1. Chapter 5 LOGARITHMIC FUNCTIONS
Section 5.1 Logarithms and Their Properties
Section 5.2 Logarithms and Exponential Models
Section 5.3 The Logarithmic Function and Its Applications
Section 5.4 Logarithmic Scales
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

2. LOGARITHMS AND THEIR PROPERTIES
5.1
LOGARITHMS
AND THEIR PROPERTIES
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

3. What is a Logarithm?
If x is a positive number, log x is the exponent of 10 that gives x. In other words, if y = log x then 10y = x.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

4. Logarithms to Exponents
Example 1
Rewrite the following statements using exponents instead of logs.
log 100 = 2 (b) log 0.01 = −2 (c) log 30 = 1.477
Solution
We use the fact that if y = log x then 10y = x.
(a) 2 = log 100 means that 102 = 100.
(b) −2 = log 0.01 means that 10−2 = 0.01.
(c) = log 30 means that = 30. (Actually, this is only an approximation. Using a calculator, we see that = and that log 30 = )
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

5. Exponents to Logarithms
Example 2
Rewrite the following statements using logs instead of exponents.
(a) 105 = 100,000 (b) 10−4 = (c) =
Solution
We use the fact that if 10y = x, then y = log x.
(a) 105 = 100,000 means that log 100,000 = 5.
(b) 10−4 = means that log = −4.
(c) = means that log = 0.8. (This, too, is only an approximation because 10.8 actually equals )
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

6. Logarithms Are Exponents
Example 3
Without a calculator, evaluate the following, if possible:
(a) log 1 (c) log 1,000,000
(d) log (f) log(−100)
Solution
(a) We have log 1 = 0, since 100 = 1.
(c) Since 1,000,000 = 106, the exponent of 10 that gives 1,000,000 is 6. Thus, log 1,000,000 = 6.
(d) Since = 10−3, the exponent of 10 that gives is −3. Thus, log = −3.
(f) Since 10 to any power is positive, −100 cannot be written as a power of 10. Thus, log(−100) is undefined.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

7. Logarithmic and Exponential Functions are Inverses
For any N, log(10N) = N and for N > 0, 10logN = N.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

8. Properties of Logarithms
Properties of the Common Logarithm • By definition, y = log x means 10y = x. • In particular, log 1 = 0 and log 10 = 1. • The functions 10x and log x are inverses, so they “undo” each other: log(10x) = x for all x, 10log x = x for x > 0. • For a and b both positive and any value of t, log(a b) = log a + log b log(a/b) = log a − log b log(bt) = t · log b.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

9. Applying Properties of Logarithms
Example 5 Solve 100・2t = 337,000,000 for t. Solution Dividing both sides of the equation by 100 gives 2t = 3,370,000. Taking logs of both sides gives log 2t= log(3,370,000). Since log(2t) = t · log 2, we have t log 2 = log(3,370,000), so, solving for t, we have t = log(3,370,000)/log 2 =
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

10. The Natural Logarithm
For x > 0, ln x is the power of e that gives x or, in symbols, ln x = y means ey = x, and y is called the natural logarithm of x.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

11. Same Properties as for Common Logarithm
Properties of the Natural Logarithm • By definition, y = ln x means ey = x. • In particular, ln 1 = 0 and ln e = 1. • The functions ex and ln x are inverses, so they “undo” each other: ln(ex) = x for all x, eln x = x for x > 0. • For a and b both positive and any value of t, ln (a b) = ln a + ln b ln (a/b) = ln a − ln b ln (bt) = t · ln b.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

12. Applying Properties of Natural Logarithms
Example 6 (a)
Solve for x: 5e2x = 50
Solution (a)
We first divide both sides by 5 to obtain
e2x = 10.
Taking the natural log of both sides, we have
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

13. Misconceptions and Calculator Errors Involving Logs
• log(a + b) is not the same as log a + log b • log(a − b) is not the same as log a − log b • log(a b) is not the same as (log a)(log b) • log (a/b) is not the same as (log a) / (log b) • log (1/a) is not the same as 1 / (log a).
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

