3.08 Applications of logarithmic functions
Practical problems with exponential and logarithms
In science class, we have discussed radiocarbon dating. In social studies class, we have studies population growth. In economics, we study compounding annual interest rates. Did you know we were studying math in those classes? We were learning about exponential and logarithmic functions.
Modeling exponential growth and decay
In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:
$y=A_0e^{kt}$y=A0ekt
where $A_0$A0 is equal to the value at time zero (the starting point), $e$e is Euler’s constant, and $k$k is a positive constant that determines the rate (percentage) of growth. We may use the exponential growth function in applications involving doubling time, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time.
An exponential function with the form $y=A_0e^{kt}$y=A0ekt has the following characteristics:
- one-to-one function
- horizontal asymptote: $y=0$y=0
- domain: $(-\infty,\infty)$(−∞,∞)
- range: $(0,\infty)$(0,∞)
- $x$x -intercept: none
- $y$y-intercept: $\left(0,A_0\right)$(0,A0)
- increasing if $k>0$k>0
- decreasing if $k<0$k<0
Worked example
Question 1
A population of bacteria doubles every hour. If the culture started with $10$10 bacteria, what is the constant that determines the rate (percentage) of growth.
Think: To find the $k$k value, we must substitute the known values into the exponential equation. To find $A_0$A0 we use the fact that $A_0$A0 is the amount at the starting point; therefore, $A_0=10$A0=10. We know what happens after one hour; the culture doubles. So, the $t=1$t=1 hour and $y=20$y=20.
Do:
| $y$y | $=$= | $A_0e^{kt}$A0ekt |
(Given) |
| $20$20 | $=$= | $(10)e^{k*1}$(10)ek*1 |
(Substitute known values) |
| $2$2 | $=$= | $e^k$ek |
(Divide by $10$10 on both sides) |
| $\ln2$ln2 | $=$= | $k$k |
(Take the natural logarithm) |
question 2
The half-life of carbon-$14$14 is $5730$5730 years. Express the amount of carbon-$14$14 remaining as a function of time, $t$t.
Think: To find the half-life of a function describing exponential decay, solve the following equation:
$\frac{1}{2}A_0=A_0e^{kt}$12A0=A0ekt
Do:
| $\frac{1}{2}A_0$12A0 | $=$= | $A_0e^{kt}$A0ekt |
(Given) |
| $\frac{1}{2}A_0$12A0 | $=$= | $A_0e^{5730k}$A0e5730k |
(Substitute: $t=5730$t=5730) |
| $\frac{1}{2}$12 | $=$= | $e^{5730k}$e5730k |
(Divide $A_0$A0 by both sides) |
| $\ln\left(\frac{1}{2}\right)$ln(12) | $=$= | $5730k$5730k |
(Take the natural log) |
| $k$k | $=$= | $\frac{\ln\left(\frac{1}{2}\right)}{5730}$ln(12)5730 |
(Divide $5730$5730 by both sides) |
To express the remaining as a function of time, rewrite the formula replacing the $k$k with value above:
$A=\frac{1}{2}A_0$A=12A0
$A=A_0e^{\frac{\ln\left(\frac{1}{2}\right)}{5730}}$A=A0eln(12)5730
Reflect: When we are measuring decay, it is expected that the $k$k value be negative. Lets find the approximate value of the $k$k using the calculator: $k$k is approximately $-1.2097$−1.2097 X $10^{-4}$10−4
Given the half-life, find the decay rate:
1. Write $A=A_0e^{kt}$A=A0ekt
2. Replace $A$A by $\frac{1}{2}A_0$12A0 and replace $t$t by the given half-life.
3. Solve to find $k$k. Express $k$k as an exact value.
Note: It is also possible to find the decay rate using $k=\frac{\ln\frac{1}{2}}{t}$k=ln12t
Question 3
How long will it take for $10%$10% of a $1000$1000-gram sample of uranium-$235$235 to decay? Uranium-$235$235's half-life is $703800000$703800000 years.
Think: We are looking for the time it will take for $10%$10% of the substance to decay. Therefore, the starting point $A_0=1000$A0=1000 and the ending point is $A=0.9\times1000$A=0.9×1000 which is $A=900$A=900.
