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GENERAL

MATHEMATICS

Logarithmic Functions

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Learning Outcome(s): At the end of the lesson, the learner is able to

represent reallife situations using logarithmic functions and solve problems

involving logarithmic functions.

Lesson Outline:

1. Logarithms, including common and natural logarithms.
2. Applications (Richter scale, decibels, pH levels)

Definition : Let a, b, and c be positive real numbers such that b ≠ 1. The

logarithm of a with base b is denoted by logba, and is defined as

c = log b

a if and only if a = b

c

Reminders.

1. In both the logarithmic and exponential forms, b is the base. In the

exponential form, c is an exponent; this implies that the logarithm is

actually an exponent. Hence, logarithmic and exponential functions are

inverses.

2. In the logarithmic form log b

x, x cannot be negative.

3. The value of log b

x can be negative.

Definition: Common logarithms are logarithms with base 10; logx is a

short notation for log 10

x.

Definition: Natural logarithms are logarithms to the base e

(approximately 2.71828), and are denoted by “ln”. In other words, lnx is

another way of writing log e

x.

Example 1. Rewrite the following exponential equations in logarithmic

form, whenever possible.

a. 5

3

= 125 b. 7

• 2

= 1/49 c. 10

2

= 100 d. (2/3)

2

e. (0.1)

• 4

= 10000 f. 4

0

= 1 g. 7

b

= 21 h. e

2

= x i. (–2)

2

Solution.

a. log 5

125 = 3 b. log 7

(1/49) = – 2 c. log100 = 2 d. log 2/3(

e. log

10000 = – 4 f. log 4

1 = 0 g. log 7

21 = b h. lnx = 2

i. cannot be written in logarithmic form

Example 4. Suppose that an earthquake released approximately 10

12

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.

(a) Since E = 10

12

, then R =

2

3

log

10

12

10

4. 40

2

3

log 10

7. 6

Since by definition, log 10

is the exponent by which 10 must be raised

to obtain 10

, then log 10

Thus, R =

2

3

(b) This earthquake releases 10

12

 39810717 times more

energy than the reference earthquake.

Sound Intensity

In acoustics, the decibel (dB) level of a sound is

D = 10 log

𝐼

10

− 12

where I is the sound intensity in watts/m

2

(the quantity 10

• 12

watts/m

2

is

least audible sound a human can hear).

Example 5. The decibel level of sound in a quiet office is 10

• 6

watts/m

2

. (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.

(a) D = 10 log

10

− 6

10

− 12

= 10 log 10

6

Since by definition, log 10

6

is the exponent by

which 10 must be raised to obtain 10

6

, then log 10

6

Thus, D = 10(6) = 60 decibels.

(b) This sound is 10

• 6

• 12

6

= 1,000,000 times more intense than the

least audible sound a human can hear.

Acidity and the pH scale

The pH level of a water-based solution is defined as

pH = – log[H

],

where [H+] is the concentration of hydrogen ions in moles per liter. Solutions

with a pH of 7 are defined neutral ; those with pH < 7 are acidic, and those

with pH > 7 are basic.

Example 6. A 1-liter solution contains 0.00001 moles of hydrogen ions.

Find its pH level.

Solution. Since there are 0.00001 moles of hydrogen ions in 1 liter, then the

concentration of hydrogen ions is 10

• 5

moles per liter.

The pH level is – log 10

• 5 . Since log 10 - 5

is the exponent by which 10 must

be raised to obtain 10

• 5

, then log 10

• 5

Thus, pH = – log 10

• 5

Solved Examples

In numbers 1-3, find the value of the following logarithmic expressions.

1. log 3

81 2. log 169

13 3. log

5

1

5

Answers. 1. 4 2. 2 3. – 1

In numbers 4-6, rewrite the following expressions in logarithmic form,

whenever possible.

