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Introductory And Intermediate Algebra Through Applications 3 Chapter 11
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March 12, 2025
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Chapter 11
FUNCTIONS
18.
()
8
hx
⎛⎞
=
⎟
⎜
⎟
⎜
⎝⎠
a.
()
22
18
26
4
81
h
−
⎛⎞
⎛⎞
⎟⎟
⎜⎜
−=
=
=
⎟⎟
⎜⎜
⎟⎟
⎜⎜
⎝⎠
⎝⎠
(
)
a.
()
41
3
3
11
44
4
64
4
g
−+
−
−=
=
=
=
(
)
(
)
28.
(
)
fx
e
+
=
(
)
2
148.413
fe
e
==
≈
(
)
()
(
)
2
4,
11
24
2
,
16
16
x
xf
x
x
y
−
=
⎛⎞
⎟
⎜
−=
−
⎟
⎜
⎟
⎜
⎝⎠
222
32.
()
3
fx
⎛⎞
⎟
⎜
=
⎟
⎜
⎟
⎜
⎝⎠
()
()
3
2
1
,
3
1
11
1
11
,
33
3
11
1
22
,
39
9
x
xf
x
x
y
−
⎛⎞
⎟
⎜
=
⎟
⎜
⎟
⎜
⎝⎠
⎛⎞
⎛⎞
⎛
⎞
⎟⎟
⎜⎜
=
⎟⎟
⎜⎜
⎟⎟
⎜⎜
⎝⎠
⎝
⎠
⎛⎞
⎛
⎞
⎟⎟
⎜⎜
=
⎟⎟
⎜⎜
⎟⎟
⎜⎜
⎝⎠
⎝
⎠
34.
(
)
3
x
fx
=−
()
(
)
()
()
3
3
2
3
3,
11
1
33
3
,
27
27
3
23
92
,
9
3
3
27
3,
27
x
xf
x
x
y
−
=−
⎛⎞
⎟
⎜
−−
=
−=
−
−
−
⎟
⎜
⎟
⎜
⎝⎠
−=
−
−
−=
−
−
36.
(
)
3
x
fx
+
=
()
(
)
1
31
2
2
3,
11
1
33
3
3
,
99
3
x
xf
x
x
y
+
−+
−
=
⎛⎞
⎟
⎜
−=
=
=
−
⎟
⎜
⎟
⎜
⎝⎠
Section 11.1 Exponential Functio
ns
()
()
1
1
3,
64
3
61
1
44
4
11
,
4
11
4
x
xf
x
x
y
−
⎛⎞
⎟
⎜
=−
⎟
⎜
⎟
⎜
=−
=
⎛⎞
⎜
⎝⎠
⎟
⎜
−
⎟
⎜
⎟
⎜
⎝⎠
=−
64
⎟
⎜
−=
−
=−
40.
4
10
10,000
10
4
x
x
==
⇒
=
44.
3
125
5
5
3
3
==
⇒
=
−
⇒
=
−
()
25
33
x
=
48.
3
35
23
2
x
x
−
−
=
50.
2
61
x
+
=
52.
()
()
()
31
4
33
31
4
x
x
−
=
−=
54.
()
()
3
52
3
32
8
22
x
−+
−
=
=
()
()
()
()
51
8
22
85
1
x
x
xx
+
=
=+
224
58.
()
()
125
625
=
1
1
1
2
11
1
x
−
⎛⎞
⎟
⎜
⎟
⎜
⎝⎠
⎛⎞
⎛
⎞
62.
Expo
nential funct
ion: it has a va
riable in
the exponent.
64.
(
)
()
51
15
1
1
fx
f
−
=−
+
⎟
⎜
−=
−
+
=
−
+
=
−
+=
⎟
⎜
⎟
66.
(
)
(
)
(
)
51
2
5
0
0
.
8
5
5
5
5
v
=≈
0
1250
0.
85
1250
1
1250
v
==
=
(
)
(
)
82
1
.
1
4
.
3
h
=≈
72.
()
30
2
At
⎛⎞
⎜
⎟
⎜
⎝⎠
130
65
1
(
)
76.
(
)
(
)
500
3
t
mt
=
(
)
()
()
5
53
500
3
13,500
33
t
t
=
=
225
1. a.
23
22
4
3
9
x
==
xx
xx
2.
The
restriction
0
b
>
guarantees t
hat the
3.
Domain
of
(
)
2.
The
term
log
b
x
may be read as “the logarithm of
4.
