Transformations into Linear Functions

Section 1.2 Problems

1.3 Graphing

1.3.3 Transformations into Linear Functions

When you look through a biology textbook, you very likely find graphs like the ones in Figures 1.52 and 1.53. In either graph, you see a straight line (with data points scattered about it). In Figure 1.52, the vertical axis is logarithmically transformed and the horizontal axis is on a linear scale; in Figure 1.53, both axes are logarithmically transformed. Why do we display data like this, and what do these graphs mean?

4 2 0 2 4 1000

100

10

1

0.1

x y

Figure 1.52 A straight line when the vertical axis is logarithmically transformed.

100000 10000 1000 100 10

1 10 100 1000

0.1 0.01

x y

Figure 1.53 A straight line when both axes are logarithmically transformed.

The first question is quick to answer: Straight lines (or linear relationships) are easy to recognize visually. If transforming data results in data points lying along

a straight line, we should do the transformation, because, as we will see, this will allow us to obtain a functional relationship between quantities. Now on to the second question: What do these graphs mean?

Exponential Functions Let’s look at Figure 1.52 and redraw just the straight line, using logy(instead of y) on the vertical axis andx on the horizontal axis (Figure 1.54). SetY =^logyand forget for a moment where the graph came from. We see a linear relationship betweenYandx—a relationship of the form

Y =c+mx

wherecis theY-intercept andmis the slope. We can read these two quantities off of the graph in Figure 1.54:

c=1.5 m =0.5 That is, we have

Y =¹.⁵+⁰.⁵x Now,Y =^logy, and thus

logy=¹.⁵+⁰.⁵x Exponentiating both sides, we find that

y =^101.5+0.5x =^101.5(^100.5)^x

Since 10^1.5≈³¹.^{62 and 100.5}≈³.162, we can write the preceding equation as y=(³¹.⁶²)(³.^162x) ^(1.4) Looking at (1.4), we see that it is an exponential function.

0.5 1

Y log y 3 2.5 2 1.5 1 0.5

0.5 1

4 2 0 2 4 x

Figure 1.54 Figure 1.52 redrawn. Now the vertical axis is labeled logy.

A graph in which the vertical axis is on a logarithmic scale and the horizontal axis is on a linear scale is called a log-linear plotor asemilog plot. If we display an exponential function of the formy = ba^x on a semilog plot, a straight line results.

To see this, we take logarithms to base 10 on both sides ofy=ba^x:

logy=^log(ba^x) ^(1.5)

Using the properties of logarithms, we simplify the right-hand side to log(ba^x)=^logb+^loga^x =^logb+xloga If we setY =^logy, then (1.5) becomes

Y =^logb+(^loga)x (1.6)

Comparing this equation with the general form of a linear functionY =c+mx, we see that theY-intercept is logband the slope is loga. You do not need to memorize this statement, since you can always do the calculation that resulted in (1.6), but you should memorize the fact that an exponential function results in a straight line on a semilog plot. Ifa>1, the slope of the line is positive; if 0<a<1, the slope of the line is negative.

EXAMPLE 4 Graph

y=².⁵·^3x, x∈^R on a semilog plot.

Solution

We take logarithms first:

logy=^log(².⁵·^3x)

=log 2.5 ≈0.3979

+x log 3

≈0.4771

The graph is shown in Figure 1.55. Note that the origin of the coordinate system is wherex = ^{0 and} y = ^{1 (or log}y = 0). The labeling on the vertical axis is fory, and we see that the labels are multiples of 10. To find 2.5 on the vertical axis, we use the fact that log 2.⁵=⁰.3979 and that 2.5 is therefore 0.3979 units above thex-axis, as illustrated in Figure 1.55. (One unit on the vertical axis corresponds to a factor of 10.)

100

10

1

0.1

2 1 0 1 2 3 y 2.5 3^x

x y

Figure 1.55 The graph ofy=2.5×3^xon a semilog plot.

4 3 2 1

1 2

2 1 0 1 2 3 4 5 6 (0, log 100)

(4, log 1) Y log y

x

Figure 1.56 The graph for Example 5. The line goes through the points(0,2)and(4,0).

EXAMPLE 5 Find the functional relationship betweenxandybased on the graph in Figure 1.56.

