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Mathematics Stage 5 PAS5.3.8 Part 3 Functions and logarithms Logarithms Number: 43668 Title: Functions and Logarithms This publication is copyright New South Wales Department of Education and Training (DET), however it may contain material from other sources which is not owned by DET. We would like to acknowledge the following people and organisations whose material has been used: Extracts from Mathematics Syllabus Years 7-10 © Board of Studies, NSW 2002 Unit overview pp iii, iv, Parts 1, 2, 3 pp 3, 4 COMMONWEALTH OF AUSTRALIA Copyright Regulations 1969 WARNING This material has been reproduced and communicated to you on behalf of the New South Wales Department of Education and Training (Centre for Learning Innovation) pursuant to Part VB of the Copyright Act 1968 (the Act). The material in this communication may be subject to copyright under the Act. Any further reproduction or communication of this material by you may be the subject of copyright protection under the Act. CLI Project Team acknowledgement: Writer: Editor: Illustrator: Desktop Publishing: James Stamell Boris Handal Thomas Brown Carolina Barbieri All reasonable efforts have been made to obtain copyright permissions. All claims will be settled in good faith. Published by Centre for Learning Innovation (CLI) 51 Wentworth Rd Strathfield NSW 2135 ________________________________________________________________________________________________ Copyright of this material is reserved to the Crown in the right of the State of New South Wales. Reproduction or transmittal in whole, or in part, other than in accordance with provisions of the Copyright Act, is prohibited without the written authority of the Centre for Learning Innovation (CLI). © State of New South Wales, Department of Education and Training 2006. Contents – Part 3 Introduction – Part 3 ..........................................................3 Preliminary quiz.................................................................5 Laws of logarithms ..........................................................11 Systems of logarithms.............................................................17 Further laws of logarithms ...............................................21 Using logarithm laws .......................................................27 Solving index equations ..........................................................30 Exponentials and logarithms ...........................................33 Exponential graphs .................................................................33 Logarithmic graphs..................................................................35 Relationship between curves..................................................37 Miscellaneous exercises .................................................41 Suggested answers – Part 3 ...........................................47 Exercises – Part 3 ...........................................................55 Part 3 Logarithms 1 2 PAS5.3.8 Functions and Logarithms Introduction – Part 3 In 1935 the American seismologist Charles F. Richter (1900-1985) recognised that the seismic waves radiated by all earthquakes can provide good estimates of their magnitudes. He collected the recordings of seismic waves from a large number of earthquakes, and developed a calibrated system of measuring them for magnitude. Richter showed that, the larger the intrinsic energy of the earthquake, the larger the amplitude of ground motion at a given distance. He developed a formula to calculate this magnitude: M = log10 A + 3log10 (8∆t) − 2.92 where A is the maximum amplitude of the earthquake wave (in millimetres), and ∆t is the time difference (in seconds) between the arrival of the primary and secondary waves. ∆t (‘delta-t’) is used to compensate for the distance from the source of the earthquake where the measurement is made. For example, if the maximum wave amplitude, A, is 23 mm and there is a 25 s gap ( ∆t ) between the arrival of the two waves, then M = log10 23 + 3log10 (8 × 25) − 2.92 = 5.34 An earthquake whose magnitude is greater than 4.5 on this scale can cause damage to buildings and other structures. Severe earthquakes have magnitudes greater than 7. The famous San Francisco earthquake of 1906 recorded 7.8 on the Richter scale. An undersea earthquake caused the Boxing Day tsunami of 2004. It killed more than 283 100 people, making it one of the deadliest disasters in modern history. The magnitude of the earthquake, the second largest earthquake ever recorded, was estimated at 9.15 on the Richter scale. This is just one example of the use of logarithms. Part 3 Logarithms 3 Students learn about: • deducing a number of laws and relationships of logarithms • applying the laws of logarithms to evaluate simple expressions • simplifying expressions using the laws of logarithms • drawing the graph of y = log x and relating it to its inverse function • solving simple equations that contain exponents or logarithms. Students learn to: • compare and contrast a set of exponential and logarithmic graphs drawn on the same axes to determine similarities and differences. Source: 4 Extracts from Mathematics Years 7–10 syllabus © Board of Studies NSW, 2002. PAS5.3.8 Functions and Logarithms Preliminary quiz Before you start this part, use this preliminary quiz to revise some skills you will need. Activity – Preliminary quiz Try these. 