14. Justification of Log Properties Using Exponents
log(a · b) = log a + log b
If a and b are both positive, we can write a = 10m and b = 10n, so log a = m and log b = n. Then, the product a · b can be written
a · b = 10m · 10n = 10m+n.
Therefore m+ n is the power of 10 needed to give a · b, so
log(a · b) = m + n,
which gives log(a · b) = log a + log b.
log(bt) = t · log b Suppose that b is positive, so we can write b = 10k for some value of k. Then bt = (10k)t. Using a property of exponents, we can write (10k)t as 10kt, so bt = (10k)t = 10kt and log(bt) = kt. But since b = 10k, we know k = log b. This means log(bt) = (log b)t = t · log b.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

15. LOGARITHMS AND EXPONENTIAL MODELS
5.2
LOGARITHMS AND EXPONENTIAL MODELS
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

16. Using Logarithms to Undo Exponents
Example 2
The population of City A begins with 50,000 people and grows at 3.5% per year. The population of City B begins with a larger population of 250,000 people but grows at the slower rate of 1.6% per year. Assuming that these growth rates hold constant, will the population of City A ever catch up to the population of City B? If so, when?
Solution
If t is time measured in years and PA and PB are the populations of these two cities, then PA = 50,000(1.035)t and PB = 250,000(1.016)t.
We want to solve the equation 50,000(1.035)t = 250,000(1.016)t.
Dividing both sides by 50,000 (1.016)t gives (1.035/1.016)t = 5.
Solving using logs: t log(1.035/1.016) = log 5, giving t ≈
Thus, the cities’ populations will be equal in just under 87 years. To check this, letting t = , PA ≈ 992,575 and PB ≈ 992,572
The answers are not exactly equal because we rounded off the value of t.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

17. Doubling Time
Example 3(a) Find the time needed for the turtle population described by the function P = 175(1.145)t to double its initial size. Solution (a) The initial size is 175 turtles; doubling this gives 350 turtles. We need to solve the following equation for t: 175(1.145)t = t = 2 log 1.145t = log 2 t · log = log 2 t = log 2/log ≈ years. Note that 175(1.145)5.119 ≈ 350
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

18. Half-Life
Example 6 The quantity, Q, of a substance decays according to the formula Q = Q0e−kt, where t is in minutes. The half-life of the substance is 11 minutes. What is the value of k? Solution We know that after 11 minutes, Q = ½ Q0. Thus, solving for k, we get Q0e−k·11 = ½ Q0 e−11k = ½ −11k = ln ½ k = (ln ½) / (−11) ≈ , so k = per minute. This substance decays at the continuous rate of 6.301% per minute.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

19. Converting Between Q = a bt and Q = a ekt
Any exponential function can be written in either of the two forms: Q = a bt or Q = a ekt. If b = ek, so k = ln b, the two formulas represent the same function.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

20. Converting Between Q = a bt and Q = a ekt
Example 7
Convert the exponential function P = 175(1.145)t to the form P = aekt.
Solution
The parameter a in both functions represents the initial population. For all t, 175(1.145)t = 175(ek)t,
so we must find k such that ek =
Therefore k is the power of e which gives By the definition of ln, we have k = ln ≈
Therefore, P = 175e0.1354t.
Example 8
Convert the formula Q = 7e0.3t to the form Q = a bt.
Solution
Using the properties of exponents, Q = 7e0.3t = 7(e0.3)t.
Using a calculator, we find e0.3 ≈ , so Q = 7(1.3499)t.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

21. Exponential Growth Problems That Cannot Be Solved By Logarithms
Example 11 With t in years, the population of a country (in millions) is given by P = 2(1.02)t, while the food supply (in millions of people that can be fed) is given by N = 4+0.5t. Determine the year in which the country first experiences food shortages. Solution Setting P = N, we get 2(1.02)t = 4+0.5t, which, after dividing by 2 and taking the log of both sides can simplify to t log 1.02 = log( t). We cannot solve this equation algebraically, but we can estimate its solution numerically or graphically: t ≈ So it will be nearly 200 years before shortages are experienced.
Finding the intersection of linear and exponential graphs
Population
(millions)
Shortages start →
N = 4+0.5t
P = 2(1.02)t
t, years
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

