Logarithmic Functions Topic 1: Evaluating Logarithmic Equations.
Lesson 14: Solving Logarithmic Equations
Transcript of Lesson 14: Solving Logarithmic Equations
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
193
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Lesson 14: Solving Logarithmic Equations
Student Outcomes
Students solve simple logarithmic equations using the definition of logarithm and logarithmic properties.
Lesson Notes
In this lesson, students solve simple logarithmic equations by first putting them into the form logπ(π) = πΏ, where π is
either 2, 10, or π, π is an expression, and πΏ is a number, and then using the definition of logarithm to rewrite the
equation in the form ππΏ = π. Students are able to evaluate logarithms without technology by selecting an appropriate
base; solutions are provided with this in mind. In Lesson 15, students learn the technique of solving exponential
equations using logarithms of any base without relying on the definition. Students need to use the properties of
logarithms developed in prior lessons to rewrite the equations in an appropriate form before solving
(A-SSE.A.2, F-LE.A.4). The lesson starts with a few fluency exercises to reinforce the logarithmic properties before
moving on to solving equations.
Classwork
Opening Exercise (3 minutes)
The following exercises provide practice with the definition of the logarithm
and prepare students to solve logarithmic equations using the methods
outlined later in the lesson. Encourage students to work alone on these
exercises, but allow students to work in pairs if necessary.
Opening Exercise
Convert the following logarithmic equations to equivalent exponential equations
a. π₯π¨π (ππ, πππ) = π πππ = ππ, πππ
b. π₯π¨π (βππ) =ππ
ππππ = βππ
c. π₯π¨π π(πππ) = π ππ = πππ
d. π₯π¨π π(πππ) = π ππ = πππ
e. π₯π§(π) = π ππ = π
f. π₯π¨π (π + π) = π π + π = πππ
logπ(π₯π¦) = logπ(π₯) + logπ(π¦);
logπ (π₯
π¦) = logπ(π₯) β logπ(π¦) ;
logπ(π₯π) = π β logπ(π₯);
logπ (1
π₯) = βlog(π₯).
Scaffolding:
Remind students of the main properties
that they use by writing the following
on the board:
logπ(π₯) = πΏ means ππΏ = π₯;
Consistently using a visual display of
these properties throughout the module
is helpful.
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
194
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Examples (6 minutes)
Students should be ready to take the next step from converting logarithmic equations to an equivalent exponential
expression to solving the resulting equation. Decide whether or not students need to see a teacher-led example or can
attempt to solve these equations in pairs. Anticipate that students neglect to check for extraneous solutions in these
examples. After the examples, lead the discussion to the existence of an extraneous solution in Example 3.
Examples
Write each of the following equations as an equivalent exponential equation, and solve for π.
1. π₯π¨π (ππ + π) = π
π₯π¨π (ππ + π) = π
πππ = ππ + π
π = ππ + π
π = βπ
2. π₯π¨π π(π + π) = π
π₯π¨π π(π + π) = π
ππ = π + π
ππ = π + π
π = ππ
3. 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) = πππ
ππ + ππ + ππ = ππ
ππ + ππ = π
π(π + π) = π
π = π or π = βπ
However, if π = βπ, then (π + π) = βπ, and (π + π) = βπ, so both logarithms in the equation are undefined.
Thus, βπ is an extraneous solution, and only π is a valid solution to the equation.
Discussion (4 minutes)
Ask students to volunteer their solutions to the equations in the Opening Exercise. This line of questioning is designed to
allow students to determine that there is an extraneous solution to Example 3. If the class has already discovered this
fact, opt to accelerate or skip this discussion.
What is the solution to the equation in Example 1?
β2
What is the result if you evaluate log(3π₯ + 7) at π₯ = β2? Did you find a solution?
log(3(β2) + 7) = log(1) = 0, so β2 is a solution to log(3π₯ + 7) = 0.
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
195
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What is the solution to the equation in Example 2?
11
What is the result if you evaluate log2(π₯ + 5) at π₯ = 11? Did you find a solution?
log2(11 + 5) = log2(16) = 4, so 11 is a solution to log2(π₯ + 5) = 4.
What is the solution to the equation in Example 3?
There were two solutions: 0 and β7.