Do:
| $A$A | $=$= | $A_0e^{kt}$A0ekt |
(Given) |
| $900$900 | $=$= | $1000e^{kt}$1000ekt |
(Substitute) |
| $k$k | $=$= | $\frac{\ln(\frac{1}{2})}{t}$ln(12)t |
(Find $k$k) |
| $k$k | $=$= | $\frac{\ln(\frac{1}{2})}{703800000}$ln(12)703800000 |
(Substitute the half-life time $t=703800000$t=703800000) |
| $900$900 | $=$= | $1000e^{\frac{\ln(\frac{1}{2})}{703800000}*t}$1000eln(12)703800000*t |
(Substitute the value of $k$k) |
| $0.9$0.9 | $=$= | $e^{\frac{\ln(\frac{1}{2})}{703800000}*t}$eln(12)703800000*t |
(Divide $1000$1000 on both sides) |
| $\ln(0.9)$ln(0.9) | $=$= | $\frac{\ln(\frac{1}{2})}{703800000}*t$ln(12)703800000*t |
(Take the natural log)
|
| $t$t | $=$= | $\frac{(703800000)*\ln(0.9)}{\ln(\frac{1}{2})}$(703800000)*ln(0.9)ln(12) |
(Multiply $\frac{703800000}{\ln(\frac{1}{2})}$703800000ln(12) on both sides) |
| $t$t | $\approx$≈ | $106979777$106979777 |
(Simplify) |
Choosing an Appropriate Model for Data
Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.
Exploration
- Consider a number plate that has $n$n characters where each character can be any one of the $26$26 letters or the $10$10 digits. How can we construct an equation that expresses $n$n in terms of the number of possible combinations, $x$x?
Well, in each position there are $36$36 possible characters available. Consider a number plate with $7$7 characters, like the one shown below. There are $36\times36\times36\times36\times36\times36\times36$36×36×36×36×36×36×36 possible combinations, which we can write more simply as $36^7$367 combinations.
$7$7 digit number plate
In general, we can say that there are $36^n$36n total possible combinations for a number plate of length $n$n, and so we have the equation$x=36^n$x=36n
We can make $n$n the subject by rewriting the equation in logarithmic form:
$n=\log_{36}x$n=log36x
This equation may be valuable when forecasting where we might need to increase the number of characters for a growing population.
- How many characters would be needed for a country with a population of $2$2 million?
A quick substitution will tell us the number of characters needed for $2$2 million different possible number plates:
Since we can only have a whole number of characters, we want to round our number up. This will account for more than $2$2 million possible number plates - if we had rounded down instead, there wouldn't be enough unique number plates for everyone in the population.$n$n $=$= $\log_{36}x$log36x (Writing down the equation) $n$n $=$= $\log_{36}2000000$log362000000 (Substituting) $n$n $\approx$≈ $4.0487...$4.0487... (Evaluating) $n$n $=$= $5$5 (Rounding up)
Now consider the graph of the logarithm function given below.
Graph of $n=\log_{36}x$n=log36x - If the population doubles every year, how many years will it take until they have to increase the number of characters used?
Well after one year, the population will be $4$4 million, then it will become $8$8 million and so on. One strategy is to first use a point on the graph to identify when $n$n first exceeds $5$5. Well we can see that graph roughly crosses $n=5$n=5 when $x=60$x=60 million.
Point indicated at $x=60$x=60 million
If we substitute $x=60$x=60 million into the equation to check, we get a value of $n=4.9978\dots$n=4.9978… which is very close to our value of $n=5$n=5, but falls a little short.
Since the population increases in powers of $2$2, we know that after $6$6 years, the population will be $2^6=64$26=64 million. Substituting $x=64$x=64 million into the equation we get$n$n $=$= $\log_{36}64000000$log3664000000 (Substituting) $n$n $\approx$≈ $5.015\dots$5.015… (Simplifying) $n$n $=$= $6$6 (Rounding up)
So after $6$6 years, the population will increase to $x=64$x=64 million. At this point, license plates that are only $5$5 digits long will not be enough to account for everyone in the population, and so the country will have to start issuing $6$6 digit license plates.
Exploration
The Trinket and Widget Corporation (TWC) is exploring ways to increase the productivity of its workforce. Initial research shows that the number of widgets, $N$N, manufactured by the average TWC warehouse each month is related to the average monthly employee salary, $S$S, by the equation $N_1=40\log_2\left(S+1\right)$N1=40log2(S+1).
 |
| Graph of $N_1=40\log_2\left(S+1\right)$N1=40log2(S+1). |
The graph of the equation passes through the origin, which suggests that the employees are not willing to build any widgets if they have no salary. Notice that as the average monthly salary increases, the number of widgets produced also increases, but this number grows less and less with every increase in salary.
