πππ₯π¨π (π.πππ)π = (πππ₯π¨π (π.πππ))π
β πππ.πππππ, Gregor is
right that π(π) = π. πππ(πππ.πππππ) is a reasonable model for his data.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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b. Gregor also wants to find a function to help him to calculate the number of trials his function π predicts it
should take to accumulate πππ, π, πππ, and ππ, πππ beans. Let the positive number π represent the number
of beans that Gregor wants to have. Solve the equation π. πππ(πππ.πππππ) = π for π.
π. πππ(πππ.πππππ) = π
πππ.πππππ =π
π. πππ
π. πππππ = π₯π¨π (π
π. πππ)
π =π₯π¨π (π) β π₯π¨π (π. πππ)
π. ππππ
c. Your answer to part (b) can be written as a function π΅ of the number of beans π, where π > π. Explain what
this function represents.
The function π΅(π) =π₯π¨π (π)βπ₯π¨π (π.πππ)
π.ππππ calculates the number of trials it will take for Gregor to accumulate π
beans.
d. When does Gregorβs model predict that he will accumulate β¦
i. πππ beans?
π΅(πππ) =π₯π¨π (πππ) β π₯π¨π (π. πππ)
π. ππππβ ππ. ππ
According to his model, it will take Gregor ππ trials to accumulate πππ beans.
ii. π, πππ beans?
π΅(πππ) =π₯π¨π (ππππ) β π₯π¨π (π. πππ)
π. ππππβ ππ. ππ
According to his model, it will take Gregor ππ trials to accumulate π, πππ beans.
iii. ππ, πππ beans?
π΅(πππππ) =π₯π¨π (πππππ) β π₯π¨π (π. πππ)
π. ππππβ ππ. ππ
According to his model, it will take Gregor ππ trials to accumulate ππ, πππ beans.
4. Helena and Karl each change the rules for the bean experiment. Helena started with four beans in her cup and
added one bean for each that landed marked-side up for each trial. Karl started with one bean in his cup but added
two beans for each that landed marked-side up for each trial.
a. Helena modeled her data by the function π(π) = π. πππ(π. ππππ). Explain why her values of π = π. πππ and
π = π. πππ are reasonable.
Since Helena starts with four beans, we should expect that π β π, so a value π = π. πππ is reasonable.
Because she is using the same rule for adding beans to the cup as we did in Lesson 23, we should expect that
π β π. π. Thus, her value of π = π. πππ is reasonable.
b. Karl modeled his data by the function π(π) = π. πππ(π. ππππ). Explain why his values of π = π. πππ and
π = π. πππ are reasonable.
Since Karl starts with one bean, we should expect that π β π, so a value π = π. πππ is reasonable. Because
Karl adds two beans to the cup for each that lands marked-side up, we should expect that the number of
beans roughly doubles with each trial. That is, we should expect π β π. Thus, his value of π = π. πππ is
reasonable.
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ALGEBRA II
Lesson 24: Solving Exponential Equations
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c. At what value of π do Karl and Helena have the same number of beans?
We need to solve the equation π(π) = π(π) for π.
π. πππ(π. ππππ) = π. πππ(π. ππππ)
π₯π¨π (π. πππ(π. ππππ)) = π₯π¨π (π. πππ(π. ππππ))
π₯π¨π (π. πππ) + π₯π¨π (π. ππππ) = π₯π¨π (π. πππ) + π₯π¨π (π. ππππ)
π₯π¨π (π. πππ) + π π₯π¨π (π. πππ) = π₯π¨π (π. πππ) + π π₯π¨π (π. πππ)
π π₯π¨π (π. πππ) β π π₯π¨π (π. πππ) = π₯π¨π (π. πππ) β π₯π¨π (π. πππ)
π(π₯π¨π (π. πππ) β π₯π¨π (π. πππ)) = π₯π¨π (π. πππ) β π₯π¨π (π. πππ)
π (π₯π¨π (π. πππ
π. πππ)) = π₯π¨π (
π. πππ
π. πππ)
π(π. ππβπππ) β π. ππβπππ
π β π. πβπππ
Thus, after trial number π, Karl and Helena have the same number of beans.
d. Use a graphing utility to graph π = π(π) and π = π(π) for π < π < ππ.
e. Explain the meaning of the intersection point of the two curves π = π(π) and π = π(π) in the context of this
problem.
