Fall17 HW04 — Semilog and double log plots

Fall17 HW04 -- Semilog and double log plots

1. (Problem # 43, p. 53) When log y is graphed as a function of x, a straight line results. Graph the straight line given by the following two points

(x1, y1) = (0, 5) (x2, y2) = (3, 1)

on a log-linear plot. The functional relationship between x and y is: y =

.

(Note: The original x-y coordinates are given.)

Answer to Problem # 1: The slope of the line in the log-linear plane is

m

=

log 1

-

log 5

=

log 5 -

=

-1

log 5

=

log(5-1/3).

3-0

3

3

Using the original point (3, 1), the point-slope form of the line in the log-linear plane is

Y - log 1 = log(5-1/3)(x - 3)

where Y = log y. The above equation becomes now

log y = log(5-1/3)x - 3 log(5-1/3)

log y = log(5-x/3) + log(5) = log(5-x/3 ? 5).

Thus the functional relationship is: y = 5 ? 5-x/3 = 5 ? 0.58x.

Alternative answer to Problem # 1: Since we have a straight line in a log-linear plot,

a linear relationship in a log-linear plot corresponds to an exponential relationship between the original quantities. That is, x and y satisfy a relationship of the form y = b ? ax, for some a and b.

If we substitute the given values (0, 5) and (3, 1) in the expression y = b ? ax, we obtain

5 = b ? a0 and 1 = b ? a3.

Thus b = 5 and the second expression above becomes

1 = 5 ? a3

a3 = 1

5

so that a = 5-1/3 = 0.58. Now b = 1/a3 = 5.

Thus the functional relationship is: y = 5 ? 5-x/3 = 5 ? 0.58x.

1

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2. (Problem # 45, p. 53) When log y is graphed as a function of x, a straight line results. Graph the straight line given by the following two points

(x1, y1) = (-2, 3) (x2, y2) = (1, 1)

on a log-linear plot. The functional relationship between x and y is: y =

.

(Note: The original x-y coordinates are given.)

Answer to Problem # 2: The slope of the line in the log-linear plane is

m

=

log 1

-

log 3

=

log 3 -

=

-1

log 3

=

log(3-1/3).

1 - (-2)

3

3

Using the original point (1, 1), the point-slope form of the line in the log-linear plane is

Y - log 1 = log(3-1/3)(x - 1)

where Y = log y. The above equation becomes now

log y = log(3-1/3)x - log(3-1/3)

log y = log(3-x/3) + log(31/3) = log(3-x/3 ? 31/3).

Thus the functional relationship is: y = 31/3 ? 3-x/3 = 31/3 ? (3-1/3)x = 1.44 ? 0.69x.

Alternative answer to Problem # 2: Since we have a straight line in a log-linear plot,

a linear relationship in a log-linear plot corresponds to an exponential relationship between the original quantities. That is, x and y satisfy a relationship of the form y = b ? ax, for some a and b.

If we substitute the given values (-2, 3) and (1, 1) in the expression y = b ? ax, we obtain

3 = b ? a-2 and 1 = b ? a.

If we solve for b in both equations and we equate the expressions, we obtain

31 =

a-2 a

3a2 = 1 a

a3 = 1 3

1 a=

= 3-1/3 = 0.69.

33

Now b = 1/a = 3 3 = 31/3 = 1.44.

Thus the functional relationship is: y = 1.44 ? 0.69x.

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3. When log y is graphed as a function of x, a straight line results. Graph the straight line given by the following two points

(x1, y1) = (0, 50) (x2, y2) = (2, 800)

on a log-linear plot. The functional relationship between x and y is: y =

.

(Note: The original x-y coordinates are given.)

Answer to Problem # 3: The slope of the line in the log-linear plane is

log 800 - log 50 log(800/50) log 16 log(42) 2 log 4

m=

=

=

=

=

= log 4.

2-0

2

2

2

2

Using the original point (0, 50), the point-slope form of the line in the log-linear plane is

Y - log 50 = (log 4)(x - 0)

where Y = log y. The above equation becomes now

log y = (log 4)x + log 50

log y = log(50) + log(4x) = log(50 ? 4x).

Thus the functional relationship is: y = 50 ? 4x.

Alternative answer to Problem # 3: Since we have a straight line in a log-linear plot, a linear relationship in a log-linear plot corresponds to an exponential relationship between the original quantities. That is, x and y satisfy a relationship of the form y = b ? ax, for some a and b.

If we substitute the given values (0, 50) and (2, 800) in the expression y = b ? ax, we obtain 50 = b ? a0 and 800 = b ? a2.

Thus b = 50 and the second expression above becomes 800 = 50 ? a2 a2 = 16

so that a = 4. Thus the functional relationship is: y = 50 ? 4x.

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4. (Problem # 47, p. 53) Consider the relationship y = 3 ? 10-2x between the quantities x and y. Use a logarithmic transformation to find a linear relationship of the form Y = mx + b between the given quantities.

Y=

m=

b=

.

Graph the resulting linear relationship on a log-linear plot.

Answer to Problem # 4: Starting with y = 3 ? 10-2x we take logarithms of both sides. We obtain

log y = log(3 ? 10-2x) = log 3 + log(10-2x)

= -2x + log 3.

