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Solve each equation.
1.
SOLUTION:
2.
SOLUTION:
7-4 Solving Logarithmic Equations and Inequalities
2
Solve each equation.
MULTIPLE CHOICE
3. Solve log5 (x − 10) = log5
3x.
1.
A 10
B
SOLUTION: 2
C
5
D 2, 5
SOLUTION:
2.
SOLUTION:
Substitute each value into the original equation.
The domain of a logarithmic function cannot be 0, so
log5 (–6) is undefined and –2 is an extraneous
2 solution.
MULTIPLE CHOICE
3. Solve log5 (x − 10) = log5
3x. C is the correct option.
A 10
B Solve each inequality.
2
C 4. log x > 3
5 5
D 2, 5
SOLUTION:
SOLUTION:
Thus, solution set is {x | x > 125}.
5. log8 x ≤ −2
SOLUTION:
Substitute each value into the original equation.
eSolutions Manual - Powered by Cognero Page1
Thus, solution set is .
The domain of a logarithmic function cannot be 0, so
log5 (–6) is undefined and –2 is an extraneous
solution. 6. log4 (2x + 5) ≤ log4 (4x − 3)
SOLUTION:
C is the correct option.
Solve each inequality.
4. log5 x > 3
SOLUTION:
Thus, solution set is {x | x ≥ 4}.
7. log8 (2x) > log8 (6x − 8)
SOLUTION:
Thus, solution set is {x | x > 125}.
5. log8 x ≤ −2
SOLUTION:
Exclude all values of x for which
So,
Thus, solution set is .
Thus, solution set is .
6. log4 (2x + 5) ≤ log4 (4x − 3)
CCSS STRUCTURE Solve each equation.
SOLUTION:
8.
SOLUTION:
Thus, solution set is {x | x ≥ 4}.
7. log8 (2x) > log8 (6x − 8)
SOLUTION:
9.
SOLUTION:
Exclude all values of x for which
So,
Thus, solution set is .
CCSS STRUCTURE Solve each equation.
10.
8.
SOLUTION:
SOLUTION:
11.
9.
SOLUTION:
SOLUTION:
12.
SOLUTION:
10.
SOLUTION:
13.
11.
SOLUTION:
SOLUTION:
12. 2
14. log3 (3x + 8) = log3 (x + x)
SOLUTION:
SOLUTION:
Substitute each value into the original equation.
13.
SOLUTION:
Thus, x = –2 or 4.
2
15. log12 (x − 7) = log12 (x + 5)
SOLUTION:
2
14. log3 (3x + 8) = log3 (x + x)
SOLUTION:
Substitute each value into the original equation.
Thus, x = –3 or 4.
Substitute each value into the original equation. 2
16. log6 (x − 6x) = log6 (−8)
SOLUTION:
Thus, x = –2 or 4.
2
15. log12 (x − 7) = log12 (x + 5)
SOLUTION:
Substitute each value into the original equation.
Substitute each value into the original equation. log6 (–8) is undefined, so 4 and 2 are extraneous
solutions.
Thus, no solution.
2
17. log9 (x − 4x) = log9 (3x − 10)
Thus, x = –3 or 4. SOLUTION:
2
16. log6 (x − 6x) = log6 (−8)
SOLUTION:
Substitute each value into the original equation.
Substitute each value into the original equation.
log9 (–4) is undefined and 2 is extraneous solution.
Thus, x = 5.
2
18. log4 (2x + 1) = log4 (10x − 7)
log6 (–8) is undefined, so 4 and 2 are extraneous
solutions. SOLUTION:
Thus, no solution.
2
17. log9 (x − 4x) = log9 (3x − 10)
SOLUTION:
Substitute each value into the original equation.
Substitute each value into the original equation. Thus, x = 1 or 4.
2
19. log7 (x − 4) = log7 (− x + 2)
SOLUTION:
log (–4) is undefined and 2 is extraneous solution.
9
Thus, x = 5.
2
18. log4 (2x + 1) = log4 (10x − 7)
SOLUTION:
Substitute each value into the original equation.
Since you can not have a log of 0, x = 3 is the
Substitute each value into the original equation. solution.
SCIENCEThe equation for wind speed w, in
miles per hour, near the center of a tornado is w
= 93 log d + 65, where d is the distance in
10
Thus, x = 1 or 4. miles that the tornado travels.
20. Write this equation in exponential form.
2
SOLUTION:
19. log7 (x − 4) = log7 (− x + 2)
SOLUTION:
21. In May of 1999, a tornado devastated Oklahoma City
with the fastest wind speed ever recorded. If the
tornado traveled 525 miles, estimate the wind speed
near the center of the tornado.
Substitute each value into the original equation.
SOLUTION:
Substitute 525 for d in the equation and simplify.
Since you can not have a log of 0, x = 3 is the Solve each inequality.
solution.
22. log6 x < −3
SCIENCEThe equation for wind speed w, in
miles per hour, near the center of a tornado is w SOLUTION:
= 93 log d + 65, where d is the distance in
10
miles that the tornado travels.
20. Write this equation in exponential form.
SOLUTION:
The solution set is .
23. log4 x ≥ 4
SOLUTION:
21. In May of 1999, a tornado devastated Oklahoma City
with the fastest wind speed ever recorded. If the
tornado traveled 525 miles, estimate the wind speed
near the center of the tornado.
The solution set is {x | x ≥ 256}.
SOLUTION:
Substitute 525 for d in the equation and simplify. 24. log3 x ≥ −4
SOLUTION:
Solve each inequality.
22. log6 x < −3
The solution set is .
SOLUTION:
25. log2 x ≤ −2
SOLUTION:
The solution set is .
23. log4 x ≥ 4
The solution set is .
SOLUTION:
26. log5 x > 2
SOLUTION:
The solution set is {x | x ≥ 256}.
24. log3 x ≥ −4
SOLUTION: The solution set is .
27. log7 x < −1
SOLUTION:
The solution set is .
25. log2 x ≤ −2
The solution set is .
SOLUTION:
28. log2 (4x − 6) > log2 (2x + 8)
SOLUTION:
The solution set is .
26. log5 x > 2
SOLUTION: The solution set is .
29. log7 (x + 2) ≥ log7 (6x − 3)
SOLUTION:
The solution set is .
27. log7 x < −1
SOLUTION:
Exclude all values of x for which
So,
The solution set is .
The solution set is .
28. log2 (4x − 6) > log2 (2x + 8)
SOLUTION: 30. log3 (7x – 6) < log3 (4x + 9)
SOLUTION:
The solution set is .
Exclude all values of x for which
29. log7 (x + 2) ≥ log7 (6x − 3)
SOLUTION:
So,
The solution set is .
Exclude all values of x for which
31. log5 (12x + 5) ≤ log5 (8x + 9)
SOLUTION:
So,
The solution set is .
Exclude all values of x for which
30. log3 (7x – 6) < log3 (4x + 9)
SOLUTION:
So,
The solution set is .
Exclude all values of x for which
32. log11 (3x − 24) ≥ log11 (−5x − 8)
SOLUTION:
So,
The solution set is .
The solution set is {x | x ≥ 2}.
31. log5 (12x + 5) ≤ log5 (8x + 9)
33. log (9x + 4) ≤ log (11x − 12)
SOLUTION: 9 9
SOLUTION:
Exclude all values of x for which
The solution set is {x | x ≥ 8}.
So, 34. CCSS MODELINGThe magnitude of an
earthquake is measured on a logarithmic scale called
the Richter scale. The magnitude M is given by M =
The solution set is . log10 x, where x represents the amplitude of the
seismic wave causing ground motion.
a. How many times as great is the amplitude caused
32. log11 (3x − 24) ≥ log11 (−5x − 8) by an earthquake with a Richter scale rating of 8 as
an aftershock with a Richter scale rating of 5?
SOLUTION: b. In 1906, San Francisco was almost completely
destroyed by a 7.8 magnitude earthquake. In 1911,
an earthquake estimated at magnitude 8.1 occurred
along the New Madrid fault in the Mississippi River
Valley. How many times greater was the New
Madrid earthquake than the San Francisco
earthquake?
The solution set is {x | x ≥ 2}.
SOLUTION:
a.
33. log9 (9x + 4) ≤ log9 (11x − 12) The amplitude of the seismic wave with a Richter
8 5
SOLUTION: scale rating of 8 and 5 are 10 and 10 respectively.
Divide 108 by 105.
The solution set is {x | x ≥ 8}. 3
The scale rating of 8 is 10 or 1000 times greater
34. CCSS MODELINGThe magnitude of an than the scale rating of 5.
earthquake is measured on a logarithmic scale called
the Richter scale. The magnitude M is given by M = b.
log10 x, where x represents the amplitude of the The amplitudes of San Francisco earthquake and
seismic wave causing ground motion. New Madrid earthquake were 107.8 and 108.1
a. How many times as great is the amplitude caused respectively.
by an earthquake with a Richter scale rating of 8 as
an aftershock with a Richter scale rating of 5? Divide 108.1 by 107.8.
b. In 1906, San Francisco was almost completely
destroyed by a 7.8 magnitude earthquake. In 1911,
an earthquake estimated at magnitude 8.1 occurred
along the New Madrid fault in the Mississippi River
Valley. How many times greater was the New
Madrid earthquake than the San Francisco
earthquake? The New Madrid earthquake was 100.3 or about 2
times greater than the San Francisco earthquake.
SOLUTION:
a.
MUSICThe first key on a piano keyboard
The amplitude of the seismic wave with a Richter 35.
scale rating of 8 and 5 are 108 and 105 respectively. corresponds to a pitch with a frequency of 27.5
8 5 cycles per second. With every successive key, going
Divide 10 by 10 . up the black and white keys, the pitch multiplies by a
constant. The formula for the frequency of the pitch
sounded when the nth note up the keyboard is played
is given by
a. A note has a frequency of 220 cycles per second.
