Mathematics Learning Centre Introduction to Eponents and Logarithms Christopher Thomas c 998 University of Sydney
Acknowledgements Parts of section of this booklet rely a great deal on the presentation given in the booklet of the same name, written by Peggy Adamson for the Mathematics Learning Centre in 987. The remainder is new. Jackie Nicholas, Sue Gordon and Trudy Weibel read pieces of earlier drafts of this booklet. I should like to thank them for their etremely helpful comments on the contents and layout. In addition, Duncan Turpie performed the laborious task of final proof reading. Thanks Duncan. Christopher Thomas December 99 This booklet was revised in 998 by Jackie Nicholas.
Contents Eponents. Introduction......2 Eponents with the Same Base....3 Eponents with Different Bases... 7.4 Scientific Notation..... 8.5 Summary...... 9.6 Eercises... 0 2 Eponential Functions 2. The Functions y =2 and y =2... 2.2 The functions y = b and y = b... 2 2.3 The Functions y = e and y = e... 3 2.4 Summary...... 4 2.5 Eercises... 4 3 Logarithms 5 3. Introduction..... 5 3.2 Logarithms to Base 0 (Common Logarithms)... 5 3.3 Logarithms to Base b.... 20 3.4 Logarithms to Base e (Natural Logarithms).... 22 3.5 Eponential Functions Revisited... 22 3.6 Summary...... 23 3.7 Eercises... 23 4 Solutions to Eercises 24 4. Solutions to Eercises from Section.... 24 4.2 Solutions to Eercises from Section 2.... 27 4.3 Solutions to Eercises from Section 3..... 29
Mathematics Learning Centre, University of Sydney Eponents. Introduction Whenever we use epressions like 7 3 or 2 5 we are using eponents. The symbol 2 5 means 2 2 2 2 2. }{{} This symbol is spoken as two raised to the 5 factors power five, two to the power five or simply two to the five. The epression 2 5 is just a shorthand way of writing multiply 2 by itself 5 times. The number 2 is called the base, and 5 the eponent. Similarly, if b is any real number then b 3 stands for b b b. Here b is the base, and 3 the eponent. If n is a whole number, b n stands for b b b. }{{} Wesay that b n is written in n factors eponential form, and we call b the base and n the eponent, power or inde. Special names are used when the eponent is 2 or 3. The epression b 2 is usually spoken as b squared, and the epression b 3 as b cubed. Thus two cubed means 2 3 =2 2 2=8..2 Eponents with the Same Base We will begin with a very simple definition. If b is any real number and n is a positive integer then b n means b multiplied by itself n times. The rules for the behaviour of eponents follow naturally from this definition. First, let s try multiplying two numbers in eponential form. For eample 2 3 2 4 = (2 2 2) (2 2 2 2) = 2 2 2 2 2 2 2 }{{} 7 factors = 2 7 = 2 3+4. Eamples like this suggest the following general rule. Rule : b n b m = b n+m. That is, to multiply two numbers in eponential form (with the same base), we add their eponents. Let s look at what happens when we divide two numbers in eponential form. For eample, 3 6 = 3 3 3 3 3 3 3 4 3 3 3 3 = 3 3 3 3 3 3 3 3 3 3 = 3 3 = 3 2 = 3 6 4.