What is the result if you evaluate log(π₯ + 2) + log(π₯ + 5) at π₯ = 0? Did you find a solution?
log(2) + log(5) = log(2 β 5) = log(10) = 1, so 0 is a solution to log(π₯ + 2) + log(π₯ + 5) = 1.
What is the result if you evaluate log(π₯ + 2) + log(π₯ + 5) at π₯ = β7? Did you find a solution?
log(β7 + 2) and log(β7 + 5) are not defined because β7 + 2 and β7 + 5 are negative. Thus, β7 is
not a solution to the original equation.
What is the term we use for an apparent solution to an equation that fails to solve the original equation?
It is called an extraneous solution.
Remember to look for extraneous solutions and to exclude them when you find them.
Exercise 1 (4 minutes)
Allow students to work in pairs or small groups to think about the exponential equation below. This equation can be
solved rather simply by an application of the logarithmic property logπ(π₯π) = π logπ(π₯). However, if students do not
see to apply this logarithmic property, it can become algebraically difficult.
Exercises
1. Drew said that the equation π₯π¨π π[(π + π)π] = π cannot be solved because he expanded
(π + π)π = ππ + πππ + πππ + ππ + π and realized that he cannot solve the equation
ππ + πππ + πππ + ππ + π = ππ. Is he correct? Explain how you know.
If we apply the logarithmic properties, this equation is solvable.
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 = ππ
π = π
Check: If π = π, then π₯π¨π π[(π + π)π] = π π₯π¨π π(π) = π β π = π, so π is a solution to the original equation.
MP.3
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
196
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Exercises 2β4 (6 minutes)
Students should work on these three exercises independently or in pairs to help develop fluency with these types of
problems. Circulate around the room, and remind students to check for extraneous solutions as necessary.
Solve the equations in Exercises 2β4 for π.
2. π₯π§((ππ)π) = ππ
π β π₯π§(ππ) = ππ
π₯π§(ππ) = π
ππ = ππ
π =ππ
π
Check: Since π (ππ
π) > π, we know that π₯π§ ((π β
ππ
π)
π
) is defined. Thus, ππ
π is the solution to the equation.
3. π₯π¨π ((ππ + π)π) = π
π β π₯π¨π (ππ + π) = π
π₯π¨π (ππ + π) = π
πππ = ππ + π
πππ = ππ + π
ππ = ππ
π =ππ
π
Check: Since π (πππ
) + π β π, we know that π₯π¨π ((π βπππ
+ π)π
) is defined.
Thus, πππ
is the solution to the equation.
4. π₯π¨π π((ππ + π)ππ) = ππ
ππ β π₯π¨π π(ππ + π) = ππ
π₯π¨π π(ππ + π) = π
ππ = ππ + π
π = ππ + π
π = ππ
π =π
π
Check: Since π (ππ
) + π > π, we know that π₯π¨π π (π βππ
+ π) is defined.
Thus, ππ
is the solution to this equation.
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
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Example 4 (4 minutes)
Students encountered the first extraneous solution in Example 3, but there were no extraneous solutions in
Exercises 2β4. After working through Example 4, debrief students to informally assess their understanding, and provide
guidance to align their understanding with the concepts. Remind students that they generally need to apply logarithmic
properties before being able to solve a logarithmic equation. Some sample questions are included with likely student
responses. Remember to have students check for extraneous solutions in all cases.
log(π₯ + 10) β log(π₯ β 1) = 2
log (π₯ + 10
π₯ β 1) = 2
π₯ + 10
π₯ β 1= 102
π₯ + 10 = 100(π₯ β 1)
99π₯ = 110
π₯ =10
9
Is 10
9 a valid solution? Explain how you know.
Yes; log (109
+ 10) and log (109
β 1) are both defined, so 10
9 is a valid solution.
Why could we not rewrite the original equation in exponential form using the definition of the logarithm
immediately?
The equation needs to be in the form logπ(π) = πΏ before using the definition of a logarithm to rewrite
it in exponential form, so we had to use the logarithmic properties to combine terms first.
Example 5 (3 minutes)
Make sure students verify the solutions in Example 5 because there is an extraneous solution.
log2(π₯ + 1) + log2(π₯ β 1) = 3
log2((π₯ + 1)(π₯ β 1)) = 3
log2(π₯2 β 1) = 3
23 = π₯2 β 1
0 = π₯2 β 9
0 = (π₯ β 3)(π₯ + 3)
Thus, π₯ = 3 or π₯ = β3. These solutions need to be checked to see if they are valid.