In an effort to produce more widgets, the management team at TWC send out an employee survey to find ways they can improve the working conditions. They find that providing free fruit for the employee lunchroom leads to a slight increase in productivity, which is modeled by the equation $N_2=60\log_2\left(S+1\right)$N2=60log2(S+1).
Let's compare the two equations on the same set of axes. The graph of the equation $N_2=60\log_2\left(S+1\right)$N2=60log2(S+1) corresponds to a stretching in the vertical direction of the graph of $N_1=40\log_2\left(S+1\right)$N1=40log2(S+1).
 |
| Graph of $N_1=40\log_2\left(S+1\right)$N1=40log2(S+1) and graph of $N_2=60\log_2\left(S+1\right)$N2=60log2(S+1). |
The management team can only go so far with free fruit. They want to consider the long term future of their workforce and have decided to trial a new fleet of robots to work on the widget assembly line.
The robots don't need a salary, and can reliably produce a large quantity of the most technically advanced components of the widget each month. The new productivity relationship is described by the equation $N_3=60\log_2\left(S+16\right)$N3=60log2(S+16).
On the axes below we show the graphs of the equation $N_2=60\log_2\left(S+1\right)$N2=60log2(S+1) and the equation $N_3=60\log_2\left(S+16\right)$N3=60log2(S+16). We can see that the graph describing the productivity with the robots is a horizontal translation of the graph describing the productivity resulting from only the free lunchroom fruit.
 |
| Graph of $N_2=60\log_2\left(S+1\right)$N2=60log2(S+1) and graph of $N_3=60\log_2\left(S+16\right)$N3=60log2(S+16). |
In general, a horizontal translation of the graph of $y=\log_bx$y=logbx is represented in the form $y=\log_b\left(x-h\right)$y=logb(x−h), where the resulting graph is translated $h$h units to the right, or by the form $y=\log_b\left(x+h\right)$y=logb(x+h), where the graph is translated $h$h units to the left.
In the example above, the graph of $N_2=60\log_2\left(S+1\right)$N2=60log2(S+1) is translated by $15$15 units to the left to produce the graph of $N_3=60\log_2\left(S+16\right)$N3=60log2(S+16).
Now suppose that new industry standards require the TWC to perform quality assurance tests on their widgets. Each month the engineering department will take $50$50 widgets at random off the production line and run through strength and durability assessments. These widgets will not be able to be sold to consumers, so this will decrease the output from the warehouse.
The productivity relationship that accounts for the quality assurance testing is described by the equation $N_4=60\log_2\left(S+16\right)-50$N4=60log2(S+16)−50. The graph of $N_3$N3 and the graph of $N_4$N4 are shown below.
 |
| Graph of $N_3=60\log_2\left(S+16\right)$N3=60log2(S+16) and graph of $N_4=60\log_2\left(S+16\right)-50$N4=60log2(S+16)−50. |
Notice that the graph of $N_4=60\log_2\left(S+16\right)-50$N4=60log2(S+16)−50 is the result of a vertical translation of the graph of $N_3=60\log_2\left(S+16\right)$N3=60log2(S+16) by $50$50 units downward.
Question 4
Following the productivity research, the Trinket and Widget Corporation has decided to implement all the recommendations of the management team. The widget productivity and employee salary relationship is now modeled by the equation $N=60\log_2\left(S+16\right)-50$N=60log2(S+16)−50.
- How many widgets will be produced if no salary is paid to employees?
If no salary is paid, then the value of S will be zero. When we substitute this into the equation we have
So there will still be $190$190 widgets produced each month.$N$N $=$= $60\log_216-50$60log216−50 $=$= $60\times4-50$60×4−50 $=$= $240-50$240−50 $=$= $190$190 - If the average salary is $\$500$$500 per month, what is the expected number of widgets produced?
By substituting $S=500$S=500 into the equation we have
The warehouse will produce around $490$490 widgets if the employees are given $\$500$$500 per month.$N$N $=$= $60\log_2\left(500+16\right)-50$60log2(500+16)−50 $=$= $60\log_2516-50$60log2516−50 $=$= $491$491 (to nearest whole widget) - How many more widgets will be produced if the monthly salary is doubled to $\$1000$$1000?
