1
Lesson 10.1: An Introduction to Logarithms
Learning Goals:
1) What is a logarithm?
2) How do we convert between logarithmic and exponential form?
Do Now: Find the inverse of each of the following: switch π₯ & y values!
1. π(π₯) = 3π₯ + 5 2. π(π₯) = βπ₯ + 53
β 3 3. π(π₯) = 2π₯
π¦ = 3π₯ + 5 π¦ = βπ₯ + 53
β 3 π¦ = 2π₯
π₯ = 3π¦ + 5 π₯ = βπ¦ + 53 β 3 π₯ = 2π¦
Solve for π¦! Solve for π¦! Solve for π¦!
π₯ β 5 = 3π¦ π₯ β 3 = βπ¦ + 53
π¦ =π₯β5
3= πβ1(π₯) (π₯ β 3)3 = (βπ¦ + 5
3)3
(π₯ β 3)3 = π¦ + 5
π¦ = (π₯ β 3)3 β 5 = πβ1(π₯)
What happens when you try to find the inverse of problem 3? Currently we donβt
know how to solve for π¦!
Defining Logarithmic Functions
The function π¦ = logππ₯ is the name we give the inverse of π¦ = ππ₯.
π¦ = logππ₯ is the same as ππ¦ = π₯
A logarithm gives as its output (π¦-value) the exponent we must raise π to in order
to produce its input (π₯-value).
How would you write the inverse to problem 3 using some form of the word
logarithm? π₯ = 2π¦ so π¦ = log2π₯
2
What are logarithms? Logarithms are another way to write an exponential
function!
logππ = π₯ means ππ₯ = π, where π > 0 and π β 1
Parts of a logarithm: base, anti-logarithm, and exponent
log πβbaseβ
π₯βantiβlog
= π¦βexponent
Trick to remember is to βloopβ:
log8π₯ =1
3 logπ₯125 = β
3
2
81
3 = π₯ π₯β3
2 = 125
π₯ = 2 (π₯β3
2)β2
3= (125)β
2
3
π₯ =1
25
Exponential Form Logarithmic Form π₯
2π₯ = 64 log264 = π₯ 2π₯ = 26 π₯ = 6
π₯4 = 81 logπ₯81 = 4 (π₯4)
14 = (81)
14 π₯ = Β±3 so π₯ = +3
52 = π₯ log5π₯ = 2 π₯ = 25
72 = π₯ log7π₯ = 2 π₯ = 49
π₯3 = 27 logπ₯27 = 3 (π₯3)
13 = (27)
13 π₯ = 3
3
Part 1: Write each exponential equation in logarithmic form.
1. 24 = 16 2. 81
3 = 2
log216 = 4 log82 =1
3
Part 2: Write each logarithmic equation in exponential form.
3. log432 =5
2 4. logπ
1
π= β1
45
2 = 32 πβ1 =1
π
Part 3: Solve each equation for π₯:
5. log2π₯ = 4 6. log81π₯ =1
2
24 = π₯ 811
2 = π₯
π₯ = 16 π₯ = 9
7. log6π₯ = β2 8. log64π₯ =2
3
6β2 = π₯ 642
3 = π₯
π₯ =1
36 π₯ = 16
Part 4: Evaluate each expression:
9. log25125 = π₯ set it = π₯ 10. log51
25= π₯ set it = π₯
25π₯ = 125 5π₯ =1
25
(52)π₯ = 53 5π₯ = 25β1
2π₯ = 3 5π₯ = (52)β1
π₯ =3
2 π₯ = β2
4
11. log491
7= π₯ set it = π₯ 12. log1000.1 = π₯ set it = π₯
49π₯ =1
7 100π₯ = 0.1
49π₯ = 7β1 100π₯ =1
10
(72)π₯ = 7β1 100π₯ = 10β1
2π₯ = β1 (102)π₯ = 10β1
π₯ = β1
2 2π₯ = β1
π₯ = β1
2
Part 5: Solve each equation for π.
13. logπ27 = 3 14. logπ1
2= β1
π3 = 27 πβ1 =1
2
βπ33
= β273
(πβ1)β1 = (1
2)β1
π = 3 π = 2
15. logπβ5 =1
4 16. logπ4 =
1
2
π1
4 = β5 π1
2 = 4
(π1
4)4
= (51
2)4
(π1
2)2
= (4)2
π = 52 = 25 π = 16
17. If π(π₯) = log3π₯, find π(1)
π(1) means π₯ = 1!
5
π(1) = log31 β "logbase" in calc!
π¦ = log31
3π¦ = 1
π¦ = 0
18. Verify the following by evaluating the logarithms:
log2(8)β + log2(4)β = log2(32)β log2(8)β + log2(4)β = log2(32)β
2π₯ = 8 2π₯ = 4 2π₯ = 32 3 + 2 = 5
2π₯ = 23 2π₯ = 22 2π₯ = 25 5 = 5 β
π₯ = 3 π₯ = 2 π₯ = 5
6
Homework 10.1: An Introduction to Logarithms
Write each of the following in logarithmic form.
1. 216 = 63 2. 4β2 = 0.0625 3. 6253
4 = 125
Write each of the following in exponential form.
4. 5 = log3243 5. β2 = log50.04 6. log49343 =3
2
Evaluate each of the following:
7. log6216 8. log232 9. 5 β log88
Solve each equation for the variable.