The two curves intersect at the π-value where Helena and Karl have the same number of beans. The π-value
indicates the number of beans they both have after five trials.
f. Which student reaches ππ beans first? Does the reasoning used in deciding whether Gregor or Fiona would
get πππ beans first hold true here? Why or why not?
Helena reaches ππ beans first. Although the function modeling Helenaβs beans has a smaller base, Karlβs does
not catch up to Helena until after five trials. After five trials, Karlβs will always be greater, and he will reach
πππ beans first. The logic we applied to comparing Gregorβs model and Fionaβs model does not apply here
because Helena and Karl do not start with the same initial number of beans.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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Debrief students after they complete Exercises 1β4 to ensure understanding of the exercises and strategies used to solve
the exercises before continuing. In Exercises 5β10, students solve exponential functions using what they know about
logarithms. After completing Exercises 5β10, debrief students about when it is necessary to use logarithms to solve
exponential equations and when it is not. Exercises 7, 8, and 9 are examples of exercises that do not require logarithms
to solve but may be appropriate to solve with logarithms depending on the approach used by students.
Exercise 5β10 (7 minutes)
For the following functions π and π, solve the equation π(π) = π(π). Express your solutions in terms of logarithms.
5. π(π) = ππ(π. π)π+π, π(π) = π(π. π)π
ππ(π. π)π+π = π(π. π)π
π(π. π)π+π = π. ππ
π₯π¨π (π) + π₯π¨π (π. ππ+π) = π₯π¨π (π. ππ)
π₯π¨π (π) + (π + π) π₯π¨π (π. π) = π π₯π¨π (π. π)
π₯π¨π (π) + π π₯π¨π (π. π) + π₯π¨π (π. π) = π π₯π¨π (π. π)
π₯π¨π (π) + π₯π¨π (π. π) = π(π₯π¨π (π. π) β π₯π¨π (π. π))
π₯π¨π (π. π) = π π₯π¨π (π. π
π. π)
π₯π¨π (π. π) = π π₯π¨π (π)
π =π₯π¨π (π. π)
π₯π¨π (π)
6. π(π) = πππ(π)ππ+π, π(π) = ππ(π)πβππ
πππ(π)ππ+π = ππ(π)πβππ
π(π)ππ+π = π(π)πβππ
π₯π¨π (π) + (ππ + π)π₯π¨π (π) = π₯π¨π (π) + (π β ππ)π₯π¨π (π)
π π₯π¨π (π) + ππ π₯π¨π (π) + π₯π¨π (π) = π₯π¨π (π) + π π₯π¨π (π) β ππ π₯π¨π (π)
ππ(π₯π¨π (π) + π₯π¨π (π)) = π π₯π¨π (π) β π π₯π¨π (π)
ππ π₯π¨π (ππ) = π π₯π¨π (π)
π =π π₯π¨π (π)
π π₯π¨π (ππ)
7. π(π) = πππππ+ππβππ, π(π) = πππππβππ
πππππ+ππβππ = πππππβππ
(πππ)ππ+ππβππ = πππππβππ
π(ππ + ππ β ππ) = πππ β ππ
ππ + ππ β ππ = ππ β ππ
ππ β π = π
π(ππ β π) = π
π(π + π)(π β π) = π
π = π, π = βπ, or π = π
Scaffolding:
Challenge advanced
students to solve Exercise
6 in more than one way,
for example, by using first
the logarithm base 5 and
then the logarithm base 3,
and comparing the results.