First we can find the number of widgets produced when the salary is $\$1000$$1000, then we can subtract from this the $491$491 widgets expected when the salary is only $\$500$$500.
By doubling the salary, the warehouse will produce and extra $549-491=58$549−491=58 widgets.$N$N $=$= $60\log_2\left(1000+16\right)-50$60log2(1000+16)−50 $=$= $60\log_21016-50$60log21016−50 $=$= $549$549 (to nearest whole widget)
Practice question
Question 5
pH is a measure of how acidic or alkaline a substance is, and the pH scale goes from $0$0 to $14$14, $0$0 being most acidic and $14$14 being most alkaline. Water in a stream has a neutral pH of about $7$7. The pH $\left(p\right)$(p) of a substance can be found according to the formula $p=-\log_{10}h$p=−log10h, where $h$h is the substance’s hydrogen ion concentration.
Store-bought apple juice has a hydrogen ion concentration of about $h=0.0002$h=0.0002.
Determine the pH of the apple juice correct to one decimal place.
Is the apple juice acidic or alkaline?
Acidic
AAlkaline
BA banana has a pH of about $8.3$8.3.
Solve for $h$h, its hydrogen ion concentration, leaving your answer as an exact value.
Question 6
A major communications company found that the more they spend on advertising, the higher their revenue. Their sales revenue, in thousands of dollars, is given by $R=10+20\log_4\left(x+1\right)$R=10+20log4(x+1), where $x$x represents the amount they spend on advertising (in thousands of dollars).
Determine their sales revenue if they spend no money on advertising.
Determine their sales revenue if they spend $\$14000$$14000 on advertising.
Give your answer to the nearest thousand $dollars$dollars.
Would you say that every extra $\$1000$$1000 spent on advertising becomes more or less effective in terms of raising revenue?
more effective: sales revenue increases as advertising spending increases.
Aless effective: every extra $\$1000$$1000 spent on advertising raises the sales revenue by less and less
B
Motivation for a log scale
Logarithm scales are often used when there is a large range of values involved with the variables under consideration. Here is a simple example to motivate the idea of a log scale:
Suppose we consider the set of five ordered pairs shown here:
| $x$x | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
|---|---|---|---|---|---|
| $y$y | $20$20 | $200$200 | $3631$3631 | $52481$52481 | $250000$250000 |
A plot of the five points would be difficult to manage because of the range of the $y$y values.
See the following graph plot, the scale on the $y$y-axis is so huge, that we lost a lot of the information from the first $3$3 points.
One way forward would be to develop a strategy that enables the reader to access the information indirectly. For example, we could plot the base $10$10 logarithm of $y$y against $x$x, with values shown in a new table.
| $x$x | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
|---|---|---|---|---|---|
| $\log_{10}y$log10y | $1.301$1.301 | $2.301$2.301 | $3.560$3.560 | $4.720$4.720 | $5.398$5.398 |
Even though the $y$y values are far more manageable in this form, we need to remember that the actual data points are those in the first table. That is to say the actual $y$y values, correct to $3$3 decimal places at least, are given by $10^{1.301},10^{2.301},10^{3.560},10^{4.720}$101.301,102.301,103.560,104.720 and $10^{5.398}$105.398.
Using the logarithm of the $y$y values gives us the following graph.
But of course, we can only retrieve the original data values by using a formula.
The idea that scientists and others struck upon was to leave the numbers alone (keep the $y$y values as they originally were in the first table) and simply change the spacings between numbers on the $y$y axis. That is, make the spacings between numbers proportional to the logarithms of the $y$y values.
Suppose for example we rule up the $y$y axis in the following way:
The first interval ( say of arbitrary length of $4$4 cm) starts from the origin, and covers the $y$y values from $1$1 to $10$10 ($10^0-10^1$100−101). The next $4$4 cm covers $y$y values from $10$10 to $100$100 ($10^1-10^2$101−102). The next $4$4 cm covers $y$y values from $100$100 to $1000$1000 ($10^2-10^3$102−103). The pattern continues with each $4$4 cm interval covering the $y$y values from $10^k$10k to $10^{k+1}$10k+1.
