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Chapter 4. Section 4.2 Logarithmic Functions and Models. Logarithmic Functions. What Is a Logarithm ? Recall the exponential function a x We define a new function, the logarithm with base a , as where log a x is the power of the base a For example.
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Chapter 4 Section 4.2 Logarithmic Functions and Models
Logarithmic Functions • What Is a Logarithm ? • Recall the exponential function ax • We define a new function, the logarithm with basea , as where logax is the power of the base a • For example with basea , a > 0 , a ≠ 1 logax = y ay = x that equals x 25 = 32 log232 = 5 since 103 = 1000 log101000 = 3 since and Section 4.7 v5.2.1
ay = x ax alog ay = Logarithmic Functions loga x = y • So we can think of the logarithm as an exponent • Since yIS the logarithm, we could also write So, acting on variable x or y or … with either function logaay = logax = y = x and … and then on that result with the other function… yields the original variable value Section 4.7 v5.2.1
= x a logax Logarithmic Functions • Consider the exponential function f(x) = ax,0 < a≠ 1 • We can show that f(x) is a 1–1 function • Hence f(x)doeshave an inversefunctionf–1(x) f–1(f(x)) =f–1(ax) = x • Since we showed that we call this inverse the logarithmfunction with basea • Since logax is an inverse ofaxthen loga(ax) = x and = alogax = y logaay = logax What are the domain and range of ax ? Question: What are the domain and range of logax ? Section 4.7 v5.2.1
y x Logarithmic Functions • A Second Look Graphs of f(x) = axand f–1(x) = logax are mirror images with respect to the line y = x Note that y = f–1(x)if and only if f(y) = ay = x y = logaxiffi ay = x f(x) = ax , for a > 1 (1, a) f–1(x) = logax ● (0, 1) ● ● (a, 1) ● (1, 0) y = x Remember: logax is the power of a that yields x Section 4.7 v5.2.1
y f(x) = ax , for a > 1 (1, a) f–1(x) = logax ● (0, 1) ● ● ● x y = x Logarithmic Functions • A Third Look • loga x = ymeansay = x What if x = a ? and ay = a Sofor any basea > 0, a ≠ 1, logaa = 1 • What if x =1 ? Then loga1 = y so ay = 1 and thus y = 0 So for any basea > 0 , a ≠ 1 , loga1 = 0 Then logaa = y = a1 so that y = 1 WHY? (a, 1) (1, 0) WHY? Can logax = ax ? Question: Section 4.7 v5.2.1
y x Logarithmic Functions • A Fourth Look • What if 0 < a < 1 ? Then what is f–1(x) ? Are the graphs still symmetricwith respect to line y = x ? YES ! Is it still true that f(1) = aandf–1(x) = 1 ? YES ! • Is it possible that loga x = ax ? YES ! on the liney = x f(x) = ax , for a< 1 (a, 1) ● ● (0, 1) ● (1, a) ● ● (1, 0) y = x f–1(x) = logax For what x does logax = ax ? Question: Section 4.7 v5.2.1
y x Logarithmic Functions • Example • Let a = ½ Then f(x) = ax Then what is f–1(x) ? f–1(x) = logax = log½x It is still true that f(1) =(½)1=½ and f–1(½) = log½(½) = 1 Where do the graphs intersect? y= (½)x = log½x = x f(x) = ax , for a = ½ = (½)x (½,1) ● ● (0,1) ● (1,½) ● ● (1,0) y = x f–1(x) = log1/2x WHY ? Section 4.7 v5.2.1
Exponential/Logarithmic Comparisons • Compare ax with logax ExponentialInverse Function 102 = 100 34 = 81 (½)–5 = 32 25 = 32 log232 = 5 5–3 = 1/125 log5(1/125) = –3 e3 ≈ 20.08553692 • Common Logarithm • Base 10 • Log10 x = Log x • Natural Logarithm • Base e • Loge x = ln x 2 log10100 = log381 = 4 log1/232 = –5 loge(20.08553692) ≈ 3 Section 4.7 v5.2.1
One-to-One Property Review • Since ax and loga x are 1-1 • 1. If x = y then ax = ay • 2. If ax = ay then x = y • 3. If x = y then loga x = loga y • 4. If log xa = loga y then x = y • These facts can be used to solve equations Example: Solve 10x – 1 = 100 10x – 1 = 102 x – 1 = 2 x = 3 { 3 } Solution set : Section 4.7 v5.2.1