Advanced students should
be able to solve Exercises
7β9 without logarithms by
expressing each function
with a common base, but
using logarithms may be a
more reliable approach for
students struggling with
the exponential
properties.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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8. π(π) = ππ(πππ+ππ), π(π) = π(πππ+ππ+π)
ππ(πππ+ππ) = π(πππ+ππ+π)
ππ(πππ+ππ) = πππ+ππ+π
ππ((ππ)ππ+ππ) = (ππ)ππ+ππ+π
ππππ+ππ+π = ππππ+πππ+ππ
πππ + ππ + π = πππ + πππ + ππ
ππ + ππ + π = π
(π + π)(π + π) = π
π = βπ or π = βπ
9. π(π) = ππ¬π’π§π(π), π(π) = πππ¨π¬π(π)
ππ¬π’π§π(π) = πππ¨π¬π(π)
π¬π’π§π(π) = ππ¨π¬π(π)
π¬π’π§(π) = ππ¨π¬(π) or π¬π’π§(π) = βππ¨π¬(π)
π =π
π+ ππ or π =
ππ
π+ ππ for all integers π
10. π(π) = (π. ππ)ππ¨π¬(π)+π¬π’π§(π), π(π) = (π. π)π π¬π’π§(π)
(π. ππ)ππ¨π¬(π)+π¬π’π§(π) = (π. π)π π¬π’π§(π)
π₯π¨π ((π. ππ)ππ¨π¬(π)+π¬π’π§(π)) = π₯π¨π (π. π)π π¬π’π§(π))
(ππ¨π¬(π) + π¬π’π§(π))π₯π¨π (π. ππ) = π π¬π’π§(π)π₯π¨π (π. π)
(ππ¨π¬(π) + π¬π’π§(π))π₯π¨π (π. ππ) = π π¬π’π§(π) π₯π¨π (π. π)
π(ππ¨π¬(π) + π¬π’π§(π))π₯π¨π (π. π) = π π¬π’π§(π) π₯π¨π (π. π)
π ππ¨π¬(π) + π π¬π’π§(π) = π π¬π’π§(π)
ππ¨π¬(π) = π
π =π
π+ ππ for all integers π
Closing (3 minutes)
Ask students to respond to the following prompts either in writing or orally to a partner.
Describe two different approaches to solving the equation 2π₯+1 = 32π₯. Do not actually solve the equation.
You could begin by taking the logarithm base 10 of both sides, or the logarithm base 2 of both sides.
You could even take the logarithm base 3 of both sides.
Could the graphs of two exponential functions π(π₯) = 2π₯+1 and π(π₯) = 32π₯ ever intersect at more than one
point? Explain how you know.
No. The graphs of these functions are always increasing. They intersect at one point, but once they
cross once they cannot cross again. For large values of π₯, the quantity 32π₯ is always greater than 2π₯+1,
so the graph of π ends up above the graph of π after they cross.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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Discuss how the starting value and base affect the graph of an exponential function and how this can help you
compare exponential functions.
The starting value determines the π¦-intercept of an exponential function, so it determines how large or
small the function is when π₯ = 0. The base is ultimately more important and determines how quickly
the function increases (or decreases). When comparing exponential functions, the function with the
larger base always overtakes the function with the smaller base no matter how large the value when
π₯ = 0.
If π(π₯) = 2π₯+1 and π(π₯) = 32π₯, is it possible for the equation π(π₯) = π(π₯) to have more than one solution?
No. Solutions to the equation π(π₯) = π(π₯) correspond to π₯-values of intersection points of the graphs
of π¦ = π(π₯) and π¦ = π(π₯). Since these graphs can intersect no more than once, the equation can have
no more than one solution.