1 Solve the following. a 3m − 2 = 27 ___________________________________________________ ___________________________________________________ ___________________________________________________ b 251− y = 1251+ y ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 2 Write the following in index form. a log 7 49 = 2 ___________________________________________________ b 2 3 = log 8 4 ___________________________________________________ ___________________________________________________ Part 3 Logarithms 5 3 Write the following in logarithmic form. a 161.5 = 64 ___________________________________________________ 81 = 9 b ___________________________________________________ ___________________________________________________ c 4 625 = 5 ___________________________________________________ d 3−2 = 1 9 ___________________________________________________ ___________________________________________________ 4 Evaluate. a log 2 8 ___________________________________________________ ___________________________________________________ ___________________________________________________ b log 8 4 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ c log 9 9 ___________________________________________________ ___________________________________________________ ___________________________________________________ 6 PAS5.3.8 Functions and Logarithms d log 8 1 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 5 Find the value of the pronumeral. a log x 2 = 1 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ b log m 32 = 5 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ c log n 1 = −3 27 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ Part 3 Logarithms 7 d log v 2 1 =− 3 4 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ e log w 9 = 2 3 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ f log 5 d = 4 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ g log10 j = 2.5 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 8 PAS5.3.8 Functions and Logarithms 6 Evaluate. a log 36 3 6 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ b log 49 49 7 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ Check your response by going to the suggested answers section. Part 3 Logarithms 9 10 PAS5.3.8 Functions and Logarithms Laws of logarithms You have already learned the laws for indices. When you remember that logarithms are, in fact, indices: the logarithm of a number to a given base is the index to which you raise that base to get the number then you can realise that the laws for logarithms are the laws for indices written in logarithmic form. Index law 1 Logarithm law 1 a0 = 1 log a 1 = 0 Any base to the power of 0 is equal to 1. (a ≠ 0) The logarithm of 1, to any base, is equal to zero. Examples Examples 20 = 1 n0 = 1 log 2 1 = 0 log 5 1 = 0 log n 1 = 0 Index law 2 Logarithm law 2 a = a1 log a a = 1 Any number raised to the power of 1 has the same value. The logarithm of a number to the same base is always equal to 1. Examples Examples 50 = 1 3 = 31 10 = 10 1 x = x1 log 3 3 = 1 log10 10 = 1 log x x = 1 You should now be able to use these two laws to write down the answers to the following. Part 3 Logarithms 11 Activity – Laws of logarithms Try these. 1 Write down the value of x in each of the following. a log 4 x = 0 __________________________________________ b log 3 1 = x ___________________________________________ c log 4 x = 1 __________________________________________ d log x 8 = 1 __________________________________________ Check your response by going to the suggested answers section. Could you immediately write down the value of x in each of these without needing to work them out? Index law 3 Logarithm law 3 am × an = am + n log a (M × N ) = log a M + log a N To multiply powers of the same base, add the indices. The log of a product of two numbers is the sum of the logs of those numbers to the same base. Now study carefully how Index law 3 leads you to the third law for logarithms. Let M = a m andN = a n . By the definition of logarithms, m = log a M andn = log a N . M × N = am × an ∴MN = a m + n 12 PAS5.3.8 Functions and Logarithms This means, log a (MN ) = m + n ∴log a (MN ) = log a M + log a N Examples log 3 35 = log 3 (5 × 7) = log 3 5 + log 3 7 log a 40 = log a (4 × 10) = log a 4 + log a 10 log a 40 = log a (5 × 8) = log a 5 + log a 8 Follow through the steps in this example. Do your own working in the margin if you wish. If log a 2 = 0.301,log a 3 = 0.477andlog a 5 = 0.699 , calculate the value of a log a 6 b log a 15 Solution a log a 6 = log a (2 × 3) = log a 2 + log a 3 = 0.301 + 0.477 = 0.778 b log a 15 = log a (3 × 5) = log a 3 + log a 5 = 0.477 + 0.699 = 1.176 As you are only given the logs of 2, 3, and 5, you need to write 6 and 15 as products of these numbers to work them out. Part 3 Logarithms 13 Activity – Laws of logarithms Try these. 2 Given log a 2 = 0.301,log a 3 = 0.477andlog a 5 = 0.699 , calculate the value of a log a 4 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ b log a 20 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 3 Show that log a 9 is double log a 3 . Hence calculate its value. _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ Check your response by going to the suggested answers section. In these examples it does not matter what the base is, as you were provided with the value of the logarithm. 14 PAS5.3.8 Functions and Logarithms From these examples you should realise that you can extend logaMN = logaM + logaN to logaMNP = logaM + logaN + logaP and so on. And remember, these rules only apply if the bases remain the same. Law 3: log a (M × N ) = log a M + log a N can also be written as log a M + log a N = log a (M × N ) . This states that The sum of the logarithms of two numbers, to the same base, is the logarithm of the product of the numbers to the same base. This rule can be extended to more than just two numbers. Examples log10 3 + log10 8 = log10 (3 × 8) = log10 24 log 2 6 + log 2 7 = log 2 (6 × 7) = log 2 42 log a 2 + log a 5 + log a 7 = log a (2 × 5 × 7) = log a 70 Sometimes you arrive at a result where you might be able to evaluate the logarithm using either of laws 1 or 2. Follow through the steps in this example. Do your own working in the margin if you wish. Evaluate by first writing as a single logarithm. a log 6 2 + log 6 3 b log15 c log10 16 + log10 5 + log10 125 1 + log15 30 2 Solution Part 3 Logarithms 15 a log 6 2 + log 6 3 = log 6 (2 × 3) = log 6 6 = 1[Logarithm law 2] b log15 c log10 16 + log10 5 + log10 125 1 ⎛1 ⎞ + log15 30 = log15 ⎜ × 30 ⎟ ⎝2 ⎠ 2 = log15 15 = 1[Logarithm law 2] = log10 (16 × 5 × 125) = log10 10000 Now this single logarithm can be expanded log10 10000 = log10 (10 × 10 × 10 × 10) = log10 10 + log10 10 + log10 10 + log10 10 = 1+1+1+1 =4 Make sure you are familiar with the logarithm laws so far before attempting the next activity. Activity – Laws of logarithms Try these. 