10. π = log416 11. log8π₯ =1
2 12. logπ64 = 6
7
13. If π(π₯) = log3π₯, find π(81).
14. When $1 is invested at 6% interest, its value, π΄, after π‘ years is π΄ = 1.06π‘.
Express π‘ in terms of π΄.
15. Can the value of log2(β4) be found? What about the value of log20? Why
or why not? What does this tell you about the domain of logππ₯?
16. Which of the following is equivalent to π¦ = log7π₯?
(1) π¦ = π₯7 (2) π₯ = π¦7 (3) π₯ = 7π¦ (4) π¦ = π₯1
7
17. Verify the following by evaluating the logarithms:
log4(4) + log4(16) = log4(64)
8
Lesson 10.2: Natural Logarithms
Learning Goals:
1) What is a logarithm?
2) What is the number π?
3) What is a natural logarithm?
4) What is a common logarithm?
5) How do we convert between logarithmic and exponential form?
Do Now: Which of the following are valid examples of logarithms?
log24 log14 log04 log327 log5(β25) log70
Valid Examples of Logarithms Invalid Examples of Logarithms
log327
log24
log04
log5(β25)
log14
log70
Definition of the Logarithm base π:
If three numbers, πΏ, π, and π₯ with 0 < π < 1 or π > 1 are related by ππΏ = π₯, then πΏ
is the logarithm base π of π₯, and we write logπ(π₯) = πΏ.
9
Compound Interest: You deposit $3500 in an account that pays 4% annual
interest compounded continuously. What is the balance after 1 year?
Solution: π = 3500 π = .04 π‘ = 1
π = amount you invest(principal) r = interest rate (%
100) π‘ = time
Use the formula for continuously compounded interest.
π΄ = π ππ π‘ Write the formula
= 3500 π(.04β1) Substitute for π, π, and π‘
π΄ = 3642.84 The balance at the end of 1 year is $3642.84
Complete the following example:
You deposit $4800 in an account that pays 6.5% annual interest compounded
continuously. What is the balance after 3 years?
When it says compounded continuously use base of π!
Solution: π = 4800 π = .065 π‘ = 3
π = amount you invest(principal) r = interest rate (%
100) π‘ = time
Use the formula for continuously compounded interest.
The Natural Base π
The natural base π is irrational. It is defined as follows:
As π approaches +β, (1 +1
π)π
approaches
π β 2.718. 2nd β Γ· β π
Natural Base Functions
A function of the form π¦ = π ππ π₯ is called a natural base exponential function.
If π > 0 and π > 0, the function is an exponential GROWTH function.
If π > 0 and π < 0, the function is an exponential DECAY function.
10
π΄ = π ππ π‘ Write the formula
= 4800 π(.065β3) Substitute for π, π, and π‘
π΄ = 5833.49 The balance at the end of 3 years is $5833.49
The natural logarithms, gives an exponent as its output. In fact, it gives the
power that we must raise π to in order to get the input.
π¦ = lnπ₯ lnπ₯ = logππ₯ lnπ₯ β lnππ₯
Without using your calculator, determine the values of each of the following.
1. ln(π) = π₯ 2. ln1 = π₯
logππ = π₯ logπ1 = π₯
ππ₯ = π1 ππ₯ = 1
π₯ = 1 π₯ = 0
3. ln(π5) = π₯ 4. lnβπ = π₯
logππ5 = π₯ logπβπ = π₯
ππ₯ = π5 ππ₯ = βπ
π₯ = 5 π₯ =1
2
Rewrite each of the following in logarithmic form.
5.. π5 = π₯ 6. ππ₯ = 15
logππ₯ = 5 logπ15 = π₯
lnπ₯ = 5 ln15 = π₯
The Natural Logarithm
The inverse of π¦ = ππ₯: π¦ = lnπ₯ π¦ = logeπ₯
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Common and Natural Logarithms {Calculator Information}
Common Logarithm: A logarithm with a base of 10.
On the graphing calculator, the Log key is used to display the common logarithm
of a number.
Examples: Find the common log of each number to four decimal places.
1. 279 = log(279) = 2.4456 2. 0.0005 = log(. 0005) = β3.3010
When I find the common log of a number, it means that
102.4456 β 279 and 10β3.3010 β 0.0005
How can we find the number if we know the common log of the number?
log(π₯) = log10(π₯)
Examples: Find to the nearest thousandth each number π whose common
logarithm is given.
3. logπ = 3.9294 4. logπ = β1.7799
log10π = 3.9294 log10π = β1.7799
103.9294 = π 10β1.779 = π
12
On the graphing calculator, the ln key is used to display the natural logarithm of a
number.
Find the natural log of each number to four decimal places.
5. ln52 = 3.9512 6. 2ln16 = 5.5452
When I find the natural log of a number, it means that π3.9512 β 52
How can we find the number if we know the natural log of the number?
ln(π₯) = logπ(π₯)
Examples: Find to four decimal places the antilogarithm of the given logarithm.
7. lnπ₯ = β.5373 8. lnπ₯ = .8297
πβ.5373 = π₯ π .8297 = π₯
π₯ = .5843 π₯ = 2.2926
Compute the value of each logarithm. Verify your answers using an exponential
statement.