Solution

Figure 1.56 shows a semilog plot. We setY = ^logy. Then, in anx–Y graph, theY- intercept is log 100=2, and, using the two points(⁰,^{log 100})^and(⁴,^{log 1})^{, we find} that the slope of the line is(log 1−log 100)/(4−0)=(0−2)/(4−0)= −0.5. Hence,

Y =c+mx=²−⁰.⁵x

SinceY =^logy, after exponentiating the linear equation, we obtain y=^102−0.5x =¹⁰²(^10−0.5)^x =(¹⁰⁰)(⁰.³¹⁶²)^x

Power Functions Let’s look back at Figure 1.53. There, both axes are logarithmically transformed. We redraw just the straight line, usingY =^logyon the vertical axis and X =^logxon the horizontal axis (Figure 1.57).

5 4 3 2 1

0.5 0 0.5 1 1.5 2 2.5 3

2 1

X = log x 2

1 Y log y

Figure 1.57 Figure 1.53 redrawn. Now the vertical axis is labeled logyand the horizontal axis is labeled logx.

Using the linear equationY =c+m X, wherecis theY-intercept andmis the slope, we find, from Figure 1.57, that

c=^{4 and} m = −² WithY =^logyandX=^logx, the linear equation becomes

logy=⁴−^{2 log}x Exponentiating both sides, we get

y =^{104−2 logx} =¹⁰⁴(^10logx−2)=¹⁰⁴x⁻² The functiony=¹⁰⁴x⁻²is a power function.

A graph in which both the vertical and the horizontal axis are logarithmically scaled is called a log-log plot or double-log plot. If we display a power function y=bx^r in a double-log plot, a straight line results. To see this, we take logarithms to base 10 on both sides ofy=bx^r:

logy=^log(bx^r) ^(1.7)

Using the properties of logarithms on the right-hand side of (1.7), we get log(bx^r)=^logb+^logx^r =^logb+rlogx

If we setY =^logyand X=^logx, then (1.7) becomes Y =^logb+r X

Comparing this equation with the general form of a linear function,Y = c+m X, we see that theY-intercept is logband the slope isr. Ifr > 0, the slope is positive.

Ifr <0, the slope is negative.

EXAMPLE 6 Graph

y=¹⁰⁰x^−2/3, x >⁰ on a double-log plot.

Solution

We take logarithms first:

logy =^log(¹⁰⁰x^−2/3)=^{log 100}− ² 3logx

We setY =^logyandX =^logx. Then, with log 100=2, we find that Y =²−²

3X

This is the equation of a straight line withX-intercept 3 andY-intercept 2 (and thus slope−²/3). We graph this function in Figure 1.58, where we haveX andY on the two axes. If we use x and y on the two axes (Figure 1.59), the labels change: The y-intercept is now 100 (corresponding to log 100 = 2) and the x-intercept is 1000 (corresponding to log 1000 = 3). Note that the origin in Figure 1.58 is X = 0 and Y =0; the origin in Figure 1.59 isx =^{1 and}y=^1.

log x log y

3 2 1

1

1 0 1 2 3 4

Figure 1.58 The graphy=100x^−2/3 on a double-log plot where the axes are labeledX=logxandY=logy.

0.1 0.1

10 100 1000

10 100 1000 10000

1

x y

Figure 1.59 The graphy=100x^−2/3on a double-log plot where the axes are labeledxandy.

EXAMPLE 7 Find the functional relationship betweenxandyon the basis of the graph in Figure 1.60.

1

2.5 3 2 1.5 1 0.5 1 0.5

0 1 2 3 4

(log 1, log 0.01) (0, 2)

(log 1000, log 1) (3, 0)

X log x Y log y

Figure 1.60 The graph of the function for Example 7: The two points on the double-log plot used for finding the relationship are(log 1,log 0.01)and(log 1000,log 1).

Solution

IfY =^logyandX=^logx, then, in an X–Y graph, theY-intercept is log 0.⁰¹= −² and, using the two points(^{log 1},^{log 0}.⁰¹)^and(^{log 1000},^{log 1}), we calculate the slope as

log 1−log 0.01

log 1000−^{log 1} = ⁰−(−2) 3−⁰ = ²

3 Hence, the equation is

Y = −²+² 3X

WithY =^logyandX=^logx, we find that logy = −²+ ²

3logx and, after exponentiating both sides of this equation, we get

y=^{10−2+23logx} =^{10−210logx2/3} =(⁰.⁰¹)x^2/3

Thus, the functional relationship betweenxandyis a power function of the form y=(⁰.⁰¹)x^2/3

Applications

EXAMPLE 8 When growing plants at sufficiently high initial densities, we often observe that the number of plants decreases as the size of the plants grows. This property is called self-thinning. When the per-plant dry weight of the aboveground biomass is plotted on a log-log plot as a function of the density of survivors, we frequently find that the data lie along a straight line with slope−3/2. Assume that, for a particular plant, such a relationship holds for plant densities between 10²and 10⁴plants per square meter and that, at a density of 100 plants per square meter, the dry weight per plant is about 10 grams. Find the functional relationship between dry weight and plant density, and graph this function on a log-log plot.