4 5 Write as a single logarithm. a log 4 7 + log 4 3 = b log b 10 + log b 9 = ____________________________________ c log 5 3 + log 5 2 + log 5 11 = ____________________________________ _____________________________ Evaluate by first writing as a single logarithm. a log18 6 + log18 3 = ___________________________________________________ ___________________________________________________ 16 PAS5.3.8 Functions and Logarithms b log 24 2 + log 24 3 + log 24 4 = ___________________________________________________ ___________________________________________________ c log10 2 + log10 50 = ___________________________________________________ ___________________________________________________ ___________________________________________________ Check your response by going to the suggested answers section. Systems of logarithms Any positive number, other than 1, can be used as the base for a system of logarithms (to that base). There are two systems in wide current use. • Natural logarithms Sometimes called Naperian logarithms, after John Napier (15501617). These are important in senior and higher mathematics and have as the base the irrational number e = 2.71828... . You will not be using these logarithms in this course. • Common logarithms These are also called Briggsian logarithms and use the base 10. They are convenient for computation in our decimal number system. Part 3 Logarithms 17 Both of these two systems may be found on your calculator. The log key gives logarithms to the base 10, and the ln key is used for logarithms to the base e. (ln stands for logarithme naturale, natural logarithm). So if you wanted to find log10 27 , you would press log 27. The answer your calculator should give is 1.431363764 When you use the log key your answer won’t always be a nice whole number. But for now you will not need your calculator. By convention, if you are working with common logarithms, that is in base 10, you can leave out the base. For example, log10 n can be simply written as log n. (log 9 means log10 9 and log 4 means log10 4 , and so on.) Follow through the steps in this example. Do your own working in the margin if you wish. Given that log 9 = 0.9542, find the value of log 90. Solution log 90 = log10 90 = log10 (9 × 10) = log10 9 + log10 10 = 0.9542 + 1[Notelog10 10 = 1] = 1.9542 You can check this answer with your calculator. 18 PAS5.3.8 Functions and Logarithms Activity – Laws of logarithms Try these. 6 Given log 15 = 1.1761, find log 150. _______________________________________________________ _______________________________________________________ 7 Given log 2 = 0.3010 and log 3 = 0.4771, find log 60. _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ Check your response by going to the suggested answers section. You have been practising using some laws of logarithms. Now check that you can solve these kinds of problems by yourself. Go to the exercises section and complete Exercise 3.1 – Laws of logarithms. Part 3 Logarithms 19 20 PAS5.3.8 Functions and Logarithms Further laws of logarithms You are familiar with this index law: a m ÷ a n = a m − n . Using M = a m andN = a n (that is, m = log a M;n = log a N ) then M = M ÷ N = am−n . N ⎛M⎞ And from the definition of logarithms, log a ⎜ ⎟ = m − n . ⎝N⎠ ⎛M⎞ That is, log a ⎜ ⎟ = log a M − log a N ⎝N⎠ Index law 4 Logarithm law 4 am ÷ an = am − n ⎛M⎞ log a ⎜ ⎟ = log a M − log a N ⎝N⎠ To divide powers of the same base, subtract the indices. The log of the quotient of two numbers is the difference of the logs of those numbers to the same base. Examples 15 = log a 15 − log a 3 3 log b (20 ÷ 7) = log b 20 − log b 7 log a log p 0.75 = log p 3 = log p 3 − log p 4 4 Conversely, you can also say The difference in the logarithms of two numbers to the same base is the logarithm of the quotient of the two numbers to the same base. Example log a 7 − log a 2 = log a Part 3 Logarithms 7 = log a 3.5 2 21 Follow through the steps in this example. Do your own working in the margin if you wish. a Given log10 2 = 0.301andlog10 5 = 0.699 , calculate log10 0.5 . b Simplify log 3 24 − log 3 8 Solution a b log10 0.5 5 = log10 10 = log10 5 − log10 10 = 0.699 − 1 = −0.301 log10 0.5 1 = log10 2 = log10 1 − log10 2 = 0 − 0.301 = −0.301 OR ⎛ 24 ⎞ log 3 24 − log 3 8 = log 3 ⎜ ⎟ ⎝ 8⎠ = log 3 3 =1 You should realise that just as loga MN = loga M + loga N then loga M = loga M – loga N loga N 22 PAS5.3.8 Functions and Logarithms Activity – Further laws of logarithms Try these. 1 You are given that log 3 = 0.477, log 4 = 0.602 and log 5 = 0.699. [As no base is given, assume it to be 10.] Calculate a log 3 = ____________________________________________ 4 ___________________________________________________ ___________________________________________________ ___________________________________________________ b log 0.8 = ____________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 2 How can you use log 3 to help you find the value of log 75? Show 4 your working. _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ Check your response by going to the suggested answers section. There is just one more logarithm law you need to know. Part 3 Logarithms 23 Consider log a x n = log a (x × x × x × x × ....... × x) = log a x + log a x + ....... + log a x = n log a x [x is a factor n times] [ log a x is a term added n times] Index law 5 Logarithm law 5 a n = a× a × a × ... ×a log a x n = n log a x there are n factors altogether To raise a number to a power, multiply the number by itself the index number of times. The logarithm of the nth power of a number is n times the logarithm of the number to the same base. For instance, log 2 5 8 = 8 log 2 5 . Follow through the steps in this example. Do your own working in the margin if you wish. Given log a 3 = 0.5 , find the value of a log a 81 b log a 1 9 Solution log a 81 a b = log a 34 = 4 log a 3 = 4 × 0.5 =2 1 log a 9 = log a 3−2 = −2 log a 3 = −2 × 0.5 = −1 Because you were given the value of loga 3, you needed to write loga 81 and loga 1 2 in terms of loga 3 to evaluate them. 24 PAS5.3.8 Functions and Logarithms Activity – Further laws of logarithms Try these. 