Mathematics Learning Centre, University of Sydney 2 This leads us to another general rule. Rule 2: bn b m = b n m. In words, to divide two numbers in eponential form (with the same base), we subtract their eponents. We have not yet given any meaning to negative eponents, so n must be greater than m for this rule to make sense. In a moment we will see what happens if n is not greater than m. Now look at what happens when a number in eponential form is raised to some power. For eample, (2 2 ) 3 = (2 2) (2 2) (2 2) = 2 6 = 2 2 3. This suggest another general rule. Rule 3: (b m ) n = b mn That is, to raise a number in eponential form to a power, we multiply the eponents. Eamples 5 2 5 4 =5 2+4 =5 6 = 5625 2.5 4 2.5 3 2.5 =2.5 4+3 =2.5 6 = 244.40625 ( π5 π 3 )3 =(π 5 3 ) 3 =(π 2 ) 3 = π 2 3 = π 6 96.389 3 n+4 3 m+ simplifies to 3n m+3. 3 y 4 y 2 simplifies to 2 y 2. Eercises Evaluate the following epressions using a calculator where necessary.. 3 4 2. 4 2 3 3 3..57 4. (2.7 2 ) 3 5. (3+2.2 3 ) 4.5 4 Simplify these, or at least change them around a bit. 6. 22+y 7. 7 9.5 2.2 9. (3 ) z 0. 3+2z 3 z 4 8. 4 +3y
Mathematics Learning Centre, University of Sydney 3 Until now we have only considered eponents which are positive integers, such as 7 or 89. Our intention is to etend this notation to cover eponents which are not necessarily positive integers, for eample 5, or 3,ornumbers such as π 3.459. Just as we 3 can make sense of epressions like 5 89,wewant to be able to make sense of epressions such as 5 3 3. But more than this, we want to make sense of these epressions in such a way that rules, 2 and 3 remain valid. It is not at all obvious how we should interpret an epression 5 3 3. Itdoes not really make sense to think of it as 5 multiplied by itself 3 times. 3 Our plan is this: if we want rules, 2 and 3 to hold for general eponents then we will try defining epressions like 5 3 3 to be whatever they must be in order that rules, 2 and 3 remain valid. In other words, we will insist that rules, 2 and 3 remain valid for these more general eponents, and hope that this requirement will tell us what the definitions of epressions like 5 3 3 must be. Let us begin by etending the notation to include an eponent equal to 0. We want to make sense of the epression b 0 in such a way that rules, 2 and 3 hold. What happens to rule 2 when n = m? Rule 2 gives b n b n = b n n or = b 0. Until now we have not attached any meaning to the epression b 0.Itdoesn t make sense to talk about a number being multiplied by itself 0 times. However, if we want rule 2 to continue to be valid when n = m then we must define the epression b 0 to mean the number. If b 0then we define b 0 to be equal to. We do not attempt to give any meaning to the epression 0 0.Itremains undefined. Using this definition we can check that rules and 3 also remain valid. For eample, to check that rule still holds, if n is a whole number and m =0then rule gives b n b 0 = b n which is okay because b 0 =. To be strictly correct we should also check that rule remains valid in the case that m =0 and n =0. You should check that this is true and that rule 3 also remains valid under this definition of b 0. We initially had no idea of how to etend our notation to cover a zero eponent, but if we wish rules, 2 and 3 to remain valid for such an eponent then the definition b 0 = is forced on us. We have no choice. Okay, we have come up with a sensible definition of b 0 by taking m = n in rule 2 and seeing what b 0 must be if rule 2 is to remain valid. To come up with a suitable meaning for negative eponents we can take n<min rule 2. For eample, let s try n =2and m =3.
Mathematics Learning Centre, University of Sydney 4 Rule 2 gives b 2 = b or b 3 = b. b This suggests that we should define b to be equal to. This definition, too, makes sense b for all values of b ecept b =0. In a similar way we can see that we should define b n to mean, ecept when b =0, b n in which case it is undefined. You should convince yourself of this by showing that the requirement that rule 2 remains valid forces on us the definitions b 2 = b 2 and b 3 = b 3. If n is a positive integer (for eample n =7or n = 78) then we define b n to be equal to. This definition makes sense for all values of b ecept b =0,inwhich case the b n epression b n remains undefined. You should check that, with this definition, rules and 3 also remain valid. Eamples 3 2 3 4 =3 0 = 2 = 2 = 2 2 3. 4+2 =2 3. 2 =8 0.82644628 = 6.6570248 ( + 3 ) = = = = + 3 + 3 2 + 3 3 2 + Eercises Evaluate the following epressions.. 5 3 2. 2 3. 5 6 9 4. 2.2 4 2 3 5. (6 2 ) 2 2 3 Simplify the following epressions. 6. ( 2 + y 2 )( 2 y 2 ) 7. ( 4 y 4 )( 3 4 + 2 y 4 + 4 y 2 + y 3 4 ) 8. 3 2 +y 2 2 +y 2 20. ( 2 ) 2 9. ( 2 ( y) ( 2 y 2 )( 2 +y 2 ) ) 2
