Document 1804579

Chapter 7,
5.
continued
y 5 385(1.04)x
9. When P 5 1500, r 5 0.07
550 5 385(1.04)x
n 5 365, and t 5 2:
x
1.43 5 1.04
r nt
A 5 P 1 1 1 }n 2
log 1.43
log 1.04
}5x
9.12 ø x
365.07 730
5 1500 1 }
365 2
The expenditures reached $550 billion in 1996 1 9,
or 2005.
ø 1725.39
A 5 Pe rt
6.
0.07 365 + 2
A 5 1500 1 1 1 }
365 2
The balance after 2 years is $1725.39.
4000 1 1000 5 4000e0.02t
y
10.
1.25 5 e0.02t
y
11.
ln 1.25 5 ln e0.02t
1
0.223 ø 0.02t
1 x
3
1 2
is y 5 5.
Domain: all real numbers
Domain: all real numbers
Range: y > 0
Range: y > 24
12.
2. The decay factor in the model y 5 7.2(0.89) x is 0.89.
base b, and it is equal to x if and only if b x 5 y.
5. y 5 (1.4) is an exponential function because it has the
form y 5 ab .
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
8.
Range: y > 0
Range: y > 0
(
)
(22, 2 )
11
4
(3, 2.72)
(2, 1)
y 5 ex 2 2
21
21
x
x
Domain: all real numbers
(0, 23)
(21, 25)
y
(0, 1)
f(x)
1
Range: y > 0
(1, 2.72)
Domain: all real numbers
f(x) 5 23 ? 4x
Domain: all real numbers
Range: y > 3
y 5 ex
x
Domain: all real numbers
1
Domain: all real numbers
14.
x
3
x
x
1
21
y
1
21, 2 4
21
(1, 1.6)
x
21
1
f(x) 5 2(0.8)x
(0, 2)
x
2
y
(2, 4.6)
4. A logarithm with base e is called a natural logarithm.
7.
13.
f(x)
f(x) 5 2(0.8)x 2 1 1 3
(1, 5)
3. Sample answer: logb y represents the logarithm of y with
21
x
11
3
(0, 23)
1 x11
1. The asymptote of the function y 5 22 }
15
4
6.
1
3
(1, 2 )
y5( ) 24
x
Chapter 7 Review (pp. 539–542)
y
(1, )
2
21
You will earn $1000 in interest in about 11 years.
1 x
3
(0, 1)
1
11.2 ø t
()
y5
Range: y > 0
15.
f(x)
(22, 7)
f(x) 5 e20.4(x 1 2) 1 6
(21, 6.67)
f(x) 5 23 ? 4x 1 1 2 2
Domain: all real numbers
2
Range: y < 22
21
(0, 1)
f(x) 5 e20.4x
(1, 0.67)
x
Domain: all real numbers
Range: y > 6
Algebra 2
Worked-Out Solution Key
409
continued
31. 2 ln 3 1 5 ln 2 2 ln 8 5 ln 32 1 ln 25 2 ln 8
16. When t 5 10: N 5 6.9e20.0695t
20.0695(10)
5 ln 9 1 ln 32 2 ln 8
20.695
5 ln 9 + 32 2 ln 8
N 5 6.9e
N 5 6.9e
N ø 3.44
288
About 3.44 picograms of nitrogen-13 remain after
10 minutes.
32.
17. 35 5 243, so log3 243 5 5.
18. 70 5 1, so log7 1 5 0.
log 32
x5}
5 2.153
log 5
1 23
19. }
5 216, so log1/6 216 5 23.
6
1
1
1
20. 12521/3 5 }, so log125 } 5 2} .
5
3
5
1 2
21.
2.153
y
y 5 log3x
1
21
1
(3, 1)
(1, 0)
x
(3, 23)
(1, 24)
y 5 log3x 2 4
23.
Domain: x > 0
Domain: x > 0
Range: all real numbers
Range: all real numbers
f(x)
f(x) 5 ln(x 2 1) 1 3
(5, 4.39)
(2, 3)
1
21
(4, 1.39)
(1, 0)
5x 5 32
0 32
5
2.153
ø 32, 32 5 32 5
33. log3 (2x 2 5) 5 2
3log3 (2x 2 5) 5 32
2x 2 5 5 9
2x 5 14
x57
Check: log3 (2x 2 5) 5 2
log3(2 + 7 2 5) 0 2
log3 9 0 2
2
3 5 9, log3 9 5 2 34. ln x 1 ln (x 1 2) 5 3
ln [x(x 1 2)] 5 3
eln [x(x 1 2)] 5 e3
x(x 1 2) 5 e3
2
x 1 2x 2 e3 5 0
Check:
22.
