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General Mathematics
Quarter 1 – Module 8:
Logarithmic Functions
What I Need to Know
This module was designed and written with you in mind. It is here to help
you master the lesson about Logarithmic Function. The scope of this module permits
it to be used in many different learning situations. The language used recognizes the
diverse vocabulary level of students. The lessons are arranged to follow the standard
sequence of the course. But the order in which you read them can be changed to
correspond with the textbook you are now using.
The module is divided into three lessons, namely:
ο‚· Lesson 1 – Introduction to Logarithms
ο‚· Lesson 2 – Logarithmic Functions, Equations, and Inequalities
ο‚· Lesson 3 – Solving Logarithmic Equations and Inequalities
After going through this module, you are expected to:
1. represent real-life situation using logarithmic functions;
2. distinguish among logarithmic function, logarithmic equation, and logarithmic
inequality;
3. apply basic properties of logarithms;
4. illustrate the laws of logarithm; and
5. solve logarithmic equations and inequalities.
What I Know
Choose the letter of the best answer. Write the chosen letter on a separate sheet of
paper.
1. Which of these statements is TRUE about Logarithmic Function?
I. The inverse of exponential function is a Logarithmic function.
II. log𝑏 𝑦 = x is a logarithm if and only if 𝑏 π‘₯ = y for b > 0, b ≠ 1.
III. The equations x = log𝑏 𝑦 and 𝑦 = 𝑏 π‘₯ are equivalent.
A. I only
C. I and III only
B. II only
D. All of these
2. What is the relationship exists between the two functions π’š = 𝒃𝒙 and
π’š = π₯𝐨𝐠 𝒃 𝒙 ?
A. Inverses
C. not inverse
B. Reciprocals
D. none of these
3. What is the value of x in πŸπ’™ = πŸ”πŸ’?
A. 2
C. 5
B. 3
D. 6
πŸ’
4. Which of the following is equivalent to πŸ‘ = πŸ–πŸ when written in logarithmic form?
A. log 3 81 = 4
C. log4 81 = 3
B. log3 4 = 81
D. log81 3 = 4
5. What exponential equation is equivalent to π₯𝐨𝐠 πŸ” πŸ‘πŸ” = 𝟐?
A. 636 = 2
C. 62 = 36
2
B. 36 = 6
D. 26 = 36
6. Without using calculator, which among these logarithms has the largest value?
A. log2 40
C. log4 20
B. log3 30
D. log5 10
7
7. To find the value of log √7 (√7) , which of the following properties must be applied?
A. log 𝑏 1 = 0
C. log 𝑏 𝑏 π‘₯ = x
B. log 𝑏 𝑏 = 1
D. 𝒃π₯𝐨𝐠𝒃 𝒙 = x
2
8. Which of the following functions is the inverse of f(x) = 𝒃𝒙 ?
A. 𝑓 −1 (π‘₯) = 𝑏 π‘₯
C. 𝑓 −1 (π‘₯) = log 𝑏 π‘₯
−1 (π‘₯)
𝑏
B. 𝑓
= π‘₯
D. 𝑓 −1 (π‘₯) = log π‘₯ 𝑏
9. What is the value when you evaluate π₯𝐨𝐠 πŸ‘ πŸπŸ’πŸ‘ − π₯𝐨𝐠 𝟐 πŸ“πŸπŸ ?
A. 4
C. -3
B. 3
D. -4
𝟏
πŸ‘
10. Evaluate the expression πŸπ’™ + π’š given that π’š = π₯𝐨𝐠 πŸ“ πŸπŸ“ and 𝒙 = π₯𝐨𝐠 𝟐 πŸ‘πŸ.
A. -5
C. -2
B. -3
D. -1
π₯𝐨𝐠 πŸ‘ πŸ— + π₯𝐨𝐠 πŸ“ √πŸ“
11. Find the value when you evaluate the expression,
.
π₯𝐨𝐠 πŸ– 𝟐
A.
B.
15
C.
2
21
D.
2
18
5
12
5
12. What do you call to a logarithm with base 10? (base 10 is usually omitted when
writing this logarithm).
A. Natural Logarithm
C. Logarithmic Function
B. Common Logarithm
D. Logarithmic Equation
13–14. What is the magnitude in the Richter scale of an earthquake that released
10 17 joules of energy?
A. 6.2
C. 8.4
B. 7.5
D. 8.9
14. How much more energy does this earthquake release than that of the reference
earthquake?
A. 3981071706000 times more energy than that by the reference
earthquake
B. 39810717060 times more energy than that by the reference earthquake
C. 3981.071706 times more energy than that by the reference earthquake
D. 3.981071706 times more energy than that by the reference earthquake
15. The decibel level of sound in a quiet room is 10−5 π‘€π‘Žπ‘‘π‘‘π‘ /π‘š2 . What is the
corresponding sound intensity in decibels?
A. 40
C. 60
B. 50
D. 70
Lesson
1
Introduction to Logarithms
What’s In
In the previous lesson, we learned that exponential function is in the form
𝑓(π‘₯) = 𝑏 , where b > 0, b ≠ 1 and x is any real number. Since the exponential function
is one to one on ℝ, then its inverse exists. To obtain its inverse, we need to solve for x
in terms of y.
π‘₯
3
Illustrative Example:
Find the inverse of 𝒇(𝒙) = πŸπ’™
To find the inverse of 𝒇(𝒙) = πŸπ’™ , we have
 Replace f(x) with y:
y = πŸπ’™
 Interchange x and y:
x = 2𝑦
 Solve for y:
y = the power to which we raise 2 to
get x
 Replace y with 𝑓 −1 (π‘₯)
𝑓 −1 (π‘₯) = the power to which we raise
2 to get x.
However, there is no algebraic method that will
allow us to do this. Hence, you can be able to
graph the inverse of y = πŸπ’™ by graphing x =
2𝑦 . (see figure at the right)
In Mathematics, the expression “the power to
which we raise 2 to get x” is replaced by the
symbol “log 2 π‘₯” read as “ the logarithm, base 2 of
x”. Thus, if 𝒇(𝒙) = πŸπ’™ then 𝑓 −1 (π‘₯) = log 2 π‘₯
For example:
𝑓 −1 (16) = log 2 16 = 4
Logarithm = Exponent
Because,
log 𝑏 𝑦 = π‘₯
4 is the power to which we raise 2 to get 16.
Common log: log y = x
Natural log :
More Examples:
ln y = x
𝑦 = 𝑏π‘₯
𝑦 = 10π‘₯
𝑦 = 𝑒π‘₯
(a) log 2 8 = 3, since 3 is the exponent to which 2 must be raised to yield 8.
1
(b) log10 10 = -1, since -1 is the exponent to which 10 must be raised to yield
1
.
10
(c) log 5 1 = 0, since 0 is the exponent to which 5 must be raised to yield 1.
What’s New
There are a wide variety of problems involving logarithmic functions. The most
common applications in real-life of logarithms are maybe encountered in areas like
biology, chemistry, physics, economics, and banking.
Have you thought of a reason why you need to study this module? What is the
use of logarithm in our daily lives? The logarithm is used in the following applications
in real-life, like Richter Scale, Sound intensity, and pH level.
What is It
We define logarithm as, log 𝑏 𝑦 = x if and only if 𝑏 π‘₯ = y for b > 0, b ≠ 1. A logarithm
is an exponent which b must have to produce y. In either equation, b is called the base
and must be a positive number, not equal to 1.
The equations,
x = π₯𝐨𝐠 𝒃 π’š and y = 𝒃𝒙 are equivalent. The first equation is in logarithmic form
and the second is an exponential form.
Common Logarithms are logarithms with base 10; log π‘₯ is a short notation for
log10 π‘₯
Natural Logarithms are logarithms to the base e (approximately 2.71828), and are
denoted by “ln”. In other words, ln x is another way of writing
log 𝑒 π‘₯.