1. log13(13) = 1 2. logβ7 (1
49) = β4
13π₯ = 13 β7π₯=
1
49
13π₯ = 131 (71
2)π₯
= 49β1
π₯ = 1 (71
2)π₯
= (72)β1
1
2π₯ = β2
π₯ = β4
13
What is wrong with the following questions?
3. log2(0) 4. log3 (β1
3)
They canβt be evaluated!
Find the value of each of the following.
5. If π₯ = log2(8) and π¦ = 2π₯, 6. If π = 26 and π = log2(π),
find the value of π¦. find the value of π .
2π₯ = 8 and π¦ = 2π₯, then π¦ = 8 2π = π sub in the value of π!
2π = 26
π = 6
Rewrite each of the following in exponential form.
7. ln25 = π₯ 8. lnπ₯ = 52 9. log45 = π₯ 10. logπ₯ = .32
ππ₯ = 25 π52 = π₯ 10π₯ = 45 10π₯ = .32
Find π₯ using your calculator. Round all answers to the nearest hundredth.
11. ln45 = π₯ 12. log75 = π₯ 13. lnπ₯ = 3.2958 14. logπ₯ = β.4881
π₯ = 3.81 π₯ = 1.88 π₯ = 27.00 π₯ = 0.33
15. Solve for π₯: 3 log(π₯ + 4) = 6 16. Solve for π₯ to the nearest
3 log(π₯+4)
3=6
3 thousandth:
log(π₯ + 4) = 2 4 + ln(π₯ β 1) = 4.8200
102 = π₯ + 4 ln(π₯ β 1) = .82
100 = π₯ + 4 π .82 = π₯ β 1
π₯ = 96 π₯ = π .82 + 1
You can always check your answers! π₯ = 3.270
14
Homework 10.2: Natural Logarithms
1. When Kyle was born, his grandparents invested $5000 in a college fund that
paid 4% compounded continuously. What is the value of the account after 18
years? {π΄ = πππ π‘}
2. A hot liquid is cooling in a room whose temperature is constant. Its
temperature can be modeled using the exponential function shown below. The
temperature, π, is in degrees Fahrenheit and it is a function of the number of
minutes, π, it has been cooling.
π(π) = 101 πβ0.03 π + 67
a) What was the initial temperature of the water at π = 0?
b) How do you interpret the statement that π(60) = 83.7?
15
Determine the value for each of the following logarithms.
3. log2 (1
64) 4. log48 5. log2β4
5
6. ln(π4) 7. lnβπ3
8. lnπ
Rewrite in logarithmic form.
9. π7 = π₯ 10. ππ₯ = 7 11. 2π₯ = 5
12. log927 13. logπ₯(π₯3)
Find π₯ in each of the following. Round answers to the nearest hundredth.
14. lnπ₯ = 3.3534 15. logπ₯ = 1.7218
16. log0.528 = π₯ 17. ln51.3 = π₯
16
Lesson 10.3: Properties of logarithms
Learning Goal: How do we use the properties to expand and simplify log
expressions?
Do Now: Answer the following questions in order to prepare for todayβs lesson.
1. Simplify: (3π₯5π¦4)
2β(2π 4π¦)
6π₯5π¦6=(32βπ₯10βπ¦8)(2π₯4π¦)
6π₯5π¦6=18π₯14π¦9
6π₯5π¦6= 3π₯9π¦3
(1) Power Law (2) Product Law (3) Quotient Law
2. Write each expression in radical form.
a. 71
2 index = 2 b. 25
3 index = 3 c. (3π)7
4 index = 4
β7 β253
β(3π)74
3. Write each expression in exponential form.
a. β23
= 21
3 b. (β10)3= 10
3
2 c. βπ₯5π¦4
= π₯5
4π¦1
4
4. What are logarithms? Logs are exponents!
5. Label the base, exponent, and anti-logarithm: logbπ = π₯
17
Using the properties of exponents, we can arrive at the properties of
logarithms.
Properties of Exponents Properties of Logarithms
Multiplication Law
ππ β ππ = ππ+π logb(π β π) = logbπ+ logbπ
Division Law ππ
ππ= ππβπ logb (
π
π) = logbπβ logbπ
Power Law
(ππ)π = ππβπ logπππ = πlogbπ
When separating logs, the power law is done last!
Directions: Rewrite each expression as sums and differences in terms of
log(π₯) , log(π¦) , and log (π§)
1. logπ₯3
π¦= 3logπ₯ β logπ¦
2. logππ₯βπ¦ = logππ₯ + logπβπ¦ = logππ₯ + logππ¦1
2 = logππ₯ +1
2logππ¦
3. logπ₯
π¦π§= logπ₯ β (logπ¦ + logπ§) = logπ₯ β logπ¦ β logπ§
4. log(βπ₯7π¦3) = log(β7 β βπ¦3) = logβ7 + logβπ¦3 = log71
2 + logπ¦3
2
=1
2log7 +
3
2logπ¦ or
1
2(7logπ₯ + 3logπ¦)
5. logβπ₯
π¦63
= logπ₯1
3 β logπ¦2 =1
3logπ₯ β 2logπ¦
6. log (βπ₯π¦73
π₯2π§) = log (π₯
1
3) + logπ¦7
3 β logπ₯2 β logπ§ =
1
3logπ₯ +
7
3logπ¦ β 2logπ₯ β logπ§
18
Directions: Use the properties of logarithms to rewrite each expression in an
equivalent form containing a single logarithm.