y
21
5x 5 32
log55x 5 log5 32
x 5 log5 32
x
}
226Ï 4 2 4(2e3)
226Ï 84.34
ø}
x 5 }}
2
2
x
f(x) 5ln x
}
ø 3.59, 25.59
Domain: x > 1
Range: all real numbers
24. d 5 22.1158 ln a 1 13.669
When a 5 25: d 5 22.1158 ln 25 1 13.669
d ø 22.1158(3.2189) 1 13.669 ø 6.8585
When a 5 50: d 5 22.1158 ln 50 1 13.669
d ø 22.1158(3.9120) 1 13.669 ø 5.3920
The average diameter of a pupil is about 6.8585
millimeters for a 25 year old and and about 5.3920
millimeters for a 50 year old.
25. log8 3xy 5 log8 3 1 log8 x 1 log8 y
26. ln 10x 3y 5 ln 10 1 ln x 3 1 ln y 5 ln 10 1 3 ln x 1 ln y
8
27. log }4 5 log 8 2 log y 4 5 log 8 2 4 log y
y
3y
28. ln }5 5 ln 3y 2 ln x 5 5 ln 3 1 ln y 2 5 ln x
x
29. 3 log 7 4 1 log 7 6 5 log 7 43 1 log 7 6 5 log 7 64 1 log 7 6
5 log 7 64 + 6 5 log 7 384
12
30. ln 12 2 2 ln x 5 ln 12 2 ln x2 5 ln }2
x
410
5 ln 36
5 ln 288 2 ln 8 5 ln }
8
Algebra 2
Worked-Out Solution Key
Check:
ln x 1 ln (x 1 2) 5 3
ln (25.59) 1 ln (25.59 1 2) 0 3
ln (25.59) 1 ln (23.59) 0 3
Because ln (25.59) and ln (23.59) are not defined,
25.59 is not a solution.
ln x 1 ln (x 1 2) 5 3
ln 3.59 1 ln (3.59 1 2) 0 3
ln 3.59 1 ln 5.59 0 3
ln 3.59 + 5.59 0 3
ln 20.07 0 3
3ø3
The solution is about 3.59.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 7,
Chapter 7,
continued
8
35. (3, 8): 8 5 ab 3 l a 5 }3
b
4.673 2 3.332
M: }}
ø 0.75
1.792 2 0
(5, 2): 2 5 ab5
ln y 2 4.673 5 0.75(ln x 2 1.792)
ln y 5 0.75 ln x 1 3.329
1 2
8
2 5 }3 b5
b
2 5 8b
ln y 5 ln x0.75 1 3.329
2
y 5 eln x
1
} 5 b2
4
0.75 1 3.329
y 5 e3.329 + eln x
y 5 27.91x0.75
1
2
}5b
8
b
Chapter 7 Test (p. 543)
8
a 5 }3 5 }3 5 64
1.
1 2
1
}
2
2.
y
y 5 2 ? 4x 2 2
1 2
2
2
b
2 5 ab22 l a 5 }
22
(1, 0.25): 0.25 5 ab
y
(1, 8) (3, 8)
1 x
2
So, an equation is y 5 64 + } .
36. (22, 2):
0.75
2
1
1b 2
(0, 2)
y 5 2 ? 4x
x
21
2 1
0.25 5 }
22 b
Domain: all real numbers
3.
0.5 5 b
f(x) 5 25 ? 2
1
2
5
2
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1b 2
9
Domain: all real numbers
Domain: all real numbers
Range: y < 3
Range: y > 0
5.
y52
65b
(
2
3
y52
()
21,
1
ln y
(0, 2)
(22, 2)
1
So, an equation is y 5 }4 + 6x.
0
6.
y
2
a 5 }2 5 }
5 }4
36
0.693 1.099 1.386 1.609 1.792
3.332 3.850 4.159 4.369 4.543 4.673
)
2
2
3
(0, 1)
x
2 x
1
3
8
3
(0, 3)
(1, )
1
1 x12
3
()
(1, )
(1, )
g(x) 5
1 x
3
2
3
1
g(x)
()
1
21
g(x) 5
()
2 x x
3
Domain: all real numbers
Domain: all real numbers
Range: y > 0
Range: y > 2
7.
8.
y
ln y
(0, 3.332)
x
21
(0, 25)
f(x) 5 25 ? 2
324 5 }2 b4
ln x
1
(21, 2 )
(4, 324): 324 5 ab4
9
x
1
(23, 22)
9
9 5 ab l a 5 }2
b
9
b
y
1
2
324 5 9b 2
4.
f(x)
13
(24, )
2
0.5
So, an equation is y 5 0.5 + 0.5x.