Is 3 a valid solution?
log2(3 + 1) + log2(3 β 1) = log2(4) + log2(2) = 2 + 1 = 3, so 3 is a valid solution.
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
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Is β3 a valid solution?
Because β3 + 1 = β2, log2(β3 + 1) = log2(β2) is undefined, so β3 not a valid solution. The value
β3 is an extraneous solution, and this equation has only one solution: 3.
What should we look for when determining whether or not a solution to a logarithmic equation is extraneous?
We cannot take the logarithm of a negative number or 0, so any solution that would result in the input
to a logarithm being negative or 0 cannot be included in the solution set for the equation.
Exercises 5β9 (8 minutes)
Have students work on these exercises individually to develop fluency with solving logarithmic equations. Circulate
throughout the classroom to informally assess understanding and provide assistance as needed.
Solve the logarithmic equations in Exercises 5β9, and identify any extraneous solutions.
5. π₯π¨π (ππ + ππ + ππ) β π₯π¨π (π + π) = π
π₯π¨π (ππ + ππ + ππ
π + π) = π
ππ + ππ + ππ
π + π= πππ
ππ + ππ + ππ
π + π= π
ππ + ππ + ππ = π + π
π = ππ + ππ + π
π = (π + π)(π + π)
π = βπ or π = βπ
Check: If π = βπ, then π₯π¨π (π + π) = π₯π¨π (π), which is undefined. Thus, βπ is an extraneous solution.
Therefore, the only solution is βπ.
6. π₯π¨π π(ππ) + π₯π¨π π(π) = π
π₯π¨π π(ππ) + π = π
π₯π¨π π(ππ) = π
ππ = ππ
π = ππ
π =π
π
Check: Since π
π> π, π₯π¨π π (π β
ππ
) is defined.
Therefore, π
π is a valid solution.
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
199
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7. π π₯π§(π + π) β π₯π§(βπ) = π
π₯π§((π + π)π) β π₯π§(βπ) = π
π₯π§ ((π + π)π
βπ) = π
π =(π + π)π
βπ
βπ = ππ + ππ + π
π = ππ + ππ + π
π = (π + π)(π + π)
π = βπ or π = βπ
Check: Thus, we get π = βπ or π = βπ as solutions to the quadratic equation. However, if π = βπ, then
π₯π§(π + π) = π₯π§(βπ), so βπ is an extraneous solution. Therefore, the only solution is βπ.
8. π₯π¨π (π) = π β 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 = ππ
Check: Since ππ > π, π₯π¨π (ππ) is defined.
Therefore, ππ is a valid solution to this equation.
9. π₯π§(π + π) = π₯π§(ππ) β 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) = ππ
ππ + ππ + π = ππ
ππ + ππ β π = π
(π β π)(π + π) = π
π = π or π = βπ
Check: If = βπ, then the expressions π₯π§(π + π) and π₯π§(π + π) are undefined.
Therefore, the only valid solution to the original equation is π.
Closing (3 minutes)
Have students summarize the process they use to solve logarithmic equations in writing. Circulate around the classroom
to informally assess student understanding.
If an equation can be rewritten in the form logπ(π) = πΏ for an expression π and a number πΏ, then
apply the definition of the logarithm to rewrite as ππΏ = π. Solve the resulting exponential equation,
and check for extraneous solutions.
If an equation can be rewritten in the form logπ(π) = logπ(π) for expressions π and π, then the fact
that the logarithmic functions are one-to-one gives π = π. Solve this resulting equation, and check for
extraneous solutions.
Exit Ticket (4 minutes)
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
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Name Date
Lesson 14: Solving Logarithmic Equations
Exit Ticket
Find all solutions to the following equations. Remember to check for extraneous solutions.
1. log2(3π₯ + 7) = 4
2. log(π₯ β 1) + log(π₯ β 4) = 1
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
201
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Exit Ticket Sample Solutions
Find all solutions to the following equations. Remember to check for extraneous solutions.
1. π₯π¨π π(ππ + π) = π
π₯π¨π π(ππ + π) = π
ππ + π = ππ
ππ = ππ β π
π = π
Since π(π) + π > π, we know π is a valid solution to the equation.