1. 2log5π+ log5π = log5π2 + log5π = log5(π
2π)
2. 1
2logπ β 3logπ = logβπ β logπ3 = log (
βπ
π3)
3. log2 + logπΌ β logπ = log (2πΌ
π)
4. 2logπ₯ β (3logπ¦ + logπ§) = logπ₯2 β (logπ¦3 + logπ§) = log (π₯2
π¦3π§)
5. 2logπ₯ + 3logπ¦ β1
2logπ§ = logπ₯2 + logπ¦3 β logβπ§ = log (
π₯2π¦3
βπ§)
6. 1
3(log(π₯) β 3 log(π¦) + log (π§)) = βlog (
π₯
π¦3π§)
3
When proving equations, you must get the left side = to the right side!
Directions: Use properties of logarithms to show that:
1. log(26) β log(13) = log (2) 2. log(3) + log(4) + log(5) β log(6) = 1
log (26
13) = log2 log (
3(4)(5)
6) = 1
log2 = log2 β log10 = 1
1 = 1 β
3. 1
2log(25) + log(4) = log (20) 4. log (
1
2β1
3) + log(2) = log (
1
3)
logβ25 + log4 = log20 log (1
6) + log2 = log (
1
3)
log5 + log4 = log20 log (1
6β 2) = log (
1
3)
log(5 β 4) = log 20 log (1
3) = log (
1
3) β
log20 = log20 β
19
5. log (1
3β1
4) + (log (
1
3) β log (
1
4)) = log (
1
9)
log (1
12) + log (
1
31
4
) = log (1
9)
log (1
12) + log (
4
3) = log (
1
9)
log (1
12β4
3) = log (
1
9)
log (1
9) = log (
1
9) β
20
Homework 10.3: Properties of logarithms
1. The expression log4π₯ is equivalent to:
(1) logπ₯4 (2) 4logπ₯ (3) log4 + logπ₯ (4) (log4)(logπ₯)
2. log βπ₯π¦
π§ is equal to:
(1) 1
2logπ₯ +
1
2logπ¦ β logπ§ (2)
1
2logπ₯ + logπ¦ β logπ§
(3) 1
2(logπ₯ + logπ¦ β logπ§) (4)
1
2logπ₯π¦
logπ§
3. The expression 3logπ₯ β1
2logπ¦ is equal to
(1) logπ₯3
π¦2 (2) log
π₯3
βπ¦ (3) logβ
3π₯
π¦ (4)
log3π₯1
2logπ¦
4. Apply properties of lograrithms to rewrite the following expressions as a single
logarithm or number.
a. 1
2ln(36) + ln (2)
b. ln(4) β 3ln (1
3) + ln (2)
c. ln(5) +3
5ln(32) β ln (4)
5. Apply properties of logarithms to rewrite each expression as a sum of terms
involving numbers, log(π₯) , log(π¦) , and log (π§), where π₯, π¦, and π§ are positive real
numbers.
a. log (π₯π¦π§3) b. log (βπ₯3
π¦π§
3) c. log(3π₯4βπ¦5)
6. Use properties of logarithms to show that:
a. log(2) β log (1
13) = log (26) b.
1
2log(16) + log(3) + log (
1
4) = log (3)
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Lesson 10.4: More Properties of Logarithms and Substitution
Learning Goals:
1) How do we use the properties to expand and simplify log expressions?
2) How do we simplify a logarithmic expression using powers of 10 rules?
Do Now: Answer the following questions in order to prepare for todayβs lesson.
1. What is a common logarithm? How do you know when you have a common
logarithm? A common log has no base visible so we know base = 10 for
example logπ₯ = log10π₯
2. Complete the following table of logarithms by using a calculator; then, answer
the questions that follow.
π₯ log (π₯) 1,000,000 106 = 6
100,000 105 = 5
10,000 104 = 4
1,000 103 = 3
100 102 = 2
10 101 = 1
a) What is log (1)? How does that follow from the definition of a base-10
logarithm?
log(1) = 0 100 = 1 count # of zeros since the base = 10
b) What is log(10π) for an integer π? How does that follow from the definition of
a base- 10 logarithm?
Write a conjecture using logarithmic notation.
log(10π) = π π is the # of zeros!
22
Logarithm Rules from Previous Lesson
Multiplication Law logπ(ππ) = logππ+ logbπ
Division Law
logπ (π
π) = logππ β logbπ
Power Law logππ
π = πlogππ
New Logarithm Rules
When expanding logarithms, you must be aware of special logs that can be
reduced!