3 Given that log a 2 = 0.3010 find the value of a log a 8 = ________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ b log a 64 = ___________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ Check your response by going to the suggested answers section. You have been practising applying further rules of logarithms. Now check that you can solve these kinds of problems by yourself. Go to the exercises section and complete Exercise 3.2 – Further laws of logarithms. Part 3 Logarithms 25 26 PAS5.3.8 Functions and Logarithms Using logarithm laws Here is a summary of the laws of logarithms and relationships you have learned so far. • log a 1 = 0 • log a a = 1 • log a (M × N ) = log a M + log a N • ⎛M⎞ log a ⎜ ⎟ = log a M − log a N ⎝N⎠ • log a x n = n log a x Make sure you understand and can use these logarithm relationships before proceeding further. The laws of logarithms help you simplify expressions such as the following. Follow through the steps in this example. Do your own working in the margin if you wish. a Simplify log x + log y – 2log z b Write 4log 2 – 2log 6 + 2log 3 – 3 log 4 as a single logarithm. Solution log x + log y − 2 log z a = log x + log y − log z 2 [2 log z = log z 2 ] x×y z2 xy = log 2 z = log Part 3 Logarithms 27 b 4 log 2 − 2 log 6 + 2 log 3 − 3log 4 = log 2 4 − log 6 2 + log 32 − log 4 3 2 4 × 32 62 × 4 3 16 × 9 = log 36 × 64 1 = log 16 = log 2 −4 = −4 log 2 = log In these two examples no base was given. You could therefore assume it to be 10. However, the base you use here doesn’t matter; the outcome is still be the same. Activity – Using logarithm laws Try these. 1 Write as a single logarithm. a log xy + log yz − log xz ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ b 4 log 27 + 2 log 2 − 2 log 3 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ Check your response by going to the suggested answers section. 28 PAS5.3.8 Functions and Logarithms You can also proceed the other way and expand logarithmic expressions. Follow through the steps in this example. Do your own working in the margin if you wish. Write log y ax in expanded form. y2 Solution log y ax = log y ax − log y y 2 2 y = log y a + log y x − 2 log y y = log y a + log y x − 2[log y y = 1] Expand these step-by-step. Activity – Using logarithm laws Try these. 2 Write these in expanded form. a log b abc ___________________________________________________ ___________________________________________________ ___________________________________________________ b log10 x y − log10 y x ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ Check your response by going to the suggested answers section. Part 3 Logarithms 29 Notice that in the second example it didn’t matter whether the base is 10 or any other base. The result is still the same. Solving index equations You are familiar with equations of the type: 8 = 2 2 x − 2 . Its solution is based on the fact that you can equalise the bases easily. 8 = 22 x − 2 23 = 22 x − 2 3 = 2x − 2[same base, therefore equate indices] 2x = 5 5 x= 2 In the equation 2 x = 3 there is no simple method of equalising the bases. Here is a way to solve such equations. If 2 x = 3 then the common logarithms of both sides are equal. That is, 2x = 3 log10 2 x = log10 3 x log10 2 = log10 3 log10 3 x= log10 2 = 1.5850[correct to 4 dp using a calculator] For this next activity you will need to use your calculator. 30 PAS5.3.8 Functions and Logarithms Activity – Using logarithm laws Try these. 3 Solve 3n +1 = 12 . Give your answer correct to 4 decimal places. _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ 4 Solve 5 y −1 = 3y . Give your answer correct to 4 significant figures. _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ Check your response by going to the suggested answers section. You have been practising using logarithm laws to simplify expressions and solving index equations with different bases. Now check that you can solve these kinds of problems by yourself. Go to the exercises section and complete Exercise 3.3 – Using logarithm laws. Part 3 Logarithms 31 32 PAS5.3.8 Functions and Logarithms Exponentials and logarithms Functions of the form y = a x are known as exponential functions. Those of the form y = log a x are known as logarithmic functions. Exponential graphs Consider graphs of the type y = a x where a is a fixed, positive number not equal to 1, and x is the index (or power or logarithm or exponent). You may recall covering these curves earlier in your course. They are called exponential graphs, or graphs of exponential functions. Activity – Exponentials and logarithms Try these. 1 Complete the table of values for y = 3x . x y Part 3 Logarithms –2 −2 3 = 1 2 3 –1 = 0 1 2 1 9 33 2 Draw the graph. y 8 6 4 2 –2 –1 0 1 2 x 3 Where does the graph cut the y-axis? ________________________ 4 Does the curve ever cut the x-axis? Give a reason. _______________________________________________________ _______________________________________________________ Check your response by going to the suggested answers section. This graph approaches the x-axis but never touches it. A line to which a curve approaches gradually but never touches is called an asymptote. The x-axis is an asymptote to this curve. Here are some other features that you ought to be aware of about exponential graphs. • When x = 0,y = a 0 = 1 . So the graph of y = a x always passes through (0, 1) • When x = 1,y = a1 = a . So (1, a) is a point on the graph. • For a point (x0 ,y0 ) on the graph, y0 = a x0 . Now if you increase the x-value by 1, the new y-value is given by 34 PAS5.3.8 Functions and Logarithms y( x0 +1) = a x0 +1 = a1 × a x0 [by the law of indices] = a × y0 So each time you add 1 to x, you multiply y by a. • Each time you subtract 1 from x, you divide y by a. From these features you should easily recognise and be able to sketch the graph of y = a x , for any values of a (≠1). y = ax y a (1, a) 1 0 1 x Logarithmic graphs y = log a x is a logarithmic function. What does a graph of this function look like? Follow through the steps in this example. Do your own working in the margin if you wish. Graph y = log 3 x Solution To draw up a table of values for this equation, it is easier to first rewrite this relation as x = 3y . In this case it is easier to give values for y and find the corresponding values of x. Part 3 Logarithms 35 y –2 3−2 = x –1 1 1 = 32 9 3−1 = 1 3 0 1 2 30 = 1 31 = 3 32 = 9 [Notice how the table is set out; y-values first, so you can then calculate the corresponding x-values.] Plotting these points onto a grid gives: y 3 2 1 –1 0 1 2 3 4 5 6 7 8 9 10 x –1 –2 –3 From this graph you can estimate values of log 3 x . For example, when x = 8, log 3 8 ≈ 1.9 . Activity – Exponentials and logarithms Try these. 