Exit Ticket (4 minutes)
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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Name Date
Lesson 24: Solving Exponential Equations
Exit Ticket
Consider the functions π(π₯) = 2π₯+6 and π(π₯) = 52π₯.
a. Use properties of logarithms to solve the equation π(π₯) = π(π₯). Give your answer as a logarithmic
expression, and approximate it to two decimal places.
b. Verify your answer by graphing the functions π¦ = π(π₯) and π¦ = π(π₯) in the same window on a calculator, and
sketch your graphs below. Explain how the graph validates your solution to part (a).
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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Exit Ticket Sample Solutions
Consider the functions π(π) = ππ+π and π(π) = πππ.
a. Use properties of logarithms to solve the equation π(π) = π(π). Give your answer as a logarithmic
expression, and approximate it to two decimal places.
ππ+π = πππ
(π + π)π₯π¨π (π) = ππ π₯π¨π (π)
ππ π₯π¨π (π) β π π₯π¨π (π) = π π₯π¨π (π)
π =π π₯π¨π (π)
π π₯π¨π (π) β π₯π¨π (π)
π =π₯π¨π (ππ)
π₯π¨π (ππ) β π₯π¨π (π)
π =π₯π¨π (ππ)
π₯π¨π (πππ
)
π β π. ππ
Any of the final three forms are acceptable, and other correct forms using logarithms with other bases (such
as base 2) are possible.
b. Verify your answer by graphing the functions π = π(π) and π = π(π) in the same window on a calculator,
and sketch your graphs below. Explain how the graph validates your solution to part (a).
Because the graphs of π = π(π) and π = π(π) intersect when π β π. ππ, we know that the equation
π(π) = π(π) has a solution at approximately π = π. ππ.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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Problem Set Sample Solutions
1. Solve the following equations.
a. π β ππ+π = ππππ
ππ+π = ππππ
ππ+π = ππ
π + π = π
π = π
b. π β πππ = πππ
πππ = πππ
πππ = ππ
ππ = π
π =π
π
c. π β πππ+π = πππππ
πππ+π = ππππ
πππ+π = πππ
ππ + π = ππ
ππ = π
π = π
d. πππβπ = ππ
πππβπ = ππ
ππβ (ππβπ) = ππ
ππ β π = π
ππ = π
π =π
π
e. π β πππ = πππ
πππ = ππ
ππ β π₯π§(π) = π₯π§(ππ)
π =π₯π§(ππ)
π β π₯π§(π)
π β π. πππ
Note: Students can also use the common logarithm to find the solution.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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f. π β ππππ = πππ
ππππ = ππ
ππ β π₯π§(ππ) = π₯π§(ππ)
π =π₯π§(ππ)
π β π₯π§(ππ)
π β π. πππ
Note: Students can also use the common logarithm to find the solution.
g. π β ππ = ππππ
ππ =πβπππ
π
π β π₯π§(π) = π₯π§ (πβπππ
π)
π =π₯π§ (
πβππππ
)
π₯π§(π)
π β π. πππ
Note: Students can also use the common logarithm to find the solution.
h. βπ β πππ = π
Solution using properties of exponents:
πππ β πππ = ππ
πππ+ππ = ππ
π
π+ ππ = π
π =π
π
i. π₯π¨π (πππ) β πππ = π₯π¨π (ππππππ)
πππ =π₯π¨π (ππππππ)
π₯π¨π (πππ)
πππ = π
πππ = πππ
ππ =π
π
π =π
ππ
2. Lucy came up with the model π(π) = π. πππ(π. πππ)π for the first bean activity. When does her model predict that
she would have π, πππ beans?
ππππ = π. πππ(π. πππ)π
π₯π¨π (ππππ) = π₯π¨π (π. πππ) + π π₯π¨π (π. πππ)
π =π₯π¨π (ππππ) β π₯π¨π (π. πππ)
π₯π¨π (π. πππ)
π β ππ. ππ
Lucyβs model predicts that it will take ππ trials to have over ππππ beans.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
399
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3. Jack came up with the model π(π) = π. πππ(π. πππ)π for the first bean activity. When does his model predict that
he would have ππ, πππ beans?