4
Example 1:
Write each exponential equation in logarithmic form.
a. 9 = 32
b. x = 5−3
c. 103 = 1,000
d. 𝑒 0 = 1
Solution:
a. 9 = 32
means log 3 9 = 2
b. x = 5−3
means log 5 π‘₯ = −3
c. 103 = 1,000
d. 𝑒 0 = 1
The base
remains
the same
means log10 1,000 = 3 π‘œπ‘Ÿ log 1,000 =3
means log 𝑒 1 = 0 or ln 1 = 0
Example 2:
Write each logarithmic equation in exponential form.
a. log 4 64 = 3
1
b. 3 = log 1 π‘₯
c. log 100 = -2
3
d. ln y = 3
Solution:
a. log 4 64 = 3
means
43 = 64
b. 3 = log 1 π‘₯
means
(3)3 = x
means
10−2 =
means
𝑒3 = y
1
3
c. log
1
100
= -2
d. ln y = 3
The logarithm
is the
exponent
1
100
Because logarithms are exponents, it is possible to evaluate some logarithms by
inspection. The logarithm of x with base b. log 𝑏 π‘₯, is the exponent to which b must be
raised to get x.
Example 3:
Evaluate each of the following logarithms.
9
16
4
a. log 5 125
b. log 3 ( )
e. log 3 1
f. log 8 8
c. log 0.001
d. log 1 16
2
g. log √2 (√2)3
h. 7log7 49
Solution:
To evaluate
logarithm,
express it to
exponential form
a. log 5 125 = πŸ‘ because 53 = 125
9
3
b. log 3 (16) = 2 because (4)2 =
4
9
16
c. log 0.001 = -3 because 10−3 = 0.001
𝟏
d. log 1 16 = -4 because (𝟐)−πŸ’ = 16
2
Because logarithms are exponents, they have properties that can be verified using
properties of exponents.
5
PROPERTIES OF LOGARITHM
e. log 3 1 = 0
π₯𝐨𝐠 𝒃 𝟏 = 0
f. log 8 8 = 1
π₯𝐨𝐠 𝒃 𝒃 = 1
g. log √2 (√2)3 = 3
π₯𝐨𝐠 𝒃 𝒃𝒙 = x
h. 7log7 49 = 49
𝒃π₯𝐨𝐠 𝒃 𝒙 = x
Applications: Some of the most common applications in real-life of logarithms are the
Richter Scale, Sound Intensity, and pH level.
The Richter magnitude scale (often shortened to Richter Scale) is the most common
standard measurement for earthquakes. It was invented in 1935 by Charles F. Richter
of the California Institute of Technology as a mathematical device to compare the size of
earthquakes. The Richter scale is used to measure the magnitude of an earthquake
that is the amount of energy released during an earthquake.
The magnitude R of an earthquake is given by
𝟐
𝑬
R = πŸ‘ π₯𝐨𝐠 πŸπŸŽπŸ’.πŸ’πŸŽ
Where E (in joules) is the energy released by the earthquake (the quantity 104.40
Joules is the energy released by a very small reference earthquake).
Example 4:
Suppose that an earthquake released approximately 1015 joules of energy (a) What is its
magnitude on a Richter scale? (b) How much more energy does this earthquake release
than the reference earthquake?
Solution:
2
1015
3
104.40
2
log 1010.6
3
(a) R = log
Then,
2
3
( 10.6)
since E is 1015
apply quotient law
by log 𝑏 𝑏 π‘₯ = x
Thus R = 7.0666… ≈ 7.1
(b) This earthquake releases
reference earthquake.
1015
104.40
= 1010.6 = 39810717060 times more energy than the
Sound Intensity
In acoustics, the decibel (dB) level for a sound is
𝑰
D = 10 π₯𝐨𝐠 𝟏𝟎−𝟏𝟐
Where I is the sound intensity in watts/π‘š2 (the quantity 10−12 watts/π‘š2 is least audible
sound a human can hear).
6
Example 5:
The decibel level of sound in a certain place is 10−5 watts/π‘š2 . (a) What is the
corresponding sound intensity in decibels? (b) How much more intense is this sound
than the least audible sound a human can hear?
Solution:
10−5
(a) D = 10 log 10−12 = 10 log 107 since I is 10−5 and by definition, log 107 is the exponent
by which 10 must be raised to obtain 107 , then 107 =
7.
Thus D = 10(7) = 70 decibels.
(b) This sound is
10−5
10−12
= 107 = 10,000,000 times more intense than the least audible
sound a human can hear.
Acidity and the pH scale
The pH level of the water-based solution is defined as
pH = - π₯𝐨𝐠[𝑯+ ]
where [𝐻 + ] is the concentration of the hydrogen ions in moles per liter. Solution with a
pH of 7 are said to be neutral; a pH below 7 indicates an acid; and a pH above 7 indicates
a base.
Example 6:
For some fruit juices, [𝐻 + ] = 3 x 10−4. Determine the pH and classify these juices as
acid or base.
Solution:
pH =
-
π₯𝐨𝐠[𝑯+ ] = -log(3π‘₯ 10−4 )
since , [𝐻 + ] = 3 x 10−4
then -log(0.0003) = 3.522878745
pH ≈ 3.5 is below 7 thus the fruit juice is an acid.
What’s More
Activity: Estimate and Evaluate
A. Without using calculator, which quantity is larger: log3 10 or log7 40?
Solution:
First, we estimate log3 10. This quantity represents the exponent to which
3 must be raised to yield 10. Since 32 =9 (less than 10), but 33 =27 (more than
10), we conclude that the quantity log3 10 lies between 2 and 3. In a similar way,
we can estimate log7 40 this quantity represents the exponent to which 7 must
be raised to yield 40. Since 71 =7(less than 40), but 72 =49 (more than 40) we can
conclude that the quantity log7 40 lies between 1 and 2. It now follows from
these two estimates that π₯π¨π πŸ‘ 𝟏𝟎 is larger than π₯π¨π πŸ• πŸ’πŸŽ.
7
B. Evaluate π₯π¨π πŸ’ πŸ‘πŸ
Solution:
Let y = log4 32. The exponential form of this equation is 4𝑦 = 32
Since both 4 and 32 are powers of 2, we can rewrite the equation 4𝑦 = 32 using
the same base on both sides:
(22 )𝑦 = 25
22𝑦 = 25
2𝑦 = 5
y=
πŸ“
𝟐
Assessment:
A. Without using calculator, estimate which quantity is larger.
1.
log 5 30 and log 8 40
2.
log10 90 and log 𝑒 𝑒 5
3.
log 2 3 and log 3 2
B. Evaluate each expression.
1.
log 9 27
2.
log 4
3.
log 5 √5
1
4. log 25 625
1
32
5. log 2 8√2
C. Complete the table by writing the correct equations in exponential/logarithmic
form.
EXPONENTIAL FORM
LOGARITHMIC FORM
34 = 81
1
2
log 1,000 = 3
4−3 =
1
64
3
4
log 1 32 = -5
√π‘₯ = y
5
2
What I Have Learned
DEFINITION: π₯𝐨𝐠 𝒃 𝒙
We define the expression π₯𝐨𝐠 𝒃 𝒙 to mean as “the exponent to which b must be raised
to yield x”. (π‘™π‘œπ‘”π‘ π‘₯ is read log base b of x or the logarithm of x to the base b)
y = log 𝑏 π‘₯ is equivalent to π‘₯ = 𝑏 𝑦
8
COMMON LOGARITHM
The Common Logarithm is the logarithm with base 10 and is usually written
in log or π‘™π‘œπ‘”10 .
NATURAL LOGARITHM
The Natural Logarithmic Function is defined by y = ln x where ln x stands for
the natural logarithm of x or the logarithm of x to the base e. Thus, ln x is real
only when x > 0, and ln 1 = 0.