Mathematics Learning Centre, University of Sydney 5 Pause for a moment and look at what has been achieved. We have been able to give a meaning to b n for all integer values of n, positive, negative, and zero, and we have done it in such a way that all three of the rules above still hold. We can give meaning to epressions like ( 35 7 )3 and π 7. Wehave come quite a way, but there are a lot of eponents that we cannot yet handle. For eample, what meaning would we give to an epression like 5 7 9? Our net task is to give a suitable meaning to epressions involving fractional powers. Let us start with b 2.Wewant to give meaning to this epression in such a way that the rules, 2 and 3 remain valid. If rule 2 is to hold then we must have b 2 b 2 = b 2 + 2 = b = b. Let s be specific and take b =4. Then, 4 2 4 2 =4,so4 2 is equal to a number whose square is 4. There are two numbers whose square is 4. They are 2 and 2. We define 4 2 to be the positive square root of 4. That is, 2. In general, b 2 is defined to be the positive square root of b, also written b.sob 2 = b. Of course, b must be positive if b 2 is to have any meaning for us, because if we take any real number and multiply itself by itself then we get a positive number. (Actually there is a way of giving meaning to the square root of a negative number. This leads to the notion of comple numbers, a beautiful area of mathematics which is beyond the scope of this booklet.) That takes care of a meaning for b 2 valid then we must have if b>0. Now have alookatb 3.Ifrule 2 is to remain b 3 b 3 b 3 = b 3 + 3 + 3 = b = b. As a concrete eample take b =8. Then 8 3 must be such that 8 3 8 3 8 3 =8. There is just one number which when multiplied by itself 3 times gives 8. That number is 2. Thus 8 3 =2. For another eample take b = 8. This time we have no trouble giving a meaning to ( 8) 3,even though 8 < 0. There is a number which when multipied by itself 3 times gives 8, namely 2, so ( 8) 3 = 2. In general if we wish we wish to give meaning to epressions like b n rule 3 holds then we must have (b n ) n = b = b. in such a way that If b is positive, b n is defined to be a positive number, the n th root of b. That is, a number whose n th power isequal to b. This number is sometimes written n b. If b is negative we need to look at separately at the cases where n is even and where n is odd. If n is even and b is negative, b n cannot be defined, because raising any number to an even power results in a positive number.
Mathematics Learning Centre, University of Sydney 6 If n is odd and b is negative, b n can be defined. It is a negative number, the n th root of b. For eample, ( 27) 3 = 3 because ( 3) ( 3) ( 3) = 27. Now we can see how to define b p q for any number of the form p, where p and q are integers. q Such numbers are called rational numbers. Notice that p = p,soifrule 3 is to hold then q q b p q =(b q ) p =(b p ) q. We know how to make sense of (b q ) p and (b p ) q, and they turn out to be equal, so this tells us how to make sense of b p q.ifwewant rules, 2 and 3 to hold then we must define b p q to be either one of (b p ) q or (b q ) p. This definition always makes sense when b is positive, but we must take care when b is negative. If q is even then we may have trouble in making sense of b p q for negative b. For eample we cannot make sense of ( 3) 3 2. This is because we cannot even make sense of ( 3) 2, let alone (( 3) 2 ) 3. Trying to take the eponents in the other order does not help us because ( 3) 3 = 27 and we cannot make sense of ( 27) 2. However it may be that the numerator and denominator of p contain common factors q which, when cancelled, leave the denominator odd. For eample we can make sense of ( 3) 4 6,even though 6 is even, because 4 = 2, and we can make sense of ( 3) 2 3. 