5 Use the graph of y = log 3 x on the previous page to answer the following. a 36 Find i log 3 3 ii log 3 1 __________________________________________ iii log 3 5 _________________________________________ _________________________________________ PAS5.3.8 Functions and Logarithms b For what value of x is i log 3 x = 0.8 ? ____________________________________ ii log 3 x = 1.6 ? ____________________________________ Check your response by going to the suggested answers section. Here are some features that you ought to be aware of about logarithmic graphs. • The curve lies totally to the right of the y-axis. • The y-axis is an asymptote to the curve y = log a x . • The curve cuts the x-axis at (1, 0), since log a 1 = 0 for all values of a. • The point (a, 1) lies on the graph y = log a x because log a a = 1 . • For x > 1, y = log a x is positive. • For 0 < x < 1, y = log a x is negative. y y= log a x 1 0 1 a x Relationship between curves The sketch of the two curves y = a x andy = log a x on the same grid shows that the figure is symmetrical. Part 3 Logarithms 37 y y = 3x 10 9 y=x 8 7 6 5 4 3 2 y = log3 x 1 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 x –1 –2 –3 In fact, each curve is a reflection of the other across the line y = x. The exponential curve y = a x is a reflection of the logarithmic curve y = log a x in the line y = x. Hence the exponential curve y = a x is the inverse function of the logarithmic curve y = log a x . Follow through the steps in this example. Do your own working in the margin if you wish. Find the area of the rectangle MNOP. 1 0 y = a M 2x 8 y P N x Solution 38 PAS5.3.8 Functions and Logarithms At the point M the value of y is 8. Substituting y = 8 into the equation y = 2 x gives y = 2 x 8 = 2 x 2 3 = 2 x ∴x = 3 The coordinates of M are (3, 8). Hence ON = 3 units, MN = 8 units. Therefore area of MNOP = 3 × 8 = 24units 2 Activity – Exponentials and logarithms Try these. 6 In the sketch RS is parallel to OT. Find the coordinates of R, S and T. y= 2x y S y= log 2 x T 0 R N x _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ Part 3 Logarithms 39 Check your response by going to the suggested answers section. You have been practising comparing the relationship between exponential and logarithmic curves. Now check that you can solve these kinds of problems by yourself. Go to the exercises section and complete Exercise 3.4 – Exponentials and logarithms. 40 PAS5.3.8 Functions and Logarithms Miscellaneous exercises You are now in a position to deal with more difficult, and abstract, examples of logarithms. Text linking to example. Follow through the steps in this example. Do your own working in the margin if you wish. Given log b 3 = mandlog b 2 = n , find the value of the following in terms of m and/or n. a log b 16 b log b 24 Solution a log b 16 = log b 2 4 = 4 log b 2 =4×n = 4n b log b 24 = log b (3 × 8) = log b 3 + log b 8 = m + log b 2 3 = m + 3log b 2 = m + 3n Part 3 Logarithms 41 Activity – Miscellaneous exercises Try these. 1 Given log b 3 = mandlog b 2 = n , find the value of the following in terms of m and/or n. a log b 8b ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ b log b 12 b2 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ Check your response by going to the suggested answers section. You should also be able to solve logarithmic equations. In these examples you will be using the idea that if log p = log q then p = q. 42 PAS5.3.8 Functions and Logarithms Follow through the steps in this example. Do your own working in the margin if you wish. a Write the value of x in terms of y if log 4x − log 3y = log 3 . b Solve the equation log 2x − log(x − 3) = log 3 . Solution log 4x − log 3y = log 3 a 4x = log 3 3y 4x =3 3y 4x = 9y 9 x= y 4 log 2x − log(x − 3) = log 3 2x = log 3 log x−3 2x =3 x−3 2x = 3(x − 3) 2x = 3x − 9 −x = −9 x=9 log b In these examples it did not matter what the base is. You can assume it to be 10, but the relationship is still true with any other base. Make sure you understand the logarithm relationships when working on these. Part 3 Logarithms 43 Activity – Miscellaneous exercises Try these. 2 Express y in terms of x without the use of logs in log 3x + log y = log 5 . _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ 3 Solve the following: log x + log(2x + 1) = log(x + 4) . _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ Check your response by going to the suggested answers section. Notice that in the last example x has to be a positive number. This is because you can only take the log of a positive number. log x can only exist when x > 0. Follow through the steps in this example. Do your own working in the margin if you wish. Simplify log 64 ÷ log16 . 44 PAS5.3.8 Functions and Logarithms Solution log 64 ÷ log16 = log 64 log16 log 2 6 log 2 4 6 log 2 = 4 log 2 6 = 4 1 =1 2 = It is important not to confuse log 64 ÷ log 16 with log 64 – log 16. log = log 64 – log 16 Activity – Miscellaneous exercises Try these. 4 Simplify log a 4 ÷ log ( a) . 3 _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ Check your response by going to the suggested answers section. Part 3 Logarithms 45 You have been practising solving a variety of logarithmic expressions and equations. Now check that you can solve these kinds of problems by yourself. Go to the exercises section and complete Exercise 3.5 – Miscellaneous exercises. 