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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