Thus the linear relationship is

Y = -2x + log 3.

That is:

Y = log y m = -2 b = log 3.

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5. (Problem # 51, p. 53) Consider the relationship y = 5 ? 24x between the quantities x and y. Use a logarithmic transformation to find a linear relationship of the form Y = mx + b between the given quantities.

Y=

m=

b=

.

Graph the resulting linear relationship on a log-linear plot.

Answer to Problem # 5: Starting with y = 5 ? 24x we take logarithms of both sides. We obtain

log y = log(5 ? 24x) = log 5 + log(24x)

= 4x log 2 + log 5

= (4 log 2)x + log 5

= (log 16)x + log 5.

Thus the linear relationship is

Y = (log 16)x + log 5.

That is:

Y = log y m = log 16 b = log 5.

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6. (Problem # 55, p. 53) When log y is graphed as a function of log x, a straight line results. Graph the straight line given by the following two points

(x1, y1) = (1, 2) (x2, y2) = (5, 1)

on a log-log plot. The functional relationship between x and y is: y =

.

(Note: The original x-y coordinates are given.)

Answer to Problem # 6: The slope of the line in the log-log plane is

m

=

log 1 log 5

- -

log 2 log 1

=

log 2 -log 5 .

Using the original point (5, 1), the point-slope form of the line in the log-log plane is

log 2 Y - log 1 = - (log x - log 5)

log 5

where Y = log y. The above equation becomes now

log 2 log y = -log 5 log x + log 2

log y = log(x-(log 2)/(log 5)) + log(2) = log(2 ? x-(log 2)/(log 5)).

Thus the functional relationship is: y = 2 ? x-(log 2)/(log 5) = 2 ? x-0.43.

Alternative answer to Problem # 6: Since we have a straight line in a log-log plot, a linear relationship in a log-log plot corresponds to a power relationship between the original quantities. That is, x and y satisfy a relationship of the form y = bxr, for some b and r.

If we substitute the given values (1, 2) and (5, 1) in the expression y = bxr, we obtain 2 = b ? 1r and 1 = b ? 5r.

Thus b = 2 and the second expression above becomes 1 = 2 ? 5r log 1 = log 2 + r log 5 r = - log 2/ log 5 = -0.43.

Thus the functional relationship is: y = 2 ? x-0.43.

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7. (Problem # 57, p. 53) When log y is graphed as a function of log x, a straight line results. Graph the straight line given by the following two points

(x1, y1) = (4, 2) (x2, y2) = (8, 8)

on a log-log plot. The functional relationship between x and y is: y =

.

(Note: The original x-y coordinates are given.)

Answer to Problem # 7: The slope of the line in the log-log plane is m = log 8 - log 2 = log(8/2) = log 4 = 2 log 2 = 2. log 8 - log 4 log(8/4) log 2 log 2

Using the original point (4, 2), the point-slope form of the line in the log-log plane is Y - log 2 = 2(log x - log 4)

where Y = log y. The above equation becomes now log y = 2 log x - 2 log 4 + log 2

log y = log(x2) + log(4-2) + log 2 = log((2x2)/16) = log(x2/8).

Thus the functional relationship is: y = 1 x2. 8

Alternative answer to Problem # 7: Since we have a straight line in a log-log plot, a linear relationship in a log-log plot corresponds to a power relationship between the original quantities. That is, x and y satisfy a relationship of the form y = bxr, for some b and r.

If we substitute the given values (4, 2) and (8, 8) in the expression y = bxr, we obtain

2 = b ? 4r and 8 = b ? 8r.

If we solve for b in both equations and we equate the expressions, we obtain

28 =

4r 8r

28 =

22r 23r

23r 8 =

22r 2

2r = 4

Thus b = 2/4r = 2/16 = 1/8.

Thus the functional relationship is: y = 1 x2. 8

r = 2.

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8. When log y is graphed as a function of log x, a straight line results. Graph the straight line given by the following two points

(x1, y1) = (1, 20) (x2, y2) = (10, 000, 20, 000)

on a log-log plot. The functional relationship between x and y is: y =

.

(Note: The original x-y coordinates are given.)

Answer to Problem # 8: The slope of the line in the log-log plane is

log 20, 000 - log 20 log(20, 000/20) log 103 3

m=

=

=

=.

log 10, 000 - log 1

log 10, 000

log 104 4

Using the original point (1, 20), the point-slope form of the line in the log-log plane is

3 Y - log 20 = (log x - log 1)

4 where Y = log y. The above equation becomes now

3/4 log y = x + log 20

log

log y = log(x3/4) + log 20 = log(20 ? x3/4).

Thus the functional relationship is: y = 20 ? x0.75.

Alternative answer to Problem # 8: Since we have a straight line in a log-log plot, a linear relationship in a log-log plot corresponds to a power relationship between the original quantities. That is, x and y satisfy a relationship of the form y = bxr, for some b and r.

If we substitute the given values (1, 20) and (104, 2 ? 104) in the expression y = bxr, we

obtain 20 = b ? 1r and 2 ? 104 = b(104)r.

Thus b = 20 and the second expression above becomes

2 ? 104 = 20 ? 104r

103 = 104r

so that r = 3/4 = 0.75. Thus the functional relationship is: y = 20 ? x0.75.

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