EXAMPLE 1

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## EXAMPLE 1

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**log**log log log a. 8 3 23 8 = = 4 12 2 1/4 b. 1 0 40 1 = = c. 12 1 121 12 = = –1 1 d. 4 –1 4 = = 4 EXAMPLE 1 Rewrite logarithmic equations Logarithmic Form Exponential Form**log**log log log 3 7 1/2 14 1. 81 4 34 81 = = 2. 7 1 71 7 = = –5 1 3. 1 0 140 1 = = 2 4. 32 –5 32 = = for Example 1 GUIDED PRACTICE Rewrite the equation in exponential form. Logarithmic Form Exponential Form**log**log log a. b. 25x 10log4 5 5 5 4 a. 10log4 = b log x = x b ( ) log 25x b. = 52 x 5 52x = 2x = bx log = x b EXAMPLE 5 Use inverse properties Simplify the expression. SOLUTION Express 25 as a power with base 5. Power of a power property**y = ln (x + 3)**log a. b. y = 6 x 6 a. From the definition of logarithm, the inverse of is x. y = 6 y = x ex = (y + 3) ex – 3 y = ANSWER The inverse ofy =ln(x + 3) isy = ex – 3. EXAMPLE 6 Find inverse functions Find the inverse of the function. SOLUTION b. y = ln (x + 3) Write original function. x = ln (y + 3) Switch x and y. Write in exponential form. Solve for y.**log**log 10. 8 log x 8 7 7 8 x = b = x log x log b 8 b 7–3x 11. = x log ax –3x 7–3x = a for Examples 5 and 6 GUIDED PRACTICE Simplify the expression. SOLUTION SOLUTION**log**log log 12. 64x 2 2 2 log 64x = 2 26x = ( ) 26 x 6x = bx log = x b 13. eln20 e e 20 = = = x log x log 20 e e for Examples 5 and 6 GUIDED PRACTICE Simplify the expression. SOLUTION Express 64 as a power with base 2. Power of a power property SOLUTION eln20**14.**Find the inverse of y = 4 x log 4 From the definition of logarithm, the inverse of y = x. y = 6 is 15. Find the inverse of y =ln(x – 5). ex = (y – 5) ex + 5 y = The inverse ofy =ln(x – 5) isy = ex + 5. ANSWER for Examples 5 and 6 GUIDED PRACTICE SOLUTION SOLUTION y =ln(x – 5) Write original function. x = ln (y – 5) Switch x and y. Write in exponential form. Solve for y.**log**a. y x = 3 EXAMPLE 7 Graph logarithmic functions Graph the function. SOLUTION 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.**log**b. y x = 1/2 EXAMPLE 7 Graph logarithmic functions Graph the function. SOLUTION 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.**Graph . State the domain and**range. y (x + 3) + 1 = log log 2 2 Sketch the graph of the parent function y = x, which passes through (1, 0), (2, 1), and (4, 2). EXAMPLE 8 Translate a logarithmic graph SOLUTION STEP 1 STEP 2 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.**log**16. y x = 5 If x = 1 y = 0, x = 5 y = 1, x = 10 y = 2 Plot several convenient points, such as (1, 0), (5, 1), and (10, 2). The y-axis is a vertical asymptote. for Examples 7 and 8 GUIDED PRACTICE Graph the function. State the domain and range. SOLUTION**for Examples 7 and 8**GUIDED PRACTICE 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.**log**17. y (x – 3) = 1/3 for Examples 7 and 8 GUIDED PRACTICE Graph the function. State the domain and range. SOLUTION domain: x > 3, range: all real numbers**log**18. y (x + 1) – 2 = 4 for Examples 7 and 8 GUIDED PRACTICE Graph the function. State the domain and range. SOLUTION domain: x > 21, range: all real numbers**log**log log log b. 0.2 a. 64 b 5 4 4 To help you find the value of y, ask yourself what power of bgives you y. a. 4 to what power gives 64? 43 64, so 64 3. = = b. 5 to what power gives 0.2? 0.2, so 0.2 –1. 5–1 log = = 5 EXAMPLE 2 Evaluate logarithms Evaluate the logarithm. SOLUTION**log**log log log log d. 125 c. 6 1/5 36 1/5 b 36 –3 1 1 To help you find the value of y, ask yourself what power of bgives you y. 5 5 c. to what power gives 125? 125, so 125 –3. = = 1 361/2 6, so 6 . d. 36 to what power gives 6? = = 2 EXAMPLE 2 Evaluate logarithms Evaluate the logarithm. SOLUTION**log 8**a. 8 100.903 8 b. ln 0.3 .3 e –1.204 0.3 EXAMPLE 3 Evaluate common and natural logarithms Expression Keystrokes Display Check 0.903089987 –1.203972804**Tornadoes**The wind speed s(in miles per hour) near the center of a tornado can be modeled by s 93logd + 65 = where dis 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. EXAMPLE 4 Evaluate a logarithmic model**s**93logd+ 65 = 93(2.342)+ 65 ANSWER The wind speed near the tornado’s center was about 283miles per hour. EXAMPLE 4 Evaluate a logarithmic model SOLUTION Write function. = 93log220+ 65 Substitute 220 for d. Use a calculator. = 282.806 Simplify.**log**log log log 5. 32 27 27 2 2 25 32, so 32 5. = = 6. 3 271/3 3, so 3 . 1 = = 3 for Examples 2, 3 and 4 GUIDED PRACTICE Evaluate the logarithm. Use a calculator if necessary. SOLUTION 2 to what power gives 32? SOLUTION 27 to what power gives 3?