Because logarithms are exponents, they have properties that can be verified using
properties of exponents
BASIC PROPERTIES OF LOGARITHMS
If b > 0 and b ≠ 1, then
1. π₯𝐨𝐠 𝒃 𝟏 = 0 because π’ƒπŸŽ = 1
4. If log 𝑏 π‘₯ = log 𝑏 𝑦, then x = y (One-toOne Property)
2. π₯𝐨𝐠 𝒃 𝒃 = 𝟏
5. log π‘Ž 𝑐 =
3. π₯𝐨𝐠𝒃 𝒃𝒙 = x
6. log π‘Ž 𝑏 =
log𝑏 𝑐
log𝑏 π‘Ž
1
log𝑏 π‘Ž
What I Can Do
Using the formula, you are now ready to solve real-life problems involving logarithmic
functions.
Problem 1: What is the magnitude in Richter scale of an earthquake which released
1013 joules of energy? How much more energy did this earthquake release than that of
the reference earthquake?
2
1013
Answer. Magnitude R = 3 log 104.40 =
1013
104.40
2
log 108.6
3
=
2
(8.6)
3
= 5.7. The earthquake released
= 108.6 = 398107107 times more energy than that by the reference earthquake.
Problem 2: An unknown substance has a hydrogen ion concentration of 10−9 . Classify
the substance as acidic, neutral, or basic.
Answer. pH = - log[𝐻 + ] = - log ( 10−9 ) = -log 0.000000001 = 9. Therefore, the solution is
basic.
9
Assessment
Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a
separate sheet of paper.
1.
Which of these statements is NOT TRUE about Logarithmic Function?
I. The inverse of exponential function is a Logarithmic function.
II. log 𝑏 𝑦 = x is a logarithm if and only if 𝑏 π‘₯ = y for b < 0, b = 1.
III. The equations, x = log 𝑏 𝑦 and 𝑦 = 𝑏 π‘₯ are equivalent
A. I only
C. II and III
B. II only
D. All of these
2. What is the relationship exists between the two functions π’š = πŸ‘π’™ and
π’š = π₯𝐨𝐠 πŸ‘ 𝒙 ?
A. Inverses
C. Not inverse
B. Reciprocals
D. None of these
𝟏
3. What is the value of x in πŸ“π’™ = πŸ”πŸπŸ“ ?
A. -4
B. -3
C. 4
D. 5
𝟐
πŸπŸ”
4. Which of the following is equivalent to (πŸ‘)πŸ’ = πŸ–πŸ when written in logarithmic form?
16
2
16
A. log 4 81 = 3
2
C. log 2 81 = 4
3
16
B. log 4 3 = 81
D. log 16 4 =
81
2
3
5. What exponential equation is equivalent to π₯𝐨𝐠 πŸ— πŸ•πŸπŸ— = πŸ‘?
A. 7293 = 9
C. 9729= 3
9
B. 3 = 729
D. 93 = 729
6. Without using calculator, which among these logarithms has the smallest value?
A. log 3 45
C. log 4 10
B. log 2 20
D. log 1 5
2
7. Applying the properties of logarithm, what is the value of 5log5 √5 ?
1
A. 5
C. 5
1
B. 52
D. -5
8. Which of the following functions is the inverse of y = 𝒃𝒙 ?
A. y = log 𝑏 π‘₯
C. y = log π‘₯ 𝑏
B. x = log 𝑏 𝑦
D. x = log 𝑦 𝑏
9. What is the value when you evaluate π₯𝐨𝐠 πŸ• πŸ‘πŸ’πŸ‘ + π₯𝐨𝐠 πŸ” πŸπŸπŸ—πŸ” ?
A. 6
C. 8
B. 7
D. 9
𝟏
10. Evaluate the expression π‘₯ 2 + 2𝑦 given that π’š = π₯𝐨𝐠 πŸ“ πŸπŸπŸ“ and 𝒙 = π₯𝐨𝐠 𝟐 πŸπŸπŸ–.
A. 120
C. 55
B. 85
D. 23
10
11. Find the value when you evaluate the expression,
A. 2.5
B. 3.5
C. 3.75
D. 4.75
π₯𝐨𝐠 πŸ‘ πŸπŸ•− π₯𝐨𝐠 𝟐 √𝟐
.
π₯𝐨𝐠 πŸ– πŸ’
12. What do you call to a logarithm with base e? (usually denoted by “ln”)
A. Natural Logarithm
C. Logarithmic Function
B. Common Logarithm
D. Logarithmic Equation
13–14. What is the magnitude in the Richter scale of an earthquake that released
10 12 joules of energy?
A. 3.5
C. 4.5
B. 3.7
D. 5.1
14. How much more energy does this earthquake release than that of the reference
earthquake?
A. 39810717 times more energy than that by the reference earthquake
B. 39810710 times more energy than that by the reference earthquake
C. 3981.071 times more energy than that by the reference earthquake
D. 3.981071 times more energy than that by the reference earthquake
15. The decibel level of sound in a quiet room is 10−6 π‘€π‘Žπ‘‘π‘‘π‘ /π‘š2 .
corresponding sound intensity in decibels?
A. 40
C. 60
B. 50
D. 70
What is the
Additional Activities
ENRICHMENT: Solve each problem and show your complete solution in a separate
sheet of paper.
1. Given that π₯𝐨𝐠 𝟐 = 𝒙, π₯𝐨𝐠 πŸ‘ = π’š 𝒂𝒏𝒅 π₯𝐨𝐠 πŸ• = 𝒛, express log 12 in terms of x, y and z.
2. Find the ERROR: Michelle wanted to find the value of x in the equation 2(3)π‘₯ =
34. She first converted the equation to log 3 2π‘₯ = 17. Next, she wrote 2x = 317 and use a
calculator to find x = 64,570,081. Is her answer correct? If not, what was her error and
what is the right answer?
3. Marc and Margaret decided to play a game in which they each selected a logarithmic
function and compare their functions to see which gave a larger value, Marc selected
the function f(x) = 10 log 2 π‘₯ and Margaret selected the function
𝑓(π‘₯) = 2 log10 π‘₯.
(a) Which of the functions has a larger value when x = 7?
(b) Which of the functions has a larger value when x = 1?
(c) Do you think the base or the multiplier is more important in determining
the value of the logarithmic function? Why?
11
What I Know
Choose the letter of the best answer. Write the chosen letter on a separate sheet
of paper.
1. Which of the following statements is/are TRUE?
1
i. If f(x) = (3)π‘₯ , then f(-1) = 3.
ii. The equation π‘Ž3 = 2 is equivalent to log π‘Ž 2 = 3.
iii. One possible solution to x in log 4 π‘₯ < 3 is -1.
A. i only
B. ii only
C. i and ii
D. all of these
2. Which of these is correctly arranged in ascending order? (you may use a calculator)
A. log 2.8, ln 2.8, 𝑒 2.8, 102.8
C. 𝑒 2.8 , ln 2.8, 102.8, log 2.8
2.8
2.8
B. ln 2.8, 10 , log 2.8, 𝑒
D. 102.8, 𝑒 2.8, ln 2.8, log 2.8
3. What should be the value of x in 4π‘₯+1 = 64?
A. 0
B. 2
C. 4
D. 8
4. Which of the graphs below represents the inverse of f(x) = 2π‘₯ ?
A.
C.
B.
D.
5. What is the reason why f(x) = log −4 π‘₯ is NOT a logarithmic function?
A. In a logarithmic function, the base must be greater than 0 and not equal to 1.
B. Cannot be written in exponential form.
C. The value of x is any real number.
D. The solution of the function is unique.
6. Which of the following expresses a relationship between two variables and can be
represented by a table or a graph.
A. Logarithmic Equation
C. Logarithmic Function
B. Logarithmic Inequality
D. None of These
7. How many possible solutions are there in solving logarithmic inequality?
A. one solution
C. infinitely many
B. 1 or 2 solutions
D. no solution
8. Which of the following is an example of a logarithmic inequality?
A. 3x < y
B. log 3 π‘₯ ≥ π‘₯
C. log 3 (π‘₯ − 1) < 3
D. log −3(π‘₯ − 1) ≤ og −3 π‘₯
9. What is the base of a logarithmic function y = log 𝑏 π‘₯ whose graph contains the
point (9, 2)?