3 How many notes up the piano keyboard is this?
The scale rating of 8 is 10 or 1000 times greater b. Another pitch on the keyboard has a frequency of
than the scale rating of 5. 880 cycles per second. After how many notes up the
keyboard will this be found?
b.
The amplitudes of San Francisco earthquake and SOLUTION:
New Madrid earthquake were 107.8 and 108.1 a.
respectively. Substitute 220 for f in the formula and solve for n.
Divide 108.1 by 107.8.
b.
Substitute 880 for f in the formula and solve for n.
The New Madrid earthquake was 100.3 or about 2
times greater than the San Francisco earthquake.
MUSICThe first key on a piano keyboard
35.
corresponds to a pitch with a frequency of 27.5
cycles per second. With every successive key, going
up the black and white keys, the pitch multiplies by a
constant. The formula for the frequency of the pitch
sounded when the nth note up the keyboard is played
MULTIPLE REPRESENTATIONSIn this
is given by 36.
problem, you will explore the graphs shown:
a. A note has a frequency of 220 cycles per second. y = log4 x and
How many notes up the piano keyboard is this?
b. Another pitch on the keyboard has a frequency of
880 cycles per second. After how many notes up the
keyboard will this be found?
SOLUTION:
a.
Substitute 220 for f in the formula and solve for n.
a. ANALYTICAL
How do the shapes of the
graphs compare? How do the asymptotes and the x-
intercepts of the graphs compare?
b. VERBAL
Describe the relationship between the
graphs.
b.
c. GRAPHICAL
Substitute 880 for f in the formula and solve for n. Use what you know about
transformations of graphs to compare and contrast
the graph of each function and the graph of y = log4
x.
1. y = log4 x + 2
2. y = log4 (x + 2)
3. y = 3 log4 x
MULTIPLE REPRESENTATIONSIn this
36. d. ANALYTICAL
problem, you will explore the graphs shown: Describe the relationship
between y = log x and y = 1(log x). What are a
y = log4 x and 4 − 4
reasonable domain and range for each function?
e.ANALYTICAL
Write an equation for a function
for which the graph is the graph of y = log3 x
translated 4 units left and 1 unit up.
SOLUTION:
a.
The shapes of the graphs are the same. The
asymptote for each graph is the y-axis and the x-
intercept for each graph is 1.
a. ANALYTICAL b.
How do the shapes of the The graphs are reflections of each other over the x-
graphs compare? How do the asymptotes and the x- axis.
intercepts of the graphs compare?
b. VERBAL c.
Describe the relationship between the 1. The second graph is the same as the first, except it
graphs. is shifted horizontally to the left 2 units.
c. GRAPHICAL
Use what you know about
transformations of graphs to compare and contrast
the graph of each function and the graph of y = log4
x.
1. y = log4 x + 2
2. y = log4 (x + 2)
3. y = 3 log x
4
2, 8] scl: 1 by [ 5, 5] scl: 1
d. ANALYTICAL [− −
Describe the relationship
between y = log x and y = 1(log x). What are a
4 − 4 2. The second graph is the same as the first, except it
reasonable domain and range for each function? is shifted vertically up 2 units.
e.ANALYTICAL
Write an equation for a function
for which the graph is the graph of y = log x
3
translated 4 units left and 1 unit up.
SOLUTION:
a.
The shapes of the graphs are the same. The
asymptote for each graph is the y-axis and the x-
intercept for each graph is 1.
b.
The graphs are reflections of each other over the x-
4, 8] scl: 1 by [ 5, 5] scl: 1
axis. [− −
c. 3. Each point on the second graph has a y-coordinate
1. The second graph is the same as the first, except it 3 times that of the corresponding point on the first
is shifted horizontally to the left 2 units. graph.
2, 8] scl: 1 by [ 5, 5] scl: 1
[− − 2, 8] scl: 1 by [ 5, 5] scl: 1
[− −
d.
2. The second graph is the same as the first, except it The graphs are reflections of each other over the x-
is shifted vertically up 2 units. axis.
D={x|x>0}; R={all real numbers}
e. where h is the horizontal
shift and k is the vertical shift. Since there is a
horizontal shift of 4 and vertical shift of 1, h = 4 and
k = 1.
y = log3 (x + 4) + 1
SOUNDThe relationship between the intensity of
37.
sound I and the number of decibels
β
4, 8] scl: 1 by [ 5, 5] scl: 1
[− − is , where I is the intensity of
3. Each point on the second graph has a y-coordinate sound in watts per square meter.
3 times that of the corresponding point on the first
graph. a. Find the number of decibels of a sound with an
intensity of 1 watt per square meter.
b. Find the number of decibels of sound with an
−2
intensity of 10 watts per square meter.
c. The intensity of the sound of 1 watt per square
meter is 100 times as much as the intensity of 10−2
watts per square meter. Why are the decibels of
sound not 100 times as great?
SOLUTION:
2, 8] scl: 1 by [ 5, 5] scl: 1 a.
[− −
d. Substitute 1 for I in the given equation and solve for
The graphs are reflections of each other over the x- .
axis. β
D={x|x>0}; R={all real numbers}
e. where h is the horizontal
shift and k is the vertical shift. Since there is a
horizontal shift of 4 and vertical shift of 1, h = 4 and
k = 1.
y = log3 (x + 4) + 1
SOUNDThe relationship between the intensity of
37.
sound I and the number of decibels β b.
is , where I is the intensity of Substitute 10−2 for I in the given equation and solve
for .
β
sound in watts per square meter.
a. Find the number of decibels of a sound with an
intensity of 1 watt per square meter.
b. Find the number of decibels of sound with an
−2
intensity of 10 watts per square meter.
c. The intensity of the sound of 1 watt per square
meter is 100 times as much as the intensity of 10−2
watts per square meter. Why are the decibels of
sound not 100 times as great?
c. Sample answer: The power of the logarithm only
changes by 2. The power is the answer to the
SOLUTION: logarithm. That 2 is multiplied by the 10 before the
a. logarithm. So we expect the decibels to change by
Substitute 1 for I in the given equation and solve for 20.
.
β
CCSS CRITIQUE
38. Ryan and Heather are solving
log x 3. Is either of them correct? Explain your
3 ≥−
reasoning.
b.
−2
Substitute 10 for I in the given equation and solve
for .
β
SOLUTION:
Sample answer: Ryan; Heather did not need to
switch the inequality symbol when raising to a
negative power.
c. Sample answer: The power of the logarithm only
CHALLENGEFind log 27 +log 27 +log 27 +
changes by 2. The power is the answer to the 39. 3 9 27
logarithm. That 2 is multiplied by the 10 before the
log81 27 + log243 27.
logarithm. So we expect the decibels to change by
20. SOLUTION:
CCSS CRITIQUE
38. Ryan and Heather are solving
log x 3. Is either of them correct? Explain your
3 ≥−
reasoning.
REASONINGThe Property of Inequality for
40.
Logarithmic Functions states that when b > 1, logb x
>logby if and only if x >y. What is the case for
when 0 < b < 1? Explain your reasoning.
SOLUTION:
Sample answer: When 0 < b < 1, logb x > logby if
and only if x 1, logb x your reasoning. All logarithmic equations are of the
>logby if and only if x >y. What is the case for
form y = logb x.
when 0 < b < 1? Explain your reasoning. a. If the base of a logarithmic equation is greater
than 1 and the value of x is between 0 and 1, then the
SOLUTION:
value for y is (less than, greater than, equal to) 0.
Sample answer: When 0 < b < 1, logb x > logby if b. If the base of a logarithmic equation is between 0
and only if x 81
SOLUTION:
53.
SOLUTION:
4a + 6 a
4 16
log 2401 57. ≤
54. 7
SOLUTION:
SOLUTION:
Solve each equation or inequality. Check your
solution.
2x + 3
5 125
55. ≤
58.
SOLUTION:
SOLUTION:
3x − 2
56. 3 >81
SOLUTION:
59.
SOLUTION:
4a + 6 a
4 16
57. ≤
SOLUTION:
60.
SOLUTION:
58.
SOLUTION:
SHIPPING The height of a shipping cylinder is 4
61.
feet more than the radius. If the volume of the
cylinder is 5 cubic feet, how tall is it? Use the
π
2
formula V = πr h.
59.
SOLUTION:
Substitute 5 for V and r + 4 for h in the formula and
π
SOLUTION: simplify.
The equation has one real root r = 1.
Thus, the height of the shipping cylinder is 1 + 4 = 5
ft.
NUMBER THEORYTwo complex conjugate
62.
60. numbers have a sum of 12 and a product of 40. Find
the two numbers
SOLUTION:
SOLUTION:
The equations that represent the situation are:
Solve equation (1).
SHIPPING The height of a shipping cylinder is 4
61.
feet more than the radius. If the volume of the
cylinder is 5 cubic feet, how tall is it? Use the
π
2
formula V = πr h.
SOLUTION: Solve equation (2).
Substitute 5 for V and r + 4 for h in the formula and
π
simplify.
The equation has one real root r = 1.
Thus, the height of the shipping cylinder is 1 + 4 = 5 Thus, the two numbers are 6 + 2i and 6 2i.
–
ft.
Simplify. Assume that no variable equals zero.
5 3
NUMBER THEORYTwo complex conjugate
62. x x
numbers have a sum of 12 and a product of 40. Find 63. ·
the two numbers SOLUTION:
SOLUTION:
The equations that represent the situation are:
a2 a6
64. ·
SOLUTION:
Solve equation (1).
2 3
65. (2p n)
SOLUTION:
Solve equation (2).
(3b3c2)2
66.
SOLUTION:
Thus, the two numbers are 6 + 2i and 6 2i. 67.
–
Simplify. Assume that no variable equals zero. SOLUTION:
x5 x3
63. ·
SOLUTION:
68.
2 6
a a SOLUTION:
64. ·
SOLUTION:
(2p2n)3
65.
SOLUTION:
(3b3c2)2
66.
