EXAMPLE 1 Rewrite logarithmic equations Logarithmic FormExponential Form 2323 = 8 a. = 2 log = 1b. 4 log 1 = 0 = c. 12 log 121 = d. 1/4 log – = 12 4 = –1 1 4

GUIDED PRACTICE for Example 1 Rewrite the equation in exponential form. Logarithmic FormExponential Form 3434 = = 3 log = log 7 = 1 = log 10 = 4. 1/2 log – = 1 32 = –5 1 2

EXAMPLE 5 Use inverse properties Simplify the expression. a. 10 log 4 b. 5 log 25 x SOLUTION Express 25 as a power with base 5. a.10 log 4 = 4 b. 5 log 25 x = ( 5252 ) x 5 log = 5 52x52x 2x2x = Power of a power property b log x b = x b log bxbx = x

EXAMPLE 6 Find inverse functions Find the inverse of the function. SOLUTION b. a. y = 6 x b. y = ln (x + 3) a. 6 log From the definition of logarithm, the inverse of y = 6 x is y = x.x. Write original function. y = ln (x + 3) Switch x and y. x = ln (y + 3) Write in exponential form. Solve for y. = exex (y + 3) = e x – 3 y ANSWER The inverse of y = ln (x + 3) is y =e x – 3.

GUIDED PRACTICE for Examples 5 and 6 Simplify the expression. SOLUTION log x 8 8 = x b log b b = x log 7 –3x SOLUTION 7 log 7 –3x = –3x a log axax = x

GUIDED PRACTICE for Examples 5 and 6 Simplify the expression. SOLUTION log 64 x Express 64 as a power with base 2. 2 log 64 x = ( 2626 ) x 2 log = 2 26x26x 6x6x = Power of a power property b log bxbx = x 13.e ln20 SOLUTION e ln20 = e log 20 e = 20 e log x e = x

GUIDED PRACTICE for Examples 5 and 6 Find the inverse of 14. y = 4 x SOLUTION From the definition of logarithm, the inverse of 4 log y = 6 y = x.x. is y = ln (x – 5). Find the inverse of 15. y = ln (x – 5) SOLUTION Write original function. Switch x and y. x = ln (y – 5 ) Write in exponential form. Solve for y. = exex (y – 5 ) = e x + 5 y ANSWER The inverse of y = ln (x – 5) is y =e x + 5.

EXAMPLE 7 Graph logarithmic functions Graph the function. SOLUTION a.y = 3 log x Plot several convenient points, such as (1, 0), (3, 1), and (9, 2). The y -axis is a vertical asymptote. From left to right, draw a curve that starts just to the right of the y -axis and moves up through the plotted points, as shown below.

EXAMPLE 7 Graph logarithmic functions Graph the function. SOLUTION b.y = 1/2 log x Plot several convenient points, such as (1, 0), (2, –1), (4, –2), and (8, –3). The y -axis is a vertical asymptote. From left to right, draw a curve that starts just to the right of the y -axis and moves down through the plotted points, as shown below.

EXAMPLE 8 Translate a logarithmic graph SOLUTION STEP 1 Graph. State the domain and range. y = 2 log (x + 3) + 1 STEP 2 Sketch the graph of the parent function y = x, which passes through (1, 0), (2, 1), and (4, 2). 2 log Translate the parent graph left 3 units and up 1 unit. The translated graph passes through (–2, 1), (–1, 2), and (1, 3). The graph’s asymptote is x = –3. The domain is x > –3, and the range is all real numbers.

GUIDED PRACTICE for Examples 7 and 8 Graph the function. State the domain and range. SOLUTION 16.y = 5 log x If x = 1 y = 0, x = 5 y = 1, x = 10y = 2 Plot several convenient points, such as (1, 0), (5, 1), and (10, 2). The y -axis is a vertical asymptote.

GUIDED PRACTICE for Examples 7 and 8 From left to right, draw a curve that starts just to the right of the y -axis and moves up through the plotted points. The domain is x > 0, and the range is all real numbers.

GUIDED PRACTICE for Examples 7 and 8 Graph the function. State the domain and range. SOLUTION 17.y = 1/3 log (x – 3) domain: x > 3, range: all real numbers

GUIDED PRACTICE for Examples 7 and 8 Graph the function. State the domain and range. SOLUTION 18.y = 4 log (x + 1) – 2 domain: x > 21, range: all real numbers

EXAMPLE 2 Evaluate logarithms 4 log a.64 b. 5 log 0.2 Evaluate the logarithm. b logTo help you find the value of y, ask yourself what power of b gives you y. SOLUTION 4 to what power gives 64 ? a. 4 log , so = 3.3. = 64 5 to what power gives 0.2 ? b. = 5 –1  0.2, so – log =

EXAMPLE 2 Evaluate logarithms Evaluate the logarithm. b logTo help you find the value of y, ask yourself what power of b gives you y. SOLUTION = – , so 1/5 log 125 = –3. c. to what power gives 125 ? 1 5 d. 36 to what power gives 6 ? 36 1/2 6, so 36 log 6 == 1 2. d. 36 log 6 c. 1/5 log 125

EXAMPLE 3 Evaluate common and natural logarithms ExpressionKeystrokesDisplay a. log 8 b. ln 0.3 Check – e –1.204

EXAMPLE 4 Evaluate a logarithmic model Tornadoes The wind speed s (in miles per hour) near the center of a tornado can be modeled by where d is the distance (in miles) that the tornado travels. In 1925, a tornado traveled 220 miles through three states. Estimate the wind speed near the tornado’s center. 93 log d + 65 s =