22. THE LOGARITHMIC FUNCTION AND ITS APPLICATIONS
5.3
THE LOGARITHMIC FUNCTION AND ITS APPLICATIONS
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

23. The Graph, Domain, and Range of the Common Logarithm- -
The domain of log x is all positive numbers. Its range is all real numbers. The log function grows very rapidly for 0 < x < 1 and very slowly for x > 1. It has a vertical asymptote at x = 0 and never touches the y-axis.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

24. Graphs of the Inverse Functions y = log x and y = 10x
Log function x
y = log x
0.01 -2
0.1 -1 1 10
100 2
1000 3
(1,10 )
y = x
y = 10x
Exponential function x
y = 10x -2
0.01 -1
0.1 1 10 2
100 3
1000
(10,1 )
(0,1 )
y = log x
(-1,0.1 )
(1,0 )
(0.1,-1 )
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

25. Graph of the Natural Logarithm
y = ln x
Like the common log, the natural log is only defined for x > 0 and has a vertical asymptote at x = 0. The graph is slowly increasing and concave down. It also passes through (1, 0).
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

26. Graph of Natural Logarithm and Its Inverse
y = ex
The functions y = ln x and y = ex are inverses of one another. Notice how they are mirror images of one another through the line y = x.
y = x
y = ln x
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

27. Chemical Acidity
In chemistry, the acidity of a liquid is expressed using pH. The acidity depends on the hydrogen ion concentration in the liquid (in moles per liter); this concentration is written [H+]. The greater the hydrogen ion concentration, the more acidic the solution. The pH is defined as: pH = −log[H+].
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

28. Chemical Acidity Example 2
The hydrogen ion concentration of seawater is [H+] = 1.1 · 10−8. Estimate the pH of seawater. Then check your answer with a calculator.
Solution
We want to estimate pH = −log(1.1 · 10−8).
Since 1.1 · 10−8 ≈ 10−8 and log 10−8 = −8, we
know that pH = −log(1.1 · 10−8) ≈ −(−8) = 8.
Using a calculator, we have
pH = −log(1.1 · 10−8) =
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

29. Logarithms and Orders of Magnitude
Logarithms are useful in measuring quantities whose magnitudes vary widely, such as acidity (pH), sound (decibels), and earthquakes (the Richter scale).
If one object is 10 times heavier than another, we say it is an order of magnitude heavier. If one quantity is two factors of 10 greater than another, we say it is two orders of magnitude greater.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

30. Comparing Orders of Magnitude
Example 4 The sound intensity of a refrigerator motor is 10−11 watts/cm2. A typical school cafeteria has sound intensity of 10−8 watts/cm2. How many orders of magnitude more intense is the sound of the cafeteria? Solution To compare the two intensities, we compute their ratio: Sound Intensitycafeteria/Sound Intensityrefrigerator = 10−8/10−11 = 103. Thus, the sound intensity of the cafeteria is 1000 times greater than the sound intensity of the refrigerator. The log of this ratio is 3. We say that the sound intensity of the cafeteria is three orders of magnitude greater than the sound intensity of the refrigerator.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

31. Decibels
The intensity of audible sound varies over an enormous range. The range is so enormous that we consider the logarithm of the sound intensity. This is the idea behind the decibel (abbreviated dB). To measure a sound in decibels, the sound’s intensity, I, is compared to the intensity of a standard benchmark sound, I0. The intensity of I0 is defined to be I0 = 10−16 watts/cm2, roughly the lowest intensity audible to humans. The comparison between a sound intensity I and the benchmark sound intensity I0 is made as follows:
Noise level in decibels = 10 ・ log(I/I0).
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

32. Comparing Orders of Magnitude
Example 5 (b) Loud music can measure 110 dB whereas normal conversation measures 50 dB. How many times more intense is loud music than normal conversation? Solution (b) Using the formula for the noise level dB, the difference is 10 log (Imusic/I0) – 10 log (Iconversation/I0) = 110 – 50 = 60. Then dividing both sides by 10 and applying laws of exponents, So Imusic = 106 Iconversation , which means that loud music is 106 times, or one million times, as intense as normal conversation.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