SOLUTION:
67.
SOLUTION:
68.
SOLUTION:
Solve each equation.
1.
SOLUTION:
2.
SOLUTION:
2
MULTIPLE CHOICE
3. Solve log5 (x − 10) = log5
3x.
A 10
B
2
C
5
D 2, 5
SOLUTION:
Solve each equation.
1.
SOLUTION:
Substitute each value into the original equation.
The domain of a logarithmic function cannot be 0, so
log5 (–6) is undefined and –2 is an extraneous
solution.
2. C is the correct option.
SOLUTION: Solve each inequality.
4. log5 x > 3
SOLUTION:
Thus, solution set is {x | x > 125}.
2
MULTIPLE CHOICE
3. Solve log5 (x − 10) = log5
5. log8 x ≤ −2
3x.
A 10
B SOLUTION:
2
C
5
D 2, 5
SOLUTION:
Thus, solution set is .
6. log4 (2x + 5) ≤ log4 (4x − 3)
SOLUTION:
Substitute each value into the original equation.
Thus, solution set is {x | x ≥ 4}.
7. log8 (2x) > log8 (6x − 8)
The domain of a logarithmic function cannot be 0, so
log (–6) is undefined and –2 is an extraneous SOLUTION:
5
solution.
C is the correct option.
Solve each inequality.
4. log5 x > 3 Exclude all values of x for which
SOLUTION:
So,
7-4 Solving Logarithmic Equations and Inequalities
Thus, solution set is .
Thus, solution set is {x | x > 125}.
CCSS STRUCTURE Solve each equation.
5. log8 x ≤ −2
8.
SOLUTION:
SOLUTION:
Thus, solution set is .
6. log4 (2x + 5) ≤ log4 (4x − 3)
SOLUTION:
9.
SOLUTION:
Thus, solution set is {x | x ≥ 4}.
7. log8 (2x) > log8 (6x − 8)
SOLUTION:
10.
SOLUTION:
Exclude all values of x for which
So,
Thus, solution set is .
CCSS STRUCTURE Solve each equation.
11.
8.
SOLUTION:
SOLUTION:
eSolutions Manual - Powered by Cognero Page2
12.
SOLUTION:
9.
SOLUTION:
13.
10.
SOLUTION:
SOLUTION:
2
14. log3 (3x + 8) = log3 (x + x)
11.
SOLUTION:
SOLUTION:
Substitute each value into the original equation.
12.
SOLUTION:
Thus, x = –2 or 4.
2
15. log12 (x − 7) = log12 (x + 5)
SOLUTION:
13.
SOLUTION:
Substitute each value into the original equation.
Thus, x = –3 or 4.
2 2
14. log3 (3x + 8) = log3 (x + x) 16. log6 (x − 6x) = log6 (−8)
SOLUTION:
SOLUTION:
Substitute each value into the original equation. Substitute each value into the original equation.
Thus, x = –2 or 4.
log6 (–8) is undefined, so 4 and 2 are extraneous
2
15. log12 (x − 7) = log12 (x + 5) solutions.
Thus, no solution.
SOLUTION:
2
17. log9 (x − 4x) = log9 (3x − 10)
SOLUTION:
Substitute each value into the original equation.
Substitute each value into the original equation.
Thus, x = –3 or 4.
2
16. log6 (x − 6x) = log6 (−8)
log9 (–4) is undefined and 2 is extraneous solution.
SOLUTION: Thus, x = 5.
2
18. log4 (2x + 1) = log4 (10x − 7)
SOLUTION:
Substitute each value into the original equation.
Substitute each value into the original equation.
log6 (–8) is undefined, so 4 and 2 are extraneous
solutions.
Thus, no solution.
2 Thus, x = 1 or 4.
17. log9 (x − 4x) = log9 (3x − 10)
2
19. log (x −4) =log (−x +2)
SOLUTION: 7 7
SOLUTION:
Substitute each value into the original equation.
Substitute each value into the original equation.
log9 (–4) is undefined and 2 is extraneous solution.
Thus, x = 5. Since you can not have a log of 0, x = 3 is the
2 solution.
18. log4 (2x + 1) = log4 (10x − 7)
SCIENCEThe equation for wind speed w, in
SOLUTION: miles per hour, near the center of a tornado is w
= 93 log d + 65, where d is the distance in
10
miles that the tornado travels.
20. Write this equation in exponential form.
SOLUTION:
Substitute each value into the original equation.
21. In May of 1999, a tornado devastated Oklahoma City
with the fastest wind speed ever recorded. If the
tornado traveled 525 miles, estimate the wind speed
Thus, x = 1 or 4. near the center of the tornado.
2 SOLUTION:
19. log7 (x − 4) = log7 (− x + 2) Substitute 525 for d in the equation and simplify.
SOLUTION:
Solve each inequality.
22. log6 x < −3
SOLUTION:
Substitute each value into the original equation.
The solution set is .
Since you can not have a log of 0, x = 3 is the 23. log4 x ≥ 4
solution.
SOLUTION:
SCIENCEThe equation for wind speed w, in
miles per hour, near the center of a tornado is w
= 93 log d + 65, where d is the distance in
10
miles that the tornado travels.
20. Write this equation in exponential form.
The solution set is {x | x ≥ 256}.
SOLUTION:
24. log3 x ≥ −4
SOLUTION:
21. In May of 1999, a tornado devastated Oklahoma City The solution set is .
with the fastest wind speed ever recorded. If the
tornado traveled 525 miles, estimate the wind speed
near the center of the tornado. 25. log2 x ≤ −2
SOLUTION:
SOLUTION:
Substitute 525 for d in the equation and simplify.
Solve each inequality. The solution set is .
22. log6 x < −3
26. log x > 2
SOLUTION: 5
SOLUTION:
The solution set is .
The solution set is .
23. log x ≥ 4
4 27. log7 x < −1
SOLUTION:
SOLUTION:
The solution set is {x | x ≥ 256}. The solution set is .
24. log3 x ≥ −4
28. log (4x − 6) > log (2x + 8)
SOLUTION: 2 2
SOLUTION:
The solution set is .
The solution set is .
25. log2 x ≤ −2
29. log (x + 2) ≥ log (6x − 3)
SOLUTION: 7 7
SOLUTION:
The solution set is .
Exclude all values of x for which
26. log5 x > 2
SOLUTION:
So,
The solution set is .
The solution set is .
30. log3 (7x – 6) < log3 (4x + 9)
27. log7 x < −1
SOLUTION:
SOLUTION:
The solution set is . Exclude all values of x for which
28. log2 (4x − 6) > log2 (2x + 8)
So,
SOLUTION:
The solution set is .
31. log5 (12x + 5) ≤ log5 (8x + 9)
The solution set is . SOLUTION:
29. log7 (x + 2) ≥ log7 (6x − 3)
SOLUTION:
Exclude all values of x for which
So,
Exclude all values of x for which
So, The solution set is .
32. log11 (3x − 24) ≥ log11 (−5x − 8)
The solution set is .
SOLUTION:
30. log3 (7x – 6) < log3 (4x + 9)
SOLUTION:
The solution set is {x | x ≥ 2}.
33. log9 (9x + 4) ≤ log9 (11x − 12)
Exclude all values of x for which SOLUTION:
So,
The solution set is . The solution set is {x | x ≥ 8}.
34. CCSS MODELINGThe magnitude of an
earthquake is measured on a logarithmic scale called
31. log5 (12x + 5) ≤ log5 (8x + 9)
the Richter scale. The magnitude M is given by M =
SOLUTION: log10 x, where x represents the amplitude of the
seismic wave causing ground motion.
a. How many times as great is the amplitude caused
by an earthquake with a Richter scale rating of 8 as
an aftershock with a Richter scale rating of 5?
b. In 1906, San Francisco was almost completely
Exclude all values of x for which destroyed by a 7.8 magnitude earthquake. In 1911,
an earthquake estimated at magnitude 8.1 occurred
along the New Madrid fault in the Mississippi River
Valley. How many times greater was the New
So, Madrid earthquake than the San Francisco
earthquake?
SOLUTION:
The solution set is . a.
The amplitude of the seismic wave with a Richter
scale rating of 8 and 5 are 108 and 105 respectively.
8 5
32. log11 (3x − 24) ≥ log11 (−5x − 8) Divide 10 by 10 .
SOLUTION:
The scale rating of 8 is 103 or 1000 times greater
than the scale rating of 5.
The solution set is {x | x ≥ 2}.
b.
33. log9 (9x + 4) ≤ log9 (11x − 12) The amplitudes of San Francisco earthquake and
New Madrid earthquake were 107.8 and 108.1
SOLUTION: respectively.
Divide 108.1 by 107.8.
The solution set is {x | x ≥ 8}.
34. CCSS MODELINGThe magnitude of an 0.3
earthquake is measured on a logarithmic scale called The New Madrid earthquake was 10 or about 2
times greater than the San Francisco earthquake.
the Richter scale. The magnitude M is given by M =
log10 x, where x represents the amplitude of the
MUSICThe first key on a piano keyboard
seismic wave causing ground motion. 35.
a. How many times as great is the amplitude caused corresponds to a pitch with a frequency of 27.5
by an earthquake with a Richter scale rating of 8 as cycles per second. With every successive key, going
an aftershock with a Richter scale rating of 5? up the black and white keys, the pitch multiplies by a
constant. The formula for the frequency of the pitch
b. In 1906, San Francisco was almost completely sounded when the nth note up the keyboard is played
destroyed by a 7.8 magnitude earthquake. In 1911, is given by
an earthquake estimated at magnitude 8.1 occurred
along the New Madrid fault in the Mississippi River a. A note has a frequency of 220 cycles per second.
Valley. How many times greater was the New How many notes up the piano keyboard is this?
Madrid earthquake than the San Francisco b. Another pitch on the keyboard has a frequency of
earthquake? 880 cycles per second. After how many notes up the
keyboard will this be found?