**log 12**7. 12 101.079 12 8. ln 0.75 .75 e –0.288 0.75 for Examples 2, 3 and 4 GUIDED PRACTICE Evaluate the logarithm. Use a calculator if necessary. Expression Keystrokes Display Check 1.079 –0.288**9.**WHAT IF?Use the function in Example 4 to estimate the wind speed near a tornado’s center if its path is 150miles long. s 93logd+ 65 = 93(2.1760)+ 65 ANSWER The wind speed near the tornado’s center is about 267 miles per hour. for Examples 2, 3 and 4 GUIDED PRACTICE SOLUTION Write function. = 93log150+ 65 Substitute 150 for d. Use a calculator. = 267 Simplify.**Use**3 0.792 and 7 1.404to evaluate the logarithm. log log log log log log log log log 4 4 4 4 4 4 4 4 4 – 7 a. 3 = and and log log 3 3 Use the given values of Use the given values of – 0.792 1.404 log log 7. 7. 4 4 4 4 3 –0.612 = 7 Write 21 as 3 7. b. (3 7) 21 = 3 7 = + + 0.792 1.404 2.196 = EXAMPLE 1 Use properties of logarithms Quotient property Simplify. Product property Simplify.**Use**3 0.792 and 7 1.404to evaluate the logarithm. log log log log log 4 4 4 4 4 c. 49 72 = Write 49 as 72 2 7 = log 7. 4 2(1.404) Use the given value of 2.808 = EXAMPLE 1 Use properties of logarithms Power property Simplify.**Use**5 0.898 and 8 1.161 to evaluate the log log 6 6 logarithm. log log log log log log 6 6 6 6 6 6 – 8 log 1. 5 = 6 and Use the given values of and log log 5 5 Use the given values of – 0.898 1.161 log log 8. 8. 6 6 6 6 5 –0.263 = 8 Write 40 as 8 5. 2. (8 5) 40 = 8 5 = + + 1.161 0.898 2.059 = for Example 1 GUIDED PRACTICE Quotient property Simplify. Product property Simplify.**Use**5 0.898 and 8 1.161 to evaluate the log log 6 6 logarithm. log log log log log log 6 6 6 6 6 6 3. 64 82 = Write 64 as 82 2 8 = log log 8. 5. 6 6 2(1.161) Use the given value of Use the given value of 2.322 = 4. 125 53 = Write 125 as 53 3 5 = 3(0.898) 2.694 = for Example 1 GUIDED PRACTICE Power property Simplify. Power property Simplify.**5x3**5x3 Expand y y log log log log log log log log log log 6 6 6 6 6 6 6 6 6 6 – 5x3 y = – 5 x3 + y = – 3 5 x + y = EXAMPLE 2 Expand a logarithmic expression SOLUTION Quotient property Product property Power property**–**– log 9 + 3log2 log 3 log 9 + log 23 log 3 = – log log 3 = (9 ) 23 log = 9 23 3 log 24 = ANSWER The correct answer is D. EXAMPLE 3 Standardized Test Practice SOLUTION Power property Product property Quotient property Simplify.**5.**Expand log 3 x4 . log 3 x4 log 3 + log x4 = log 3 + 4 log x = for Examples 2 and 3 GUIDED PRACTICE SOLUTION Product property Power property**6.**Condense ln 4 + 3 ln 3 – ln 12. – ln 4 + ln 33 ln 12 = – ln ln 12 = 4 33 ln = (4 ) 33 12 ln 9 = for Examples 2 and 3 GUIDED PRACTICE SOLUTION ln 4 + 3 ln 3 – ln 12 Power property Product property Quotient property Simplify.**Evaluate**8 8 8 using common logarithms and natural logarithms. log log log 3 3 3 log 8 0.9031 1.893 = 0.4771 log 3 ln 8 2.0794 1.893 = 1.0986 ln 3 EXAMPLE 4 Use the change-of-base formula SOLUTION Using common logarithms: Using natural logarithms:**Sound Intensity**I I I L(I) 10 log = 0 0 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 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**L(2I) – L(I)**= – 10 log 10 log = 2I 2I I I I I I I I I I I 0 0 0 0 0 0 10 – log log = – log 10 log 2 log = + 10 log 2 = 3.01 ANSWER The loudness increases by about 3decibels. EXAMPLE 5 Use properties of logarithms in real life SOLUTION Let Ibe the original intensity, so that 2Iis the doubled intensity. Increase in loudness Write an expression. Substitute. Distributive property Product property Simplify. Use a calculator.**8**14 8 14 log log log log 8 8 5 5 7. log 8 0.9031 1.292 = 0.6989 log 5 8. log 14 1.146 1.269 = 0.9031 log 8 for Examples 4 and 5 GUIDED PRACTICE Use the change-of-base formula to evaluate the logarithm. SOLUTION SOLUTION**log**log log log 12 26 26 12 9. 30 9 30 9 log 9 0.9542 0.674 = 1.4149 log 26 10. log 30 1.4777 1.369 = 1.076 log 12 for Examples 4 and 5 GUIDED PRACTICE Use the change-of-base formula to evaluate the logarithm. SOLUTION SOLUTION**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. I I L(I) 10 log = 0 for Examples 4 and 5 GUIDED PRACTICE SOLUTION Let Ibe the original intensity, so that 3Iis the tripled intensity.**Increase in loudness**L(3I) – L(I) = – 10 log 10 log = – 10 log log = 3I 3I I I I I – log 10 log 3 log = + I I I I I I 10 0 0 0 0 0 0 log 3 = 4.771 ANSWER The loudness increases by about 4.771decibels. for Examples 4 and 5 GUIDED PRACTICE Write an expression. Substitute. Distributive property Product property Simplify. Use a calculator.