6 3 A rational number p is said to be epressed in its lowest form if p and q contain no q common factors. If p, when epressed in its lowest form, has q odd then we can make q sense of b p q even for b<0. To recapitulate, we define b p q =(b q ) p =(b p ) q. This definition makes sense for all p if q b>0. If b<0 then this definition makes sense providing that p is epressed in its lowest q form and q is odd. So far, if b>0, we have been able to give a suitable meaning to b for all rational numbers. Not every number is a rational number. For eample, 2isanirrational number: there do not eist integers p and q such that 2= p.however for b>0itispossible to q etend the definition of b to irrational eponents so that rules,2 and 3 remain valid. Thus if b>0 then b is defined for all real numbers and satisfies rules, 2 and 3. We will not show how b may be defined for irrational numbers. Eamples ( 3 ) = =3 ( 3 ) (0.2) 3 = = = 25 (0.2) 3 0.008 ( 64) 2 3 =[( 64) 3 ] 2 =( 4) 2 =6or, ( 64) 2 3 =[( 64) 2 ] 3 = (4096) 3 =6 6 3 4 =( 4 6) 3 =2 3 =8
Mathematics Learning Centre, University of Sydney 7 ( 6) 3 4 is not defined. 5 3 2 =5 + 2 =5 5 2 =5 5 Eercises If the following epressions are not defined then say so. Otherwise evaluate them. 2. 25 3 2 22. ( 8) 5 4 23. 8 5 4 24. ( 27) 3 2 25. ( 27) 2 3.3 Eponents with Different Bases From the definition of eponents we know that if n is a positive integer then (ab) n = (ab) (ab) (ab) }{{} n factors = a a a } {{ } n factors = a n b n. b b b }{{} n factors (switching the order around) Just as in section.2, we can show that this equation holds true for more general eponents than integers, and we can formulate the following rule: Rule 4: (ab) = a b whenever both sides of this equation make sense, that is, when each of (ab), a and b make sense. Again, from the definition of eponents we know that if n is a positive integer then ( ) a n = a b b a b a (b 0) }{{ b} n factors n factors {}}{ a a a = b b b }{{} n factors = an b n As in section.2, we can show that this equation remains valid if the integer n is replaced by a more general eponent. We can formulate the following rule: Rule 5: ( a b ) = a whenever both sides of this equation make sense, that is, whenever b ( a b ), a and b make sense. An epression of the form a b y cannot generally be simplified, though it can be written in the form (ab y ) or (a y b) y if necessary. For eample, we cannot really make the epression a 2 b 5 any simpler than it is, though we could write it in the form (ab 5 2 ) 2 or (a 2 5 b) 5. Eamples (2 3) 3 =2 3 3 3 =8 27 =26=6 3
Mathematics Learning Centre, University of Sydney 8 (4) 2 =4 2 2 =2 2 =2 ( 40) 3 =( 8 5) 3 =( 8) 3 (5) 3 = 2 3 5 ( 2 3 )3 = 23 = 8 3 3 27 ( 4 7 ) 2 = = 72 = 49 ( 4 7 )2 4 2 6 ( 27 8 ) 3 =( 8 ) 3 = ( 8) 3 = 2 27 27 3 3 Eercises Simplify the following epressions. 26. ( 4 y) 2 27. 6 3 36 3 28. ( ) y 3 4 4 29. ( 8 6 ) 4 30. (a 3 b ) 3.4 Scientific Notation Scientific notation is a way of epressing any number, especially a very large or a very small number, in a concise and convenient fashion using powers of 0. For eample, consider the numbers 2700000000000000000 and 27000000000000000. It is plain that they are both quite large numbers. However, written down in this way it is difficult for us to see just how large these numbers are, or to compare their sizes. It is clear that if we did want to compare the size of these numbers we would probably begin by counting the numbers of zeros at the tail of each of these numbers. The idea behind scientific notation is that any number can be written as a number between and 0 multiplied by a power of 0. For eample: 45 = 4.5 0 450 = 4.5 0 2 4500 = 4.5 0 3 0.45 = 4.5 0 0.045 = 4.5 0 2 0.0045 = 4.5 0 3. Notice that the power to which 0 is raised (the eponent) indicates where to shift the the decimal point. Thus in the epression 4500 = 4.5000 0 3 the eponent is positive and decimal point has been shifted to the right by 3 places. Similarly, in 0.045 =4.5 0 2 the eponent is negative and the decimal point has been shifted to the left by 2 places. To return to the eamples given in the beginning of this section, the first number can be written as 2.7 0 8 and the second as 2.7 0 6. Not only is it easier and quicker to write these numbers in scientific notation, but written in this fashion it is clear that the second number is smaller than the first by a factor of 0 2 = 00.