38.
Range: y > 0
x13
}
a5}
22 5
22 5 0.5
36 5 b
Domain: all real numbers
Range: y > 0
0.125 5 b 3
37. (2, 9):
x
21
0.25 5 2b3
2
b
(2, 2)
y
y 5 2.5e 20.5x 1 1
(0, 3.5)
(1, 2.52)
(1.792, 4.673)
2
21
(0, 2.5)
x
(1, 1.52)
21
21
y 5 2.5e 20.5x x
0.5
0.2
ln x
Domain: all real numbers
Domain: all real numbers
Range: y > 0
Range: y > 1
Algebra 2
Worked-Out Solution Key
411
Chapter 7,
22.
h(x)
()
1
3
h(x) 5
x
4
(2, 21.09) x
h(x) 5
( )e
1
3
x21
x ø 0.8740
Check: 72x 5 30
72(0.8740) 0 30
71.748 0 30
30 ø 30 22
Domain: all real numbers
Range: y > 22
10. 52 5 25, so log5 25 5 2.
1
1
11. 225 5 }, so log2 } 5 25.
32
32
14.
y
y
y 5 ln x
1
1
x
21
(1, 0)
(2, 0.69)
x
21
(2, 22.31)
(1, 23)
y 5 ln x 2 3
Domain: x > 0
Domain: x > 0
Range: all real numbers
Range: all real numbers
Domain: x > 23
f(x)
f(x) 5 log(x 1 3) 1 2
(21, 2.30)
Range: all real numbers
(22, 2)
1
(2, 0.30)
(1, 0)
x
f(x) 5log x
49
16. 2 ln 7 2 3 ln 4 5 ln 72 2 ln 43 5 ln 49 2 ln 64 5 ln }
64
5
17. log4 3 1 5 log4 2 5 log4 3 1 log4 2
5 log4 3 1 log4 32
5 log4 3 + 32 5 log4 96
18. log 5 1 log x 2 2 log 3 5 log 5 1 log x 2 log 32
5 log 5 1 log x 2 log 9
5x
5 log 5 + x 2 log 9 5 log }
9
log 50
19. log5 50 5 } ø 2.431
log 5
log 23
20. log6 23 5 } ø 1.750
log 6
log 45
21. log9 45 5 } ø 1.732
log 9
412
Algebra 2
Worked-Out Solution Key
x 5 104
Check: 3 log (x 2 4) 5 6
3 log (104 2 4) 0 6
3 log 100 0 6
3+206
656
24. log4 x 1 log4 (x 1 6) 5 2
12. 60 5 1, so log6 1 5 0.
21
x 2 4 5 100
x5}
2 log 7
21
15.
10log (x 2 4) 5 102
log 30
1
3
13.
log (x 2 4) 5 2
2x 5 log 7 30
(0, )
5
3
23. 3 log (x 2 4) 5 6
log 7 72x 5 log 7 30
e
(1, 0.91)
(1, 2 )
72x 5 30
log4 [x(x 1 6)] 5 2
4log 4[x(x 1 6)] 5 42
x(x 1 6) 5 16
x2 1 6x 5 16
2
x 1 6x 2 16 5 0
(x 1 8)(x 2 2) 5 0
x 5 28 or x 5 2
Check: log4 x 1 log4 (x 1 6) 5 2
log4 (28) 1 log4 (28 1 6) 0 2
log4 (28) 1 log4 (22) 0 2
Because log4 (28) and log4 (22) are not defined, 28 is
not a solution.
log4 x 1 log4 (x 1 6) 5 2
log4 2 1 log4 (2 1 6) 0 2
log4 2 1 log4 8 0 2
log4 16 0 2
Because 42 5 16, log4 16 5 2. The solution is 2.
48
25. (21, 48): 48 5 ab21 l a 5 }
b21
(2, 6): 6 5 ab2
1b 2
48
2
65 }
21 b
6 5 48 + b3
1
8
} 5 b3
1
2
}5b
48
b
48
}
a5}
21 5
21 5 24
1 2
1
}
2
1 x
So, an equation is y 5 241 }2 2 .
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
9.
continued
Chapter 7,
26. (3, 8):
continued
8
3
8 5 a(3)b l a 5 }b
TAKS Practice (pp. 546–547)
1. B;
(6, 15): 15 5 a(6)b
In choice A, the two vertical lines should be horizontal.
13 2
8
15 5 }b 6b
15 5 8 + 2b
In choice C, the two horizontal lines should be on the
trapezoid, and the vertical line should be on the triangle.
