Logarithmic Functions and Graphs Section 5.3 Logarithmic Functions and Graphs

Objectives Find common logarithms and natural logarithms with and without a calculator. Convert between exponential and logarithmic equations. Change logarithmic bases. Graph logarithmic functions. Solve applied problems involving logarithmic functions.

Logarithmic Functions These functions are inverses of exponential functions. We can draw the graph of the inverse of an exponential function by interchanging x and y. To Graph: x = 2y. 1. Choose values for y. 2. Compute values for x. 3. Plot the points and connect them with a smooth curve. * Note that the curve does not touch or cross the y-axis.

Example Graph: x = 2y.

Example (continued) This curve looks like the graph of y = 2x reflected across the line y = x, as we would expect for an inverse. The inverse of y = 2x is x = 2y.

Logarithmic Function, Base a We define y = loga x as that number y such that x = ay, where x > 0 and a is a positive constant other than 1. We read loga x as “the logarithm, base a, of x.”

Example Find each of the following logarithms. a) log10 10,000 b) log10 0.01 c) log5 125 d) log9 3 e) log6 1 f) log5 5 a) The exponent to which we raise 10 to obtain 10,000 is 4; thus log10 10,000 = 4. b) We have The exponent to which we raise 10 to get 0.01 is –2, so log10 0.01 = –2.

Example (continued) c) 125 = 53: The exponent to which we raise 5 to get 125 is 3, so log5 125 = 3. d) log9 3: The exponent to which we raise 9 to get 3 is 1/2; thus log9 3 = 1/2. e) log6 1: 1 = 60. The exponent to which we raise 6 to get 1 is 0, so log6 1 = 0. f) 5 = 51. The exponent to which we raise 5 to get 5 is 1, so log5 5 = 1.

Logarithms loga 1 = 0 and loga a = 1, for any logarithmic base a. A logarithm is an exponent!

Example Convert each of the following to a logarithmic equation. a) 16 = 2x b) 10–3 = 0.001 c) et = 70 The exponent is the logarithm. a) 16 = 2x log216 = x The base remains the same. b) 10–3 = 0.001 g log10 0.001 = –3 c) et = 70 g log e 70 = t

Example Convert each of the following to an exponential equation. a) log 2 32 = 5 b) log a Q = 8 c) x = log t M The logarithm is the exponent. a) log 2 32 = 5 25 = 32 The base remains the same. b) log a Q = 8 → a8 = Q c) x = log t M → tx = M

Example Find each of the following common logarithms on a calculator. Round to four decimal places. a) log 645,778 b) log 0.0000239 c) log (3) Function Value Readout Rounded a) log 645,778 5.8101 b) log 0.0000239 –4.6216 c) log (–3) Does not exist

Natural Logarithms Logarithms, base e, are called natural logarithms. The abbreviation “ln” is generally used for natural logarithms. Thus, ln x means loge x. ln 1 = 0 and ln e = 1, for the logarithmic base e.

Example Find each of the following natural logarithms on a calculator. Round to four decimal places. a) ln 645,778 b) ln 0.0000239 c) log (5) d) ln e e) ln 1 Function Value Readout Rounded a) ln 645,778 13.3782 b) ln 0.0000239 –10.6416

Example (continued) Function Value Readout Rounded c) ln (–5) Does not exist. b) ln e 1 c) ln 1 0

Changing Logarithmic Bases The Change-of-Base Formula For any logarithmic bases a and b, and any positive number M,

Example Find log5 8 using common logarithms First, we let a = 10, b = 5, and M = 8. Then we substitute into the change-of-base formula:

Example We can also use base e for a conversion. Find log5 8 using natural logarithms Substituting e for a, 6 for b and 8 for M, we have

Example Graph: y = f (x) = log5 x. Method 1: y = log5 x is equivalent to x = 5y. Select y and compute x.

Example (continued) Graph: y = f (x) = log5 x. Method 2 Use a graphing calculator. First change bases.

Example (continued) Graph: y = f (x) = log5 x. Method 3 Calculators which graph inverses automatically. Begin with y1 = 5x, the graphs of both y1 and its inverse y2 = log 5 x will be drawn.

Example Graph each of the following. Describe how each graph can be obtained from the graph of y = ln x. Give the domain and the vertical asymptote of each function. a) f (x) = ln (x + 3) b) c) f (x) = |ln (x – 1)|

Example (continued) a) f (x) = ln (x + 3) The graph is a shift 3 units left. The domain is the set of all real numbers greater than –3, (–3, ∞). The line x = –3 is the vertical asymptote.

Example (continued) b) The graph is a vertical shrinking of y = ln x, followed by a reflection across the x-axis and a translation up 3 units. The domain is the set of all positive real numbers, (0, ∞). The y-axis is the vertical asymptote.

Example (continued) c) The graph is a translation of y = ln x, right 1 unit. The effect of the absolute is to reflect the negative output across the x-axis. The domain is the set of all positive real numbers greater than 1, (1, ∞). The line x =1 is the vertical asymptote.

Example In a study by psychologists Bornstein and Bornstein, it was found that the average walking speed w, in feet per second, of a person living in a city of population P, in thousands, is given by the function w(P) = 0.37 ln P + 0.05.

Example a. The population of Billings, Montana, is 106954. Find the average walking speed of people living in Billings. The population of Chicago, Illinois, is 2,714,856. Find the average walking speed of people living in Chicago. Graph the function. d. A sociologist computes the average walking speed in a city to the approximately 2.0 ft/sec. Use this information to estimate the population of the city.

Example (continued) a. Since P is in thousands and 106,954 = 106.954 thousand, we substitute 106.954 for P: w(106.954) = 0.37 ln 106.954 + 0.05  1.8 ft/sec. The average walking speed of people living in Billings is about 1.8 ft/sec. b. Substitute 2714.856 for P: w(2714.856) = 0.37 ln 2714.856 + 0.05  3.0 ft/sec. The average walking speed of people living in Chicago is about 3.0 ft/sec.

Example (continued) c. Graph with a viewing window [0, 600, 0, 4] because inputs are very large and outputs are very small by comparison.

Example (continued) d. To find the population for which the walking speed is 2.0 ft/sec, we substitute 2.0 for w(P), 2.0 = 0.37 ln P + 0.05, and solve for P. Use the Intersect method. Graph y1 = 0.37 ln x + 0.05 and y2 = 2. In a city with an average walking speed of 2.0 ft/sec, the population is about 194.5 thousand or 194,500.