Mathematics Learning Centre, University of Sydney 9 Remember that in scientific notation it is conventional that the epression is written as anumber between and 0 multiplied by a power of 0. For eample we would write 3.7 0 8 rather than 37 0 7 or.37 0 9. Eamples 967 =9.67 0 2.0000439 = 4.39 0 5 8.26 0 3 = 82600000000000 (.2 0 4 ) (3.7 0 2 )=.2 3.7 0 4+2 =4.44 0 6 (3.47 0 7 ) (2.3 0 23 )=3.47 2.3 0 7+23 =7.98 0 7+23 =7.98 0 6 (.3 0 2 )+(7.35 0 3 )=.03 +.00735 =.02035 = 2.035 0 2 4.28 0 28 = 4.28 2.9 0 2.9 0.4708 0 28 =.4708 0 7 3.77 0 8 = 3.77.5 0 4.5 0 4 3.278 0 8 ( 4) =3.278 0 4 Eercises Write the following numbers in scientific notation. 3. 0.0049 32. 3. 0 2 4.2 0 8 33. 2.7 0 2 +4.3 0 34. 8327 35. 2. 0 2 8.7 0 3.5 Summary If b>0 then b is defined for all numbers. Ifb<0 then b is defined for all integers and all numbers of the form p where p and q are integers, p is epressed in its lowest form and q q q is odd. The number b is called the base and is called the power, inde or eponent. Eponents have the following properties:. If n is a positive integer and b is any real number then b n = b b b. }{{} n factors 2. b n n = b, and if n is even we take this to mean the positive n th root of b. 3. If b 0then b 0 =. b 0 is undefined for b =0. 4. If p and q are integers then b p q =(b q ) p =(b p ) q. 5. b b y = b +y whenever both sides of this equation are defined. 6. b b y = b y whenever both sides of this equation are defined. 7. b = whenever both sides of this equation are defined. b 8. (ab) = a b whenever both sides of this equation are defined. 9. ( a b ) = a b whenever both sides of this equation are defined. Any real number can be written in the form a 0 n where a is a number between 0 and 0 and n is an (positive or negative) integer. This is called scientific notation.
Mathematics Learning Centre, University of Sydney 0.6 Eercises The following epressions evaluate to quite a simple number. If you leave some of your answers in fractional form you won t need a calculator. 36. 9 2 37. 6 3 4 38. ( 5 ) 39. (3 ) 2 40. ( 5 2 ) 2 4. ( 8) 3 2 42. ( 27 8 ) 2 3 43. 5 27 5 24 44. 8 2 2 2 45. ( 25) 2 3 These look a little complicated but are equivalent to simpler ones. Simplify them. Again, you won t need a calculator. 46. 3n+2 6 47. 48. (a 3 n 2 6 2 + b 2 ) 2 49. ( 2 + y 2 ) 2 2 ( 2 + y 2 ) 2 50. 2 + 2 5. (u 3 v 3 )(u 2 3 +(uv) 3 + v 2 3 ) Write these numbers in scientific notation. 52. 0.0037 53. 2.5 0 7 3.54 0 54. 3.47 0 7 7.4 0 3 55. (2.7 0 65 ) 3 56. 5.98 0 6 3.7 0 5 57. 3.8 0 27 2.45 0 8 Don t bother working these ones out, just decide whether or not the epressions are defined. 58. (.7) 8 59. ( 3) 2 9 60. ( 4.8) 6 4 6. (π) 2 62. ( π) 8 4
Mathematics Learning Centre, University of Sydney 2 Eponential Functions 2. The Functions y =2 and y =2 In the previous section we saw how, if b is a positive number, we can make sense of the epression b for all real numbers. Itturns out that functions of the type y = f() =b, where b is a positive number, are of great importance in mathematics and in all branches of the sciences. To get an indication of how these functions behave we have graphed the function f() =2 in Figure. You should be aware of several important features of this graph. y 5 0 5-4 -3-2 - 2 3 4 Figure : Graph of the function f() =2 The function f() =2 is always positive (the graph of the function never cuts the - ais), although the value of the function gets very close to zero for values of very large negative (ie a long way to the left along the -ais). For eample, when = 5 wehave 2 =0.0325. The function 2 increases very rapidly for large values of. From the rules of eponents discussed in section you should know that 2 + =2 2. Inwords, the value of 2 doubles if is increased by. The graph of y =2 intercepts the y-ais at y =.You should epect this because you know from the rules of eponents that 2 0 =. Figure 2 displays the graph of the function f() =2. How is the graph of y =2 related to the graph of y =2? Well, if we set =then 2 =2 =, which is the 2 value which would have been obtained by setting = inthe function y =2.Inthe same way we see that if we set = 7 inthe function y =2 then we obtain the same value as we would by setting =7in the function y =2. Proceeding like this we see that the graph of the function y =2 is the reflection in the y-ais of the graph of y =2. Compare Figure with Figure 2. From the rules of eponents discussed in section it follows that 2 =(2 ) =( 2 ). The function y =2 is the same as the function y =( 2 ), and so 2 (+) =( 2 )+ = 2 ( 2 ) =( 2 ) 2.