15
8
In choice D, the circle should be on the trapezoid, and the
vertical line should be on the triangle.
} 5 2b
15
log }
8
}5b
In choice B, the circle and the vertical line are positioned
correctly, so the net represents the figure in choice B.
log 2
2. G;
0.907 ø b
8
3
On their own, 8 triangles and 6 octagons have
8
a 5 }b ø }
ø 2.954
0.907
8(3) 1 6(8) 5 72 edges.
3
So, an equation is y 5 (2.954)x0.907.
In the solid, each edge is shared by exactly two polygons.
1
27. H 5 a(1 1 r)t
The initial amount is a 5 62 and the percent increase
is r 5 0.0485. So, the exponential growth model is
H 5 62(1.0485)t.
So, the number of edges is }2 (72) 5 36.
F1V5E12
14 1 V 5 36 1 2
28. When P 5 2500, r 5 0.035, and t 5 8: A 5 Pe rt
0.035(B)
A 5 2500e
0.28
5 2500e
The solid has 24 vertices.
ø 3307.82
3. C;
The balance after 8 years is $3307.82.
29. a.
ln x
21.609
ln y
22.303 20.643 20.288
1.609
2.398
2.996 3.807
0
0.405
Faces P and U are perpendicular, not parallel.
Faces R and S are parallel, not perpendicular.
4. G;
1
There are 6 triangles and 1 hexagon in a hexagonal
pyramid for a total of 7 faces. On their own, 6 triangles
and 1 hexagon have 6(3) 1 1(6) 5 24 edges. In the solid,
each edge is shared by exactly two polygons. So the
ln x
21
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Faces S and T are perpendicular, not parallel.
Faces Q and T are perpendicular, so choice C is true.
ln y
b.
V 5 24
0.405 2 (22 + 203)
M 5 }}
5 0.5
3.807 2 (21 + 609)
ln y 2 0.405 5 0.5(ln x 2 3.807)
ln y 5 0.5ln x 2 1.4985
ln y 5 ln x0.5 2 1.4985
y 5 eln x
0.5 2 1.4985
y 5 e21.4985eln x
0.5
y 5 0.2235x0.5
1
number of edges is }2 (24) 5 12.
F1V5E12
7 1 V 5 12 1 2
V57
The solid has 7 vertices.
5. D;
2
3y 5 22x 1 3 l y 5 2}3 x 1 1
3
When x 5 120:
0.5
y 5 0.2235(120)
ø 0.2235(10.9545) l y ø 2.4483
The speed of the current needed to carry a particle
with a diameter of 120 millimeters downstream is
about 2.45 meters per second.
2y 5 3x 1 8 l y 5 }2 x 1 4
2 3
Because 2}3 1 }2 2 5 21, the lines are perpendicular.
6. J;
Volume of box 5 * + w + h
5 9 + 6 + 12
5 648 in.3
Volume of cube 5 s 3 5 33 5 27 in.3
648 in.3
27 in.
Cubes in box 5 }
5 24
3
24 cubes can be placed completely inside the box.
Algebra 2
Worked-Out Solution Key
413
Chapter 7,
continued
7. B;
h 5 216t 2 1 12t 1 5
7 5 216t 2 1 12t 1 5
0 5 216t 2 1 12t 2 2
}}
212 6 Ï122 2 4(216)(22)
2(216)
t 5 }}}
212 6 4
t5}
232
t 5 0.25 or t 5 0.5
The ball has a height of 7 feet on its way up at 0.25
second, and on its way down at 0.5 second. Because
Alan hit the ball on its way down, the ball was in the air
for 0.5 second.
8. H;
5x 2 6y 5 22
26y 5 25x 2 2
5
1
y 5 }6 x 1 }3
1
y 5 mx 1 b, so b 5 }3 .
9. D;
3
2}4 (12x 2 4y) 1 (5y 2 8x) 5 29x 1 3y 1 5y 2 8x
5 8y 2 17x
10. J;
1 1 1 1 1
Sequence: }2 , }4, }6, }8 , }
,...5
10
1
1
1
1
2+1 2+2 2+3 2+4
1
2n
11. C;
The range of the function is 24 < y a 3.
12. H;
24 2 1
5
slope: }
5 }3
23 2 0
y-intercept: (0, 1)
5
y > }3x 1 1
3y > 5x 1 3
25x 1 3y > 3
13. 4s 1 6h 5 54
h 5 3s 2 2
4s 1 6(3s 2 2) 5 54
22s 5 66
s53
h 5 3(3) 2 2 5 7
There are 7 hexagonal tables in the dining room.
414
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}, }, }, }, . . . , }, . . .