A. 1
B. 3
C. 6
D. 9
12
10. Which of the following best describes a logarithmic equation?
A. An equation involving exponents.
B. An equation having 2 or more variables.
C. An equation involving rational expression.
D. An equation involving logarithms.
11. Referring to the graph of two logarithms below, what are the x-coordinates of the
points intersecting the two curves?
A.
B.
C.
D.
-1 and 0
-1 and 2
2 and 0.602
0 and 0.602
12. Which of the following will satisfy both the logarithmic equation, log 2(π‘₯ + 2) = 4 and
the logarithmic inequality log 2(π‘₯ − 2) < 4?
A. 14
B. 15
C. 17
D. 18
13–15. Determine the type of the following Logarithms. Write A if it is a Logarithmic
Function, B if Logarithmic Equation, C if Logarithmic Inequality, and D
none of these.
________ 13. ln √π‘₯ + 1 = 0
________ 14. log12(π‘₯ − 1) < 2
________ 15. x = log 𝑏 π‘₯
Lesson
2
Logarithmic Functions,
Equations, and Inequalities
What’s In
Using your idea about exponential, let us determine whether the following is an
exponential function, an exponential equation, an exponential inequality, or none of
these.
a) 10π‘₯−2 = 100
Exponential Equation
1 π‘₯
b) ( ) ≤ 2
Exponential Inequality
2
π‘₯
c) 7 = 𝑦
Exponential Function
d) f(x) = (−5)π‘₯
None of these
(base cannot be negative)
Since a, b, and c are exponential then we can write these in logarithmic form as
follows:
a) log 100 = π‘₯ − 2
b) log 1 2 ≤ π‘₯
2
c) log 7 𝑦 = π‘₯
now, share your ideas how a, b, and c are similar and how they are different.
13
What’s New
In the previous lesson, we defined logarithm as log 𝑏 𝑦 = x if and only if 𝑏 π‘₯ = y for
b > 0, b ≠ 1. A logarithm is an exponent which b must have to produce y. In either
equation, b is called the base and must be a positive number, not equal to 1. This is
usually written in the form of f(x) = log 𝑏 π‘₯, π‘™π‘–π‘˜π‘’(a) f(x) = log 2 π‘₯. But if a logarithm is
expressed as (b) log 2(π‘₯ + 3) = log 2(6 − π‘₯) + 3 π‘Žπ‘›π‘‘ (𝑐) (log 2 π‘₯)2 < 2 log 2 (π‘₯) + 3, are these
logarithms still a function? If NOT what are these called for and how they differ from
one another?
One way of describing this is to show them in a graph. Let’s start with the given
function f(x) = π₯𝐨𝐠 𝟐 𝒙
The figure at the left is the graph
value
of the function, f(x) = As
log 2the
π‘₯
of x increases
the value of y
also increases
without
bound
f(x) = log 2 π‘₯
When we try to graph the
functions,
𝑓(π‘₯) = log 2 (π‘₯ + 3) and
g(x) = log 2 (6 − π‘₯) + 3 in one set of
axes the graph shows that they
intersect at one point ( 5, 3 )
meaning that we can verify our
solution in solving logarithmic
equation by graph. Hence, the
value of x in
log 2 (π‘₯ + 3) = log 2 (6 − π‘₯) + 3 is 5.
𝑓(π‘₯) = log 2 (π‘₯ + 3)
and g(x) = log 2 (6 − π‘₯) + 3
Geometrically, we see the graphs of the
two functions at the right,
f(x) = (log 2 π‘₯)2 and g(x) = 2log 2 (π‘₯) + 3
when we try to solve for the
logarithmic inequality,
(log 2 π‘₯)2 < 2log 2 (π‘₯) + 3 for x values
1
shows a solution interval of (2 , 8)
means that to satisfy the inequality we
can replace all real numbers between
1
π‘Žπ‘›π‘‘ 8.
2
14
f(x) = (log 2 π‘₯) 2 and g(x) = 2log 2 (π‘₯) + 3
What is It
Read and study carefully on how we can determine a logarithm whether it is an
Equation, an Inequality, or a Function.
Logarithmic Equation
Logarithmic Inequality
Logarithmic Function
•can be solve for all x values
that satisfy the equation.
•can be solve for all x values
that satisfy the inequality.
•there is always an (=) sign
•consider these signs ( <, >,
≤, ≥ ) these are also known
as less than, greater than,
less than equal and greater
than equal
•Example:
•log 3 (2π‘₯ + 4)>log 3 (π‘₯ − 4)
•Example:
•log 3 8π‘₯ − 15 = 4
•expresses a relationship
between two variables and
can be represented by a
table of values or a graph.
•consider the following
involve in the given: f(x), g(x)
or y and any letter with (x).
•Example:
•f(x) = log 5 π‘₯
Remember the specific parts of each
to help you in distinguishing them:
log 3 (8π‘₯ − 15) = 4
log 3 (2π‘₯ + 4) > log 3 (π‘₯ − 4)
f(x) = log 5 π‘₯
What’s More
Determine whether the given is a logarithmic function, a logarithmic equation, a
logarithmic inequality, or none of these.
Examples:
π‘Ž.
b.
c.
d.
log 6 π‘₯ = 7776
log 7 + log π‘₯ > log 49
h(x) = log 2 2π‘₯
log −3 27 = π‘₯
Answer
[Logarithmic Equation]
[Logarithmic Inequality]
[Logarithmic Function]
[none (base cannot be negative)]
Assessment: Determine whether the given is a logarithmic function, a logarithmic
equation, a logarithmic inequality, or neither. Write A if it is a logarithmic function,
B if logarithmic equation, C if logarithmic inequality, and D if neither.
1. log 3 3π‘₯ = − log 3(2π‘₯ + 5)
6. y =log 1 π‘₯
2. g(x) = log 9 π‘₯
7. log 1 π‘₯ > log 1 (5)−3
3. ln (2x + 10) = 2 ln2
4. log 4 2π‘₯
5. log 4 ( 4π‘₯ + 1) ≤ log 4 2π‘₯
8. log 6 π‘₯ = 1 − log 6 (π‘₯ − 1)
9. 2x + y ≤ 64
10. (log π‘₯)2 + 2 log π‘₯ − 3 = 0
2
5
15
1
5
What I Have Learned
The definitions of logarithmic equations, inequalities, and functions are shown
below:
Logarithmic
Equation
Definition
An equation
involving
logarithms.
Logarithmic
Inequality
Logarithmic
Function
An inequality
involving
logarithms.
A function of the
form
f(x)= log 𝑏 π‘₯ where
b > 0 and b ≠ 1
Example
ln x > 1
log 2 128 = π‘₯
g(x) = 4 log 2 π‘₯
What is It
Read and study carefully on how we can determine a logarithm whether it is an
Equation, an Inequality, or a Function.
Logarithmic Equation
Logarithmic Inequality
•can be solve for all x values
that satisfy the equation.
•can be solve for all x values
that satisfy the inequality.
•there is always an (=) sign
•consider these signs ( <, >,
≤, ≥ ) these are also known
as less than, greater than,
less than equal and greater
than equal
•Example:
•log 3 (2π‘₯ + 4)>log 3 (π‘₯ − 4)
•Example:
•log 3 8π‘₯ − 15 = 4
Logarithmic Function
•expresses a relationship
between two variables and
can be represented by a
table of values or a graph.
•consider the following
involve in the given: f(x), g(x)
or y and any letter with (x).
•Example:
•f(x) = log 5 π‘₯
Remember the specific parts of each
to help you in distinguishing them:
log 3 (8π‘₯ − 15) = 4
log 3 (2π‘₯ + 4) > log 3 (π‘₯ − 4)
f(x) = log 5 π‘₯
16
What’s More
Determine whether the given is a logarithmic function, a logarithmic equation, a
logarithmic inequality, or none of these.