Mathematics Learning Centre, University of Sydney 2 y 5 0 5-4 -3-2 - 2 3 4 Figure 2: Graph of the function y =2 In words, the value of the function y =2 is decreased by a factor of 2 by. if is increased 2.2 The functions y = b and y = b Any function of the form y = b where b>0 and b behaves like one of the functions y =2 or y =( 2 ) =2. If b> then the function y = b is increasing and behaves like y =2. If b< then the function is decreasing and behaves like y =( 2 ) =2. If b =then y = =for all. Notice that regardless of the value of b, providing always that b>0, the function y = b intercepts the y-ais at y =. This is because b 0 =for all numbers b. Figure 3 shows the graphs of the functions y = b for various values of b. y 20 y = 0 y = e 5 0 - y =.5 = (2/3) 5 y =.5 y = = -4-3 -2-2 3 4 Figure 3: Graphs of y = b for various values of b.
Mathematics Learning Centre, University of Sydney 3 Eercises. Make a careful sketch of the graphs of the functions y =2.5 and y =5. Indicate where (if at all) these functions intercept the aes. 2. Which of the following functions are increasing and which are decreasing? You should be able to decide without graphing the functions or substituting any values, though you may do so if you wish. a. f() =2.7 b. f() =( 2.7 ) c. f() =3 d. f() =0.22 2.3 The Functions y = e and y = e There is a number called e which has a special importance in mathematics. Like the number π, the number e is an irrational number (see section.2), which is equivalent to saying that it has a non-terminating, non-repeating decimal representation. In other words we can never write down eactly what e is. To 5 decimal places it is equal to 2.7828, but this is just an approimation of the correct value. Unless you really need to write down an approimate value for e it is more convenient and accurate to leave the symbol e in epressions involving this number. For eample, it is preferable to write 2e rather than 2 2.7828 or 5.43656. In mathematics the functions e and e are particularly important. Because of this we have graphed them in figure 4. You can see how similar these functions are to the other eponential functions. The reasons for their importance are discussed briefly in section 5.5. The function y = e is often referred to as the eponential function, and is even given another special symbol, ep, so that ep() =e and ep( ) =e. y = e - y 20 y = e 5 0 5-4 -3-2 - 2 3 4 Figure 4: Graphs of e and e.
Mathematics Learning Centre, University of Sydney 4 2.4 Summary Functions of the form f() =b, where b>0 and b are called eponential functions. If b< then b is a decreasing function, and if b> then b is an increasing function. The function b is equal to the function ( b ). The number e 2.7828 and the functions e and e are of special importance in mathematics. The function e is often given the special name ep, so that ep() =e and ep( ) =e. 2.5 Eercises 3. Which of the following functions are increasing and which are decreasing? If you have understood this section fully you will be able to answer this question without graphing the functions or substituting any values. a. f() =( 5 3 ) b. f() =( 5 3 ) c. f() =( 3 5 ) d. f() =( 3 5 ) 4. Sketch the graphs of the functions f() =3 and f() =3.Onthe same diagrams mark in roughly the graphs of f() =2.9 and 2.9. 5. It is true that e.0986 3. Try it for yourself on a calculator if you have one. How do you think the functions y =3 and y = e.0986 compare? Why? If you cannot solve this otherwise, you might like to try substituting in a few numbers for in both of the functions and comparing the values.
Mathematics Learning Centre, University of Sydney 5 3 Logarithms 3. Introduction Taking logarithms is the reverse of taking eponents, so you must have a good grasp on eponents before you can hope to understand logarithms properly. Review the material in the first two sections of this booklet if necessary. We begin the study of logarithms with a look at logarithms to base 0. It is important that you realise from the beginning that, as far as logarithms are concerned, there is nothing special about the number 0. Indeed, the most natural logarithms are logarithms to base e, and they are introduced in section 3.4. Logarithms to base 0 are in common use only because we use a decimal system of counting, and this is probably a result of the fact that humans have ten fingers. We have begun with logarithms to base 0 only to be definite, and we could just as easily have started with logarithms to any other convenient base. 3.2 Logarithms to Base 0 (Common Logarithms) We will begin by considering the function y =0, graphed in Figure 5. As we know y 20 5 0 5-2 -.5 - -0.5 0.5.5 2 Figure 5: Graph of f() =0 from the discussion in section, given any number, wecan raise 0 to the power of to obtain another number which we write as 0. What of the reverse procedure? Suppose we begin with a number and we wish to find the power to which 0 must be raised to obtain that number. For eample, suppose we begin with the number 7 and we wish to find the power to which 0 must be raised to obtain 7. This number is called the logarithm to the base 0 of 7 and is written log 0 7. Similarly, log 0 5 is equal to the power to which 0 must be raised to obtain 5. Forageneral number, log 0 is equal to that power to which 0 must be raised to obtain the number.