πππππ = π. πππ(π. πππ)π
π₯π¨π (πππππ) = π₯π¨π (π. πππ) + π π₯π¨π (π. πππ)
π =π₯π¨π (πππππ) β π₯π¨π (π. πππ)
π₯π¨π (π. πππ)
π β ππ. ππ
Jackβs model predicts that it will take ππ trials to have over ππ, πππ beans.
4. If instead of beans in the first bean activity you were using fair pennies, when would you expect to have
$π, πππ, πππ?
One million dollars is πππ pennies. Using fair pennies, we can model the situation by π(π) = π. ππ.
πππ = π. ππ π = π π₯π¨π (π. π)
π =π
π₯π¨π (π. π)
π β ππ. ππ
We should expect it to take ππ trials to reach more than $π million using fair pennies.
5. Let π(π) = π β ππ and π(π) = π β ππ.
a. Which function is growing faster as π increases? Why?
The function π is growing faster due to its larger base, even though π(π) > π(π).
b. When will π(π) = π(π)?
π(π) = π(π)
π β ππ = π β ππ
π₯π§(π β ππ) = π₯π§(π β ππ)
π₯π§(π) + π π₯π§(π) = π₯π§(π) + π π₯π§(π)
π π₯π§(π) β π π₯π§(π) = π₯π§(π) β π₯π§(π)
π π₯π§ (π
π) = π₯π§ (
π
π)
π = π
Note: Students can also use the common logarithm to find the solution.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
400
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6. The growth of a population of E. coli bacteria can be modeled by the function π¬(π) = πππ(ππ. πππ)π, and the
growth of a population of Salmonella bacteria can be modeled by the function πΊ(π) = ππππ(π. πππ)π, where π
measures time in hours.
a. Graph these two functions on the same set of axes. At which value of π does it appear that the graphs
intersect?
From the graph, it appears that the two curves intersect at π β π. π.
b. Use properties of logarithms to find the time π when these two populations are the same size. Give your
answer to two decimal places.
π¬(π) = πΊ(π)
πππ(ππ. πππ)π = ππππ(π. πππ)π
ππ. ππππ = π(π. πππ)π
π π₯π¨π (ππ. πππ) = π₯π¨π (π) + π π₯π¨π (π. πππ)
π(π₯π¨π (ππ. πππ) β π₯π¨π (π. πππ)) = π₯π¨π (π)
π =π₯π¨π (π)
π₯π¨π (ππ. πππ) β π₯π¨π (π. πππ)
π β π. ππβπππ
It takes approximately π. ππ hours for the populations to be the same size.
7. Chain emails contain a message suggesting you will have bad luck if you do not forward the email to others.
Suppose a student started a chain email by sending the message to ππ friends and asking those friends to each send
the same email to π more friends exactly one day after receiving the message. Assuming that everyone that gets
the email participates in the chain, we can model the number of people who receive the email on the πth day by the
formula π¬(π) = ππ(ππ), where π = π indicates the day the original email was sent.
a. If we assume the population of the United States is πππ million people and everyone who receives the email
sends it to π people who have not received it previously, how many days until there are as many emails being
sent out as there are people in the United States?
πππ(πππ) = ππ β ππ
πππ(πππ) = ππ
π₯π¨π (πππ) + π₯π¨π (πππ) = π β π₯π¨π (π)
π₯π¨π (πππ) + π = π β π₯π¨π (π)
π =π + π₯π¨π (πππ)
π₯π¨π (π)
π β ππ. ππ
So by the ππth day, more than πππ million emails are being sent out.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
401
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b. The population of earth is approximately π. π billion people. On what day will π. π billion emails be sent out?