46 PAS5.3.8 Functions and Logarithms Suggested answers – Part 3 Check your responses to the preliminary quiz and activities against these suggested answers. Your answers should be similar. If your answers are very different or if you do not understand an answer, contact your teacher. Activity – Preliminary quiz 1 a 3m − 2 = 27 b (5 ) 3m − 2 = 33 m−2= 3 m=5 2 1− y ( ) = 53 1+ y 5 2 − 2 y = 5 3+ 3 y 2 − 2y = 3 + 3y 5y = −1 1 y=− 5 2 3 2 a 7 2 = 49 b 8 =4 3 a log16 64 = 1.5 b log 81 9 = c log 625 5 = d log 3 a Let x = log 2 8 . b Let x = log 8 4 . 4 1 4 Let x = log 9 9 . 9 x = 91 x =1 Logarithms 1 = −2 9 (2 3 )x = 2 2 2x = 23 x=3 c 1 2 8x = 4 2x = 8 Part 3 251− y = 1251+ y 2 3x = 2 2 3x = 2 2 x= 3 d Let x = log 8 1 . 8x = 1 8 x = 80 x=0 47 5 a log x 2 = 1 b m 5 = 32 x1 = 2 x=2 c log n log m 32 = 5 1 = −3 27 1 n −3 = 27 −3 n = 3−3 n=3 d m 5 = 2 5 (orm = 5 32 ) m=2 2 1 log v = − 3 4 2 − 1 v 3= 4 ⎛ − 23 ⎞ ⎜⎝ v ⎟⎠ − 3 2 ⎛ 1⎞ =⎜ ⎟ ⎝ 4⎠ − 3 2 3 −2 − 2 ( ) v= 2 = 23 =8 e log w 9 = 2 3 f log 5 d = 4 d= 2 3 w =9 3 2 3 ⎛ 23 ⎞ 2 ⎜⎝ w ⎟⎠ = ( 9 ) 3 2 2 ( ) w= 3 ( 5) ⎛ 1⎞ = ⎜ 52 ⎟ ⎝ ⎠ 4 4 = 52 = 25 = 33 = 27 g log10 j = 2.5 j = 10 2.5 = 10 2 × 10 0.5 = 100 × 10 = 100 10 48 PAS5.3.8 Functions and Logarithms 6 a Let x = log 36 3 6 . Let x = log 49 49 7 . b 36 x = 3 6 (6 ) 2 x =6 49 x = 49 7 1 3 (7 ) 2 x =7 ×7 1 62 x = 6 3 1 2x = 3 1 x= 6 72 x = 7 2 2 1 2 1 2 1 2 1 x =1 4 2x = 2 Activity – Laws of logarithms 1 a x=1 2 a log a 4 = log a (2 × 2) = log a 2 + log a 2 = 0.301 + 0.301 = 0.602 5 Part 3 x=0 c b x=4 d x=8 log a 20 = log a (2 × 2 × 5) = log a 2 + log a 2 + log a 5 = 0.301 + 0.301 + 0.699 = 1.301 log a 9 = log a (3 × 3) = log a 3 + log a 3 = 2 log a 3(which is double log a 3) = 2 × 0.477 = 0.954 3 4 b a log 4 7 + log 4 3 = log 4 (7 × 3) = log 4 21 c log 5 3 + log 5 2 + log 5 11 = log 5 (3 × 2 × 11) = log 5 66 a log18 6 + log18 3 = log18 (6 × 3) = log18 18 =1 Logarithms b log b 10 + log b 9 = log b (10 × 9) = log b 90 b log 24 2 + log 24 3 + log 24 4 = log 24 (2 × 3 × 4) = log 24 24 =1 49 c log10 2 + log10 50 = log10 (2 × 50) = log10 100 = log10 (10 × 10) = log10 10 + log10 10 6 7 = 1+1 =2 log150 = log(15 ×10) = log15 + log10 = 1.1761+1 = 2.1761 log 60 = log(2 × 3 × 10) = log 2 + log 3 + log10 = 0.3010 + 0.4770 + 1 = 1.778 Activity – Further laws of logarithms 3 = log 3 − log 4 4 = 0.477 − 0.602 = −0.125 1 a 2 Start by writing log log b 4 5 = log 4 − log 5 = 0.602 − 0.699 = −0.097 log 0.8 = log 3 75 = –0.125. = log 4 100 75 log = −0.125 100 log 75 − log100 = −0.125 log 75 − log(10 × 10) = −0.125 log 75 − (log10 + log10) = −0.125 log 75 − (1 + 1) = −0.125(sincelog10 = log10 10 = 1) log 75 = 2 − 0.125 = 1.875 3 50 a log a 8 = log a 2 3 b log a 64 = log a 2 6 = 3log a 2 = 3 × 0.3010 = 6 log a 2 = 6 × 0.3010 = 0.9030 = 1.8060 PAS5.3.8 Functions and Logarithms Activity – Using logarithm laws 1 a log xy + log yz − log xz = (log x + log y) + (log y + log z) − (log x + log z) = log x + log y + log y + log z − log x − log z = 2 log y b log 27 + 2 log 2 − 2 log 3 = log 33 + 2 log 2 − 2 log 3 = 3log 3 + 2 log 2 − 2 log 3 = log 3 + 2 log 2 = log 3 + log 2 2 = log(3 × 2 2 ) = log(3 × 4) = log12 2 a log b abc = log b a + log b b + log b c = log b a + 1 + log b c b 3 log10 x y − log10 = (log10 x − log10 y) − (log10 y − log10 x) y x = log10 x − log10 y − log10 y + log10 x = 2 log10 x − 2 log10 y 3n +1 = 12 log10 3n +1 = log10 12 (n + 1)log10 3 = log10 12 n +1= log10 12 log10 3 = 2.261859507[using a calculator] n = 1.2619[correct to 4 dp] [Notice that Part 3 Logarithms log10 12 ≠ log10 4 . Don’t confuse this!] log10 3 51 5 y −1 = 3y 4 log10 5 y −1 = log10 3y (y − 1)log10 5 = y log10 3 y log10 5 − log10 5 = y log10 3 y log10 5 − y log10 3 = log10 5 y(log10 5 − log10 3) = log10 5 log10 5 log10 5 − log10 3 = 3.151[correct to 4 sig fig] y= Activity – Exponentials and logarithms 1 Table values are: 1 1 , ,1,3,9 . 9 3 y 2 8 6 4 2 –2 –1 0 2 x 1 3 (0, 1) 4 No, the curve never cuts the x-axis. As you move to the left the yvalue gets smaller and smaller but always remains positive. 52 5 a b i i 1 2.4 6 R(1, 0), T(0, 1), S(1,2) ii ii 0 5.5 iii 1.5 PAS5.3.8 Functions and Logarithms Activity – Miscellaneous exercises 1 a b log b 8b = log b 8 + log b b = log b 2 3 + log b b 12 b2 = log b 12 − log b b 2 log b = log b (3 × 2 2 ) − 2 log b b = 3log b 2 + 1 = log b 3 + log b 2 2 − 2 × 1 = 3n + 1 = log b 3 + 2 log b 2 − 2 = m + 2n − 2 2 log 3x + log y = log 5 3x = log 5 log y 3x =5 y 3x = 5y 3 y= x 5 3 log x + log(2x + 1) = log(x + 4) log x(2x + 1) = log(x + 4) x(2x + 1) = x + 4 2x 2 + x = x + 4 2x 2 = 4 x2 = 2 x=± 2 x= 2 (since x must be positive) log a 4 ÷ log 4 = = ( a) 3 log a 4 ⎛ 1⎞ log ⎜ a 2 ⎟ ⎝ ⎠ 3 log a 4 3 2 log a 4 log a = 3 2 log a 3 2 2 =4× 3 2 =2 3 =4÷ Part 3 Logarithms 53 54 PAS5.3.8 Functions and Logarithms Exercises – Part 3 Exercises 3.1 to 3.5 Name ___________________________ Teacher ___________________________ Exercise 3.1 – Laws of logarithms 1 Solve. a y = log 5 5 ___________________________________________ b log 5 2x = 1 ___________________________________________________ 2 Given log a 2 = 0.301,log a 5 = 0.699, and log a 7 = 0.845, find the value of a log a 14 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ b log a 35 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ Part 3 Logarithms 55 c log a 25 ___________________________________________________ ___________________________________________________ ___________________________________________________ d log a 49 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 3 a Using the values in question 2, find the value of log a 10 . ___________________________________________________ ___________________________________________________ b What is the value of the base a? _________________________ Why? _____________________________________________ 4 Write as a single logarithm. a log m 6 + log m 8 ___________________________________________________ b log b 9 + log b 4 + log b 5 ___________________________________________________ ___________________________________________________ 5 Evaluate each of the following by first writing as a single logarithm. a log10 8 + log10 125 ___________________________________________________ ___________________________________________________ ___________________________________________________ 56 PAS5.3.8 Functions and Logarithms b log 6 4 + log 6 9 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ c log12 3 + log12 36 + log12 16 ___________________________________________________ ___________________________________________________ ___________________________________________________ d log10 5 + log10 8 + log10 25 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 6 Given log 2 = 0.3010, log 3 = 0.477, and log 7 = 0.8451 evaluate a log 20 ___________________________________________________ ___________________________________________________ ___________________________________________________ b log 21 ___________________________________________________ ___________________________________________________ ___________________________________________________ c log 140 ___________________________________________________ ___________________________________________________ ___________________________________________________ Part 3 Logarithms 57 d log 42 ___________________________________________________ ___________________________________________________ ___________________________________________________ 7 a Given that log 3 = 0.4771, evaluate log 5 + log 6. ___________________________________________________ ___________________________________________________ ___________________________________________________ b Given that log 6 = 0.7782, evaluate log 5 + log 3 + log 4. ___________________________________________________ ___________________________________________________ ___________________________________________________ c Given that log 4 = 0.6021, evaluate log 5 + log 8. ___________________________________________________ ___________________________________________________ ___________________________________________________ d Given that log 9 = 0.9542, evaluate log 5 + log 6 +log 3. ___________________________________________________ ___________________________________________________ ___________________________________________________ e Given that log 12 = 1.0792, evaluate log 15 + log 8. ___________________________________________________ ___________________________________________________ ___________________________________________________ 58 PAS5.3.8 Functions and Logarithms Exercise 3.2 – Further laws of logarithms 1 Given that log 2 = 0.301, log 3 = 0.477 and log 5 = 0.699 evaluate a log 0.2 ___________________________________________________ ___________________________________________________ ___________________________________________________ b log 2.5 ___________________________________________________ ___________________________________________________ ___________________________________________________ c log 1 2 3 ___________________________________________________ ___________________________________________________ ___________________________________________________ d log 7.5 ___________________________________________________ ___________________________________________________ ___________________________________________________ 2 Given that log a 2 = 0.301,log a 3 = 0.477andlog a 5 = 0.699 evaluate a log a 20 ___________________________________________________ ___________________________________________________ ___________________________________________________ b log a 24 ___________________________________________________ Part 3 Logarithms 59 ___________________________________________________ ___________________________________________________ c log a 45 ___________________________________________________ ___________________________________________________ ___________________________________________________ d log a 50 ___________________________________________________ ___________________________________________________ ___________________________________________________ e log a 2 1 4 ___________________________________________________ ___________________________________________________ ___________________________________________________ f log a 24 25 ___________________________________________________ ___________________________________________________ ___________________________________________________ g log a 6.25 ___________________________________________________ ___________________________________________________ ___________________________________________________ 60 PAS5.3.8 Functions and Logarithms 3 Evaluate a 1.5 log 3 9 ___________________________________________________ ___________________________________________________ ___________________________________________________ b 4 log 5 125 3 ___________________________________________________ ___________________________________________________ c ___________________________________________________ 2 log 2 32 5 ___________________________________________________ ___________________________________________________ d ___________________________________________________ 3 log 3 81 4 ___________________________________________________ ___________________________________________________ e ___________________________________________________ 3 1 log 3 2 9 ___________________________________________________ ___________________________________________________ f ___________________________________________________ 2 1 log 4 3 64 ___________________________________________________ ___________________________________________________ ___________________________________________________ Part 3 Logarithms 61 Exercise 3.3 – Using logarithm laws 1 Simplify these by writing as a single logarithm. a 2 log x − 3log y ___________________________________________________ ___________________________________________________ ___________________________________________________ b ___________________________________________________ 1 log x + log y − 2 log z 2 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ c log 4 − 4 log 5 + 3log 50 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ d ___________________________________________________ 1 2 log 3 − log 2 + log 9 2 