2. π₯π¨π (π β π) + π₯π¨π (π β π) = π
π₯π¨π ((π β π)(π β π)) = π
π₯π¨π (ππ β ππ + π) = π
ππ β ππ + π = ππ
ππ β ππ β π = π
(π β π)(π + π) = π
π = π or π = βπ
Check: Since the left side of the equation is not defined for π = βπ, this is an extraneous solution.
Therefore, the only valid solution is π.
Problem Set Sample Solutions
1. Solve the following logarithmic equations.
a. π₯π¨π (π) =ππ
π₯π¨π (π) =π
π
π = ππππ
π = πππβππ
Check: Since πππβππ > π, we know π₯π¨π (πππβππ) is defined.
Therefore, the solution to this equation is πππβππ.
b. π π₯π¨π (π + π) = ππ
π₯π¨π (π + π) = π
π + π = πππ
π + π = πππ
π = ππ
Check: Since ππ + π > π, we know π₯π¨π (ππ + π) is defined.
Therefore, the solution to this equation is ππ.
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
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c. π₯π¨π π(π β π) = π
π β π = ππ
π = βππ
Check: Since π β (βππ) > π, we know π₯π¨π π(π β (βππ)) is defined.
Therefore, the solution to this equation is βππ.
d. π₯π¨π π(ππππ) = π
π₯π¨π π[(ππ)π] = π
π β π₯π¨π π(ππ) = π
π₯π¨π π(ππ) = π
ππ = ππ
π =π
π
Check: Since ππ (ππ
)π
> π, we know π₯π¨π π (ππ (ππ
)π
) is defined.
Therefore, the solution to this equation is π
π.
e. π₯π¨π π(πππ + πππ + ππ) = π
π₯π¨π π[(ππ + π)π] = π
π β π₯π¨π π(ππ + π) = π
π₯π¨π π(ππ + π) = π
ππ + π = ππ
ππ + π = ππ
ππ = ππ
π =ππ
π
Check: Since π (πππ
)π
+ ππ (πππ
) + ππ = πππ, and πππ > π, π₯π¨π π (π (πππ
)π
+ ππ (πππ
) + ππ) is defined.
Therefore, the solution to this equation is ππ
π.
2. Solve the following logarithmic equations.
a. π₯π§(ππ) = ππ
π β π₯π§(π) = ππ
π₯π§(π) = π
π = ππ
Check: Since ππ > π, we know π₯π§((ππ)π) is defined.
Therefore, the only solution to this equation is ππ.
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
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b. π₯π¨π [(πππ + πππ β ππ)π] = ππ
π β π₯π¨π (πππ + πππ β ππ) = ππ
π₯π¨π (πππ + πππ β ππ) = π
πππ + πππ β ππ = πππ
πππ + πππ β πππ = π
πππ + πππ β ππ β πππ = π
ππ(π + ππ) β π(π + ππ) = π
(ππ β π)(π + ππ) = π
Check: Since πππ + πππ β ππ > π for both π = βππ and π =ππ
, we know the left side of the original
equation is defined at these values.
Therefore, the two solutions to this equation are βππ and π
π.
c. π₯π¨π [(ππ + ππ β π)π] = π
π π₯π¨π (ππ + ππ β π) = π
π₯π¨π (ππ + ππ β π) = π
ππ + ππ β π = πππ
ππ + ππ β π = π
ππ + ππ β π = π
π =βπ Β± βπ + ππ
π
= βπ Β± βπ
Check: Since ππ + ππ β π = π when π = βπ + βπ or π = βπ β βπ, we know the logarithm is defined for
these values of π.
Therefore, the two solutions to the equation are βπ + βπ and βπ β βπ.
3. Solve the following logarithmic equations.
a. π₯π¨π (π) + π₯π¨π (π β π) = π₯π¨π (ππ + ππ)
π₯π¨π (π) + π₯π¨π (π β π) = π₯π¨π (ππ + ππ)
π₯π¨π (π(π β π)) = π₯π¨π (ππ + ππ)
π(π β π) = ππ + ππ
ππ β ππ β ππ = π
(π + π)(π β π) = π
Check: Since π₯π¨π (βπ) is undefined, βπ is an extraneous solution.
Therefore, the only solution to this equation is π.