SOLUTION:
a.
The amplitude of the seismic wave with a Richter SOLUTION:
scale rating of 8 and 5 are 108 and 105 respectively. a.
8 5 Substitute 220 for f in the formula and solve for n.
Divide 10 by 10 .
The scale rating of 8 is 103 or 1000 times greater b.
than the scale rating of 5. Substitute 880 for f in the formula and solve for n.
b.
The amplitudes of San Francisco earthquake and
New Madrid earthquake were 107.8 and 108.1
respectively.
Divide 108.1 by 107.8.
MULTIPLE REPRESENTATIONSIn this
36.
problem, you will explore the graphs shown:
y = log4 x and
The New Madrid earthquake was 100.3 or about 2
times greater than the San Francisco earthquake.
MUSICThe first key on a piano keyboard
35.
corresponds to a pitch with a frequency of 27.5
cycles per second. With every successive key, going
up the black and white keys, the pitch multiplies by a
constant. The formula for the frequency of the pitch
sounded when the nth note up the keyboard is played
a. ANALYTICAL
How do the shapes of the
is given by graphs compare? How do the asymptotes and the x-
intercepts of the graphs compare?
a. A note has a frequency of 220 cycles per second. b. VERBAL
Describe the relationship between the
How many notes up the piano keyboard is this? graphs.
b.
Another pitch on the keyboard has a frequency of c. GRAPHICAL
Use what you know about
880 cycles per second. After how many notes up the transformations of graphs to compare and contrast
keyboard will this be found?
the graph of each function and the graph of y = log4
x.
SOLUTION:
a.
Substitute 220 for f in the formula and solve for n. 1. y = log4 x + 2
2. y = log4 (x + 2)
3. y = 3 log4 x
d. ANALYTICAL
Describe the relationship
between y = log x and y = 1(log x). What are a
4 − 4
b. reasonable domain and range for each function?
e.ANALYTICAL
Substitute 880 for f in the formula and solve for n. Write an equation for a function
for which the graph is the graph of y = log3 x
translated 4 units left and 1 unit up.
SOLUTION:
a.
The shapes of the graphs are the same. The
asymptote for each graph is the y-axis and the x-
intercept for each graph is 1.
b.
MULTIPLE REPRESENTATIONSIn this
36. The graphs are reflections of each other over the x-
problem, you will explore the graphs shown: axis.
y = log4 x and
c.
1. The second graph is the same as the first, except it
is shifted horizontally to the left 2 units.
a. ANALYTICAL
How do the shapes of the
graphs compare? How do the asymptotes and the x-
2, 8] scl: 1 by [ 5, 5] scl: 1
intercepts of the graphs compare? [− −
b. VERBAL
Describe the relationship between the
graphs. 2. The second graph is the same as the first, except it
is shifted vertically up 2 units.
c. GRAPHICAL
Use what you know about
transformations of graphs to compare and contrast
the graph of each function and the graph of y = log4
x.
1. y = log4 x + 2
2. y = log4 (x + 2)
3. y = 3 log x
4
4, 8] scl: 1 by [ 5, 5] scl: 1
d. ANALYTICAL [− −
Describe the relationship
between y = log x and y = 1(log x). What are a
4 − 4 3. Each point on the second graph has a y-coordinate
reasonable domain and range for each function? 3 times that of the corresponding point on the first
e.ANALYTICAL
Write an equation for a function graph.
for which the graph is the graph of y = log3 x
translated 4 units left and 1 unit up.
SOLUTION:
a.
The shapes of the graphs are the same. The
asymptote for each graph is the y-axis and the x-
intercept for each graph is 1.
b.
The graphs are reflections of each other over the x-
axis. 2, 8] scl: 1 by [ 5, 5] scl: 1
[− −
c. d.
1. The second graph is the same as the first, except it The graphs are reflections of each other over the x-
is shifted horizontally to the left 2 units. axis.
D={x|x>0}; R={all real numbers}
e. where h is the horizontal
shift and k is the vertical shift. Since there is a
horizontal shift of 4 and vertical shift of 1, h = 4 and
k = 1.
y = log3 (x + 4) + 1
SOUNDThe relationship between the intensity of
37.
sound I and the number of decibels
β
2, 8] scl: 1 by [ 5, 5] scl: 1
[− − is , where I is the intensity of
2. The second graph is the same as the first, except it sound in watts per square meter.
is shifted vertically up 2 units.
a. Find the number of decibels of a sound with an
intensity of 1 watt per square meter.
b. Find the number of decibels of sound with an
−2
intensity of 10 watts per square meter.
c. The intensity of the sound of 1 watt per square
meter is 100 times as much as the intensity of 10−2
watts per square meter. Why are the decibels of
sound not 100 times as great?
4, 8] scl: 1 by [ 5, 5] scl: 1 SOLUTION:
[− − a.
Substitute 1 for I in the given equation and solve for
3. Each point on the second graph has a y-coordinate .
3 times that of the corresponding point on the first β
graph.
b.
−2
2, 8] scl: 1 by [ 5, 5] scl: 1 Substitute 10 for I in the given equation and solve
[− −
d. for .
The graphs are reflections of each other over the x- β
axis.
D={x|x>0}; R={all real numbers}
e. where h is the horizontal
shift and k is the vertical shift. Since there is a
horizontal shift of 4 and vertical shift of 1, h = 4 and
k = 1.
y = log3 (x + 4) + 1
SOUNDThe relationship between the intensity of
37.
sound I and the number of decibels c. Sample answer: The power of the logarithm only
β changes by 2. The power is the answer to the
is , where I is the intensity of logarithm. That 2 is multiplied by the 10 before the
logarithm. So we expect the decibels to change by
sound in watts per square meter. 20.
CCSS CRITIQUE
a. Find the number of decibels of a sound with an 38. Ryan and Heather are solving
log x 3. Is either of them correct? Explain your
intensity of 1 watt per square meter. 3 ≥−
b. Find the number of decibels of sound with an reasoning.
−2
intensity of 10 watts per square meter.
c. The intensity of the sound of 1 watt per square
meter is 100 times as much as the intensity of 10−2
watts per square meter. Why are the decibels of
sound not 100 times as great?
SOLUTION:
a.
Substitute 1 for I in the given equation and solve for
.
β
SOLUTION:
Sample answer: Ryan; Heather did not need to
switch the inequality symbol when raising to a
negative power.
CHALLENGEFind log 27 +log 27 +log 27 +
b. 39. 3 9 27
Substitute 10−2 for I in the given equation and solve
log81 27 + log243 27.
for .
β
SOLUTION:
REASONINGThe Property of Inequality for
40.
c. Sample answer: The power of the logarithm only
Logarithmic Functions states that when b > 1, logb x
changes by 2. The power is the answer to the
logarithm. That 2 is multiplied by the 10 before the >logby if and only if x >y. What is the case for
logarithm. So we expect the decibels to change by when 0 < b < 1? Explain your reasoning.
20.
SOLUTION:
CCSS CRITIQUE Sample answer: When 0 < b < 1, log x > log y if
38. Ryan and Heather are solving b b
log x 3. Is either of them correct? Explain your
3 ≥−
and only if x 1, logb x
SOLUTION:
>logby if and only if x >y. What is the case for a. less than
when 0 < b < 1? Explain your reasoning. b. less than
c. no
SOLUTION: d. infinitely many
Sample answer: When 0 < b < 1, logb x > logby if
and only if x 81
SOLUTION:
51. log6 216
SOLUTION:
log 27 4a + 6 a
52. 4 16
3 57. ≤
SOLUTION:
SOLUTION:
53.
SOLUTION:
58.
SOLUTION:
log 2401
54. 7
SOLUTION:
Solve each equation or inequality. Check your
59.
solution.
2x + 3
5 125
55. ≤
SOLUTION:
SOLUTION:
3x − 2
56. 3 >81
SOLUTION:
60.
SOLUTION:
4a + 6 a
4 16
57. ≤
SOLUTION:
SHIPPING The height of a shipping cylinder is 4
61.
feet more than the radius. If the volume of the
cylinder is 5 cubic feet, how tall is it? Use the
π
2
formula V = πr h.
SOLUTION:
Substitute 5 for V and r + 4 for h in the formula and
π
simplify.
58.
SOLUTION:
The equation has one real root r = 1.
Thus, the height of the shipping cylinder is 1 + 4 = 5
ft.
NUMBER THEORYTwo complex conjugate
62.
numbers have a sum of 12 and a product of 40. Find
the two numbers
59.
SOLUTION:
SOLUTION:
The equations that represent the situation are:
Solve equation (1).
60.
SOLUTION:
Solve equation (2).
SHIPPING The height of a shipping cylinder is 4 Thus, the two numbers are 6 + 2i and 6 2i.
61. –
feet more than the radius. If the volume of the Simplify. Assume that no variable equals zero.
cylinder is 5 cubic feet, how tall is it? Use the
π 5 3
2 x x
63. ·
formula V = πr h.
SOLUTION:
SOLUTION:
Substitute 5 for V and r + 4 for h in the formula and
π
simplify.
a2 a6
64. ·
SOLUTION:
The equation has one real root r = 1.
Thus, the height of the shipping cylinder is 1 + 4 = 5
ft.
(2p2n)3
65.
NUMBER THEORYTwo complex conjugate
62.
numbers have a sum of 12 and a product of 40. Find
the two numbers SOLUTION:
SOLUTION:
The equations that represent the situation are: (3b3c2)2
66.
SOLUTION:
Solve equation (1).
67.
SOLUTION:
Solve equation (2).
68.
SOLUTION:
Thus, the two numbers are 6 + 2i and 6 2i.
–
Simplify. Assume that no variable equals zero.
x5 x3
63. ·
SOLUTION:
a2 a6
64. ·
SOLUTION:
(2p2n)3
65.
SOLUTION:
(3b3c2)2
66.
SOLUTION:
67.
SOLUTION:
68.
