Pre-Class Problems 20 for Friday, November 3
These are the type of problems that you will be working on in class.
You can go to the solution for each problem by clicking on the problem letter.
Discussion of Logarithmic Functions
Sketch of the graph of Logarithmic Functions
1.Graph the following logarithmic functions.
a. b.
c. d.
e. f.
g. h.
i. j.
k.
2.Sketch the graph of the following functions. State the domain of the function and use the sketch to state the range of the function.
a. b.
c. d.
e. f.
g. h.
i.
j.
3.Find the domain of the following functions.
a. b.
c. d.
Problems available in the textbook: Page 438 … 9 – 50, 55 – 92 and Examples 1 – 9 starting on page 428.
SOLUTIONS:
1a.Back to Problem 1.
Note that the domain of the logarithmic function g is . In order to graph the function g given by , we set and graph the equation . By the definition of logarithm, if and only if .
x y
1 0
3 1
9 2
The Drawing of this Graph
The x-intercept of the graph of the function is the point .
Note that as from the right, . Thus, the vertical line of , which is the y-axis, is a vertical asymptote of the graph of the function.
The functions and are inverse functions of one another:
We graphed the function in Pre-Class Problems 19.
The graph of is red and the graph of is blue:
The Drawing of these Graphs
Each graph is a reflection of the other through the line , which is gray.
1b.Back to Problem 1.
Note that the domain of the logarithmic function f is . In order to graph the function f given by , we set and
graph the equation . By the definition oflogarithm,
if and only if .
x y
8
4
2
1 0
1
2
3 The Drawing of this Graph
The x-intercept of the graph of the function is the point .
Note that as from the right, . Thus, the vertical line of , which is the y-axis, is a vertical asymptote of the graph of the function.
The functions and are inverse functions of one another:
We graphed the function in Pre-Class Problems 19.
The graph of is red and the graph of is blue:
The Drawing of these Graphs
Each graph is a reflection of the other through the line , which is gray.
1c.Back to Problem 1.
Note that the domain of the logarithmic function h is . In order to graph the function h given by , we set and graph the equation . By the definition oflogarithm,
if and only if .
x y
0
1
2 The Drawing of this Graph
The x-intercept of the graph of the function is the point .
Note that as from the left, . Thus, the vertical line of , which is the y-axis, is a vertical asymptote of the graph of the function.
The functions and are inverse functions of one another:
The graph of is red and the graph of is blue:
The Drawing of these Graphs
Each graph is a reflection of the other through the line , which is gray.
1d.Back to Problem 1.
Note that the domain of the logarithmic function k is . In order to graph the function k given by , we set and graph the equation . Since , then by the definition oflogarithm, if and only if .
x y
16
4
1 0
1
2 The Drawing of this Graph
The x-intercept of the graph of the function is the point .
Note that as from the right, . Thus, the vertical line of , which is the y-axis, is a vertical asymptote of the graph of the function.
The functions and are inverse functions of one another:
We graphed the function in Pre-Class Problems 19.
The graph of is red and the graph of is blue:
The Drawing of these Graphs
Each graph is a reflection of the other through the line , which is gray.
1e.Back to Problem 1.
Note that the domain of the logarithmic function is . By the definition of logarithm, if and only if .
x y
0
1
2
3 The Drawing of this Graph
The x-intercept of the graph of the function is the point .
Note that as from the left, . Thus, the vertical line of , which is the y-axis, is a vertical asymptote of the graph of the function.
The functions and are inverse functions of one another.
1f.Back to Problem 1.
Recall: , where
Note that the domain of the logarithmic function f is . In order to graph the function f given by , we set andgraph the equation . By the definition of logarithm, if and only if .
x y
1 0
1
2
3 The Drawing of this Graph
The x-intercept of the graph of the function is the point .
Note that as from the right, . Thus, the vertical line of , which is the y-axis, is a vertical asymptote of the graph of the function.
The functions and are inverse functions of one another.
1g.Back to Problem 1.
Recall:
Note that the domain of the logarithmic function g is . In order to graph the function g given by , we set and graph the equation . Since , then
by the definition of logarithm, if and only if .
x y
1 0
10 3
100 6
The Drawing of this Graph
The x-intercept of the graph of the function is the point .
Note that as from the right, . Thus, the vertical line of , which is the y-axis, is a vertical asymptote of the graph of the function.