Below are two special cases:
log(10π) = π
ln(π) = 1
π is the # of zeros or exponent of 10. ln(π) = logππ = 1 so π1 = π
Directions: Apply properties of logarithms to rewrite each expression as a sum of
terms involving numbers, log(π₯) , log(π¦) , and log (π§), where π₯, π¦, and π§ are positive
real numbers.
a. log(1000βπ₯) = log1000 + logβπ₯ = 3 +1
2logπ₯
b. log (100π₯2
π¦3) = log100 + 2logπ₯ β 3logπ¦ = 2 + 2logπ₯ β 3logπ¦
c. log (1
10π₯2π§) = log1 β (log10 + 2logπ₯ + logπ§) = 0 β (1 + 2logπ₯ + logπ§)
= β1 β 2logπ₯ β logπ§
Directions: Write each expression as a sum or difference of constants and
logarithms of simpler terms. Remember: lnπ = 1
a. ln(3π) = ln3 + lnπ = ln3 + 1
b. ln (π4
π₯π¦) = lnπ4 β (lnπ₯ + lnπ¦) = 4lnπ β lnπ₯ β lnπ¦ = 4 β lnπ₯ β lnπ¦
c. ln (β5π₯3
π2) = ln(β5π₯3) β lnπ2 = ln (5
1
2 β π₯3
2) β lnπ2 =
1
2ln5 +
3
2lnπ₯ β 2lnπ =
1
2ln5 +
3
2lnπ₯ β 2 or
1
2(ln5 + 3lnπ₯) β 2
23
Substitution with Logarithms
When asked to substitute using logarithms, you must EXPAND the
logarithms first by using the properties of logarithms.
Example 1: If log2 = π₯ and log3 = π¦, express log6 in terms of π₯ and π¦.
log6 = log(2 β 3) = log2 + log3 = π₯ + π¦
Example 2: If ln2 = π₯ and ln3 = π¦, express ln4
9 in terms of π₯ and π¦.
ln4 β ln9 = ln22 β ln32 = 2ln2 β 2ln3 = 2π₯ β 2π¦
Example 3: If log5 = π, then log250 can be expressed as:
(1) 50π (2) 2π + 1 (3) 10 + 2π (4) 25π
log250 = log(10 β 25) = log10 + log25 = 1 + log52 = 1 + 2log5 = 1 + 2π
Example 4: If logπ₯ = π, logπ¦ = π, and logπ§ = π, rewrite logπ₯2π¦
βπ§ in terms of
π, π, and π.
logπ₯2 + logπ¦ β logβπ§ = 2logπ₯+ logπ¦ β1
2logπ§ = 2π + π β
1
2π
Practice
1. Use the approximate logarithm values below to estimate the value of each of
the following logarithms.
log(2) = 0.3010
log(3) = 0.4771
log(7) = 0.8451
a. log12 = log(4 β 3) = log(22 β 3) = 2log2 + log3 = 2(. 3010) + .4771 = 1.0791
b. log (1
9) = log1 β log9 = 0 β log32 = β2log3 = β2(. 4771) = β.9542
c. log(70) = log(7 β 10) = log7 + log10 = .8451 + 1 = 1.8451
24
2. Given: logπ3 = π and logπ5 = π
Express each of the following in terms of π and π.
a. logπ45 = logπ(9 β 5) = logπ(32) + logπ5 = 2logπ3 + logπ5 = 2π + π
b. logπβ3
5= log
π
β3
β5= log
πβ3 β logπβ5 =1
2log
π3 β
1
2log
π5 =
1
2π β
1
2π or
1
2(π β π)
c. logπ27
β53 = logπ27 β logπβ5
3= logπ(3
3) β logπ (51
3) =
3logπ3 β1
3logπ5 = 3π β
1
3π
d. logπ9
β53 = logπ9 β logπβ5
3= logπ(3
2) β logπ (51
3) =
2logπ3 β1
3logπ5 = 2π β
1
3π
3. If lnπ = 2 and lnπ = 3, what is the numerical value of ln (πβπ
π3)?
lnπ +1
2lnπ β 3lnπ = 1 +
1
2(2) β 3(3) = 1 + 1 β 9 = β7
25
Homework 10.4: More Properties of Logarithms and Substitution
1. If logπ = π₯ and logπ = π¦, then logβππ is equivalent to:
(1) 1
2π₯ + π¦ (2)
1
2(π₯ + π¦) (3)
1
2π₯π¦ (4)
1
4π₯π¦
2. Given: ln2 = π₯ and ln3 = π¦
Express each of the following in terms of π₯ and π¦: ln β2
9
3. Express each of the following as a single logarithm.
(a) log(8) + log (1
4)
(b) 1
3log(8) + 2 log(3) β log (6)
(c) 3 log(π₯) β (2 log(π¦) +1
2log (π§))
4. In the following expression, π₯, π¦, and π§ represent positive real numbers. Use
properties of logarithms to rewrite the expression in an equivalent form
containing only log(π₯) , log(π¦) , log(π§), and numbers.
logβ(π₯3π¦2
10π§)
5. Rewrite the expression as sums and differences in terms of
ln(π₯) , ln(π¦) , and ln (π§).
ln (π
π¦π§3)
26
Lesson 10.5: Solving Exponential Equations Using Logarithms
Learning Goal: How do we use logarithms to solve an exponential equation?
a) Solve the following equation for π₯: 43π₯β1 = 32
Get like bases! (22)3π₯β1 = 25
2(3π₯ β 1) = 5
6π₯ β 2 = 5
6π₯ = 7
You can always check your answer! π₯ =7
6
b) What happens when you try to solve 52π₯ = 20 for π₯?
You cannot get like bases, so we will need to use logs!
How can we solve exponential equations when you cannot get like bases?
1) Look to isolate the base with the variable exponent.
2) Take the common log of both sides
3) Use the power law to move the exponent
4) Solve for the variable
1. 2 β 52π₯ = 20
2β52π₯
2=20
2
52π₯ = 10
2π₯log5 = log10
2π₯log5
log5=log10
log5
2π₯ = 1.430676558
π₯ = .715338279
27
Solve each equation for the given variable and express your answer to the
nearest hundredth.