SOLUTION:
Solve each equation.
1.
SOLUTION:
2.
SOLUTION:
2
MULTIPLE CHOICE
3. Solve log5 (x − 10) = log5
3x.
A 10
B
2
C
5
D 2, 5
SOLUTION:
Solve each equation.
1.
SOLUTION:
Substitute each value into the original equation.
The domain of a logarithmic function cannot be 0, so
log5 (–6) is undefined and –2 is an extraneous
solution.
2.
C is the correct option.
SOLUTION:
Solve each inequality.
4. log5 x > 3
SOLUTION:
2 Thus, solution set is {x | x > 125}.
MULTIPLE CHOICE
3. Solve log5 (x − 10) = log5
3x.
A 10 5. log8 x ≤ −2
B
2
C SOLUTION:
5
D 2, 5
SOLUTION:
Thus, solution set is .
6. log4 (2x + 5) ≤ log4 (4x − 3)
SOLUTION:
Substitute each value into the original equation.
Thus, solution set is {x | x ≥ 4}.
The domain of a logarithmic function cannot be 0, so 7. log8 (2x) > log8 (6x − 8)
log5 (–6) is undefined and –2 is an extraneous
solution. SOLUTION:
C is the correct option.
Solve each inequality.
4. log5 x > 3
Exclude all values of x for which
SOLUTION:
So,
Thus, solution set is {x | x > 125}.
Thus, solution set is .
5. log8 x ≤ −2
CCSS STRUCTURE Solve each equation.
SOLUTION:
8.
SOLUTION:
Thus, solution set is .
6. log4 (2x + 5) ≤ log4 (4x − 3)
SOLUTION:
9.
SOLUTION:
Thus, solution set is {x | x ≥ 4}.
7. log8 (2x) > log8 (6x − 8)
SOLUTION:
10.
Exclude all values of x for which SOLUTION:
So,
Thus, solution set is .
CCSS STRUCTURE Solve each equation.
8.
11.
SOLUTION:
SOLUTION:
12.
9.
SOLUTION:
SOLUTION:
13.
10.
SOLUTION:
SOLUTION:
7-4 Solving Logarithmic Equations and Inequalities
2
14. log3 (3x + 8) = log3 (x + x)
11.
SOLUTION:
SOLUTION:
12.
Substitute each value into the original equation.
SOLUTION:
Thus, x = –2 or 4.
2
15. log12 (x − 7) = log12 (x + 5)
SOLUTION:
13.
SOLUTION:
Substitute each value into the original equation.
2 Thus, x = –3 or 4.
14. log3 (3x + 8) = log3 (x + x)
2
16. log6 (x − 6x) = log6 (−8)
SOLUTION:
SOLUTION:
eSolutions Manual - Powered by Cognero Page3
Substitute each value into the original equation.
Substitute each value into the original equation.
Thus, x = –2 or 4.
2 log (–8) is undefined, so 4 and 2 are extraneous
15. log12 (x − 7) = log12 (x + 5) 6
solutions.
SOLUTION: Thus, no solution.
2
17. log9 (x − 4x) = log9 (3x − 10)
SOLUTION:
Substitute each value into the original equation.
Substitute each value into the original equation.
Thus, x = –3 or 4.
2
16. log6 (x − 6x) = log6 (−8)
SOLUTION: log (–4) is undefined and 2 is extraneous solution.
9
Thus, x = 5.
2
18. log4 (2x + 1) = log4 (10x − 7)
SOLUTION:
Substitute each value into the original equation.
log (–8) is undefined, so 4 and 2 are extraneous Substitute each value into the original equation.
6
solutions.
Thus, no solution.
2
17. log9 (x − 4x) = log9 (3x − 10) Thus, x = 1 or 4.
SOLUTION: 2
19. log7 (x − 4) = log7 (− x + 2)
SOLUTION:
Substitute each value into the original equation.
Substitute each value into the original equation.
log9 (–4) is undefined and 2 is extraneous solution.
Thus, x = 5.
2 Since you can not have a log of 0, x = 3 is the
18. log4 (2x + 1) = log4 (10x − 7) solution.
SOLUTION:
SCIENCEThe equation for wind speed w, in
miles per hour, near the center of a tornado is w
= 93 log d + 65, where d is the distance in
10
miles that the tornado travels.
20. Write this equation in exponential form.
SOLUTION:
Substitute each value into the original equation.
21. In May of 1999, a tornado devastated Oklahoma City
with the fastest wind speed ever recorded. If the
Thus, x = 1 or 4. tornado traveled 525 miles, estimate the wind speed
2 near the center of the tornado.
19. log7 (x − 4) = log7 (− x + 2)
SOLUTION:
Substitute 525 for d in the equation and simplify.
SOLUTION:
Solve each inequality.
22. log6 x < −3
SOLUTION:
Substitute each value into the original equation.
The solution set is .
Since you can not have a log of 0, x = 3 is the
solution. 23. log4 x ≥ 4
SOLUTION:
SCIENCEThe equation for wind speed w, in
miles per hour, near the center of a tornado is w
= 93 log d + 65, where d is the distance in
10
miles that the tornado travels.
20. Write this equation in exponential form.
The solution set is {x | x ≥ 256}.
SOLUTION:
24. log3 x ≥ −4
SOLUTION:
21. In May of 1999, a tornado devastated Oklahoma City
with the fastest wind speed ever recorded. If the The solution set is .
tornado traveled 525 miles, estimate the wind speed
near the center of the tornado.
25. log2 x ≤ −2
SOLUTION:
Substitute 525 for d in the equation and simplify. SOLUTION:
Solve each inequality.
22. log6 x < −3 The solution set is .
SOLUTION:
26. log5 x > 2
SOLUTION:
The solution set is .
The solution set is .
23. log4 x ≥ 4
27. log7 x < −1
SOLUTION:
SOLUTION:
The solution set is {x | x ≥ 256}.
24. log3 x ≥ −4 The solution set is .
SOLUTION:
28. log2 (4x − 6) > log2 (2x + 8)
SOLUTION:
The solution set is .
25. log2 x ≤ −2 The solution set is .
SOLUTION:
29. log7 (x + 2) ≥ log7 (6x − 3)
SOLUTION:
The solution set is .
26. log5 x > 2 Exclude all values of x for which
SOLUTION:
So,
The solution set is .
The solution set is .
27. log7 x < −1
30. log3 (7x – 6) < log3 (4x + 9)
SOLUTION:
SOLUTION:
The solution set is .
Exclude all values of x for which
28. log2 (4x − 6) > log2 (2x + 8)
SOLUTION: So,
The solution set is .
31. log5 (12x + 5) ≤ log5 (8x + 9)
The solution set is .
SOLUTION:
29. log7 (x + 2) ≥ log7 (6x − 3)
SOLUTION:
Exclude all values of x for which
Exclude all values of x for which So,
So,
The solution set is .
The solution set is . 32. log11 (3x − 24) ≥ log11 (−5x − 8)
SOLUTION:
30. log3 (7x – 6) < log3 (4x + 9)
SOLUTION:
The solution set is {x | x ≥ 2}.
33. log9 (9x + 4) ≤ log9 (11x − 12)
Exclude all values of x for which
SOLUTION:
So,
The solution set is .
The solution set is {x | x ≥ 8}.
34. CCSS MODELINGThe magnitude of an
31. log5 (12x + 5) ≤ log5 (8x + 9) earthquake is measured on a logarithmic scale called
SOLUTION:
the Richter scale. The magnitude M is given by M =
log10 x, where x represents the amplitude of the
seismic wave causing ground motion.
a. How many times as great is the amplitude caused
by an earthquake with a Richter scale rating of 8 as
an aftershock with a Richter scale rating of 5?
Exclude all values of x for which b. In 1906, San Francisco was almost completely
destroyed by a 7.8 magnitude earthquake. In 1911,
an earthquake estimated at magnitude 8.1 occurred
along the New Madrid fault in the Mississippi River
So, Valley. How many times greater was the New
Madrid earthquake than the San Francisco
earthquake?
The solution set is . SOLUTION:
a.
The amplitude of the seismic wave with a Richter
8 5
32. log11 (3x − 24) ≥ log11 (−5x − 8) scale rating of 8 and 5 are 10 and 10 respectively.
Divide 108 by 105.
SOLUTION:
The scale rating of 8 is 103 or 1000 times greater
The solution set is {x | x ≥ 2}. than the scale rating of 5.
33. log9 (9x + 4) ≤ log9 (11x − 12) b.
The amplitudes of San Francisco earthquake and
SOLUTION: New Madrid earthquake were 107.8 and 108.1
respectively.
Divide 108.1 by 107.8.
The solution set is {x | x ≥ 8}.
34. CCSS MODELINGThe magnitude of an
earthquake is measured on a logarithmic scale called
The New Madrid earthquake was 100.3 or about 2
the Richter scale. The magnitude M is given by M = times greater than the San Francisco earthquake.
log10 x, where x represents the amplitude of the
seismic wave causing ground motion.
MUSICThe first key on a piano keyboard
a. How many times as great is the amplitude caused 35.
by an earthquake with a Richter scale rating of 8 as corresponds to a pitch with a frequency of 27.5
an aftershock with a Richter scale rating of 5? cycles per second. With every successive key, going
b. In 1906, San Francisco was almost completely up the black and white keys, the pitch multiplies by a
destroyed by a 7.8 magnitude earthquake. In 1911, constant. The formula for the frequency of the pitch
an earthquake estimated at magnitude 8.1 occurred sounded when the nth note up the keyboard is played
along the New Madrid fault in the Mississippi River is given by
Valley. How many times greater was the New a. A note has a frequency of 220 cycles per second.
Madrid earthquake than the San Francisco How many notes up the piano keyboard is this?
earthquake? b. Another pitch on the keyboard has a frequency of
SOLUTION: 880 cycles per second. After how many notes up the
a. keyboard will this be found?