The functions and are inverse functions of one another.
1h.Back to Problem 1.
Note that the domain of the logarithmic function h is . In order to graph the function h given by , we set and graph the equation . Since
, then by the definition of logarithm, if and only if .
NOTE:
x y
0
2
4
The Drawing of this Graph
The x-intercept of the graph of the function is the point .
Note that as from the right, . Thus, the vertical line of , which is the y-axis, is a vertical asymptote of the graph of the function.
The functions and are inverse functions of one another.
1i.Back to Problem 1.
Note that the domain of the logarithmic function f is . In order to
graph the function f given by , we set and
graph the equation . Since
, then by the definition of logarithm, if and only if .
NOTE:
t y
1 0
2
4
8 The Drawing of this Graph
The t-intercept of the graph of the function is the point .
Note that as from the right, . Thus, the vertical line of , which is the y-axis, is a vertical asymptote of the graph of the function.
The functions and are inverse functions of one another.
1j.Back to Problem 1.
Note that the domain of the logarithmic function g is . In order to
graph the function g given by , we set and graph the equation . Since
, then by the definition of logarithm, if and only if .
NOTE: Since , then =
NOTE:
t y
0
1
The Drawing of this Graph
The t-intercept of the graph of the function is the point .
Note that as from the left, . Thus, the vertical line of , which is the y-axis, is a vertical asymptote of the graph of the function.
The functions and = are inverse functions of one another.
1k.Back to Problem 1.
Note that the domain of the logarithmic function h is . In order to
graph the function h given by , we set and
graph the equation . Since
, then by the definition of logarithm, if and only if .
t y
10
5
1 0
4
16
The Drawing of this Graph
The t-intercept of the graph of the function is the point .
Note that as from the right, . Thus, the vertical line of , which is the y-axis, is a vertical asymptote of the graph of the function.
The functions and are inverse functions of one another.
2a.Back to Problem 2.
Since we can only take the logarithm of positive numbers, we need that be positive. That is, we need that
Domain:
To graph the function f, we set and graph the equation
.
The graph of is the graph of shifted 3 units to the right.
The range of f is . Note that the x-intercept is the point .
2b.Back to Problem 2.
Since we can only take the logarithm of positive numbers, we need that be positive. That is, we need that
Domain:
To graph the function g, we set and graph the equation
.
The graph of is the graph of shifted 4 units downward.
The range of g is .
2c.Back to Problem 2.
Since we can only take the logarithm of positive numbers, we need that be positive. That is, we need that
Domain:
To graph the function h, we set and graph the equation
.
The graph of is the graph of shifted 5 units to the left and 8 units upward.
The range of h is .
2d.Back to Problem 2.
Since we can only take the logarithm of positive numbers, we need that be positive. That is, we need that
Domain:
To graph the function f, we set and graph the equation
.
The graph of is the graph of shifted 2 units downward.
The range of f is .
2e.Back to Problem 2.
Since we can only take the logarithm of positive numbers, we need that be positive. That is, we need that
Domain:
To graph the function g, we set and graph the equation
.
The graph of is the graph of shifted 1 units to the right and 6 units upward. The graph of is the graph of reflected through the t-axis. That is, flipped over with respect to the t-axis.
The range of g is .
2f.Back to Problem 2.
Since we can only take the logarithm of positive numbers, we need that be positive. That is, we need that
Domain:
To graph the function h, we set and graph the equation
.
The graph of is the graph of shifted 4 units to the left and 3 units downward. The graph of is the graph of reflected through the x-axis. That is, flipped over with respect to the x-axis.
The range of h is .
2g.Back to Problem 2.
Since we can only take the logarithm of positive numbers, we need that be positive. That is, we need that
Domain:
To graph the function f, we set and graph the equation
.
The graph of is the graph of shifted 2 units to the left.
The graph of :
The graph of :
The range of f is .
2h.Back to Problem 2.
Since we can only take the logarithm of positive numbers, we need that be positive. That is, we need that .
Domain:
To graph the function g, we set and graph the equation
.
The graph of is the graph of shifted 6 units to the right and 1 unit downward.
The range of g is .
2i.Back to Problem 2.
Since we can only take the logarithm of positive numbers, we need that be positive. That is, we need that
.