Mathematics Learning Centre, University of Sydney 6 When we see an epresion like log 0 29 we can think of it as a sort of a question. The question we have in mind is this: to what power must we raise 0 to get 29? Or, 0? = 29. The answer to this question is a number, and we call that number log 0 29. The definition of the logarithm to base 0 is the basis on which the remainder of this section rests, and it is etremely important that you understand it properly. Again: log 0 is equal to that power to which 0 must be raised to obtain the number. As an eample, let s calculate log 0 0 3. According to the definition, log 0 0 3 is equal to that power to which 0 must be raised to obtain 0 3.Towhat power must we raise 0 to obtain 0 3? Or, 0? =0 3. Surely the answer is 3. Notice that 0 3 = 000, so we have worked out log 0 000, and without using a calculator! We have been able to work this out because we have understood the meaning of the logarithm of a number. We will need to use a calculator to work out the logarithms of most numbers, but it is very important that we understand what it is that the calculator is working out for us when we push the buttons. Without a calculator we can work out the logarithms of many numbers. Eamples: log 0 00 = log 0 0 2 =2 log 0 0. = log 0 0 = log 0 0 0 = log 0 0.5 =.5 Eercise. By epressing these numbers as powers of 0, and without using a calculator, calculate the logarithms to base 0 of the following numbers. a. 0000 b. c. 0.00 d. 0 2.3 00 e. 0 f. 0 4 0 g. ( 000 ) 3 0 h. 0.00 Can we take the logarithm of any number? In other words, given any number can we find a power to which 0 may be raised to obtain the number? Look at the graph of y =0 in Figure 5. We see that 0 is never negative and indeed never even takes the value 0. There is no power to which we may raise 0 to obtain a number less than or equal to 0. This means that we cannot take the logarithm of a number less than or equal to zero. We say that log 0 is undefined for 0. The graph of 0 gives us another important piece of information. If >0 then there is only one power to which we may raise 0 to get. Our definition of log 0 is unambiguous. The graph of y = log 0 is shown in Figure 6. important features of this graph. You should pay attention to several The graph intercepts the -ais at =.Inother words, log 0 =0. You should epect this because you should know from section that 0 0 =.
Mathematics Learning Centre, University of Sydney 7 y 0.5 2 4 6 8 0 2 4 6-0.5 - Figure 6: Graph of f() =log 0 The graph does not etend to the left of the y-ais, and in fact never even intercepts the y-ais. We have already commented on the fact that the logarithm of a number less than or equal to zero is not defined. The function y = log 0 gets as large as we like as gets large. By this we mean that we can make log 0 as large as we choose by choosing to be sufficiently large. The graph does not stay below a certain height as gets large (it does not have a horizontal asymptote). However the function y = log 0 increases very slowly as increases. The fact that we bother to specify the base as being 0 suggests that we can take logarithms to other bases. We can, and we shall say more about this later, but for now let us stick with base 0. You should be aware that many writers may not mention the base of the logarithms they are referring to if it is obvious from the contet what that base is, or if it does not matter which base is used. They may just write the logarithm of orlog. Because logarithms to base 0 have been used so often they are called common logarithms. If you have a calculator it probably has a Log button on it. You could use it to find, for eample, log 0 7 and log 0 0.0. From the eamples above you should be able to see that if we epress a number as a power of 0 then we can read off the logarithm to base 0 of that number from the power. Let s try to make this precise. Suppose that is any real number. What is log 0 0? Well, log 0 0 is that power to which 0 must be raised to obtain the number 0. To what power must we raise 0 to obtain the number 0? Or, to put this question another way, 0? =0. The answer must be. Thuslog 0 0 =. This is our first rule of logarithms. Rule A: Forany real number, log 0 0 =.