It is important to understand that within any of these intervals, the scale is not linear. Here is the beginning of the scale showing the first two intervals and the position of the first data point.
Note carefully that the gaps are getting smaller and smaller between $1$1 and $10$10 and between $10$10 and $100$100. Each tick between $1$1 and $10$10 is the position of $1,2,3,...9,10$1,2,3,...9,10. Each tick between $10$10 and $100$100 is the position of $10,20,30,...90,100$10,20,30,...90,100.
Because $\log_{10}20=1.301$log1020=1.301, the height of the point shown using the cm ruler would be $1.301\times4=5.204$1.301×4=5.204 cm.
Types of scales
Technically speaking, the above scale is called a semi-log $y$y scale because the $x$x-axis is still a linear scale. If we had changed the $x$x axis to a log scale instead of the $y$y -axis (for example if the $x$x values rather than the $y$y values had a large range), we would call it a semi-log $x$x scale. If we put both axes to log scales, we would call it a log-log scale.
Graphs of functions on log scales
On semi-log $y$y paper, a graph of the function $y=a^x$y=ax becomes a straight line. For example, consider the curve given by $y=2^x$y=2x for $x\ge0$x≥0. First we'll create a table of values shown here:
| $x$x | $0$0 | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
|---|---|---|---|---|---|---|
| $y$y | $1$1 | $2$2 | $4$4 | $8$8 | $16$16 | $32$32 |
We can plot the points on a semi-log $y$y graph as follows:
This is because, by taking logs on both sides, we see that $\log_{10}y=\log_{10}\left(2^x\right)=x\log_{10}\left(2\right)$log10y=log10(2x)=xlog10(2).
If we then set $Y=\log_{10}y$Y=log10y and the constant $\log_{10}2=m$log102=m, then the last equation becomes the straight line given by $Y=mx$Y=mx.
Generalizing a little, the function $y=A\times2^x$y=A×2x will also be a straight line on semi-log $y$y paper since by taking logs, we have $\log_{10}y=\log_{10}A+x\log_{10}2$log10y=log10A+xlog102, which could be expressed as $Y=mx+c$Y=mx+c.
The important point being is that using semi-log paper must necessarily change familiar curve shapes to quite different shapes.
Application of a log scale
As a final note, plotting with semi-log scales is a common strategy used by scientists to verify to nature of certain collected data.
For example, it may be that a scientist looks at population data that seems to exhibit exponential growth. When the data is plotted on normal axes, it looks to rise in a way consistent with such a model.
To test the hypothesis, she might plot the data on semi-log $y$y paper to see if all the data points fall onto a straight line. If the data does, then she has verified that growth is indeed exponential.
Practice questions
question 7
Below is a table of values that shows a log scale relating $x$x and $y$y. Form an equation relating $x$x and $y$y. Express the equation in logarithmic form.
| log scale measure ($y$y) | linear measure ($x$x) | |
|---|---|---|
| $0$0 | $=$= | $1$1 |
| $1$1 | $=$= | $10$10 |
| $2$2 | $=$= | $100$100 |
| $3$3 | $=$= | $1000$1000 |
| $4$4 | $=$= | $10000$10000 |
question 8
The decibel scale, used to record the loudness of sound, is a logarithmic scale. The least audible sound, with intensity $10^{-12}$10−12 watts/m2 is assigned the value of $0$0. A sound that is $10$10 times louder than this is assigned a decibel value of $10$10. A sound $100$100 ($10^2$102) times louder is assigned a decibel value of $20$20, and so on. In general, an increase of $10$10 decibels corresponds to an increase in magnitude of $10$10.
If the sound of a normal speaking voice is $50$50 decibels, and the sound in a bus terminal is $80$80 decibels, then how many times louder is the bus terminal compared to the speaking voice?
Give your final answer as a basic numeral, not in exponential form.
question 9
The Richter Scale is a base-$10$10 logarithmic scale used to measure the magnitude of an earthquake, given by $R=\log_{10}x$R=log10x, where $x$x is the relative strength of the quake. This means an earthquake that measures $4.0$4.0 on the Richter Scale will be $10$10 times stronger than one that measures $3.0$3.0.
The aftershock of an earthquake measured $6.7$6.7 on the Richter Scale, and the main quake was $4$4 times stronger. Solve for $r$r, the magnitude of the main quake on the Richter Scale, to one decimal place.