Domain:
To graph the function h, we set and graph the equation
.
The graph of is the graph of
shifted units to the left and 12 unitsupward.
The graph of :
The graph of :
The range of h is .
2j.Back to Problem 2.
Since we can only take the logarithm of positive numbers, we need that be positive. That is, we need that .
Domain:
To graph the function f, we set and graph the equation
.
The graph of is the graph of
shifted 8 units to the right and 15 units downward. The graph of is the graph of reflected through the x-axis. That is, flipped over with respect to the x-axis.
The range of f is .
3a.Back to Problem 3.
Want (Need):
Step 1:
is defined for all real numbers x.
Step 2:Plot all the numbers found in Step 1 on the real number line.
+ Sign of
3
Step 3:Use the real number line to identify the open intervals determined by the plotted numbers. Pick a test value for each open interval.
IntervalTest ValueSign of =
0
4
Answer:
3b.Back to Problem3.
Want (Need):
Step 1:
undefined
Step 2:Plot all the numbers found in Step 1 on the real number line.
+ + Sign of
2
Step 3:Use the real number line to identify the open intervals determined by the plotted numbers. Pick a test value for each open interval.
IntervalTest ValueSign of
0
3
Answer:
3c.Back to Problem 3.
NOTE: The nonlinear expression will be defined and will be positive as long as the linear expression is positive.
Want (Need):
Answer:
3d.Back to Problem 3.
NOTE: The nonlinear expression is greater than or equal to zero for all real numbers x. Since the logarithm of zero is undefined, we need that .
Answer:
Definition The logarithmic function with base b is the function defined by , where and .
Recall that if and only if
Recall the following information about logarithmic functions:
1.The domain of is the set of positive real numbers. That is, the domain of is .
2.The range of is the set of real numbers. That is, the range of is .
3.The logarithmic function and the exponential function are inverses of one another:
,
for all x in the domain of g, which is the set of all real numbers.
, for all x in the domain of f, which is the set of real numbers in the interval .
Definition The natural logarithmic function is the logarithmic function whose base is the irrational number e. Thus, the natural logarithmic function is the function defined by , where . Recall that .
Definition The common logarithmic function is the logarithmic function whose base is the number 10. Thus, the common logarithmic function is the function defined by . Recall that .
Theorem (Properties of Logarithms)
1. =
2. = +
3. =
4.
5.
6.
7.
8.Change of Bases Formula:
Proof
1.Let . Then by the definition of logarithms, . Thus, . Writing the exponential equation in terms of a logarithmic equation, we have that . Since , then we have that .
2.Let and . Then by the definition of logarithms, and . Thus, . Writing the exponential equation in terms of a logarithmic equation, we have that . Since and , then .
3.Let and . Then by the definition of logarithms, and . Thus, . Writing the exponential equation in terms of a logarithmic equation, we have that . Since and , then .
Alternate proof: Since , we have that . Now, applying Property 2, we have that . Now, applying Property 1, we have that . Thus, we have that .
6.Let . Then by the definition of logarithms, . Since , then .
7.Follows from applying Property 1 and then Property 4.
8.Let , , and . Then by the definition of logarithms, we have that , , and . Since , then . Since and , then . Since , then . Thus, . Since , , and , then =
. Since b is the base of a logarithm, then . Since if and only if , then . So, we can solve for by dividing both sides of the equation = by . Thus, we obtain that .
Alternate proof: Let . Then by the definition of logarithms, . Taking the logarithm base a of both sides of this equation, we obtain that . By Property 1, we have that . Thus, . Since b is the base of a logarithm, then . Since if and only if , then . Solving for y, we obtain that . Since , then .
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The sketch of the graph of , where :
The x-intercept of the graph of the function is the point .
The vertical line , which is the y-axis, is a vertical asymptote of the
graph of the function.
The sketch of the graph of , where:
The x-intercept of the graph of the function is the point .
The vertical line , which is the y-axis, is a vertical asymptote of the
graph of the function.
The sketch of the graph of , where :
The x-intercept of the graph of the function is the point .
The vertical line , which is the y-axis, is a vertical asymptote of the
graph of the function.
The sketch of the graph of , where:
The x-intercept of the graph of the function is the point .
The vertical line , which is the y-axis, is a vertical asymptote of the
graph of the function.
Backto the top.