Examples:
π‘Ž.
b.
c.
d.
Answer
log 6 π‘₯ = 7776
log 7 + log π‘₯ > log 49
h(x) = log 2 2π‘₯
log −3 27 = π‘₯
[Logarithmic Equation]
[Logarithmic Inequality]
[Logarithmic Function]
[none (base cannot be negative)]
Assessment: Determine whether the given is a logarithmic function, a logarithmic
equation, a logarithmic inequality, or neither. Write A if it is a logarithmic function,
B if logarithmic equation, C if logarithmic inequality, and D if neither.
1. log 3 3π‘₯ = − log 3(2π‘₯ + 5)
6. y =log 1 π‘₯
2. g(x) = log 9 π‘₯
7. log 1 π‘₯ > log 1 ( )−3
3. ln (2x + 10) = 2 ln2
4. log 4 2π‘₯
5. log 4 ( 4π‘₯ + 1) ≤ log 4 2π‘₯
8. log 6 π‘₯ = 1 − log 6 (π‘₯ − 1)
9. 2x + y ≤ 64
10. (log π‘₯)2 + 2 log π‘₯ − 3 = 0
2
5
5
1
5
What I Have Learned
The definitions of logarithmic equations, inequalities, and functions are shown
below:
Logarithmic
Equation
Definition
An equation
involving
logarithms.
Logarithmic
Inequality
An inequality
involving
logarithms.
Logarithmic
Function
A function of the
form
f(x)= log 𝑏 π‘₯ where
b > 0 and b ≠ 1
Example
ln x > 1
log 2 128 = π‘₯
17
g(x) = 4 log 2 π‘₯
Assessment
Choose the letter of the best answer. Write the chosen letter on a separate sheet of
paper.
1. Which of the following statements is/are NOT TRUE?
1
i. If f(x) = (3)π‘₯ , then f(1) = 3.
ii. The equation π‘Ž4 = 5 is equivalent to log π‘Ž 5 = 4.
iii. One possible solution to x in log 4 π‘₯ < 3 is -64.
A. i only
B. ii only
C. i and ii
D. i and iii
2. Which of these is correctly arranged in descending order? (you may use a
calculator)
A. log 0.8, ln 0.8, 𝑒0.8 , 100.8
C.
𝑒0.8 , ln 0.8, 100.8 , log 0.8
B. ln 0.8, 100.8 , log 0.8 , 𝑒0.8
D. 100.8 , 𝑒0.8 , log 0.8, ln 0.8
3. What must be the value of x in 72π‘₯ = 2401?
A. 0
B. 2
C. 4
D. 8
4. Which of the following table of values represents the graph of f(x) = log 4 2π‘₯?
A.
C.
x
f(x)
x
f(x)
x
x
f(x)
B. .
f(x)
D.
5. What is the reason why the value of x in a logarithmic function f(x) = log 𝑏 π‘₯ must
always be positive?
A. By definition of logarithmic function the base b > 0 and b ≠ 1.
B. Can be written in exponential form.
C. The value of x is any real number.
D. The value of y or f(x) is nonnegative.
6. Which of the following expresses a relationship between two variables and can be
represented by a table or a graph?
A. Logarithmic Equation
C. Logarithmic Function
B. Logarithmic Inequality
D. None of these
18
7. How many possible solutions are there in solving logarithmic function?
A. one solution
C. infinitely many
B. 1 or 2 solutions
D. no solution
8. Which of the following is NOT an example of a logarithmic inequality?
A. 3x < y + 2
C. log 3 (π‘₯ − 1) < 3
B. log 3 π‘₯ ≥ −3
D. log 3(π‘₯ − 1) ≤ log 3 π‘₯
9. What is the base of a logarithmic function y = log 𝑏 π‘₯ whose graph contains the point
(729, 9)?
A. 1
B. 3
C. 6
D. 9
10. Which of the following best describes a logarithmic inequality?
A. An inequality involving exponents.
B. An inequality having 2 or more variables.
C. An inequality involving rational expression.
D. An inequality involving logarithms.
11. The graph of two logarithms are shown below. At what point does the two curves
intersect?
A. ( 3, 1 )
B. ( 3, 2 )
C.
( 3, 3 )
D. ( 3, 4 )
12. Which of the following will both satisfy the logarithmic equation, log 4 (π‘₯ + 2) = 4
and the logarithmic inequality log 4 (π‘₯ − 2) < 4?
A. 14
B. 16
C. 254
D. 258
13–15. Identify the following Logarithms. Write A if it is a Logarithmic Function, B if
Logarithmic Equation, C if Logarithmic Inequality, and D neither.
________ 13. log 25 53 = x
________ 14. log 5 (π‘₯ − 1) < 125
________ 15. h(x) = 2 log 2 (π‘₯ + 1)
Additional Activities
ENRICHMENT:
Answer the following and show your complete solution in a clean sheet of paper.
1. What is the value of ln 3 +
1
3
( ln 48 -
1
2
ln 36 ) when simplified?
2. (a) On the same set of axes, graph the curves y = ln (x+2) and
y=
ln (-x) – 1. According to your graph, in which quadrant do the two curves
intersect?
(b) Find the x-coordinate of the intersection point?
3. In your own understanding, discuss the similarities and differences between
Logarithmic Function, Logarithmic Equation, and Logarithmic Inequality.
19
What I Know
Multiple Choice: Choose the letter of the best answer. Write the chosen letter on a
separate sheet of paper.
1. Which of the following equations illustrates the Product law of Logarithm?
A. log π‘Ž π‘š + log π‘Ž 𝑛 = log π‘Ž π‘šπ‘›
B. log π‘Ž π‘š + log π‘Ž 𝑛 = log π‘Ž (π‘š + 𝑛)
C. log π‘Ž π‘š + log π‘Ž 𝑛 = m log π‘Ž 𝑛 + n log π‘Ž π‘š
D. log π‘Ž π‘š + log π‘Ž 𝑛 = log π‘Ž (π‘š − 𝑛)
2. What is the perimeter of a rectangle with π₯𝐞𝐧𝐠𝐭𝐑 π₯𝐨𝐠 𝒙 π’–π’π’Šπ’•π’” 𝐚𝐧𝐝 𝐰𝐒𝐝𝐭𝐑 π₯𝐨𝐠 π’š π’–π’π’Šπ’•π’”?
A. 2 log(π‘₯ + 𝑦)
B. 2 + log π‘₯𝑦
C. log π‘₯ + log 𝑦
D. 2 log π‘₯𝑦
3. Which of the following expressions is equal to π₯𝐨𝐠 𝐚 𝐦𝐧 ?
A. 𝑛 log π‘Ž π‘š
B. (log π‘Ž π‘š)( log π‘Ž 𝑛) C. log π‘Ž π‘š + log π‘Ž 𝑛
D. log π‘Ž (π‘š + 𝑛)
4. If b > 1 and 𝒃𝒙 > π’ƒπ’š , then what is the relationship between x and y?
A. x > y
C. x = y
B. x < y
D. not enough information
5. Which does NOT help to explain why you cannot use the laws of logarithms to expand
or simplify, π₯𝐨𝐠 πŸ’ (πŸ‘π’š − πŸ’)?
A. The expression 3y − 4 cannot be factored.
B. The expression 3y − 4 is not raised to a power.
C. 3y and 4 are neither multiplied together, nor are they divided into each other.
D. Each term in the expression does not have the same variable.
6. Describe the strategy you would use to solve π₯𝐨𝐠 πŸ” 𝒙 = π₯𝐨𝐠 πŸ” πŸ’ + π₯𝐨𝐠 πŸ” πŸ–
A. Use the product rule to turn the right side of the equation into a single logarithm.
Recognize the resulting value is equal to x.
B. Express the equation in exponential form, set the exponents equal to each
other, and solve.
C. Use the fact that the logs have the same base to add the expressions on the
right side of the equation together. Express the results in exponential form,
set the exponents equal to each other and solve.