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
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b. π₯π§(ππππ) β π π₯π§(π) = π
π₯π§(ππππ) β π₯π§(ππ) = π
π₯π§ (ππππ
π) = π
πππ = ππ
ππ =ππ
π
π =βππ
π or π = β
βππ
π
Check: Since the value of π in the logarithmic expression is squared, π₯π§(ππππ) is defined for any nonzero
value of π.
Therefore, both βππ
π and β
βππ
π are valid solutions to this equation.
c. π₯π¨π (π) + π₯π¨π (βπ) = π
π₯π¨π (π(βπ)) = π
π₯π¨π (βππ) = π
βππ = πππ
ππ = βπ
Since there is no real number π so that ππ = βπ, there is no solution to this equation.
d. π₯π¨π (π + π) + π₯π¨π (π + π) = π
π₯π¨π ((π + π)(π + π)) = π
(π + π)(π + π) = πππ
ππ + ππ + ππ β πππ = π
ππ + ππ β ππ = π
π =βπ Β± βππ + πππ
π
= βπ Β± βπππ
Check: The left side of the equation is not defined for π = βπ β βπππ, but it is for π = βπ + βπππ.
Therefore, the only solution to this equation is π = βπ + βπππ.
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NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
205
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
e. π₯π¨π (πππ + π) β π = π₯π¨π (π β π)
π₯π¨π (πππ + π) β π₯π¨π (π β π) = π
π₯π¨π (πππ + π
π β π) = π
πππ + π
π β π= πππ
πππ + π
π β π= ππππ
πππ + π = πππππ β ππππ
ππππ = ππππ
π =ππ
ππ
Check: Both sides of the equation are defined for π =ππππ
.
Therefore, the solution to this equation is ππ
ππ.
f. π₯π¨π π(π) + π₯π¨π π(ππ) + π₯π¨π π(ππ) + π₯π¨π π(ππ) = π
π₯π¨π π(π β ππ β ππ β ππ) = π
π₯π¨π π(ππππ) = π
π₯π¨π π[(ππ)π] = π
π β π₯π¨π π(ππ) = π
π₯π¨π π(ππ) = π
ππ = ππ
π =π
π
Check: Since π
π> π, all logarithmic expressions in this equation are defined for π =
ππ
.
Therefore, the solution to this equation is π
π.
![Page 14: Lesson 14: Solving Logarithmic Equations](https://reader030.fdocuments.us/reader030/viewer/2022020916/61ab646ff7b38f06217f3981/html5/thumbnails/14.jpg)
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 14
ALGEBRA II
Lesson 14: Solving Logarithmic Equations
206
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
4. Solve the following equations.
a. π₯π¨π π(π) = π
ππ
b. π₯π¨π π(π) = π
π
c. π₯π¨π π(π) = βπ
π
ππ
d. π₯π¨π βπ(π) = π
π
e. π₯π¨π βπ(π) = π
πβπ
f. π₯π¨π π(ππ) = π
π, βπ
g. π₯π¨π π(πβπ) = ππ
π
ππ
h. π₯π¨π π(ππ + π) = π
π
i. π = π₯π¨π π(ππ β π)
π
j. π₯π¨π π(π β ππ) = π
π
k. π₯π§(ππ) = π
ππ
π
l. π₯π¨π π(ππ β ππ + π) = π
π, βπ
m. π₯π¨π ((ππ + π)π) = ππ
πβπ, βπβπ
n. π₯π¨π (π) + π₯π¨π (π + ππ) = π
π
o. π₯π¨π π(π β π) + π₯π¨π π(ππ) = π
π
p. π₯π¨π (π) β π₯π¨π (π + π) = βπ
π
π
q. π₯π¨π π(π + π) β π₯π¨π π(π β π) = π
ππ
ππ
r. π₯π¨π (π) + π = π₯π¨π (π± + π)
π
s. π₯π¨π π(ππ β π) β π₯π¨π π(π + π) = π
π
t. π β π₯π¨π π(π β π) = π₯π¨π π(ππ)
π
u. π₯π¨π π(ππ β ππ) β π₯π¨π π(π β π) = π
No solution
v. π₯π¨π (β(π + π)π) =ππ
π
w. π₯π§(πππ β π) = π
βπ
π, β
βπ
π
x. π₯π§(π + π) β π₯π§(π) = π
ππ β π