Solution

Since the relationship between density (x) and dry weight (y) follows a straight line with slope−³/2 on a log-log plot, we set

logy=C− ³

2logx for 10² ≤x ≤¹⁰⁴

where C is a constant. To find C, we use the fact that when x = ^100, y = ^10.

Therefore,

log 10=C −³ 2log 100 or

1=C− ³

2·2, which implies that C =4 Hence,

logy=⁴− ³ 2logx

Exponentiating both sides (and remembering that “log” denotes the logarithm to base 10), we find that

y=¹⁰⁴x^−3/2 for 10² ≤x ≤¹⁰⁴

The graph of this function on a log-log scale is shown in Figure 1.61.

EXAMPLE 9 Polonium 210 (Po²¹⁰) is a radioactive material. To determine the half-life of Po²¹⁰ experimentally, we measure the amount of radioactive material left after timet for various values oft. When we plot the data on a semilog plot, we find that we can fit a straight line to the curve. The slope of the straight line is−⁰.0022/day. Find the half-life of Po²¹⁰.

Solution

Radioactive decay follows the equation

W(t)=W(⁰)e^−λt fort ≥⁰

0.001100 0.01 0.1

y 10⁴ x^3/2

1 10 100

1000 10000

y

x Figure 1.61 The graph of the function for Example 8: The line has slope−³/2 and goes through the point(log 100,log 10)on a double-log plot.

whereW(t)is the amount of radioactive material left after timet. If we log transform this equation, we obtain

logW(t)=^logW(⁰)−λtloge

Note that we use logarithms to base 10. If we plot W(t) as a function oft on a semilog plot, we obtain a straight line with slope−λ^loge. Matching this slope with the number given in the example, we obtain

λloge=0.0022/day Solving forλ^yields

λ= ¹

loge⁰.⁰⁰²²/^day

To find the half-lifeT_h, we use the formula (see Subsection 1.2.5) T_h = ^{ln 2}

λ = ^{ln 2}

0.⁰⁰²²(^loge)^days

≈¹³⁶.^{8 days}

Note that in the preceding example we used logarithms to base 10 to do the log transformation. The radioactive law was given in terms of the natural exponente. The slope therefore contained the factor loge≈⁰.^4343.

EXAMPLE 10 Light intensity in lakes decreases with depth. Denote by I(z)the light intensity at depthz, withz =0 representing the surface. Then the percentage surface radiation at depthz, denoted by PSR(z), is computed as

PSR(z)=¹⁰⁰I(z) I(0)

When we graph the percentage surface radiation as a function of depth on a semilog plot, a straight line results. An example of such a curve is given in Figure 1.62, where the coordinate system is rotated clockwise by 90^◦so that the depth axis points downward. Derive an equation forI(z)on the basis of the graph.

Solution

We see that the dependent variable, 100I(z)/I(⁰), is logarithmically transformed, whereas the independent variable,z, is on a linear scale. The graph is a straight line.

We thus find that

log 100I(z)

I(⁰) =c+mz

wherecis the intercept on the percentage surface radiation axis andmis the slope.

We see that

c=^{log 100} and, using the points(⁰,¹⁰⁰)^and(³⁰,¹)^{, we get}

m= ^{log 100}−^{log 1} 0−³⁰ = −²

30 = −¹ 15 Hence,

log 100I(z)

I(0) =^{log 100}− ¹ 15z

The left-hand side simplifies to log 100+log^I(z)_I(0). After canceling log 100 on both sides and exponentiating both sides, we find that

I(z)

I(⁰) =^10−(1/15)z =^exp[^{ln 10−(1/15)z}] Thus

I(z)= I(⁰)e⁻⁽

1 15ln 10)z

The number ₁₅¹ ln 10 is called thevertical attenuation coefficient. The magnitude of 0

5 10 15 20 25 30

z

35 40 45 50

100 10

0.01 0.1 1

100I(z) I(0)

Figure 1.62 The graph of percentage surface radiation as a function of depth.

this number tells us how quickly light is absorbed in a lake.