D. Use the fact that since both sides of the equations have logarithms with the
same base to set the expressions equal to each other and solve.
7. Use properties of logarithm to condense the given expression as a single logarithm.
1
log10 π‘₯ + 4 log10(π‘₯ + 2).
3
3
A. log10 [ √π‘₯ /(π‘₯ + 2)4 ]
C. log10 π‘₯ 3 (π‘₯ + 2)
3
B. log10 [ √π‘₯ (π‘₯ + 2)4 ]
D. log π‘₯ 3 - (π‘₯ + 2)4
1
8. Which of the following is NOT a solution to the logarithmic inequality,
log 3(x + 4) < log 3 3π‘₯ ?
A. 2
B. 3
C. 4
D. 5
9. Find x: log 2 π‘₯ − log 2 (√π‘₯ − 1 ) = 2.
A. 2
B. 4
C. 6
D. 8
10. A bacteria culture starts with 10,000 bacteria and the number doubles every
40 minutes. After how many minutes will there be 50,000 bacteria?
A. 10.22 minutes
C. 92.88 minutes
B. 50.82 minutes
D. 98.20 minutes
20
Lesson
3
Solving Logarithmic
Equations and Inequalities
What’s In
In the previous lessons, we defined logarithmic equations and the logarithmic
inequalities their similarities and differences as well as their basic properties. But before
we find and solve logarithmic equations and inequalities, we need to know first the laws
of logarithm.
LAWS OF LOGARITHM: Let b > 0, b ≠ 1 and let n ∈ ℝ for u > 0, v > 0, then
1. PRODUCT LAW: π₯𝐨𝐠 𝒃(𝒖𝒗) = π₯𝐨𝐠 𝒃 𝒖 + π₯𝐨𝐠 𝒃 𝒗
Example: log 2 (4 ∗ 2) = log 2 4 + log 2 2 = 3
𝒖
2. QUOTIENT LAW: π₯𝐨𝐠 𝒃 𝒗 = π₯𝐨𝐠 𝒃 𝒖 − π₯𝐨𝐠 𝒃 𝒗
4
Example: log 2 ( 2) = log 2 4 − log 2 2 = 1
3. POWER LAW: π₯𝐨𝐠 𝒃 𝒖𝒏 = n π₯𝐨𝐠 𝒃 𝒖
Example: log 2 42 = 2 log 2 4 = 2 * 2 = 4
We use these properties of logarithms to expand each expression in terms of the
logarithms of the factors (Assume each factor is positive) and to condense the
expressions as a single logarithm.
Example 1: Expand log 4 5π‘₯𝑦
log 4 5π‘₯𝑦 = log 4 5 + log 4 π‘₯ +log 4 𝑦
using Product Law
5
Example 2: Expand log 5 ( π‘₯)3
5
log 5 ( π‘₯)3 = 3 ( log 5 5 − log 5 π‘₯ )
using Power Law and Quotient Law
Example 3: Express 4 ln x – ln y to a single logarithm
ln π‘₯ 4 − ln y = ln
π‘₯4
𝑦
CHANGE OF BASE FORMULA
Any logarithmic expression can be expressed as a quotient of two logarithmic
expressions with a common base. Let a, b, and x be positive real numbers with a ≠
1, b ≠ 1.
π₯𝐨𝐠 𝒃 𝒙 =
π₯𝐨𝐠 𝒂 𝒙
π₯𝐨𝐠 𝒂 𝒃
Example: Use the change of base formula to rewrite the logarithmic expressions
to the indicated base like log 6 4 (π‘β„Žπ‘Žπ‘›π‘”π‘’ π‘‘π‘œ π‘π‘Žπ‘ π‘’ 2)
log 4
log 6 4 = log2 6 =
2
2
log2 6
21
Applying this laws may help you to find the solution in solving logarithmic
equations and logarithmic inequalities.
Problems about exponential growth and decay are some applications of solving
exponential functions, likewise, these kind of problems can be solve using the idea of
logarithmic equation. Let’s take this as an example.
A particular bacterial colony doubles its population every 15 hours. A scientist
running an experiment is stating with 100 bacteria cells. She expects the number of cells
𝑑
to be given by the formula, C = 100((2)15 , where t is the number of hours, since the
experiment started. After how many hours would the scientist expect to have 300 bacteria
cells? Give your answer to the nearest hour.
Before you answer this problem, read and study how to solve logarithmic
equation.
What is It
SOLVING LOGARITHMIC EQUATIONS
Generally, there are 2 types of logarithmic equations. Study each case carefully
before you start looking at the worked examples below
If log 𝑏 𝑀 = log 𝑏 𝑁 if and only if M = N (One-to-One Property)
If you have a single logarithm on each side of the equation having the same base,
then you can set the arguments equal to each other and solve. The arguments here are
the algebraic expressions represented by M and N.
The second type look like this,
log 𝑏 𝑀 = N
M = 𝑏 𝑁 (writing equation to exponential form)
If you have a single logarithm on one side of equation, then you can express it as
an exponential equation and solve.
CASE
1
Solve π₯𝐨𝐠 πŸ‘ (πŸ–π’™ − πŸπŸ“) = 4
Solution: log 3 (8π‘₯ − 15) = 4
original equation
34 = 8x – 15
write in exponential form
81 = 8x – 15
add 15 on both side of the equation
81 + 15 = 8x
96 = 8x
x = 12
simplify
divide both side by 8
solution
Check: log 3 (8π‘₯ − 15) = 4
log 3 (8(12) − 15) = 4
log 3 (96 − 15) = 4
log 3 (81) = 4
34 = 81
22
2
CASE
:
1Solution:
Find the value of x: ln (2x+10) = 2 ln 2
ln (2x+10) = 2 ln 2
original equation
2x + 10 = 22
one-to-one property and power law
2x + 10 = 4
simplify
2x = 4 – 10
add -10 on both side
2x = -6
divide both side by 2
x = -3
solution
Check: ln (2x+10) = 2 ln 2
ln (2(-3)+10) = 2 ln 2
ln 4 = ln 22
ln 4 = ln 4
3
CASE
1
Solve the logarithmic equation: π₯𝐨𝐠 𝟐(𝒙 + 𝟐) + π₯𝐨𝐠 𝟐 πŸ‘ = π₯𝐨𝐠 𝟐 πŸπŸ•
Solution: log 2 (π‘₯ + 2) + log 2 3 = log 2 27
log 2 3(π‘₯ + 2) = log 2 27
3x + 6 = 27
3x = 21
x=7
use product law to condense logarithm
one-to-one property
divide both side by 3
solution
Check: log 2(π‘₯ + 2) + log 2 3 = log 2 27
log 2 (7 + 2) + log 2 3 = log 2 27
log 2 9(3) = log 2 27
log 2 27 = log 2 27
ALWAYS check your solved
values with the original
logarithmic equation
CASE
4
Solve for x: π₯𝐨𝐠 𝟐 (π’™πŸ − πŸ’π’™) = π₯𝐨𝐠 𝟐 (πŸ’ − 𝒙) + πŸ“
Solution: log 2 (π‘₯ 2 − 4π‘₯) = log 2 (4 − π‘₯) + 5
1
log 2 (π‘₯ 2 − 4π‘₯) − log 2 (4 − π‘₯) = 5
log 2
π‘₯ 2 −4π‘₯
4−π‘₯
25 =
x2 −4x
4−x
32 =
original logarithmic inequality
combine terms with log
apply Quotient Law
=5
express in exponential form
x2 −4x
4−x
multiply both side by 4-x
π‘₯ 2 – 4x = 128 – 32x
simplify
π‘₯ 2 + 28x – 128 = 0
set equation to zero
(x- 4)(x+32) = 0
solve by factoring
x=4
extraneous root
x = -32
solution
23
Check: if x = -32
Check: if x = 4
log 2
(π‘₯ 2
log 2
log 2 ((−32)2 − 4(−32)) = log 2 (4 − (−32) + 5
− 4π‘₯) = log 2 (4 − π‘₯) + 5
(42
log 2 (1024 + 128) = log 2 (4 − (−32)) + 5
− 4(4)) = log 2 (4 − 4)
+5
log 2 1152 = log 2 36 + log 2 32
log 2 1152 = log 2 36(32)
log 2 0 = undefined
π₯𝐨𝐠 𝟐 πŸπŸπŸ“πŸ = π₯𝐨𝐠 𝟐 πŸπŸπŸ“πŸ
CAUTION: the logarithm of a negative number and the logarithm of zero are both
NOT DEFINED
Example: log 𝑏 (π‘›π‘’π‘”π‘Žπ‘‘π‘–π‘£π‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ) = 𝑒𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
log 𝑏 0 = 𝑒𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
You have solved logarithmic equations algebraically. You can also solve logarithmic
equation by graphing or by using a table.