4

1

9

• 2

Answers. 4. log

2

16 = 4 5.log

81

1

2

6. log

3

1

9

In numbers 7-9, rewrite the following logarithmic equations in exponential

form, whenever possible.

7. log

3

9 = 2 8. log

16

1

2

9. ln x =

Answers. 7. 3

2

1/

= 4 9. e

1

= x

A logarithmic equation or inequality can be solved for all x values that

satisfy the equation or inequality. A logarithmic function expresses a

relationship between two variables (such as x and y), and can be

represented by a table of values or a graph.

Solved Examples

Determine whether the given is a logarithmic function, a logarithmic

equation, a logarithmic inequality or neither.

1. g(x) = log 5

𝑥 (Answer: Logarithmic Function)

2. y = 2 log

4

𝑥 (Answer: Logarithmic Function)

3. log( 4 𝑥) = - log( 3 𝑥 + 5 ) (Answer: Logarithmic Equation)
4. x log 𝑥

(𝑥) – 1 > 0 (Answer: Logarithmic Inequality)

5. log 𝑥(𝑥 − 3 ) = log 4 (Answer: Logarithmic Equation)

Supplementary Exercises

Determine whether the given is a logarithmic function, a logarithmic

equation, a logarithmic inequality or neither.

1. log

3

( 2 𝑥 − 1 ) > log

3

2. h(x) = log

3. 25

3. 2 + y = log

3

4. log

3

5. log 𝑥

2

Basic Properties of Logarithms

Learning Outcome(s) : At the end of the lesson, the learner is able to apply

basic properties of logarithms and solve problems involving logarithmic

equations.

Lesson Outline:

1. Basic properties of logarithms.
2. Simplifying logarithmic expressions.

Definition : Let b and x be real numbers such that b > 0 and b ≠ 1, the basic

properties of logarithms are as follows:

1. log

𝑏

2. log

𝑏

𝑥

= x

3. If x > 0, then 𝑏

log

𝑏

𝑥

= x

Example 1. Use the basic properties of logarithms to find the value of the

following logarithmic expressions.

a. log 10 b. ln 𝑒

3

c. log

4

64 d. log

5

1

125

e. 5

log

5

2

f. log 1

Solution.

a. log 10 = log

10

1

= 1 (Property 2)

b. ln 𝑒

3

= log

𝑒

3

= 3 (Property 2)

c. log

4

64 = log

4

3

= 3 (Property 2)

d. log

5

1

125

) = log

5

− 3

= - 3 (Property 2)

e. 5

log 5

2

= 2 (Property 3)

f. log 1 = 0 (Property 1)

EXAMPLE 2. Suppose you have seats to a concert featuring your favorite

musical artist. Calculate the approximate decibel level associated if a typical

concert’s sound intensity is 10

• 2

W/m

2

Solution.

D = 10log(

𝐼

𝐼

0

D = 10log(

10

− 2

10

− 12

D = 10log( 10

10

D = 10 • 10 (Property 2)

D = 100 dB

Answer. A concert’s decibel level is 100dB.

EXAMPLE 3. Calculate the hydrogen ion concentration of vinegar that has a

pH level of 3.0.

Solution.

pH = – log[H

]

• 3.0 = – log[H

]

b. log

3

3

𝑥

3

Solution.

log

3

3

𝑥

3

= 3log

3

3

𝑥

= 3 (log

3

3 - log

3

= 3 (1- log

3

= 3 – 3 log

3

c. ln[𝑥(𝑥 − 5 )]

Solution.

ln

[

)]

= ln 𝑥 + ln(𝑥 − 5 )

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:

log

𝑏

log

𝑎

𝑥

log

𝑎

𝑏

Example. Use the change-of-base formula to rewrite the following

logarithmic expressions to the indicated base.

a. log 6

4 (change to base 2)

log

6

log

2

4

log

2

6

2

log

2

6

b. log 1/

2 (change to base e)

log

1/

ln 2

ln(

1

2

)

ln 2

ln 1 − ln 2

ln 2

0 − ln 2

ln 2

− ln 2

Supplementary Exercises

1. Use the properties of logarithms to expand the expressions as a sum,

difference or multiple of logarithms.

a. log

𝑏

2

b. ln

𝑥

3

𝑦

2

c. log[𝑥

]

Solving Logarithmic Equations and Inequalities

Learning Outcome(s): At the end of the lesson, the learner is able to solve

logarithmic equations and inequalities and solve problems involving

logarithmic functions, equations, and inequalities.

Lesson Outline:

1. Solve logarithmic equations.
2. Solve logarithmic inequalities.
3. Applications to problems in real-life contexts.

Property of Logarithmic Equations

If b > 1, then the logarithmic function y = log b

x is increasing for all x. If 0 < b

< 1, then the logarithmic function y = log b

x is decreasing for all x. This

means that log b

u = log b

v if and only if u = v.

Techniques. Some strategies for solving logarithmic equations:

1. Rewriting to exponential form;
2. Using logarithmic properties;
3. Applying the one-to-one property of logarithmic functions;
4. The Zero Factor Property: If ab = 0, then a = 0 or b = 0.

Example 1. Find the value of x in the following equations.

a. log 4

(2x) = log 4

Solution.

log 4

(2x) = log 4

2x = 10 (one-to-one property)

x = 5

Check: 5 is a solution since log 4

(2·5) = log 4

(10) is defined.

x = – 10, 10

Check: Both are solutions since log(–10)

2

and log(10)

2

are defined.

Solution B.

logx

2

logx

2

= log

2

→ 2 = 2(1) = 2(log10) = log

2

x

2

x

2

(x + 10)(x – 10) = 0

x = – 10, 10

Check: Both are solutions since log(–10)

2

and log(10)

2

are defined.

Incorrect Method. (using log b

u n = n·log b

u immediately)

logx

2

2logx = 2 (This is not a valid conclusion because log x

2

= 2log x only if

x > 0).

Property of Logarithmic Inequalities

If 0 < b < 1, then x 1

< x 2

if and only if log b

x 1

log b

x 2

If b > 1, then x 1

< x 2

if and only if log b

x 1

< log b

x 2

Example 3. Solve the following logarithmic inequalities.

a. log 3

(2x – 1) > log 3

(x + 2)

Solution.

Step 1: Ensure that the logarithms are defined.

Then 2x – 1 > 0 and x + 2 > 0 must be satisfied.

2x – 1 > 0 implies x > 1/2 and x + 2 > 0 implies x > – 2.

To make both logarithms defined, then x > 1/2. (If x > 1/2, then x is surely

greater than – 2.)

Step 2: Ensure that the inequality is satisfied.

The base 3 is greater than 1.

Thus, since log3(2x – 1) > log3(x + 2), then:

2x – 1 > x + 2

x > 3 (subtract x from both sides; add 1 to both sides)

∆ x > 3

Hence, the solution is (3, +∞).

b. – 2 < logx < 2

Solution.

Step 1: Ensure that the logarithms are defined.

This means that x > 0.

Step 2: Ensure that the inequality is satisfied.

Rewrite – 2 and 2 as logarithms to the base 10, which are log

• 2

and log

2

respectively, obtaining the inequality: log

• 2

< logx

< log

2

We split the compound inequality into two simple inequalities:

log

• 2

< logx and logx < log

2

Since the base 10 is greater than 1, simplify both inequalities as

• 2

< x and x < 1 0

2

Thus obtaining 1/100 < x < 100, which automatically satisfies

the condition in Step 1.

Hence, the solution is (1/100, 100).

Graphing Logarithmic Functions

Learning Outcome(s) : At the end of the lesson, the learner is able to represent

a logarithmic function through its table of values, graph, and equation, find

the domain and range of a logarithmic function, and graph logarithmic

functions

Lesson Outline :

1. Graph of y = log b

x for b > 1 and for 0 < b < 1

2. Domain, range, intercepts, zeroes, and asymptotes of logarithmic functions
3. Graphs of transformations of logarithmic functions

In the following examples, the graph is obtained by first plotting a few points.

Results will be generalized later on.

Example 1. Sketch the graph of y = log 2

x.

Solution.

Step 1: Construct a table of values of ordered pairs for the given function. A

table of values for y = log2x is as follows:

It can be observed that the function is defined only for x > 0. The function is

strictly decreasing, and attains all real values. As x approaches 0 from the

right, the function increases without bound, i.e., the line x = 0 is a vertical

asymptote.

In general, the graphs of y = log b

x, where b > 0 and b  1 are shown below.

PROPERTIES OF LOGARITHMIC FUNCTIONS:

1. The domain is the set of all positive numbers, or {x | x > 0}.
2. The range is the set of all positive real numbers.
3. It is a one-to-one function. It satisfies the Horizontal Line Test.
4. The x-intercept is 1. There is no y-intercept.
5. The vertical asymptote is the line x = 0 (or the y-axis). There is no

horizontal asymptote.

Relationship Between the Graphs of Logarithmic and Exponential

Functions

Since logarithmic and exponential functions are inverses of each other, their

graphs are reflections of each other about the line y = x, as shown below.

Example 3. Sketch the graphs of y = 2log2x. Determine the domain, range,

vertical asymptote, x-intercept, and zero.

Solution.

The graph of y = 2log 2

x can be obtained from the graph of y = log 2

x by

multiplying each y-coordinate by 2, as the following table of signs shows.

The graph is shown below.

Analysis:

a. Domain: {x | x Є ℝ, x >

b. Range : {y | y Є ℝ}

c. Vertical Asymptote: x = 0

d. x-intercept: 1

e. Zero: 1

Example 4. Sketch the graph of y = log 3

x – 1.

Solution.

Sketch the graph of the basic function y = log 3

x. Note that the base 3 > 1.

The “–1” means vertical shift downwards by 1 unit.

Some points on the graph of y = log 3

x are (1,0), (3,1), and (9,2).

Shift these points 1 unit down to obtain (1, – 1), (3,0), and (9,1). Plot these

points.

The graph is shown below.

Analysis:

a. Domain: {x | x Є ℝ, x >–2}

(The expression x+2 should be greater than 0 for log0.25(x+2) to be

defined. Hence, x must be greater than – 2.)

b. Range : {y | y Є ℝ}

c. Vertical Asymptote: x = – 2

d. x-intercept: – 1

e. Zero: – 1

The examples above can be generalized to form the following guidelines for

graphing transformations of logarithmic functions:

Solved Examples

Analyze each of the following functions by (a) using the transformations to

describe how the graph is related to a logarithmic function y = log b

x, (b)

identifying the x-intercept, vertical asymptote, domain and range. (c) Sketch

the graph of the function.

a.) F(x) = log 2

(x-3)

Solution.

The graph of F(x) is shifted 3 units to the right from the graph of f(x) = log 2

(x).

Domain: { x | x Є ℝ > 3 }

Range: all real numbers

Vertical Asymptote: x = 3

x-intercept: (4, 0)

b.) G(x) = log

(x) - 3

Solution.

The graph of G(x) is a vertical shift of 3 units downwards from the graph of

g(x) = log

(x).

Domain: { x | x Є ℝ > 0 }

Range: all real numbers

Vertical Asymptote: x = 0

x-intercept: (0.125, 0)

c.) H(x) = 3log 2 x

Solution.

The graph of H(x) is a stretch by a factor of 3 from the graph of h(x) = log 2

x.

Domain: { x Є ℝ | x > 0 }

Range: all real numbers

Vertical Asymptote: x = 0

x-intercept: (1, 0)