The amplitude of the seismic wave with a Richter
scale rating of 8 and 5 are 108 and 105 respectively. SOLUTION:
8 5 a.
Divide 10 by 10 . Substitute 220 for f in the formula and solve for n.
The scale rating of 8 is 103 or 1000 times greater
than the scale rating of 5. b.
Substitute 880 for f in the formula and solve for n.
b.
The amplitudes of San Francisco earthquake and
7.8 8.1
New Madrid earthquake were 10 and 10
respectively.
Divide 108.1 by 107.8.
MULTIPLE REPRESENTATIONSIn this
36.
problem, you will explore the graphs shown:
y = log4 x and
0.3
The New Madrid earthquake was 10 or about 2
times greater than the San Francisco earthquake.
MUSICThe first key on a piano keyboard
35.
corresponds to a pitch with a frequency of 27.5
cycles per second. With every successive key, going
up the black and white keys, the pitch multiplies by a
constant. The formula for the frequency of the pitch
sounded when the nth note up the keyboard is played
a. ANALYTICAL
is given by How do the shapes of the
graphs compare? How do the asymptotes and the x-
a. A note has a frequency of 220 cycles per second. intercepts of the graphs compare?
How many notes up the piano keyboard is this? b. VERBAL
Describe the relationship between the
b. Another pitch on the keyboard has a frequency of graphs.
880 cycles per second. After how many notes up the
c. GRAPHICAL
keyboard will this be found? Use what you know about
transformations of graphs to compare and contrast
the graph of each function and the graph of y = log4
SOLUTION:
a. x.
Substitute 220 for f in the formula and solve for n.
1. y = log4 x + 2
2. y = log4 (x + 2)
3. y = 3 log4 x
d. ANALYTICAL
Describe the relationship
between y = log x and y = 1(log x). What are a
b. 4 − 4
Substitute 880 for f in the formula and solve for n. reasonable domain and range for each function?
e.ANALYTICAL
Write an equation for a function
for which the graph is the graph of y = log x
3
translated 4 units left and 1 unit up.
SOLUTION:
a.
The shapes of the graphs are the same. The
asymptote for each graph is the y-axis and the x-
intercept for each graph is 1.
MULTIPLE REPRESENTATIONSIn this
36. b.
problem, you will explore the graphs shown: The graphs are reflections of each other over the x-
y = log4 x and axis.
c.
1. The second graph is the same as the first, except it
is shifted horizontally to the left 2 units.
a. ANALYTICAL
How do the shapes of the
graphs compare? How do the asymptotes and the x-
intercepts of the graphs compare?
2, 8] scl: 1 by [ 5, 5] scl: 1
b. VERBAL [− −
Describe the relationship between the
graphs.
c. GRAPHICAL 2. The second graph is the same as the first, except it
Use what you know about
transformations of graphs to compare and contrast is shifted vertically up 2 units.
the graph of each function and the graph of y = log4
x.
1. y = log4 x + 2
2. y = log4 (x + 2)
3. y = 3 log4 x
d. ANALYTICAL
Describe the relationship
4, 8] scl: 1 by [ 5, 5] scl: 1
[− −
between y = log x and y = 1(log x). What are a
4 − 4
reasonable domain and range for each function?
3. Each point on the second graph has a y-coordinate
e.ANALYTICAL
Write an equation for a function 3 times that of the corresponding point on the first
for which the graph is the graph of y = log3 x graph.
translated 4 units left and 1 unit up.
SOLUTION:
a.
The shapes of the graphs are the same. The
asymptote for each graph is the y-axis and the x-
intercept for each graph is 1.
b.
The graphs are reflections of each other over the x-
axis.
c. 2, 8] scl: 1 by [ 5, 5] scl: 1
[− −
1. The second graph is the same as the first, except it d.
is shifted horizontally to the left 2 units. The graphs are reflections of each other over the x-
axis.
D={x|x>0}; R={all real numbers}
e. where h is the horizontal
shift and k is the vertical shift. Since there is a
horizontal shift of 4 and vertical shift of 1, h = 4 and
k = 1.
y = log3 (x + 4) + 1
SOUNDThe relationship between the intensity of
37.
sound I and the number of decibels
2, 8] scl: 1 by [ 5, 5] scl: 1 β
[− −
is , where I is the intensity of
2. The second graph is the same as the first, except it
is shifted vertically up 2 units. sound in watts per square meter.
a. Find the number of decibels of a sound with an
intensity of 1 watt per square meter.
b. Find the number of decibels of sound with an
−2
intensity of 10 watts per square meter.
c. The intensity of the sound of 1 watt per square
meter is 100 times as much as the intensity of 10−2
watts per square meter. Why are the decibels of
sound not 100 times as great?
4, 8] scl: 1 by [ 5, 5] scl: 1
[− −
SOLUTION:
a.
3. Each point on the second graph has a y-coordinate Substitute 1 for I in the given equation and solve for
3 times that of the corresponding point on the first .
graph. β
b.
2, 8] scl: 1 by [ 5, 5] scl: 1
[− − −2
d. Substitute 10 for I in the given equation and solve
The graphs are reflections of each other over the x- for .
axis. β
D={x|x>0}; R={all real numbers}
e. where h is the horizontal
shift and k is the vertical shift. Since there is a
horizontal shift of 4 and vertical shift of 1, h = 4 and
k = 1.
y = log3 (x + 4) + 1
SOUNDThe relationship between the intensity of
37.
sound I and the number of decibels
β
c. Sample answer: The power of the logarithm only
is , where I is the intensity of changes by 2. The power is the answer to the
logarithm. That 2 is multiplied by the 10 before the
sound in watts per square meter. logarithm. So we expect the decibels to change by
20.
a. Find the number of decibels of a sound with an
CCSS CRITIQUE
intensity of 1 watt per square meter. 38. Ryan and Heather are solving
log x 3. Is either of them correct? Explain your
b. Find the number of decibels of sound with an 3 ≥−
−2 reasoning.
intensity of 10 watts per square meter.
c. The intensity of the sound of 1 watt per square
meter is 100 times as much as the intensity of 10−2
watts per square meter. Why are the decibels of
sound not 100 times as great?
SOLUTION:
a.
Substitute 1 for I in the given equation and solve for
.
β
SOLUTION:
Sample answer: Ryan; Heather did not need to
switch the inequality symbol when raising to a
negative power.
b.
−2
CHALLENGEFind log 27 +log 27 +log 27 +
Substitute 10 for I in the given equation and solve 39. 3 9 27
for .
β log81 27 + log243 27.
SOLUTION:
c.
Sample answer: The power of the logarithm only REASONINGThe Property of Inequality for
40.
changes by 2. The power is the answer to the
logarithm. That 2 is multiplied by the 10 before the Logarithmic Functions states that when b > 1, logb x
logarithm. So we expect the decibels to change by >logby if and only if x >y. What is the case for
20.
when 0 < b < 1? Explain your reasoning.
SOLUTION:
CCSS CRITIQUE
38. Ryan and Heather are solving
log x 3. Is either of them correct? Explain your Sample answer: When 0 < b < 1, log x > log y if
3 ≥− b b
reasoning.
and only if x 1, logb x (s) for b in the equation y = logb 1.
>logby if and only if x >y. What is the case for
SOLUTION:
when 0 < b < 1? Explain your reasoning. a. less than
b. less than
SOLUTION:
c. no
Sample answer: When 0 < b < 1, logb x > logby if d. infinitely many
and only if x 81
SOLUTION:
51. log6 216
SOLUTION:
log 27
52. 3
4a + 6 a
4 16
57. ≤
SOLUTION:
SOLUTION:
53.
SOLUTION:
58.
SOLUTION:
log 2401
54. 7
SOLUTION:
Solve each equation or inequality. Check your
solution.
2x + 3
5 125 59.
55. ≤
SOLUTION:
SOLUTION:
3x − 2
56. 3 >81
SOLUTION:
60.
SOLUTION:
4a + 6 a
4 16
57. ≤
SOLUTION:
SHIPPINGThe height of a shipping cylinder is 4
61.
feet more than the radius. If the volume of the
cylinder is 5 cubic feet, how tall is it? Use the
π
2
formula V = πr h.
SOLUTION:
Substitute 5 for V and r + 4 for h in the formula and
π
58. simplify.
SOLUTION:
The equation has one real root r = 1.
Thus, the height of the shipping cylinder is 1 + 4 = 5
ft.
NUMBER THEORYTwo complex conjugate
62.
59. numbers have a sum of 12 and a product of 40. Find
the two numbers
SOLUTION:
SOLUTION:
The equations that represent the situation are:
Solve equation (1).
60.
SOLUTION:
Solve equation (2).
SHIPPING The height of a shipping cylinder is 4
61.
feet more than the radius. If the volume of the Thus, the two numbers are 6 + 2i and 6 2i.
–
cylinder is 5π cubic feet, how tall is it? Use the Simplify. Assume that no variable equals zero.
2
formula V = πr h. 5 3
63. x · x
SOLUTION:
Substitute 5 for V and r + 4 for h in the formula and SOLUTION:
π
simplify.
a2 a6
64. ·
The equation has one real root r = 1. SOLUTION:
Thus, the height of the shipping cylinder is 1 + 4 = 5
ft.
NUMBER THEORYTwo complex conjugate
62. (2p2n)3
numbers have a sum of 12 and a product of 40. Find 65.
the two numbers
SOLUTION:
SOLUTION:
The equations that represent the situation are:
(3b3c2)2
66.
SOLUTION:
Solve equation (1).
67.
SOLUTION:
Solve equation (2).
68.
SOLUTION:
Thus, the two numbers are 6 + 2i and 6 2i.
–
Simplify. Assume that no variable equals zero.
x5 x3
63. ·
SOLUTION:
a2 a6
64. ·
SOLUTION:
(2p2n)3
65.
SOLUTION:
(3b3c2)2
66.
SOLUTION:
67.