π. π(πππ) = ππ(ππ)
π. π(πππ) = ππ
π₯π¨π (π. π(πππ)) = π β π₯π¨π (π)
π =π + π₯π¨π (π. π)
π₯π¨π (π)
π β ππ. ππππ
By the ππth day, more than π. π billion emails will be sent.
8. Solve the following exponential equations.
a. ππ(ππβπ) = ππ
ππππβπ = ππ
ππ β π = π π₯π¨π (π)
π(π β π₯π¨π (π)) = π
π =π
π β π₯π¨π (π)
b. ππ
π = πππβπ
πππ = πππβπ
π
π π₯π¨π (π) = (ππ β π)π₯π¨π (π)
ππ π₯π¨π (π) β ππ₯π¨π (π)
π= π π₯π¨π (π)
π (π π₯π¨π (π) βπ₯π¨π (π)
π) = π π₯π¨π (π)
π =π π₯π¨π (π)
π π₯π¨π (π) βπ₯π¨π (π)
π
c. ππππ+π = ππππππ+π+π
ππππ+π = ππππππ+π+π
ππ + π = (πππ + π + π)π₯π¨π (πππ)
ππ + π = πππ + ππ + π
πππ + ππ β π = π
(ππ β π)(π + π) = π
π =π
π or π = βπ
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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d. πππβππ+π = πππβπ
πππβππ+π = πππβπ
(ππ β ππ + π) π₯π¨π π(π) = (ππ β π) π₯π¨π π(π)
π(ππ β ππ + π) = ππ β π
πππ β ππ + π = ππ β π
πππ β πππ + ππ = π
(ππ β π)(π β π) = π
π =π
πor π = π
9. Solve the following exponential equations.
a. (ππ)π = ππ
πππ= ππ
πππ₯π¨π π(π) = π π₯π¨π π(π)
ππ = ππ
ππ β ππ = π
π(π β π) = π
π = π or π = π
b. (ππ)π = ππ
πππ= ππ
ππ π₯π¨π (π) = π₯π¨π (ππ)
ππ =π₯π¨π (ππ)
π₯π¨π (π)
π = βπ₯π¨π (ππ)
π₯π¨π (π) or π = ββ
π₯π¨π (ππ)
π₯π¨π (π)
10. Solve the following exponential equations.
a. πππ+π β πππβπ = ππππ
πππ+π β πππβπ = ππππ
πππ(πππβπ) β πππβπ = ππππ
πππβπ(πππ β π) = ππππ
ππ(πππβπ) = ππππ
πππβπ = ππ
π β π = π₯π¨π (ππ)
π = π₯π¨π (ππ) + π
b. π(ππ) + ππ+π = πππ
π(ππ) + ππ+π = πππ
π(ππ) + π(ππ) = πππ
π(ππ) = πππ
ππ = ππ
π = π₯π¨π π(ππ) =π₯π¨π (ππ)
π₯π¨π (π)=
π
ππ₯π¨π π(ππ)
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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11. Solve the following exponential equations.
a. (πππ)π β π(πππ) + π = π Hint: Let π = πππ, and solve for π before solving for π.
Let π = πππ. Then
ππ β ππ + π = π
(π β π)(π β π) = π
π = π or π = π
If π = π, we have π = πππ, and then π = π₯π¨π (π).
If π = π, we have π = πππ, and then π = π.
Thus, the two solutions to this equation are π and π₯π¨π (π).
b. (ππ)π β π(ππ) β π = π
Let π = ππ.
ππ β ππ β π = π
(π β π)(π + π) = π
π = π or π = βπ
If π = π, we have ππ = π, and then π = π.
If π = βπ, we have ππ = βπ, which has no solution.
Thus, the only solution to this equation is π.
c. π(ππ)π β π(ππ) β π = π
Let π = ππ.
πππ β ππ β π = π
(π β π)(ππ + π) = π
π = π or π = βπ
π
If π = π, we have ππ = π, and then π = π₯π§(π).
If π = βππ
, we have ππ = βππ
, which has no solution because ππ > π for every value of π.