= 3–3 = 3–3 3log x 3log x 3 3 1 1 1 1 4 27 27 27 x = x = { } Solution set: x log36 = By definition the logarithm (i.e. x) is the power of the base that yields = 36x 4 4 4 6 6 6 = 61/4 62x 2x = 1 {} 1 x Solution set: = 8 8 Solving Equations • Solve • 1.log3 x = –3 With inverse function: • 2. With definition: 3–3 = x This is already solved for x , so simplify Section 4.7 v5.2.1
logx 5 = x4 x 4 x4 x = = 4 x = x , for x > 0 5 = { } –x , for x < 0 4 4 4 5 5 5 5 5 Since x2 = 5 yields x = doesn’t x4 = 5 yield x = ? – If x = Solving Equations • Solve • 3.logx 5 = 4 By inverse function: 5 = x4 By definition: ( loga x = b iffi ab = x ) x4 = 5 Solution set: Question: , then why Note: then logx 5 is defined with a negativebase !! Section 4.7 v5.2.1
log3 (x2+ 5) = = 32 3 x2+ 5 x = 4 x 2 = { 2 } Solution set: Solving Equations • Solve • 4.log3 (x2+ 5) = 2 By inverse function: 9 – 5 = x2 • 5. 12 = 4x log4 12 = log4 (4x) By definition: ( 2 is the power of 3 that yields x2+ 5 ) 32 = 9 = x2+ 5 x2 = 4 x = 2 = x Question: Now, how do we find log4 12 ? Section 4.7 v5.2.1
Solving Equations • Remember:ax and logax are inverses • 1. To remove a variable from an exponent, find the logarithm of the exponential form • 2. To remove a variable from a logarithm, exponentiate Section 4.7 v5.2.1
R = log10() R = log10() so > 0 > 1 x x x x0 x0 x0 The Richter Scale • For fixed intensity x0 (as measured with a seismometer) the ratio of the seismic intensityx of an earthquake,relative to x0 , is measured by • This is the famous Richter Scale for measuring the relative “strength” of earthquakes • For x > x0 note that How fast does R grow ? Question: If R increases by 1 what is the change in x ? Section 4.7 v5.2.1
N and RN = log10 x0 107.3 = N L L x0 x0 x0 Richter Scale Comparisons • In 1992 the Landers earthquake produced a Richter scale value of 7.3 compared with the 1994 Northridge earthquake which hit 6.7 on the Richter scale • How much more powerful was the Landers earthquake, expressed as a ratio ? • Let RL = Landers intensity and RN = Northridge intensity So RL = log10 • From the definition of logarithm Thus L = x0107.3 and N = x0106.7 • Now all we need is the ratio of L to N ... = 6.7 = 7.3 and 106.7 = Section 4.7 v5.2.1
x0 107.3 x0 107.3 L L 107.3 – 4.8 107.3 – 6.7 100.6 102.5 = = = = = = N S x0 106.7 x0 104.8 The Richter Scale (continued) • The ratio is Thus L = 3.981N , i.e. 3.981times as strong as Northridge • How much stronger was Landers than the smallest recorded earthquake with Rs = 4.8 ? Landers was 316times stronger than the smallest quake ≈ 3.981 ≈ 316 Section 4.7 v5.2.1
2.00 3.16 1.58 Earthquake Comparison Ratios Richter Scale Relative Strength Ratios • 2010 Haiti • 1994 Northridge • 2010 Haiti • 1992 Landers • 2010 Chile • 2004 Indonesia • 1960 Chile 6.1 3.98 6.7 7.94 2.00 7.0 3.98 501.2 7.3 63.10 ? 31.62 199.53 8.8 100.0 9.3 9.5 Section 4.7 v5.2.1
R R 10.0 9.0 8.0 8.0 7.0 7.0 6.0 6.0 5.0 4.0 x 3.0 20 10 2.0 x 1.0 4,000 1,000 3,000 2,000 Earthquake Comparison Graph ● ● Chile9.5 Indonesia9.3 ● Chile8.8 ● ● ● Landers7.3 ● Haiti7.0 Landers7.3 ● Haiti7.0 ● Northridge6.7 ● Haiti6.1 Intensityx 106 Intensityx 106 Section 4.7 v5.2.1
398.1 1.585 1.995 Earthquake Comparison Ratios Richter Scale Relative Strength Ratios • 2011 Hawaii • 2011 Hawaii • 2010 Haiti • 2010 Chile • 2011 Japan • 2004 Indonesia • 1960 Chile 4.5 79.43 6.4 316.2 3.981 7.0 31,622.8 63.10 8.8 100.0 ? 1.585 1,258.9 9.0 9.3 3.162 9.5 Section 4.7 v5.2.1
125.9 1.995 1.585 Earthquake Comparison Ratios Richter Scale Relative Strength Ratios • 2011 Hawaii • 1989 California • 2010 Haiti • 2010 Chile • 2011 Japan • 1964 Alaska • 1960 Chile 6.4 3.162 6.9 3.981 1.259 7.0 398.1 63.10 8.8 ? 100.0 1.585 398.1 9.0 9.2 3.162 9.5 Section 4.7 v5.2.1
Think about it ! Section 4.7 v5.2.1