1. 72π€ + 8 = 34 2. π2π₯ = 5
72π€ = 26 2π₯lnπ = ln5 or 2π₯logπ = log5
2π€log7 = log26 2π₯ = ln5 or 2π₯ =log5
logπ
2π€log7
log7=log26
log7 π₯ = .80
2π€ = 1.67433041
π€ = 0.84
3. 52+π₯ β 5π₯ = 10 4. 32π₯β3 = 2π₯+4
GCF: 5π₯ (2π₯ β 3)log3 = (π₯ + 4)log2
5π₯(52 β 1) = 10 2π₯ β 3 = (π₯ + 4)log2
log3
5π₯(25 β 1) = 10 2π₯ β 3 = (π₯ + 4)(.6309297536)
5π₯(24) = 10 2π₯ β 3 = .63π₯ + 2.52
5π₯ =5
12 1.37π₯ = 5.52
π₯log5 = log5
12 π₯ = 4.03
π₯ = β0.54
Solve each equation for the given variable and express your answer to the
nearest hundredth.
5. 5π3π₯+4 = 15 6. 5π₯+3 = 42π₯β1
π3π₯+4 = 3 (π₯ + 3)log5 = (2π₯ β 1)log4
(3π₯ + 4)lnπ = ln3 π₯ + 3 = (2π₯ β 1)log4
log5
3π₯ + 4 = ln3 (π₯ + 3) = (2π₯ β 1)(.8613531161)
28
π₯ =ln (3)β4
3 π₯ + 3 = 1.72π₯ β .86
π₯ = β.9671292371 β.72π₯ = β3.86
π₯ = β0.97 π₯ = 5.36
7. 2(4π₯) + 4π₯+1 = 342 8. β3 β 33π₯ = 9
GCF: 4π₯ 31
2 β 33π₯ = 32
4π₯(2 + 41) = 342 31
2+3π₯ = 32
4π₯(6) = 342 1
2+ 3π₯ = 2
4π₯ = 57 3π₯ =3
2
π₯log4 = log57 π₯ =1
2
π₯ = 2.92
Applications of Logarithms
9. You are investing $600 at an interest rate of 7.5% compounded continuously.
Using the formula, π΄ = πππ π‘, where π΄ is your final amount, π, is your starting
amount, π is the interest rate, and π‘ is the time in years, determine how long it will
take you to reach $1000, to the nearest tenth of a year?
1000 = 600 π .075βπ‘ 50 = 80(0.98)π‘,
5
3= π .075βπ‘
5
8= (0.98)π‘
ln (5
3) = .075π‘(lnπ) log (
5
8) = π‘ log(0.98)
π‘ = 6.8 π‘ = 23.26438837
1998 + 23 = 2021
10. A population of wolves in a county is represented by the equation
π(π‘) = 80(0.98)π‘, where π‘ is the number of years since 1998. In what year will
50 wolves be reached? See above to right for answer
29
Homework 10.5: Solving Exponential Equations Using Logarithms
1. Solve the following exponential equation and express your answer to the
nearest hundredth: (1
2)
π₯
3+1=1
6.
2. Solve the following equation for π₯ and express your answer to three decimal
places:
2π₯+1 = 31βπ₯
3. Find the solution to the exponential equation below to the nearest hundredth:
4(2)π₯ β 3 = 17
30
4. The logarithmic expression ln (βπ
π¦3) can be rewritten:
(1) 3lnπ¦ β 2 (2) 1β6lnπ¦
2 (3)
lnπ¦β6
2 (4) βlnπ¦ β 3
5. The savings bank account can be modeled using π(π‘) = 1250 π0.045 π‘, where π‘ is the number of years the money has been in the account. Determine, to the
nearest tenth of a year, how long it will take for the amount of savings to double
from the initial amount deposited of $1250.
6. Matt places $1200 in an investment account earning an annual rate of 6.5%,
compounded continuously. Using the formula π΄ = πππ π‘, how long will it take for
his initial investment to quadruple in value? Round to the nearest year.
31
Lesson 10.6: Solving Logarithmic Equations
Learning Goals:
1) How do we solve logarithmic equations?
2) How do we check for extraneous roots?
Warm Up:
Using what we have learned about solving for an exponent, answer each of the
following questions:
a) Solve the following equation for π‘: π β ππ π‘ = π
πβππ π‘
π=π
π
ππ π‘ =π
π
ππ‘(logπ) = log (π
π)
π π‘ (logπ)
π(logπ)=
log(π
π)
π(logπ)
π‘ =log(
π
π)
π(logπ)
RECALL
ππΏ = π₯ can be written in the form logπ(π₯) = πΏ where
0 < π < 1 or π > 1 and π₯ > 0
*This means the base cannot be 0, 1, or a negative number and the
anti-logarithm cannot be 0 or a negative number.
32
b) Solve the following equation for π: 3 + π₯π+π = π
π₯π+π = π β 3
(π + π)logπ₯ = log (π β 3)
π + π =log (πβ3)
logπ₯
π =log (πβ3)
logπ₯β π
There are 3 types of log equations:
Type Procedure Example
1 log βloopβ Cannot loop until you isolate!
log((2π₯ + 5)2) = 4
104 = (2π₯ + 5)2 3 β log2(π₯ β 1) = 0 βlog2(π₯ β 1) = β3 log2(π₯ β 1) = 3
23 = π₯ β 1
1
2log(π₯ + 2) = 2
log(π₯ + 2)12 = 2
log(π₯ + 2)12β2 = 22
log(π₯ + 2) = 4
104 = π₯ + 2
1 log on each side
βcancelβ Logs must have like bases!
2 ln(π₯ + 2) = ln (βπ₯) ln(π₯ + 2)2 = ln (βπ₯) (π₯ + 2)2 = βπ₯
2 or more logs on one side
βcombineβ logπ + logπ = log (π β π)
logπ β logπ = log (π
π)
log2(3π₯) + log2(4) = 4 log2(12π₯) = 4 24 = 12π₯
2log5π₯ β log55 = log5125
log5π₯2 β log5 = log5125
π₯2
5= 125
33
Practice! Determine which of the three methods would be most appropriate
before solving each problem. Be sure to check for extraneous solutions.
1. log2(9π₯2 + 30π₯ + 25) = 8 2. ln(π₯ + 2) = ln(12) β ln(π₯ + 3)
28 = 9π₯2 + 30π₯ + 25 π₯ + 2 =12
π₯+3 Cross Multiply!
256 = 9π₯2 + 30π₯ + 25 π₯(π₯ + 2)(π₯ + 3) = 12
0 = 9π₯2 + 30π₯ β 231 π₯2 + 5π₯ + 6 = 12
0 = 3(3π₯2 + 10π₯ β 77) π₯2 + 5π₯ β 6 = 0
Quadratic Formula or AC Method (π₯ β 1)(π₯ + 6) = 0
π₯ =11
3 & β 7 β π₯ = 1 & β 6
Always check! Omit β6 so the only answer is π₯ = 1
π₯ + 2 > 0 π₯ + 3 > 0
β6 + 2 > 0 β6 + 3 > 0
β4 > 0 no β3 > 0 no
3. log(π₯2 + 7π₯ + 12) β log(π₯ + 4) = 0 4. log7(π₯ β 2) + log7(π₯ + 3) = log714
π₯2+7π₯+12
π₯+4= 100 (π₯ β 2)(π₯ + 3) = 14
(π₯+4)(π₯+3)
(π₯+4)= 1 π₯2 + π₯ β 6 = 14
π₯ + 3 = 1 π₯2 + π₯ β 20 = 0
π₯ = β2 β (π₯ β 4)(π₯ + 5) = 0
Always check! π₯ = 4 & π₯ = β5(omit after checking)
5. ln((4π₯)5) = 15
5 ln(4π₯) = 15
ln4π₯ = 3
π3 = 4π₯ π₯ = 5.021384231 β Always check!
34
6. log2((5π₯ + 7)19) = 57
19log2(5π₯ + 7) = 57
log2(5π₯ + 7) = 3
5π₯ + 7 = 23
5π₯ + 7 = 8
5π₯ = 1
π₯ =1
5 β Always check!
7. log(π₯+2)(π₯2 + 5π₯ + 18) β log(π₯+2)(2) = 2
π₯2+5π₯+18
2= (π₯ + 2)2
π₯2+5π₯+18
2= (π₯ + 2)(π₯ + 2)
π₯2+5π₯+18
2= π₯2 + 4π₯ + 4
2π₯2 + 8π₯ + 8 = π₯2 + 5π₯ + 18
π₯2 + 3π₯ β 10 = 0
(π₯ β 2)(π₯ + 5) = 0
π₯ = 2 & β 5 Always Check!
Omit β5 so only π₯ = 2 will work!
35
Homework 10.6: Solving Logarithmic Equations
1. Solve for π₯: ln(π₯ + 2) + ln(π₯ β 2) = ln(9π₯ β 24)
2. Solve for π₯: log(10π₯ + 5) β 3 = log (π₯ β 5)
3. Solve for π₯: ln(32π₯2) β 3 ln(2) = 3
4. Solve for π₯: log2(π₯) + log2(2π₯) + log2(3π₯) + log2(36) = 6
5. Jenn claims that because log(1) + log(2) + log(3) = log (6), then
log(2) + log(3) + log(4) = log (9). Is she correct? Explain how you know.
36
Lesson 10.7: Change of Base Rule
Learning Goals:
1) What is the change of base rule for logarithms?
2) How do we solve logarithmic equations using change of base rule?
Warm Up:
Evaluate the following:
a) log28 use βlogbaseβ in calc b) log418
2π₯ = 8 4π₯ = 18
π₯log2 = log8 π₯log4 = log18
π₯ =log8
log2= 3 π₯ =
log18
log4= 2.1
1. Letβs look at the two examples from the warm-up. Use the change of base
rule to rewrite the logarithms with base 10 and then use the LOG key in your
calculator.
a) log28 =log8
log2= 3 b) log418 =
log18
log4= 2.1
2. Use the change of base property to rewrite each logarithmic function in terms
of the common logarithm function.
a) π1(π₯) = log14
(π₯) b) π2(π₯) = log12
(π₯) c) π3(π₯) = log2(π₯)
logπ₯
log1
4
logπ₯
log1
2
logπ₯
log2
Change of Base Formula for Logarithms
If π₯, π, and π are all positive real numbers with π β 1 and π β 1, then
logπ(π₯) =logπ(π₯)
logπ(π)
37
3. Using the change of base rule, rewrite the given logarithmic expressions in
base 7.
a) log211 b. log5π₯2 c. logπ(π₯ + 3)
log711
log72
log7π₯2
log75
log7(π₯+3)
log7π
Push yourself!