EXAMPLE 4 Evaluate a logarithmic model SOLUTION = 93 log Write function. 93(2.342) + 65 = Substitute 220 for d. Use a calculator. Simplify. The wind speed near the tornado’s center was about 283 miles per hour. ANSWER 93 log d + 65 s =

GUIDED PRACTICE for Examples 2, 3 and 4 Evaluate the logarithm. Use a calculator if necessary. 2 to what power gives 32 ? , so = 5.5. = 2 log to what power gives 3 ? /3 3, so 27 log 3 == SOLUTION 27 log 6.3 SOLUTION

GUIDED PRACTICE for Examples 2, 3 and 4 Evaluate the logarithm. Use a calculator if necessary. ExpressionKeystrokesDisplay 7. log ln 0.75 Check – e –0.288

GUIDED PRACTICE for Examples 2, 3 and 4 WHAT IF? Use the function in Example 4 to estimate the wind speed near a tornado’s center if its path is 150 miles long. 9. = 93 log Write function. 93(2.1760) + 65 = 267 Substitute 150 for d. Use a calculator. Simplify. 93 log d + 65 s = SOLUTION The wind speed near the tornado’s center is about 267 miles per hour. ANSWER

EXAMPLE 1 Use properties of logarithms b. 4 log 21 a. 4 log 3 7 = 3 – = –0.612 = 4 log (3 7) = 3 4 log = – log Use and 4 log to evaluate the logarithm. Quotient property Simplify. Use the given values of 3 4 log and Write 21 as 3 7. Product property Use the given values of 3 4 log and Simplify.

EXAMPLE 1 Use properties of logarithms c. 4 log log Use and 4 log to evaluate the logarithm = 4 log = (1.404) Write 49 as 7272 Power property Use the given value of log Simplify. 4 log = 27

GUIDED PRACTICE for Example log 40 = 5 – 6 log 8 6 = –0.263 = 6 log (8 5) = 8 6 log = Quotient property Simplify – Use the given values of 5 6 log and Write 40 as 8 5. Product property Use the given values of 5 6 log and Simplify log 5 6 Use and to evaluate the logarithm. 6 log

GUIDED PRACTICE for Example log Use and to evaluate the logarithm. 6 log = 6 log = = 28 6 log 2(1.161) Write 64 as 8282 Power property Use the given value of log Simplify log = 6 log = = 35 6 log 3(0.898) Write 125 as 5353 Power property Use the given value of log Simplify.

EXAMPLE 2 Expand a logarithmic expression Power property Expand 6 log 5x35x3 y 6 5x35x3 y Quotient property = 5 6 log x3x3 y 6 – 6 + = 5 6 x y 6 – Product property SOLUTION = 5x35x3 y 6 log – 6

EXAMPLE 3 Standardized Test Practice Quotient property Product property Simplify. SOLUTION – log log 2 log 3 = – log 9 + log 2323 log 3 Power property = 24 log = (9 )2323 – log 3 = log The correct answer is D. ANSWER

GUIDED PRACTICE for Examples 2 and 3 Expand 5. log 3 x4x4. SOLUTION = log 3 + log x 4 log 3 x4x4 = log log x Power property Product property

GUIDED PRACTICE for Examples 2 and 3 SOLUTION Quotient property Product property Simplify. = – ln 4 + ln3 ln 12 Power property = ln (4 )3 – ln 12 = ln = ln 9 Condense ln ln 3 – ln ln ln 3 – ln 12

EXAMPLE 4 Use the change-of-base formula SOLUTION 3 log 8 Evaluate using common logarithms and natural logarithms. Using common logarithms: Using natural logarithms: 3 log 8 = log 8 log log 8 = ln 8 ln

EXAMPLE 5 Use properties of logarithms in real life For a sound with intensity I (in watts per square meter), the loudness L(I) of the sound (in decibels) is given by the function = log L(I)L(I)10 I 0 I Sound Intensity 0 I where is the intensity of a barely audible sound (about watts per square meter). An artist in a recording studio turns up the volume of a track so that the sound’s intensity doubles. By how many decibels does the loudness increase? 10 –12

EXAMPLE 5 Use properties of logarithms in real life Product property Simplify. SOLUTION Let I be the original intensity, so that 2I is the doubled intensity. Increase in loudness = L(2I) – L(I) = log 10 I 0 I log 10 2I2I 0 I – I 0 I 2I2I 0 I = log – = 210 log I 0 I – I 0 I + ANSWERThe loudness increases by about 3 decibels. 10 log 2 = 3.01 Write an expression. Substitute. Distributive property Use a calculator.

GUIDED PRACTICE for Examples 4 and 5 Use the change-of-base formula to evaluate the logarithm. 5 log 87. SOLUTION 5 log 8 = log 8 log log 148. SOLUTION 8 log 14 = log 14 log

GUIDED PRACTICE for Examples 4 and 5 Use the change-of-base formula to evaluate the logarithm. 26 log 99. SOLUTION = log 30 log log log log 30 = log 9 log 26

GUIDED PRACTICE for Examples 4 and 5 WHAT IF? In Example 5, suppose the artist turns up the volume so that the sound’s intensity triples. By how many decibels does the loudness increase? 11. SOLUTION Let I be the original intensity, so that 3I is the tripled intensity. = log L(I)L(I)10 I 0 I

GUIDED PRACTICE for Examples 4 and 5 Product property Simplify. = Increase in loudness L(3I) – L(I) = log 10 I 0 I log 10 3I3I 0 I – = 3 log I 0 I – I 0 I + ANSWER The loudness increases by about decibels. 10 log 3 = Write an expression. Substitute. Distributive property Use a calculator. = 10 log I 0 I 3I3I 0 I –