Chapter 6
Exponential and Logarithmic Functions
and Applications
Section 6.2
Section 6.2
Logarithmic Functions and Their Graphs
• Definition of Logarithm
• Common Logarithms
• Evaluating Logarithms
• Basic Logarithmic Properties
• Logarithmic Functions: Definition, Characteristics, and Graphs
• Transformations of Graphs of Logarithmic Functions
• Applications
Introduction to Logarithms
A logarithm (denoted by “log”) is basically another name for an
exponent.
52 = 25
log525 = 2
In this example, the logarithm (or “exponent”) is 2.
“log525 = ?” means we are asking the following:
What is the exponent to which we must raise base 5 to obtain
25?
52 = 25 is the exponential form, and log525 = 2 is the
logarithmic form. Both are equivalent.
Basically, both are telling us that the outcome of raising base 5
to the exponent 2 will be 25.
Thus, a logarithm is simply the exponent or power to which a
base must be raised to produce a given number.
General Definition of Logarithms
Let b > 0, and b  1,
logbx = y
if and only if
x = by
for x > 0 and for every real number y.
We read logbx = y as "the logarithm base b of x is y."
Examples:
log216 = 4: “The logarithm base 2 of 16 is 4” because 2 raised
to the exponent 4 is 16, or 24 = 16.
log101000 = 3: “The logarithm base 10 of 1000 is 3” because
10 to the power of 3 is 1000, or 103 = 1000.
If the logarithm has a base b = 10, we do not need to write the
base; we call it a common logarithm.
Convert from exponential form to logarithmic form:
a. 9 = 32
b. 10–3 = 1/1000
c. 41/2 = 2
d. t = rs
a. We know x = by is equivalent to logbx = y.
The base in “9 = 32” is 3; the exponent is 2.
Hint: When converting from exponential to log form, the base of
the exponent becomes the base of the log; the exponent of the
exponential form becomes the exponent (“answer”) of the log.
Base, 3, of the exponential form becomes base, 3, of the
log. Exponent 2 is the exponent (“answer”) of the log.
The logarithmic form of 9 = 32 is log39 = 2.
(continued on the next slide)
(Contd.)
Convert from exponential form to logarithmic form:
a. 9 = 32
b. 10–3 = 1/1000
c. 41/2 = 2
d. t = rs
b. The base in “10–3 = 1/1000” is 10; the exponent is –3.
The base, 10, of the exponential form becomes the base,
10, of the log. Exponent –3 is the exponent (“answer”) of
the log.
The logarithmic form is log10(1/1000) = –3, which we can
also write as log(1/1000) = –3.
c. The logarithmic form of “41/2 = 2” is log42= 1/2.
Note: Remember that 41/2 is equivalent to “the square root of 4.”
d. The logarithmic form is logr t = s.
Convert from logarithmic form to exponential form:
a. log749 = 2
b. log(100) = 2
c. log5(1/25) = –2
a. We know logbx = y is equivalent to x = by.
The exponential form of log749 = 2 is 72 = 49.
Hint:
Start at the base of the log and move counterclockwise:
log749 = 2
is equivalent to 72 = 49.
(continued on the next slide)
(Contd.)
Convert from logarithmic form to exponential form:
a. log749 = 2
b. log(100) = 2
c. log5(1/25) = –2
b. We know log(100) = 2 is the same as log10(100) = 2.
log10(100) = 2
is equivalent to 102 = 100.
c. The exponential form of log5(1/25) = –2 is 5–2 = 1/25.
Evaluate each logarithm.
a. log416
Let log416 = y. The exponential form is 4y = 16.
To what exponent y must we raise 4 to get 16?
42 = 16
Therefore, log416 = 2.
b. log7 7
Let log7 7  y.
The exponential form is 7y  7.
71/2  7
Therefore, log7 7  1/2.
Evaluate each logarithm.
c. log2(0.25)
Let log2(0.25) = y. The exponential form is 2y = 0.25.
To what exponent y must we raise 2 to get 0.25?
1 1
22  2   0.25
2 4
Therefore, log2(0.25) = –2.
d. log 1
This is equivalent to log10(1).
Let log10(1) = y. The exponential form is 10y = 1.
We know a0 =1 for all a ≠ 0.
Therefore, 100 = 1, and log10(1) = 0.
Basic Properties of Logarithms
For b > 0, and b  1,
1. logb 1 = 0
2. logb b = 1
3. logb bx = x
4. blog x  x
b
Apply the basic properties to simplify each logarithm.
1. log1/2 (1/2)
log1/2 (1/2) = 1
Note that (1/2)1 = 1/2
2. log7 1
log7 1 = 0
Note that (7)0 = 1
3. 5log (13)
5log (13)  13
We know log5(13) = log5(13), changed to exponential form,
5
5
5log (13)  13.
5
4. log8 82x
log8 82x = 2x
Changing log8 82x = 2x to exponential form, we have
(8)2x = 82x.
Logarithmic Functions
Let x > 0, b > 0, and b  1. The inverse function of the
exponential function f(x) = bx is called the logarithmic
function with base b, and it is denoted by f –1(x) = logb x.
y = logbx
is equivalent to
by = x
Since x > 0, the domain of f –1(x) = logb x is the set of all
positive real numbers.
Examples:
The inverse function of y = 0.3x is y = log0.3 x.
The inverse function of y = 10x is y = log10 x, or simply
y = log x.
If f(x) = 5x, find the following:
1. f(1)
f(1) = 5(1) = 5
2. f–1(x)
f–1(x) = log5 x
3. f–1(1)
f–1(1) = log5 (1) = 0
 Note that 5(0) = 1
4. f–1(–25)
f–1(–25) = log5 (–25) which is undefined.
Note that the definition of logarithmic function states that
x > 0.
(Recall that if y = log5 x, base 5 raised to any exponent
will not yield a negative value.)
Graphs of Exponential and Logarithmic Functions
We know that the logarithmic function is the inverse of the
exponential function, thus the graph of f–1(x) = logb x is the
graph of f(x) = bx reflected about the line y = x.
The domain of f–1 is the same as the range of f, and the range
of f–1 is the same as the domain of f.
For any point (a, b) on the graph of f(x) = bx , the point (b, a)
lies on the graph of f–1(x) = logb x.
Graphs of Logarithmic Functions
For the logarithmic function f(x) = logb x, b > 0, b  1:
The domain is the set of all positive real numbers, or
(0, ∞) and the range is the set all real numbers, or (–∞, ∞).
 The x-intercept is (1, 0).
 The y-axis (that is, the line x = 0) is a vertical
asymptote.
The function is one-to-one.
The graph passes though the point (b, 1)
(continued on the next slide)
Graphs of Logarithmic Functions
 The graph has one of the following shapes:
 If b > 1, the function is increasing.
 If 0 < b < 1, the function is decreasing.
Transformations of Graphs of Logarithmic Functions
Transformations of graphs of logarithmic functions are
similar to other basic graphs we have studied.
That is, we will shift the graphs vertically, horizontally,
stretch or compress them, and reflect them across the
axes.
Next, we will see some examples.
Transformations of Graphs of Logarithmic Functions (I)
Apply transformations to the graph of f(x) = log x to construct
the graph of each function. Determine the domain and
asymptote for each function.
a. f(x) = log x + 4
b. f(x) = log x – 4
a. Vertical shift, up 4 units
Vertical asymptote: x = 0
Domain: (0, ∞)
b. Vertical shift, down 4 units
Vertical asymptote: x = 0
Domain: (0, ∞)
Transformations of Graphs of Logarithmic Functions (II)
Apply transformations to the graph of f(x) = log x to construct
the graph of each function. Determine the domain and
asymptote for each function.
a. f(x) = log (x + 4)
b. f(x) = log (x – 4)
a. Horizontal shift, left 4 units
Vertical asymptote: x = –4
Domain: (–4, ∞)
b. Horizontal shift, right 4 units
Vertical asymptote: x = 4
Domain: (4, ∞)
Transformations of Graphs of Logarithmic Functions (III)
Apply transformations to the graph of f(x) = log x to construct
the graph of f(x) = –log x – 5. Determine the domain and
asymptote.
Reflection about the x-axis
Vertical shift, down 5 units
Vertical asymptote: x = 0
Domain: (0, ∞)
Application: Measuring Earthquake Intensity
The Richter Scale is used to measure the magnitude of
earthquakes, and is given by the formula
I
R  log  
 I0 
where I and I0 are numbers that represent the wave energy
of the earthquake.
I = intensity of the current earthquake
I0 = intensity of a standard earthquake (reference intensity)
The Richter scale grows by powers of 10. This means that
an increase of 1 point results in an earthquake 10 times
greater than the previous level.
Magnitude measures the energy released at the source of
the earthquake. Intensity measures the strength of shaking
produced by the earthquake at a certain location.
On January 12, 2010, an earthquake of magnitude 7.0 caused
major destruction in the country of Haiti. On December 26, 2004,
the ocean floorof Sumatra, Indonesia was the epicenter of an
earthquake of magnitude of 9.3. This earthquake caused a
disastrous tsunami that washed away whole cities in Indonesia.
Sources: www.earthquake.usgs.gov; www.mapsofworld.com
a. Find the intensity of the 2004 earthquake. Express its intensity
in terms of the intensity of a standard earthquake, that is, I0.
Applying the Richter formula, we let R = 9.3 and solve for I in
terms of I0.
I
I
9.3  log  
R  log  
 I0 
 I0 
I
 10 9.3
I0
I  10 9.3 I0
The intensity of the Sumatra earthquake was 109.3 times that of a
standard earthquake.
(continued on the next slide)
(Contd.)
b. Find the intensity of the 2010 earthquake. Express its intensity
in terms of the intensity of a standard earthquake, that is, I0.
Applying the Richter formula, we let R = 7 and solve for I in terms
of I0.
I
7  log  
 I0 
I
 10 7
I0
I  107 I0
The intensity of the 2010 earthquake was 107 times that of a
standard earthquake.
(continued on the next slide)
(Contd.)
c. Compare the intensity of the 2004 earthquake with the one in
2010.
We now calculate:
I2004
I2010
10 9.3 I0
2.3

10
 199.53
7
10 I0
The Indonesian earthquake was approximately 199.53 times
more intense than the Haitian earthquake.
Application: Acidity or Alkalinity of a Solution (pH values)
Potential of hydrogen (pH) is a measure to determine the
relative alkalinity, acidity, or neutrality of a solution. It is
measured on a scale of 0 to 14.
The lower the pH value, the more acidic the solution; the higher
the pH value the more alkaline (or base) the solution. A value of
7 represents a neutral solution, such as distilled water.
The formula for calculating potential of hydrogen is given by
pH = – log [H+]
where [H+] represents the hydrogen ion concentration
measured in moles of hydrogen per liter.
Note: A mole is simply a unit of measurement that provides a consistent
method to convert between atoms or molecules and grams.
Many dentists recommend their patients to try to limit soda
consumption. The threshold pH for enamel dissolution is 5.5.
At a pH of less than 5.5, the acid begins to dissolve the hard
enamel of teeth. The pH level of a soda is, on average, 3.
Find its hydrogen ion concentration.
Sources: www.healthenlightenment.com; www.dentalgentlecare.com/diet_soda.htm
To find the hydrogen ion concentration, we will use the
formula for calculating potential of hydrogen, replacing pH
with 3 and solving for [H+].
pH = – log [H+]
3 = – log [H+]
–3 = log [H+]
10–3 = [H+]
The hydrogen ion concentration of a soda is 10–3 or 0.001.
Using your textbook, practice the
problems assigned by your instructor to
review the concepts from Section 6.2.