Show that for any positive numbers π and π with π β 1 and π β 1,
logπ(π) β logπ(π) = 1.
logb
logaβlogπ
logπ= 1
1 = 1 β
How can we use the change of base formula to solve a logarithmic equation?
Rewrite the logarithmic expresssion with the larger base in terms of the
smaller base.
Cross multiply and drop the logs on both sides.
1. Solve for π₯: log(π₯) = log100(π₯2 β 2π₯ + 6)
logπ₯
1=log (π₯2β2π₯+6)
log100
logπ₯
1=log (π₯2β2π₯+6)
2
2logπ₯ = log (π₯2 β 2π₯ + 6)
logπ₯2 = log (π₯2 β 2π₯ + 6) Cancel like bases!
π₯2 = π₯2 β 2π₯ + 6
2π₯ = 6
π₯ = 3 β Always check! log(π₯) = log100(π₯2 β 2π₯ + 6)
π₯ > 0 (π₯2 β 2π₯ + 6) > 0
3 > 0 9 > 0
38
2. Solve for π₯: log2(π₯ + 1) = log4β(π₯2 + 3π₯ + 4)
Change to base 2!
log2(π₯ + 1) =log2(π₯
2+3π₯+4)
log24
log2(π₯ + 1) =log2(π₯
2+3π₯+4)
2
2log2(π₯ + 1) = log2(π₯2 + 3π₯ + 4)
log2(π₯ + 1)2 = log2(π₯
2 + 3π₯ + 4) Cancel like bases!
(π₯ + 1)2 = (π₯2 + 3π₯ + 4)
(π₯ + 1)(π₯ + 1) = π₯2 + 3π₯ + 4
π₯2 + 2π₯ + 1 = π₯2 + 3π₯ + 4
βπ₯ = 3
π₯ = β3 Reject so βNo Solutionβ!
3. Solve for π₯: log9β(π₯2 + 2π₯ + 6) = log3(π₯ + 2)
Change to base 3!
log3(π₯2+2π₯+6)
log39= log3(π₯ + 2)
log3(π₯2+2π₯+6)
2= log3(π₯ + 2)
2log3(π₯ + 2) = log3(π₯2 + 2π₯ + 6)
log3(π₯ + 2)2 = log3(π₯
2 + 2π₯ + 6) Cancel like bases!
(π₯ + 2)2 = π₯2 + 2π₯ + 6
(π₯ + 2)(π₯ + 2) = π₯2 + 2π₯ + 6
π₯2 + 4π₯ + 4 = π₯2 + 2π₯ + 6
2π₯ = 2 Always check! log9(π₯2 + 2π₯ + 6) = log3(π₯ + 2)
π₯ = 1 β π₯2 + 2π₯ + 6 > 0 π₯ + 2 > 0
9 > 0 3 > 0
39
4. Solve for π₯: log(π₯ β 2) = log100β (14 β π₯)
Change to base 10!
log(π₯ β 2) =log (14βπ₯)
log100
log(π₯ β 2) =log (14βπ₯)
2
2 log(π₯ β 2) = log (14 β π₯)
log(π₯ β 2)2 = log (14 β π₯) Cancel like bases!
(π₯ β 2)2 = 14 β π₯
(π₯ β 2)(π₯ β 2) = 14 β π₯
π₯2 β 4π₯ + 4 = 14 β π₯
π₯2 β 3π₯ β 10 = 0
(π₯ β 5)(π₯ + 2) = 0
π₯ = 5 & β 2 Always check! log(π₯ β 2) = log100(14 β π₯)
π₯ β 2 > 0 14 β π₯ > 0
Must omit the β2 β4 > 0 16 > 0
So only π₯ = 5 β π₯ β 2 > 0 14 β π₯ > 0
3 > 0 9 > 0
40
Homework 10.7: Change of Base Rule
1. Use the change of base rule to solve for π₯ in the given logarithmic equation:
log(π₯ + 1) = log100(π₯2 + 4π₯)
2. Solve the given logarithmic equation: log4(π₯) = log16(3π₯2 β 3π₯ + 1)
3. Which expression could be used to determine the value of π¦ in the equation
log6π₯ = π¦? (1) logπ₯
6 (2)
log6
logπ₯ (3)
6
logπ₯ (4)
logπ₯
log6
4. Solve for π₯: ln(π₯ + 2) + ln(π₯ β 2) = ln (β2π₯ β 1)
41
5. Solve for π₯: log (β(π₯ + 3)3) =3
2
6. Drew said that the equation log2[(π₯ + 1)4] = 8 cannot be solved because he
expanded (π₯ + 1)4 = π₯4 + 4π₯3 + 6π₯2 + 4π₯ + 1 and realized that he cannot solve
the equation π₯4 + 4π₯3 + 6π₯2 + 4π₯ + 1 = 28. Is he correct? Explain how you
know.
7. Write ln (β5π₯3
π2) as a sum or difference of constants and logarithms of simpler
terms.
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