Thus, the only solution to this equation is π₯π§(π).
d. ππ + π(ππ) + ππ = π
Let π = ππ.
(ππ)π + π(ππ) + ππ = π
ππ + ππ + ππ = π
(π + π)(π + π) = π
π = βπ or π = βπ
But ππ > π for every value of π, thus there are no solutions to this equation.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
404
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e. (πππ)π β π(πππ) β π = π
Let π = πππ.
ππ β ππ β π = π
π = π + βπ or π = π β βπ
If π = π + βπ, we have πππ = π + βπ, and then π = π₯π¨π (π + βπ).
If π = π β βπ, we have πππ = π β βπ, which has no solution because π β βπ < π.
Thus, the only solution to this equation is π₯π¨π (π + βπ).
12. Solve the following systems of equations.
a. ππ+ππ = π
πππ+π = π
ππ+ππ = ππ
πππ+π = ππ
π + ππ = π ππ + π = π
π + ππ = π ππ + ππ = π
π = π π = βπ
b. πππ+πβπ = ππ
ππβππ = π πππ+πβπ = ππ
(ππ)πβππ = ππ
ππ + π β π = π π(π β ππ) = π
ππ + π = π ππ β ππ = π
π = π
π =π
π
c. πππ = πππ+π
πππ = πππβπ
πππ = (ππ)ππ+π (ππ)ππ = πππβπ
ππ = π(ππ + π) π(ππ) = (ππ β π)
ππ β ππ = π ππ β ππ = π
π = π π = π
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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13. Because π(π) = π₯π¨π π(π) is an increasing function, we know that if π < π, then π₯π¨π π(π) < π₯π¨π π(π). Thus, if we take
logarithms of both sides of an inequality, then the inequality is preserved. Use this property to solve the following
inequalities.
a. ππ >ππ
ππ >π
π
π₯π¨π (ππ) > π₯π¨π (π
π)
π π₯π¨π (π) > π₯π¨π (π) β π₯π¨π (π)
π >π₯π¨π (π) β π₯π¨π (π)
π₯π¨π (π)
b. (ππ
)π
> π
(π
π)
π
> π
π π₯π¨π (π
π) > π₯π¨π (π)
But, remember that π₯π¨π (ππ
) < π, so we need to divide by a negative number. We then have
π <π₯π¨π (π)
π₯π¨π (π)βπ₯π¨π (π).
c. ππ > ππβπ
(ππ)π > (ππ)πβπ
πππ > πππβπ
ππ > ππ β π
π > π
d. ππ+π > ππβππ
ππ+π > ππβππ
(π + π)π₯π¨π (π) > (π β ππ)π₯π¨π (π)
ππ π₯π¨π (π) + π π₯π¨π (π) > π π₯π¨π (π) β π π₯π¨π (π)
π >π π₯π¨π (π) β π π₯π¨π (π)
π π₯π¨π (π) + π₯π¨π (π)
π >π₯π¨π (
ππππ
)
π₯π¨π (ππ)
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 24
ALGEBRA II
Lesson 24: Solving Exponential Equations
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e. (ππ
)π
> (ππ
)π+π
(π
π)
π
> (π
π)
π+π
π π₯π¨π (π
π) > (π + π)π₯π¨π (
π
π)
π (π₯π¨π (π
π) β π₯π¨π (
π
π)) > π₯π¨π (
π
π)
But, π₯π¨π (ππ
) = βπ₯π¨π (ππ
), so we have
π (βπ₯π¨π (π
π) β π₯π¨π (
π
π)) > π₯π¨π (
π
π)
π (βπ π₯π¨π (π
π)) > π₯π¨π (
π
π) .
But, βπ π₯π¨π (ππ
) < π, so we need to divide by a negative number, so we have
π <π₯π¨π (
ππ
)
βπ π₯π¨π (ππ
)
π < βπ
π.