T
he graph of a lo
garithmic
function is s
ymmetric
8.
2
4
log
16
2
4
16
=⇒
=
−
16.
13
64
18.
15
30.
12
15
226
32.
(
)
22
log
64
log
2
6
==
(
)
11
0
11
0
48.
3
16
4
16
4
11
log
log
46
4
=
5
3
1
x
x
−
=
=
x
60.
log
81
2
x
=
()
81
x
x
=
=
()
14
1
16
x
=
227
()
21
33
x
=
70.
()
16
23
44
x
=
=
72.
()
12
5
22
55
x
xx
−
=
−=
⇒
=
−
74.
(
)
4
log
4
y
fx
y
x
x
==
⇒=
()
()
2
2
4,
111
2
4
16
16,
2
y
yx
x
y
−
=
⎛⎞
=
()
()
0
2
4
4
1
01
1
,
0
11
1
1
yx
x
y
⎟
⎜
=
⎟
⎜
⎟
⎜
⎝⎠
⎜
⎟
⎜
⎝⎠
⎛⎞
⎟
⎜
=
⎟
⎜
⎟
⎛⎞
⎛
⎞
64
64
78.
(
)
2
log
2
y
fx
y
x
x
−
==
−
⇒=
()
()
()
()
()
()
()
()
3
3
2
2
1
1
0
2,
32
2
8
8
,
3
22
2
4
4
,2
12
2
2
2
,
1
02
1
1
,
0
y
yx
x
y
−
−−
−−
−−
−
=
−=
=
−
−=
=
−
−=
=
−
=
228
2
2
25
64
25
x
=
x
=
82.
(
)
4
log
4
y
fx
y
x
x
−
==
−
⇒=
()
()
()
22
4,
24
4
1
6
1
6
,2
y
yx
x
y
−
−−
=
−=
=
−
90.
10
0
⎜
⎟
⎜
⎝⎠
()
()
()
10
6
10
20
log
10
20
6
=
=
92.
2
⎜
⎟
⎜
⎝⎠
2
1
1600
log
8
1
⎛⎞
⎟
⎜
=−
⎟
⎜
⎟
⎜
⎝⎠
⎛⎞
⎟
⎜
94.
10
10
log
tN
=−
10
1
10
log
10
t
⎛⎞
⎟
⎜
=−
⎟
⎜
⎟
⎜
⎝⎠
229
1.
T
he value of
y
can never be 0 because
(
)
1000
−
40.
3
54.
7
Since ,
7
20.
bx
==
58.
log
log
23
Since ,
23.
bx
n
==
10
10
log
27
log
2
=+
62.
23
33
3
3
33
log
64
log
125
=−
log
log
b
=+
()
log
4
log
6
log
2
log
64
log
36
log
16
log
64
log
36
log
16
xxx
xxx
xx
x
=−
−
=−
−
=−
+
68.
()
77
1
6l
o
g
lo
g
u
70.
(
)
()
(
)
22
2
54
2
4l
o
g
lo
g
3
log
3
nn
nn
++
=+
72.
(
)
()
22
2
ab
a
b
−−
−
⎢
⎥
⎣
⎦
1
log
⎢
⎥
⎣
⎦
74.
888
4
log
5
log
2
log
bac
−+
−
76.
77
7
33
+−
[]
()
7
77
7
77
1
2
log
l
og
log
3
uv
w
=+
−
82.
33
33
3
3
log
3
log
3
log
log
uv
u
v
=++
2
55
3
1
log
9
2
log
v
⎡
⎤
=+
⎢
⎥
⎣
⎦
231
86.
55
5
4
log
log
log
b
=−
2
22
33
1
log
3
x
xy
⎡⎤
⎡⎤
=−
⎢⎥
⎣⎦
zz
(
)
()
14
32
32
1
log
log
5
log
log
log
5log
44
nn
nn
n
ac
ac
b
=+
−
55
55
44
100.
4
102.
55
5
22
2
log
log
log
4
1
log
log
yy
y
==
⎜
⎟
⎜
⎢⎥
⎟
⎟
⎜
⎝⎠
⎣⎦
104.
55
5
25
()
(
)
10
2.5
8
20
=−
−
=
()
10
10
I
I
⎢
⎥
⎣
⎦
110. a.
(
)
10
10
10
log
log
log
1.05
t
AP
t
=+
b.
(
)
(
)
1000
1.05
1000
1.05
1050
A
==
=
232
is log
.
b
yx
=
So log
.
x
b
bx
=
Alternate
proof:
2.