33. Asymptotes and Limit Notation
Let y = f(x) be a function and let a be a finite number. • The graph of f has a horizontal asymptote of y = a if • The graph of f has a vertical asymptote of x = a if
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

34. Graphs of y = log x and y = 10x
The graph of y = log x has a vertical asymptote of x = 0 since y →-∞ as x → 0+
y = 10x
The graph of y = 10x has a horizontal asymptote of y= 0 since y → 0 as x → -∞
y = log x
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

35. 5.4
LOGARITHMIC SCALES
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

36. The Solar System and Beyond Linear Versus Log Scales
Distances from the Sun
To represent all the distances from the sun would require a huge linear scale, but one could represent the log base 10 of these distances, although the planets are closely cramped. This is called a log scale.
Object
Km from Sun
Log of Distance
Mercury
5.8∙107
Venus
1.08∙108
Earth
1.49∙108
Mars
2.28∙108
Jupiter
7.78∙108
Saturn
1.426∙109
Uranus
2.869∙109
Neptune
4.495∙109
Pluto
5.900∙109
Proxima Centauri
4.1∙1013
Andromeda Galaxy
2.4∙1019
Andromeda Galaxy
Proxima Centauri
Planets
Sun
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

37. Representing Distances on a Log Scale
Ages of events in earth’s history and logarithms of the ages The history of the world, like the distance to the stars and planets, involves numbers of vastly different sizes. Ages of certain events (from man emerging to the earth forming) have been converted to the logarithms of their ages and plotted below.
Notice how the powers of the logarithms are equally spaced on the scale.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

38. Representing Distances on a Log Scale
Another Way to Label a Log Scale
In the previous figure, the log scale has been labeled so that exponents are evenly spaced.
Another way to label a log scale is with the values themselves instead of the exponents.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

39. Log-Log Relationship Between Weight and Metabolic Rate
Animal
Wt (lbs)
Rate (k cal/day)
Log
(Wt)
(Rate)
Rat 1 35
1.544
Cat 8
166
0.903
2.220
Human
150
2000
2.176
3.301
Horse
1750
9470
3.243
3.976
It is not practical to plot the weight and rate data on an ordinary set of axes. Both input and output values span too broad a range.
However, we can plot the data using log scales for both the horizontal (weight)
axis and the vertical (rate) axis. Note the linear relationship.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

40. Log-Log Relationship Between Weight and Metabolic Rate
Labeling the log scale with the values themselves gives the following graph:
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

41. Fitting Log Functions to Data
Observing Non-Linear Rates of Growth The graph below shows the total number of iTunes sales (in millions of songs) over time, where t is the number of days since the iTunes Store opened on April 28, We see that sales have grown at an ever increasing rate. The trend is not linear.
The number of songs sold, N, as a function of
days, t, since iTunes store opened
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

42. Fitting Log Functions to Data
Using a Log or Log-Log Scale to Linearize Data How do we decide what functions fit the data in the figure below? A common method is to take logs of one or both variables and see if the resulting data is close to linear. In this case, taking the natural log of both sales and time in days works best.
A calculator or computer gives the equation of the line as
y = − x.
To transform back to our original variables, N and t, we substitute y = lnN and x = ln t, giving
lnN = − ln t.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

43. Fitting Exponential and Power Functions to Data
To fit an exponential function, N = a ekt, to a set of data of the form (t, N), we use three steps. First, we take the natural log of both sides and make the substitution y = ln N. This leads to the equation
y = ln N = ln (a ekt) = ln a + ln ekt = ln a + kt.
Setting b = ln a gives a linear equation with k as the slope and b as the y-intercept: y = b + k t.
To fit a power function, N = k tp, to a set of data of the form (t, N), we take similar steps. We first transform the relationship by the substitutions x = ln t and y = ln N. This leads to the equation
y = ln N = ln (k tp) = ln k + ln tp = ln k + p ln t.
Setting b = ln k and m = p gives a linear equation with p as the slope and b as the y-intercept: y = b + p ln t = b + p x.
Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally