115.3 Solutions to some odd numbered problems from section 1.3-1
y
1.3-1 Sketch the graph of y = x + 1 without using a graphing calculator.
2
Solution:
Starting with the graph of y = x 2 in black,
and take that graph shifted up 1 unit in red: y = x 2 + 1.
1.3-11 Sketch the graph of y = exp(x − 2) = ex−2 without using a graphing calculator.
Starting with the graph of y = ex in black,
we get that shifted 2 units to the right in red: y = e−x .
Solution:
y
10
9
8
7
6
5
4
3
2
1
0
-4-3-2-10 1 2 3 4
x
10
9
8
7
6
5
4
3
2
1
0
-4-3-2-10 1 2 3 4
Sa
115.3 Solutions to some odd numbered problems from section 1.3
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x
115.3 Solutions to some odd numbered problems from section 1.3-2
Solution:
0.0002
1
-4
10
2
3 4 5 67891
-3
10
2
3 4 5 67891
0.02
-2
10
2
1
3 4 5 67891
-1
10
2
3 4 5 67891
0
10
5
2
1
3 4 5 6789
1
10
Sa
1.3-33 Find the following numbers on a number line that is on a logarithmic scale (base 10):
0.0002,0.02,1,5,50,100,1000, 8000, and 20000.
sk
3 4 5 67891
2
10
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50 100
2
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1000
2
3 4 5 67891
3
10
8000 20000
2
3 4 5 67891
4
10
2
3 4 5 6789
5
10
115.3 Solutions to some odd numbered problems from section 1.3-3
(x1 , y1 ) = (0, 5),
Sa
1.3-43 When log y is graphed as a function of x on log-linear paper, a straight line results. Graph straight lines,
each given by two points, on a log-linear plot, and determine the functional relationship.
sk
Solution:
Since the graph is a straight line on log-linear graph paper, we have Y = mx + b for
constants m and b which we must find.
At (x1 , y1 ) = (0, 5) we have log 5 = m(0) + b, so b = log 5.
have the equation Y = −
log 5
3 x
y = 10Y = 10log y = 10−
log 5
−0.23 and we
3
+ log 5.
Exponentiating, we get
log 5
3 x+log 5
log 5 x
= 10log 5 × 10− 3
5 × (0.58)x
Y=log y
9
8
7
6
5
4
3
2
1
0
1
2
3
4
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(x2 , y2 ) = (3, 1)
At (x2 , y2 ) = (3, 1) we have log 1 = m(3) + b, so 0 = 3m + log 5, and thus m = −
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x
115.3 Solutions to some odd numbered problems from section 1.3-4
(x1 , y1 ) = (−2, 3),
Sa
1.3-45 When log y is graphed as a function of x on log-linear paper, a straight line results. Graph straight lines,
each given by two points, on a log-linear plot, and determine the functional relationship.
sk
(x2 , y2 ) = (1, 1)
Since the graph is a straight line on log-linear graph paper, we have Y = mx + b for
constants m and b which we must find.
At (x1 , y1 ) = (−2, 3) we have log 3 = m(−2) + b, and
at (x2 , y2 ) = (1, 1) we have log 1 = m(1) + b, or 0 = m + b, so b = −m. Substituting this in the first
log 3
log 3
equation, we get log 3 = −2m − m = −3m, so m = −
and b =
3
3
log 3
log 3
and we have the equation Y = −
x+
.
3
3
Exponentiating, we get
1 x
log 3
log 3
log 3
log 3 x
1
y = 10Y = 10log y = 10− 3 x+ 3 = 10 3 × 10− 3
= 3 3 × 3− 3
Y=log y
9
8
7
6
5
4
3
2
1
-1
0
1
2
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Solution:
-2
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3
x
115.3 Solutions to some odd numbered problems from section 1.3-5
y = 2 × 10−2x .
Solution:
Sa
1.3-47 Use a logarithmic transformation to find a linear relationship between the given quantities and graph the
resulting linear relationship on a log-linear plot:
sk
We have Y = log(4 × 10
) = log 4 + 5x
9
8
7
6
5
4
3
2
1
-1
0
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Y=log y
5x
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1
x
115.3 Solutions to some odd numbered problems from section 1.3-6
Solution:
We have Y = log y = mlogx + b,
so when (x1 , y1 ) = (1, 2), we have log 2 = m log 1 + b = m(0) + b = b, so b = log 2,
and when (x2 , y2 ) = (5, 1), we have log 1 = m log 5 + log 2, so 0 = m log 5 + log 2 and thus m = −
so we have the equation Y = −
log 2
X + log 2.
log 5
Exponentiating, we have
10Y = 10 + log y = y = 10
log 2
− log 5 X+log 2
= 10log 2 10
log 2
− log 5 log x
= 2×x
log 2
− log 5
.
y
9
8
7
6
5
4
3
2
1
1
2
3
4
5
6
7 8 9 10
log 2
,
log 5
Sa
1.3-55 When log y is graphed as a function of log x a straight line results. Graph straight lines, each given by two
points on a log-log plot, and determine the functional relationship. (The original x-y coordinates are given.)
(x1 , y1 ) = (1, 2), (x2 , y2 ) = (5, 1)
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115.3 Solutions to some odd numbered problems from section 1.3-7
y = 2x 5
Solution:
Y = log y = log(2x 5 ) = log 2 + 5 log x = log 2 + 5X, where X = log x.
y
9
8
7
6
5
4
3
2
1
1
2
3
4
5
6
7 8 9 10
Sa
1.3-59 Use a logarithmic transformation to find a linear relationship between the given quantities and graph the
resulting linear relationships on a log-log plot.
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115.3 Solutions to some odd numbered problems from section 1.3-8
f (x) = 3x 1.7
Solution: F = log f (x) = log 3x 1.7 = log 3 + log x 1.7 = log 3 + 1.7 log x = log 3 + 1.7X which we
plot on log-log paper:
y
9
8
7
6
5
4
3
2
1
1
2
3
4
5
6
7 8 9 10
Sa
1.3-67 Use a logarithmic transformation to find a linear relationship between the given quantities and determine
whether a log-log or a log-linear plot should be used to graph the resulting relationship
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115.3 Solutions to some odd numbered problems from section 1.3-9
N(t) = 130 × 21.2t
Solution: log N(t) = log 130 × 21.2t = log 130 + log 21.2t = log 130 + (1.2 log 2)t which we plot
on log-linear paper:
log L
900
800
700
600
500
400
300
200
-1
100
0
1
c
Sa
1.3-69 Use a logarithmic transformation to find a linear relationship between the given quantities and determine
whether a log-log or a log-linear plot should be used to graph the resulting relationship
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115.3 Solutions to some odd numbered problems from section 1.3-10
x and y
x
1
2
4
10
20
y
1.8
2.07
2.38
2.85
3.28
Solution:
Sa
1.3-73 The following table is based on a functional relationship between x and y, which is either an exponential
or a power function. Use an appropriate logarithmic transformation and a graph to decide whether the
table comes from a power function or an exponential function and find the functional relationship between
sk
log-log paper:
Y=log y
9
8
7
6
5
4
3
9
8
7
6
5
4
3
2
2
x
1
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20
1
1
2
3 4 5 67891
2
3 4 5 678910
We conclude that the data displayed on loglog paper is closest to being a straight line, so we take a line
Y = mX + b, or log y = m log x + b, through the points corresponding to (1,1.8) and (20,3.28):
At (1,1.8) we have Y = log y = log 1.8 = m log 1 + b = m(0) + b = b, so b = log 1.8.
log 3.28
log 3.28 − log 1.8
1.8
At (20,3.28) we have Y = log 3.28 = m log 20 + b = m log 20 + log 1.8, so m =
=
.
log 20
log 20
Thus we have Y =
log 3.28
1.8
log 20
X + log 1.8 0.2X + log 1.8
Exponentiating, we get
10Y = y = 10
log 3.28
1.8
log 20
X+log 1.8
= 10log 1.8 10
log 3.28
1.8
log 20
X
= 1.8 × 10X
3.28
1.8
log 20
log
1.8 × x 0.2
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First we graph the data on semi-log and
Y=log y
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X=log x
115.3 Solutions to some odd numbered problems from section 1.3-11
x and y
x
−1
−0.5
0
0.5
1
y
0.398
1.26
4
12.68
40.18
Solution:
We only graph the data on semi-log,
because negative values cannot be represented on log-log paper:
Y
9
8
7
6
5
4
3
2
1
9
8
7
6
5
4
3
2
-2.0
-1.5
-1.0
-0.5
1
9
8
7
6
5
4
3
2
1
0.0
0.5
1.0
1.5
2.0
Sa
1.3-75 The following table is based on a functional relationship between x and y, which is either an exponential
or a power function. Use an appropriate logarithmic transformation and a graph to decide whether the
table comes from a power function or an exponential function and find the functional relationship between
sk
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115.3 Solutions to some odd numbered problems from section 1.3-12
At (-1,0.398) we have Y = log y = log 0.398 = m(−1) + b = −m + b, and
at (1,40.18) we have Y = log 40.18 = m(1) + b = m + b, so adding the two equations, we get
2b = log 0.398 + log 40.18 = log(0.398 × 40.18) = log 15.99,
and subtracting the first equationfrom the
second gives
40.18
2m = log 40.18 − log 0.398 = log
= log 100.95,
0.398
1
1
So m = 12 log 40.18
0.398 and b = 2 log(0.398 × 40.18) 2 log 15.99 0.60
40.18
1
Thus we have Y = 2 log 0.398 x + 12 log(0.398 × 40.18) 50.43x + 0.60
Exponentiating, we get
Y
10 = y = 10
1
2
log
40.18
0.398
1
x+ 2 log(0.398×40.18)
= 10
1
2
log(0.398×40.18)
1
2
log
40.18
0.398
x
10
=

1 x
x
1
1
1
40.18 2 
log 40.18
0.398
= 15.99164 2 × 
4 × 10x
(0.398 × 40.18) 2 × 10 2
0.398

Sa
We conclude that the data displayed on semilog paper is close to being a straight line, so we take a line
Y = mx + b, or log y = mx + b, through the points corresponding to (-1,0.398) and (1,40.18):
sk
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115.3 Solutions to some odd numbered problems from section 1.3-13
x and y
x
0.1
0.5
1
1.5
2
y
0.045
1.33
5.7
13.36
24.44
Solution:
First we graph the data on semi-log and
log-log paper:
Y
9
8
7
6
5
4
3
2
9
8
7
6
5
4
3
2
1
9
8
7
6
5
4
3
2
1
9
8
7
6
5
4
3
2
1
9
8
7
6
5
4
3
2
1
0.0 0.5 1.0 1.5 2.0
1
1
9
8
2 3 45677
891
6
5
4
3
2
1
9
8
7
6
5
4
3
2
1
2 3 45678910
Sa
1.3-77 The following table is based on a functional relationship between x and y, which is either an exponential
or a power function. Use an appropriate logarithmic transformation and a graph to decide whether the
table comes from a power function or an exponential function and find the functional relationship between
sk
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2003 Doug MacLean
115.3 Solutions to some odd numbered problems from section 1.3-14
At (0.1,0.045) we have Y = log y = log 0.045 = m log(0.1) + b = m(−1) + b = −m + b, and
at (2,24.44) we have Y = log 24.44 = m(log 2) + b = (log 2)m + b, so on subtracting the second equation
from the first, we get log 0.45 − log 24.44 = −(1 + log 2)m,
so m =
log 24.44−log 0.045
1+log 2
2.1 and b = m + log 0.045 2.1 + (−1.346) = 0.76
Thus we have Y = 2.1X + 0.76
Exponentiating, we get
10Y = y = 102.1X+0.76 = 100.76 102.1 log x = 5.7 × x 2.1
Sa
We conclude that the data displayed on loglog paper is close to being a straight line, so we take a line
Y = mX + b, or log y = m log x + b, through the points corresponding to (0.1,0.045) and (2,24.44):
sk
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2003 Doug MacLean