33
3.3
=−
3.2
=
and Change of Base
Exercises
2.
T
he inverse of the natural exponential
log
0.000001
log10
6
==
−
e
1.2691
≈
1.2619
≈−
66.
(
)
80
8
ln
1
sx
=−
+
c
35
7
b
c
234
(
)
(
)
()
5
2l
o
g
3
2
53
69
3
x
x
xx
x
+
=+
++
=
+
()
()
5
?
2l
o
g
2
3
?
52
3
−+
=−
+
5
?
2l
o
g
0
50
=
Logarithmic Equations
2.
65
6
log
6
log
56
x
=
=
4.
1
15
3
1
2.4650
x
x
⎟
⎜
⎟
⎜
⎝⎠
⎛⎞
6.
6
7
7
x
=
Section 11.5 Exponential and
Logarithmic Equations
235
10.
2
57
.
5
=
log1000
log
95
3
log
95
x
x
−=
−⋅
=
14.
1
26
8
log
2
log
68
x
+
=
=
16.
35
63
2
4
log
6
log
324
x
−
=
=
log
6
x
(
)
2
24.
(
)
3
2
13
x
−
−=
(
)
3
24
2
6
log
12
6
x
=
30.
55
log
log
2
1
x
−=
236
32.
(
)
()
33
3
log
6
log
3
log
6
3
xx
xx
++
=
⎡⎤
+=
⎣⎦
Check 9.
x
=−
()(
)
(
)
(
)
?
33
?
33
log
9
6
log
9
3
log
3
log
9
3
−+
+
−
=
−+
−=
Check 3.
x
=
?
()
()
7
log
3
3
1
xx
⎡⎤
−+
=
⎣⎦
(
)()
77
log
4
3
log
4
3
1
−−
+
−+
=
Check
4.
x
=
()
()
?
77
log
4
3
log
4
3
1
−+
+=
36.
(
)
(
)
()
()
log
3
4
log
5
9
1
log
3
4
5
9
1
xx
xx
++
−=
⎡⎤
+−
=
⎣⎦
15
x
=−
23
Check
.
15
x
=−
log
4
log
9
1
53
⎟⎟
⎜⎜
−+
+
−−
=
⎟⎟
⎜⎜
⎟⎟
⎜⎜
⎝⎠
⎝⎠
()
()
?
38.
(
)
22
10
4
16
10
4
16
x
x
x
xx
+
=
+=
237
(
)
(
)
55
5
1
log
2
1
log
8
1
21
log
1
8
21
xx
x
x
x
−−
−
=
−
=
−
−
=
()
()
9
32
log
6
2
69
xx
xx
⎡⎤
−=
⎣⎦
−=
Check
3.
x
=−
?
3
Check 9.
x
=
The solution is
9.
44.
(
)
(
)
33
log
9
log
2
3
2
xx
−−
+
=
−
3
9
log
2
23
9
3
x
x
x
−
=−
+
−
=
46.
44
4
log
log
3
2
log
2
3
x
x
=
()
(
)
(
)
44
24
0
xx
+−
=
Check 2.
x
=−
Check
4.
x
=
()
?
44
?
44
3
log
4
2
log
4
2
3
log
2
log
4
2
−+
=
+=
238
(
)
25
log
4
log
75
2
5
log
4
log75
x
x
+
=
+=
9
t
≈
54.
0
VV
e
=
31,000
2l
n
31
k
e
k
=
=
The car will be worth approximately $11,000.
56.
0
⎜
⎟
⎜
⎝⎠
12
140
10log
10
I
−
⎛⎞
⎟
⎜
=
⎟
⎜
⎟
⎜
⎝⎠
N
when
0.
y
=
log
3
1.6
y
=≈
62.
()
0
k
=−
k
is
0.231
;
−
it represents the rate at which
239
1. a.
Enter the expression
3
2
x
into Y1 and
6
in Y2, and then, graph. Use the TRACE or
INTERSECT feature to find the point(s) of
intersection of the graphs. The
x
-coordinate
of any point of intersection is the solution.
The solution is approximately
0.8617.
3l
o
g
2
ln
2
ln
ln
2
e
t
r
=
=
0.6931
69
72
72
9 yr
.
[
)
2
2,
x
≥∞
b.
(
)
(
)
2
1
log
1
1
12
3
3,
x
x
x
−>
−>
>∞
()
()
Let
4.
Let
0.
=−
=
()
?
Let
4.
log17
1
x
=
≤