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 62 PAS5.3.8 Functions and Logarithms 2 Write these in expanded form. a log 2 4 p2 q ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ b log x x2 y ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 3 Solve, giving answers correct to 3 decimal places. a 5 = 3x ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ b 2 m = 11 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ c 2 2 k +1 = 7 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ Part 3 Logarithms 63 d 5 y = 3y + 3 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ e 6 2 x −1 = 3x ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 4 Solve 4 t +1 = 1 8 2 _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ 5 Solve log 27 3 = x _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ 64 PAS5.3.8 Functions and Logarithms Exercise 3.4 – Exponentials and logarithms 1 Complete the table of values for y = log 2 x . a x y b –3 –2 –1 0 1 2 3 1 Draw the graph of y = log 2 xfor ≤ x ≤ 8 . 8 y 4 3 2 1 –1 1 0 2 3 4 5 6 7 8 x –1 –2 –3 –4 2 Two curves y = a x andy = log a x are shown. y 7 6 5 4 3 2 1 –1 0 1 2 3 4 5 6 7 x –1 –2 –3 a What is the value of a? ________________________________ How did you work it out? Explain. ___________________________________________________ ___________________________________________________ Part 3 Logarithms 65 ___________________________________________________ b On this graph draw in the line y = x. c What is the important link between y = a x andy = log a x , and this line? ___________________________________________________ ___________________________________________________ 3 For the diagram below, find a the co-ordinates of Q _________________________________ b the area of the rectangle OPQR. ___________________________________________________ ___________________________________________________ y g2 x lo y= P Q R 0 4 x 4 In the figure below, NM is drawn parallel to OX. Find the length of MN. y= 2x y N y= log 2 x Q R 0 4 x _______________________________________________________ _______________________________________________________ _______________________________________________________ 66 PAS5.3.8 Functions and Logarithms Find the co-ordinates of A and B, and hence calculate the length of AB. 2x 5 y= y 4 0 A B x= 2y x _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ Part 3 Logarithms 67 Exercise 3.5 – Miscellaneous exercises 1 If log a 2 = pandlog a 3 = q , give the values of the following in terms of p and/or q. a log a 32 ___________________________________________________ ___________________________________________________ ___________________________________________________ b log a 81 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ c log a 18 ___________________________________________________ ___________________________________________________ ___________________________________________________ d log a 72 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 2 If log 5 a = p , what is the value of a log 5 25a ___________________________________________________ ___________________________________________________ ___________________________________________________ 68 PAS5.3.8 Functions and Logarithms b log 5 5a 2 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 3 Express y in terms of x without the use of logs. a log 5x – log 2y = log 8 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ b log 6x + log y = 2 [Hint: log 100 = 2] ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ c 2log x – 3log y = log 9 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ Part 3 Logarithms 69 4 Solve the following equations. a log 3x + log 2 = log (x + 1) ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ b log (x + 3) – log (x – 1) = log 4 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ c log y + log (y – 1) = log (5 – y) ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ d log (z + 5) – log 3z = log (z + 1) ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ 70 PAS5.3.8 Functions and Logarithms 5 Simplify a log 81 ÷ log 27 ___________________________________________________ ___________________________________________________ ___________________________________________________ b log 16 ÷ log 128 ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ c log 4 ÷ log 1 4 ___________________________________________________ ___________________________________________________ ___________________________________________________ d log a 4 ÷ log a ___________________________________________________ ___________________________________________________ ___________________________________________________ ___________________________________________________ e log m 3 ÷ log 3 m ___________________________________________________ ___________________________________________________ ___________________________________________________ f log b ÷ log 1 b ___________________________________________________ ___________________________________________________ Part 3 Logarithms 71 ___________________________________________________ ___________________________________________________ The following is part of the graph for the function y = log 7 x . 6 y 0.5 0 0.5 1 1.5 2 2.5 3x –0.5 Use the graph to answer the following questions, to an accuracy allowed by the graph. a Find i log 7 2 ii log 7 1.4 iii log 7 0.6 ___________________________________________________ ___________________________________________________ b Find the value of x for which i log 7 x = 0 ii log 7 x = 0.5 iii log 7 x = −0.7 ___________________________________________________ ___________________________________________________ 7 8 True or false. a The log of a number between 0 and 1 is negative.____________ b The log of a negative number is less than zero. ______________ c Double the number and its logarithm is doubled too. _________ d The line x = 0 is an asymptote to y = log 7 x .________________ What is the equation of the inverse function to y = log 7 x ? _______________________________________________________ _______________________________________________________ 72 PAS5.3.8 Functions and Logarithms