SOLUTION:
68.
SOLUTION:
Solve each equation.
1.
SOLUTION:
2.
SOLUTION:
2
MULTIPLE CHOICE
3. Solve log5 (x − 10) = log5
3x.
A 10
B
2
C
5
D 2, 5
SOLUTION:
Solve each equation. Substitute each value into the original equation.
1.
SOLUTION:
The domain of a logarithmic function cannot be 0, so
log5 (–6) is undefined and –2 is an extraneous
solution.
C is the correct option.
Solve each inequality.
4. log5 x > 3
2.
SOLUTION:
SOLUTION:
Thus, solution set is {x | x > 125}.
5. log8 x ≤ −2
SOLUTION:
2
MULTIPLE CHOICE
3. Solve log5 (x − 10) = log5
3x.
A 10
B
2
C
5
D 2, 5 Thus, solution set is .
SOLUTION:
6. log4 (2x + 5) ≤ log4 (4x − 3)
SOLUTION:
Thus, solution set is {x | x ≥ 4}.
Substitute each value into the original equation.
7. log8 (2x) > log8 (6x − 8)
SOLUTION:
The domain of a logarithmic function cannot be 0, so
log5 (–6) is undefined and –2 is an extraneous
solution.
Exclude all values of x for which
C is the correct option.
Solve each inequality. So,
4. log5 x > 3
SOLUTION:
Thus, solution set is .
CCSS STRUCTURE Solve each equation.
8.
Thus, solution set is {x | x > 125}.
SOLUTION:
5. log8 x ≤ −2
SOLUTION:
Thus, solution set is .
9.
6. log4 (2x + 5) ≤ log4 (4x − 3)
SOLUTION:
SOLUTION:
Thus, solution set is {x | x ≥ 4}.
7. log8 (2x) > log8 (6x − 8)
10.
SOLUTION:
SOLUTION:
Exclude all values of x for which
So,
11.
Thus, solution set is .
SOLUTION:
CCSS STRUCTURE Solve each equation.
8.
SOLUTION:
12.
SOLUTION:
9.
SOLUTION:
13.
SOLUTION:
10.
SOLUTION:
2
14. log3 (3x + 8) = log3 (x + x)
SOLUTION:
11.
SOLUTION:
Substitute each value into the original equation.
12.
Thus, x = –2 or 4.
SOLUTION: 2
15. log12 (x − 7) = log12 (x + 5)
SOLUTION:
13.
Substitute each value into the original equation.
SOLUTION:
Thus, x = –3 or 4.
2
16. log6 (x − 6x) = log6 (−8)
SOLUTION:
2
14. log3 (3x + 8) = log3 (x + x)
SOLUTION:
Substitute each value into the original equation.
Substitute each value into the original equation.
log6 (–8) is undefined, so 4 and 2 are extraneous
solutions.
Thus, no solution.
Thus, x = –2 or 4. 2
17. log9 (x − 4x) = log9 (3x − 10)
2
15. log (x −7) =log (x + 5)
12 12 SOLUTION:
SOLUTION:
Substitute each value into the original equation.
Substitute each value into the original equation.
log9 (–4) is undefined and 2 is extraneous solution.
7-4 Solving Logarithmic Equations and Inequalities
Thus, x = –3 or 4. Thus, x = 5.
2 2
16. log6 (x − 6x) = log6 (−8) 18. log4 (2x + 1) = log4 (10x − 7)
SOLUTION: SOLUTION:
Substitute each value into the original equation. Substitute each value into the original equation.
log (–8) is undefined, so 4 and 2 are extraneous Thus, x = 1 or 4.
6
solutions. 2
Thus, no solution. 19. log7 (x − 4) = log7 (− x + 2)
2 SOLUTION:
17. log9 (x − 4x) = log9 (3x − 10)
SOLUTION:
Substitute each value into the original equation.
Substitute each value into the original equation.
Since you can not have a log of 0, x = 3 is the
solution.
log9 (–4) is undefined and 2 is extraneous solution. SCIENCEThe equation for wind speed w, in
Thus, x = 5. miles per hour, near the center of a tornado is w
= 93 log d + 65, where d is the distance in
2 10
miles that the tornado travels.
18. log4 (2x + 1) = log4 (10x − 7) 20. Write this equation in exponential form.
SOLUTION:
SOLUTION:
eSolutions Manual - Powered by Cognero Page4
21. In May of 1999, a tornado devastated Oklahoma City
Substitute each value into the original equation. with the fastest wind speed ever recorded. If the
tornado traveled 525 miles, estimate the wind speed
near the center of the tornado.
SOLUTION:
Substitute 525 for d in the equation and simplify.
Thus, x = 1 or 4.
2
19. log7 (x − 4) = log7 (− x + 2)
Solve each inequality.
SOLUTION:
22. log6 x < −3
SOLUTION:
The solution set is .
Substitute each value into the original equation.
23. log4 x ≥ 4
SOLUTION:
Since you can not have a log of 0, x = 3 is the
solution.
The solution set is {x | x ≥ 256}.
SCIENCEThe equation for wind speed w, in
miles per hour, near the center of a tornado is w
= 93 log d + 65, where d is the distance in 24. log3 x ≥ −4
10
miles that the tornado travels.
20. Write this equation in exponential form. SOLUTION:
SOLUTION:
The solution set is .
25. log2 x ≤ −2
21. In May of 1999, a tornado devastated Oklahoma City SOLUTION:
with the fastest wind speed ever recorded. If the
tornado traveled 525 miles, estimate the wind speed
near the center of the tornado.
SOLUTION:
Substitute 525 for d in the equation and simplify. The solution set is .
26. log5 x > 2
Solve each inequality. SOLUTION:
22. log6 x < −3
SOLUTION:
The solution set is .
27. log7 x < −1
The solution set is .
SOLUTION:
23. log4 x ≥ 4
SOLUTION:
The solution set is .
28. log2 (4x − 6) > log2 (2x + 8)
The solution set is {x | x ≥ 256}.
SOLUTION:
24. log3 x ≥ −4
SOLUTION:
The solution set is .
The solution set is . 29. log7 (x + 2) ≥ log7 (6x − 3)
SOLUTION:
25. log2 x ≤ −2
SOLUTION:
Exclude all values of x for which
The solution set is .
So,
26. log5 x > 2
SOLUTION: The solution set is .
30. log3 (7x – 6) < log3 (4x + 9)
SOLUTION:
The solution set is .
27. log7 x < −1
SOLUTION:
Exclude all values of x for which
So,
The solution set is .
The solution set is .
28. log2 (4x − 6) > log2 (2x + 8)
SOLUTION:
31. log5 (12x + 5) ≤ log5 (8x + 9)
SOLUTION:
The solution set is .
29. log7 (x + 2) ≥ log7 (6x − 3) Exclude all values of x for which
SOLUTION:
So,
The solution set is .
Exclude all values of x for which
32. log11 (3x − 24) ≥ log11 (−5x − 8)
SOLUTION:
So,
The solution set is .
30. log (7x – 6) < log (4x + 9)
3 3 The solution set is {x | x ≥ 2}.
SOLUTION:
33. log9 (9x + 4) ≤ log9 (11x − 12)
SOLUTION:
Exclude all values of x for which
The solution set is {x | x ≥ 8}.
So,
34. CCSS MODELINGThe magnitude of an
earthquake is measured on a logarithmic scale called
the Richter scale. The magnitude M is given by M =
The solution set is . log10 x, where x represents the amplitude of the
seismic wave causing ground motion.
a. How many times as great is the amplitude caused
31. log5 (12x + 5) ≤ log5 (8x + 9) by an earthquake with a Richter scale rating of 8 as
an aftershock with a Richter scale rating of 5?
SOLUTION: b. In 1906, San Francisco was almost completely
destroyed by a 7.8 magnitude earthquake. In 1911,
an earthquake estimated at magnitude 8.1 occurred
along the New Madrid fault in the Mississippi River
Valley. How many times greater was the New
Madrid earthquake than the San Francisco
Exclude all values of x for which earthquake?
SOLUTION:
a.
So, The amplitude of the seismic wave with a Richter
scale rating of 8 and 5 are 108 and 105 respectively.
Divide 108 by 105.
The solution set is .
32. log11 (3x − 24) ≥ log11 (−5x − 8)
SOLUTION: The scale rating of 8 is 103 or 1000 times greater
than the scale rating of 5.
b.
The amplitudes of San Francisco earthquake and
New Madrid earthquake were 107.8 and 108.1
respectively.
The solution set is {x | x ≥ 2}.
8.1 7.8
33. log9 (9x + 4) ≤ log9 (11x − 12) Divide 10 by 10 .
SOLUTION:
The New Madrid earthquake was 100.3 or about 2
times greater than the San Francisco earthquake.
The solution set is {x | x ≥ 8}.
MUSICThe first key on a piano keyboard
35.
34. CCSS MODELINGThe magnitude of an corresponds to a pitch with a frequency of 27.5
earthquake is measured on a logarithmic scale called cycles per second. With every successive key, going
up the black and white keys, the pitch multiplies by a
the Richter scale. The magnitude M is given by M = constant. The formula for the frequency of the pitch
log10 x, where x represents the amplitude of the sounded when the nth note up the keyboard is played
seismic wave causing ground motion.
a. How many times as great is the amplitude caused is given by
by an earthquake with a Richter scale rating of 8 as a. A note has a frequency of 220 cycles per second.
an aftershock with a Richter scale rating of 5? How many notes up the piano keyboard is this?
b. In 1906, San Francisco was almost completely b. Another pitch on the keyboard has a frequency of
destroyed by a 7.8 magnitude earthquake. In 1911, 880 cycles per second. After how many notes up the
an earthquake estimated at magnitude 8.1 occurred keyboard will this be found?
along the New Madrid fault in the Mississippi River
Valley. How many times greater was the New
Madrid earthquake than the San Francisco SOLUTION:
earthquake? a.