5
CASE
Solve π₯𝐨𝐠 𝟐(𝒙 + πŸπŸ•) = π₯𝐨𝐠 𝟐 (πŸ‘π’™ + πŸπŸ‘) by graphing.(may use graphic calculator)
1
Use the intersect feature on the graph to see the ordered pair of the point at which the
curves cross. Since the x-coordinates of the intersection is -3, therefore, the value of x
in this logarithmic equation is -3.
SOLVING LOGARITHMIC INEQUALITIES
If a > 1 and x > y, then π₯𝐨𝐠 𝒂 𝒙 > π₯𝐨𝐠 𝒂 π’š,
π₯𝐨𝐠 𝒂 π’š.
otherwise,
if 0 < a < 1 then π₯𝐨𝐠 𝒂 𝒙 <
Of course, the base of a logarithm cannot be 1 or negative. More importantly, the
converse is true as well.
If a > 1 and π₯𝐨𝐠 𝒂 𝒙 > π₯𝐨𝐠 𝒂 π’š , 𝒕𝒉𝒆𝒏 𝒙 > π’š otherwise 0 < a < 1, then x < y
The argument of a logarithm MUST be positive, thus, it is also necessary to take into
account any inequality resulting from the arguments being positive.
For example, as an inequality involving the term π₯𝐨𝐠 𝟐(πŸπ’™ − πŸ‘) immediately
x>
πŸ‘
.
𝟐
Case 1: Base greater than 1 ( b > 1 )
(a) Find the solution for the inequality: π₯𝐨𝐠 πŸ–(πŸ‘π’™ − πŸ“) < 2
24
requires
Solution: log 8 (3π‘₯ − 5) < 2
3x – 5 > 0, then x >
πŸ“
πŸ‘
ensure that the argument is positive
3x -5 < 82
write in exponential form
3x < 64 + 5
add 5 in both side of the inequality
3x < 69
divide both side by 3 and simplify
Combine the inequalities, so the
x < 23 and x >
πŸ“
πŸ‘
solution to x must be between
5
3
and 23 and solution maybe express
using interval notation.
πŸ“
Check: If we will assign x = 2, since πŸ‘ < 𝟐 < πŸπŸ‘. Substitute 2 in x from the given
πŸ“
πŸ‘
πŸ“
< 𝒙 < πŸπŸ‘ or ( πŸ‘, 23 )
logarithmic inequality log 8 (3π‘₯ − 5) < 2
log 8 (3(2) − 5) < 2
log 8 (6 − 5) < 2
simplify
log 8 1 < 2
apply log 𝑏 1=0 property
0<2
True
(b) Solve for x: π₯𝐨𝐠 πŸ‘ (πŸπ’™ + πŸ’) > π₯𝐨𝐠 πŸ‘ (𝒙 − πŸ’)
Solution: 2x + 4 > 0 and x – 4 > 0
x > -2 and x > 4
ensure that the arguments are positive
2x + 4 > x – 4
base is greater than 1, retain the inequality sign
2x – x > -4 – 4
combine similar terms
x > -8
( 4, ∞ )
Check: Let x = 5 ,
get the intersection of the 3 inequalities
values greater than 4 are also greater than -2 and -8
π₯𝐨𝐠 πŸ‘(πŸπ’™ + πŸ’) > π₯𝐨𝐠 πŸ‘(𝒙 − πŸ’)
π₯𝐨𝐠 πŸ‘ (𝟐(πŸ“) + πŸ’) > π₯𝐨𝐠 πŸ‘ (πŸ“ − πŸ’)
substitute 5 in x
π₯𝐨𝐠 πŸ‘ πŸπŸ’ > π₯𝐨𝐠 πŸ‘ 𝟏
simplify
2.402 > 0
True
(c) Solve the inequality π₯𝐨𝐠 𝟏𝟎 𝒙 + π₯𝐨𝐠 𝟏𝟎 (𝒙 − 𝟐) ≤ π₯𝐨𝐠 𝟏𝟎 πŸπŸ’
As you know, the logarithm function requires positive inputs. So for the
expression log10 π‘₯, we require x > 0, while for the expression log10 (π‘₯ − 2), we require x2>0, and therefore, x > 2.
Solution: π₯𝐨𝐠 𝟏𝟎 𝒙 + π₯𝐨𝐠 𝟏𝟎 (𝒙 − 𝟐) ≤ π₯𝐨𝐠 𝟏𝟎 πŸπŸ’
π₯𝐨𝐠 𝟏𝟎 𝒙(𝒙 − 𝟐) ≤ π₯𝐨𝐠 𝟏𝟎 πŸπŸ’
condense logarithm using product law
π‘₯ 2 – 2x ≤ 24
apply one-to-one property
π‘₯ 2 – 2x – 24 ≤ 0
set inequality to zero
(x-6) (x+4) ≤ 0
solving quadratic expression by factoring
[ -4, 6 ]
closed interval
( 2, 6 ]
solution set
25
Figure below shows a graphical view of this result:
Check:
a. Let x = 3
log10 3 + log10(3 − 2) ≤ log10 24
log10 3(1) ≤ log10 24 TRUE
b. Let x = 6
log10 6 + log10(6 − 2) ≤ log10 24
log10 6(4) ≤ log10 24 TRUE
Case 2: Base less than 1 ( 0 < b < 1 )
What values of x satisfy the following inequality?
𝟏
(a) π₯𝐨𝐠 𝟏 𝒙 > π₯𝐨𝐠 𝟏 ( πŸ“)−πŸ‘
πŸ“
πŸ“
Solution:
x>0
1
log 1 π‘₯ > log 1 ( 5)−3
5
5
1
log 1 π‘₯ < log 1 ( 5)−3
5
5
1
( )−3
5
x<
x < 125
0 < x < 125 or ( 0, 125)
ensure that the argument is positive
original logarithmic inequality
since base is less than 1, reverse the
inequality sign
applying one-to-one property
solution set
Check: assign x = 1, since 0 < 1 < 125
1
log 1 π‘₯ > log 1 ( )−3
original logarithmic inequality
5
log 1 1 >
5
5
1 −3
log 1 ( 5)
5
5
0 > -3
substitute x as 1
TRUE
(b) For what values of x satisfy the inequality, π₯𝐨𝐠 𝟏 πŸ‘π’™ > π₯𝐨𝐠 𝟏(πŸπ’™ + πŸ‘)
Solution: log 1 3π‘₯ < log 1 (2π‘₯ + 3)
2
2
3x > 0
2x + 3 > 0
−3
x>0
x> 2
3x < 2x + 3
3x – 2x < 3
x<3
0 < x < 3 or ( 0, 3 )
𝟐
𝟐
reverse the sign since the base is less than
arguments must be positive
applying one-to-one
combine like terms
solution set, since all real numbers greater
−3
than 0 are also greater than - 2
Check: Let x = 2, since 0 < 2 < 3
log 1 3π‘₯ < log 1 (2π‘₯ + 3)
original logarithmic inequality
2
2
log 1 3(2) < log 1 [2(2) + 3]
2
2
log 1 6 < log 1 7
2
substituting x as 2
TRUE
2
26
What’s More
Application: A particular bacterial colony doubles its population every 15 hours. A
scientist running an experiment is stating with 100 bacteria cells. She expects the number
𝑑
of cells to be given by the formula C = 100((2)15 , where t is the number o
f hours, since the experiment started. After how many hours would the scientist expect to
have 300 bacteria cells? Give your answer to the
nearest hour.