Substitute 220 for f in the formula and solve for n.
SOLUTION:
a.
The amplitude of the seismic wave with a Richter
scale rating of 8 and 5 are 108 and 105 respectively.
Divide 108 by 105.
b.
Substitute 880 for f in the formula and solve for n.
The scale rating of 8 is 103 or 1000 times greater
than the scale rating of 5.
b.
The amplitudes of San Francisco earthquake and
New Madrid earthquake were 107.8 and 108.1
MULTIPLE REPRESENTATIONSIn this
respectively. 36.
problem, you will explore the graphs shown:
8.1 7.8 y = log4 x and
Divide 10 by 10 .
The New Madrid earthquake was 100.3 or about 2
times greater than the San Francisco earthquake.
MUSICThe first key on a piano keyboard
35. a. ANALYTICAL
corresponds to a pitch with a frequency of 27.5 How do the shapes of the
cycles per second. With every successive key, going graphs compare? How do the asymptotes and the x-
up the black and white keys, the pitch multiplies by a intercepts of the graphs compare?
b. VERBAL
constant. The formula for the frequency of the pitch Describe the relationship between the
sounded when the nth note up the keyboard is played graphs.
c. GRAPHICAL
Use what you know about
is given by transformations of graphs to compare and contrast
a. A note has a frequency of 220 cycles per second. the graph of each function and the graph of y = log4
How many notes up the piano keyboard is this? x.
b. Another pitch on the keyboard has a frequency of
880 cycles per second. After how many notes up the 1. y = log4 x + 2
keyboard will this be found?
2. y = log4 (x + 2)
SOLUTION:
3. y = 3 log4 x
a.
Substitute 220 for f in the formula and solve for n.
d. ANALYTICAL
Describe the relationship
between y = log x and y = 1(log x). What are a
4 − 4
reasonable domain and range for each function?
e.ANALYTICAL
Write an equation for a function
for which the graph is the graph of y = log3 x
translated 4 units left and 1 unit up.
b. SOLUTION:
Substitute 880 for f in the formula and solve for n. a.
The shapes of the graphs are the same. The
asymptote for each graph is the y-axis and the x-
intercept for each graph is 1.
b.
The graphs are reflections of each other over the x-
axis.
c.
1. The second graph is the same as the first, except it
is shifted horizontally to the left 2 units.
MULTIPLE REPRESENTATIONSIn this
36.
problem, you will explore the graphs shown:
y = log4 x and
2, 8] scl: 1 by [ 5, 5] scl: 1
[− −
2. The second graph is the same as the first, except it
a. ANALYTICAL
How do the shapes of the is shifted vertically up 2 units.
graphs compare? How do the asymptotes and the x-
intercepts of the graphs compare?
b. VERBAL
Describe the relationship between the
graphs.
c. GRAPHICAL
Use what you know about
transformations of graphs to compare and contrast
the graph of each function and the graph of y = log4
x.
1. y = log4 x + 2 4, 8] scl: 1 by [ 5, 5] scl: 1
[− −
2. y = log (x + 2)
4
3. Each point on the second graph has a y-coordinate
3. y = 3 log4 x 3 times that of the corresponding point on the first
graph.
d. ANALYTICAL
Describe the relationship
between y = log x and y = 1(log x). What are a
4 − 4
reasonable domain and range for each function?
e.ANALYTICAL
Write an equation for a function
for which the graph is the graph of y = log3 x
translated 4 units left and 1 unit up.
SOLUTION:
a.
The shapes of the graphs are the same. The
2, 8] scl: 1 by [ 5, 5] scl: 1
asymptote for each graph is the y-axis and the x- [− −
intercept for each graph is 1. d.
b. The graphs are reflections of each other over the x-
axis.
The graphs are reflections of each other over the x-
axis. D={x|x>0}; R={all real numbers}
e. where h is the horizontal
c. shift and k
1. The second graph is the same as the first, except it is the vertical shift. Since there is a
is shifted horizontally to the left 2 units. horizontal shift of 4 and vertical shift of 1, h = 4 and
k = 1.
y = log3 (x + 4) + 1
SOUNDThe relationship between the intensity of
37.
sound I and the number of decibels
β
is , where I is the intensity of
sound in watts per square meter.
a. Find the number of decibels of a sound with an
2, 8] scl: 1 by [ 5, 5] scl: 1
[− − intensity of 1 watt per square meter.
b. Find the number of decibels of sound with an
2. The second graph is the same as the first, except it −2
is shifted vertically up 2 units. intensity of 10 watts per square meter.
c. The intensity of the sound of 1 watt per square
−2
meter is 100 times as much as the intensity of 10
watts per square meter. Why are the decibels of
sound not 100 times as great?
SOLUTION:
a.
Substitute 1 for I in the given equation and solve for
.
β
4, 8] scl: 1 by [ 5, 5] scl: 1
[− −
3. Each point on the second graph has a y-coordinate
3 times that of the corresponding point on the first
graph.
b.
Substitute 10−2 for I in the given equation and solve
for .
β
2, 8] scl: 1 by [ 5, 5] scl: 1
[− −
d.
The graphs are reflections of each other over the x-
axis.
D={x|x>0}; R={all real numbers}
e. where h is the horizontal
shift and k
is the vertical shift. Since there is a
horizontal shift of 4 and vertical shift of 1, h = 4 and
c. Sample answer: The power of the logarithm only
k = 1. changes by 2. The power is the answer to the
y = log3 (x + 4) + 1 logarithm. That 2 is multiplied by the 10 before the
logarithm. So we expect the decibels to change by
20.
SOUNDThe relationship between the intensity of
37.
sound I and the number of decibels
β
CCSS CRITIQUE
38. Ryan and Heather are solving
log x 3. Is either of them correct? Explain your
is , where I is the intensity of 3 ≥−
sound in watts per square meter. reasoning.
a. Find the number of decibels of a sound with an
intensity of 1 watt per square meter.
b. Find the number of decibels of sound with an
−2
intensity of 10 watts per square meter.
c. The intensity of the sound of 1 watt per square
−2
meter is 100 times as much as the intensity of 10
watts per square meter. Why are the decibels of
sound not 100 times as great?
SOLUTION:
a.
Substitute 1 for I in the given equation and solve for
.
β
SOLUTION:
Sample answer: Ryan; Heather did not need to
switch the inequality symbol when raising to a
negative power.
CHALLENGEFind log 27 +log 27 +log 27 +
39. 3 9 27
log81 27 + log243 27.
SOLUTION:
b.
Substitute 10−2 for I in the given equation and solve
for .
β
REASONINGThe Property of Inequality for
40.
Logarithmic Functions states that when b > 1, logb x
>logby if and only if x >y. What is the case for
when 0 < b < 1? Explain your reasoning.
SOLUTION:
c. Sample answer: The power of the logarithm only
changes by 2. The power is the answer to the Sample answer: When 0 < b < 1, logb x > logby if
logarithm. That 2 is multiplied by the 10 before the
logarithm. So we expect the decibels to change by and only if x 1, logb x d. infinitely many
>logby if and only if x >y. What is the case for
WRITING IN MATHExplain why any logarithmic
44.
when 0 < b < 1? Explain your reasoning.
function of the form y = logb x has an x-intercept of
SOLUTION: (1, 0) and no y-intercept.
Sample answer: When 0 < b < 1, logb x > logby if
SOLUTION:
and only if x 81
50.
SOLUTION:
SOLUTION:
log 216
51. 6
4a + 6 a
4 16
SOLUTION: 57. ≤
SOLUTION:
log 27
52. 3
SOLUTION:
58.
53.
SOLUTION:
SOLUTION:
log 2401
54. 7
SOLUTION:
59.
SOLUTION:
Solve each equation or inequality. Check your
solution.
2x + 3
5 125
55. ≤
SOLUTION:
60.
SOLUTION:
3x − 2
56. 3 >81
SOLUTION:
SHIPPING The height of a shipping cylinder is 4
61.
4a + 6 a feet more than the radius. If the volume of the
4 16 cylinder is 5 cubic feet, how tall is it? Use the
57. ≤ π
2
formula V = πr h.
SOLUTION:
SOLUTION:
Substitute 5 for V and r + 4 for h in the formula and
π
simplify.
58. The equation has one real root r = 1.
SOLUTION: Thus, the height of the shipping cylinder is 1 + 4 = 5
ft.
NUMBER THEORYTwo complex conjugate
62.
numbers have a sum of 12 and a product of 40. Find
the two numbers
SOLUTION:
The equations that represent the situation are:
59.
SOLUTION:
Solve equation (1).
Solve equation (2).
60.
SOLUTION:
Thus, the two numbers are 6 + 2i and 6 2i.
–
Simplify. Assume that no variable equals zero.
x5 x3
63. ·
SHIPPING
61. The height of a shipping cylinder is 4 SOLUTION:
feet more than the radius. If the volume of the
cylinder is 5 cubic feet, how tall is it? Use the
π
2
formula V = πr h.
2 6
SOLUTION: a a
Substitute 5 for V and r + 4 for h in the formula and 64. ·
π
simplify. SOLUTION:
(2p2n)3
65.
The equation has one real root r = 1.
Thus, the height of the shipping cylinder is 1 + 4 = 5 SOLUTION:
ft.
NUMBER THEORYTwo complex conjugate
62. (3b3c2)2
numbers have a sum of 12 and a product of 40. Find 66.
the two numbers
SOLUTION:
SOLUTION:
The equations that represent the situation are:
67.
SOLUTION:
Solve equation (1).
68.
Solve equation (2). SOLUTION:
Thus, the two numbers are 6 + 2i and 6 2i.
–
Simplify. Assume that no variable equals zero.
x5 x3
63. ·
SOLUTION:
a2 a6
64. ·
SOLUTION:
(2p2n)3
65.
SOLUTION:
(3b3c2)2
66.
SOLUTION:
67.
SOLUTION:
68.
SOLUTION:
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