𝑑
Solution: , C = 100((2)15
𝑑
300 = 100((2)15
3=2
𝑑
log 3 = log 215
log 3 =
log 3
log 2
substitute the given
𝑑
15
𝑑
15
=
log 2
𝑑
15
express in logarithm
apply power law
divide both side by log 2
𝑑
1.585 =
15
23.77 = t
t ≈ πŸπŸ’ 𝒉𝒐𝒖𝒓𝒔
simplify
after 24 hours are expected to have 300 bacteria
Activity: Discover the secret word below by solving each logarithmic equation:
______
______
_______
_______
______
X = 16
X = 10
X = 110.5
X=7
X = -3
A. log 4 2π‘₯ = log 4 10
P. log 4 π‘₯ = 2
E. log 3 (2π‘₯ − 5) = 2
R. ln (2x+10) = 2 ln 2
F. log π‘₯ 16 = 2
T. log 2 (5π‘₯ + 1) = 3
M. log 2 (π‘₯ + 1) log 2 (π‘₯ − 1) = 3
W. log 6 (2π‘₯ − 5) = 3
O. log π‘₯ 2 = 2
Y. log 6 (π‘₯ − 2) = 3
27
What I Have Learned
Logarithmic Inequality
A logarithmic inequality is an inequality that involves logarithm.
Properties of Logarithmic Inequalities
Given the logarithmic expression log 𝑏 π‘₯
If 0 < b < 1, then π‘₯1 < π‘₯2 if and only if log 𝑏 π‘₯1 > log 𝑏 π‘₯2
If b > 1, then π‘₯1 < π‘₯2 if and only if log 𝑏 π‘₯1 < log 𝑏 π‘₯2
Guidelines in Solving Logarithmic Inequalities
1. Ensure that the logarithm is defined (arguments must be greater than 0)
2. If b > 1, retain the inequality symbol, while if 0 < b < 1, reverse the inequality
sign.
3. Solve for the variable.
4. Evaluate
5. Determine the domain. (interval notation)
6. Check the solution with the original logarithmic inequality.
Logarithmic Equation
A logarithmic equation is an equation containing a variable in a logarithmic
expression.
Properties of Logarithmic Equation
One-to-One Property: In log 𝑏 π‘₯1 = log 𝑏 π‘₯2 if and only if π‘₯1 = π‘₯2
Zero Factor Property: If ab = 0 then a = 0 or b = 0
Guidelines in Solving Logarithmic Equations
1.
2.
3.
4.
Isolate the logarithmic term on one side of equation.
Write the equation in exponential form.
Solve for the variable.
Check to make sure you do not have extraneous solution.
28
Assessment
Multiple Choice: Choose the letter of the best answer. Write the chosen letter on a
separate sheet of paper.
1. Which of the following equations illustrates the Quotient law of Logarithm?
A. log π‘Ž π‘š + log π‘Ž 𝑛 = log π‘Ž π‘šπ‘›
B. logπ‘Ž π‘š − logπ‘Ž 𝑛 = logπ‘Ž (π‘š + 𝑛)
C. log π‘Ž π‘š + log π‘Ž 𝑛 = m log π‘Ž 𝑛 + n log π‘Ž π‘š
π‘š
D. log π‘Ž π‘š − log π‘Ž 𝑛 = log π‘Ž ( )
𝑛
2. What is the perimeter of a rectangle with π₯𝐞𝐧𝐠𝐭𝐑 π₯𝐨𝐠 πŸπ’™ π’–π’π’Šπ’•π’” 𝐚𝐧𝐝 𝐰𝐒𝐝𝐭𝐑 π₯𝐨𝐠 π’š π’–π’π’Šπ’•π’”?
A. 2 log(2π‘₯ + 𝑦)
B. 2 (log 2π‘₯𝑦)
C. log 2π‘₯ + log 𝑦
D. 2 log π‘₯𝑦
3. Which of the following expression is equal to π₯𝐨𝐠 𝐚 𝐱 πŸ‘ ?
A. 3 log π‘Ž π‘₯
C. log π‘Ž π‘₯ + log π‘Ž 3
B. (log π‘Ž π‘₯)( log π‘Ž 3)
D. log π‘Ž (π‘₯ + 3)
4. If 0<b < 1 and 𝒃𝒙 > π’ƒπ’š , then what is the relationship between x and y?
A. x > y
B. x < y
C. x = y
D. not enough information
5. Which does NOT help to explain why you cannot use the laws of logarithms to expand
or simplify π₯𝐨𝐠 πŸ’ (π’š + πŸ’)?
A. The expression y + 4 cannot be factored.
B. The expression y + 4 is not raised to a power.
C. y and 4 are neither multiplied together, nor are they divided into each other.
D. Each term in the expression does not have the same variable.
6. Describe the strategy that you should use to solve π₯𝐨𝐠 πŸ” 𝒙 = π₯𝐨𝐠 πŸ” πŸ’ − π₯𝐨𝐠 πŸ” πŸ–
A. Use the quotient rule to turn the right side of the equation into a single logarithm.
Recognize the resulting value is equal to x.
B. Express the equation in exponential form, set the exponents equal to each other,
and solve
C. Use the fact that the logs have the same base to subtract the expressions on the
right side of the equation together. Express the results in exponential form, set
the exponents equal to each other, and solve.
D. Use the fact that since both sides of the equations have logarithms with the same
base to set the expressions equal to each other and solve.
7. Use properties of logarithm to expand the given expression log 𝑏 π‘₯ 2 𝑦.
A. 2log 𝑏 ( π‘₯𝑦)
C. 2log 𝑏 π‘₯ + log 𝑏 𝑦
B. log 𝑏 2π‘₯ + log 𝑣 𝑦
D. log 𝑏 2π‘₯ − log 𝑏 𝑦
8. Which of the following are the solutions to the logarithmic inequality
log 3 (x − 4) < log 3 2π‘₯ ?
A. (-4, ∞)
B. ( 4, ∞)
C. [ 4, ∞)
D. ( -∞, 4 ]
9. Find x∢ ln (3x) = ln (4x+1) + ln (1-x)
1
A. -2
B. -4
C. 4
29
D.
1
2
10. Bacteria grown in a certain culture increase according to the formula y = 𝐢𝑒 π‘˜π‘‘ , where
y is the number of bacteria present after t hours and k is constant. If there are 1,000
bacteria present now and the number will double in 20 minutes, determine how long
it will take for the number of bacteria to reach 500,000?
A. 3 hours
B. 4 hours
C. 5 hours
D. 6 hours
Additional Activities
Take the Challenge!
Solve and analyze the following. Show your complete solution.
1. If 2 log 2 + log a + log b = 2 log ( a + b ), then which of the following is TRUE?
(a). a + b = 0 ;
(b). a – b = 0 ;
(c).
π‘Ž
𝑏
= 2, or (d) ab = 4
2. Find the value of π’π’π’ˆπŸ π’π’π’ˆπŸ π’π’π’ˆπŸ‘ π₯𝐨𝐠 πŸ‘ πŸπŸ•πŸ‘ .
3. Explain why log 𝑏 0.008 = -3 log 𝑏 5.
4. Given log 5 2 ≈ 0.431. Use this fact and the laws of logarithm to approximate the
value of ;
𝟏
π₯𝐨𝐠 πŸ“ πŸ– ; π₯𝐨𝐠 πŸ“
; π₯𝐨𝐠 πŸ“ √𝟐 ; π₯𝐨𝐠 πŸπŸ“ 𝟐; π₯𝐨𝐠 πŸπŸ“ πŸ–
πŸπŸ”
30
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