Mathematics Learning Centre, University of Sydney 8 Eamples log 0 0 3.7 = 3.7 log 0 0.000 = log 0 0 4 = 4 log 0 0 4 5 0 3 = log 0 0 4 (0 3 ) 5 = log0 0 4+ 3 23 5 = 5 Rule A tells us what happens if we first raise 0 to the power and then take the logarithm to base 0 of the result. We end up with what we started with. What happens if we do things in the reverse order? Consider the number log 0 7. If you have a calculator with a Log button on it you can see that this number is approimately 0.845. Now suppose we raise 0 to the power log 0 7. What do you think the result is? In symbols, what is 0 log 0 7? Well, remember that log 0 7isequal to that power to which 0 must be raised to give the number 7. So if we raise 0 to that power then we must get 7. The same reasoning applies to show that if >0 then 0 log 0 =. The number log 0 is that power towhich 0 must be raised to obtain. Soifweraise 0 to this power we must get. Wewill write this down as the second of our rules of logarithms. Rule B: Forany real number >0, 0 log 0 =. Eamples 0 log 0 π = π 0 log 0 (2 +y 2 ) = 2 + y 2 0 log 0 033 = 0 33 Eercises 2. Simplify the following epressions. a. 0 log 0 37.23 b. log 0 0 2 y c. 0 log 0 (0 ) d. log 0 0 2y e. 0 0log 0 f. log 0 0 +y z g. 0 log 0 ( 3y z ) h. log 0 0 02 Rules A and B epress the fact that the functions y =0 and y = log 0 are inverse functions of one another. If you have not come across the concept of inverse functions before then do not worry about what this means. If you have, then you will probably remember that the graph of an inverse function is obtained by reflecting the graph of the original function in the line y =, that is the line which runs in the north-east and south-west direction. Take another look at Figures 5 and 6. We can use the rules of eponents discussed in section to work out more rules for logarithms. If and y are numbers greater than zero then, by rule B, =0 log 0 and y =0 log 0 y, so y = 0 log 0 0 log 0 y = 0 log 0 +log 0 y (by the rules for eponents).
Mathematics Learning Centre, University of Sydney 9 This equation tell us that if we raise 0 to the power log 0 +log 0 y then we get the number y. Inother words it tells us that log 0 + log 0 y is the answer to the question 0? = y. But the answer to this question is also log 0 y. Thus log 0 y = log 0 + log 0 y. This we will call our third rule of logarithms. Rule C: Forany real numbers >0 and y>0, log 0 y = log 0 + log 0 y. So much for multiplication. What of division? If >0 and y>0 then y = 0log 0 0 log 0 y (by rule B) = 0 log 0 log 0 y (by the rules for eponents). This equation tells us that if we raise 0 to the power log 0 log 0 y then we get the number y.inother words, log 0 y = log 0 log 0 y. This is our fourth rule of logarithms. Rule D: Forany real numbers >0 and y>0, log 0 ( y )=log 0 log 0 y. If is a number, >0, and n is any number at all then: n = (0 log 0 ) n (by rule B) = 0 n log 0 (by the rules for eponents). This equation tells us that if we raise 0 to the power n log 0 then we get the number n.inother words, log 0 n = n log 0. This is our fifth rule of logarithms. Rule E: For real numbers and n, with >0, log 0 n = n log 0 Eamples log 0 y z = log 0 + log 0 y log 0 z log 0 3 y 2 = 3log 0 2 log 0 y 2 log 0 y 4 log 0 ( 2 z 3 ) = log 0 y 2 ( 2 z 3 ) 4 Eercises 3. Rewrite the following epressions so that they involve just one logarithm. a. log 0 3 2.5 log 0 y b. log 0 6+log 0 2 c. 5 log 0 3 4 log 0 (y + z 2 ) d. 2 log 0 y +3log 0 (z 2 y 2 ) e. log 0 ( + y) 3 log 0 4 f. log 0 y.7 log 0 y 2
Mathematics Learning Centre, University of Sydney 20 3.3 Logarithms to Base b As we mentioned above, we can take logarithms to other bases. If b is a real number, b>, and if is a real number, >0, then we define the logarithm to base b of to be that power to which b must be raised to obtain the number. Youmay also think of log b as the answer to the question b? =. You should notice that if b =0then this definition agrees with the one given earlier for log 0. Again: the logarithm to base b of a number >0 (written log b ) is that power to which b must be raised to obtain the number. Eamples: log 5 25 = log 5 5 3 =3 log 6 2 = log 6 6 4 = 4 log 7 49 = log 7 7 2 = 2 y 2 y = log 2 y = log e 0 y = log 0 2 3 4 5 6 7 8 - -2 Figure 7: Graph of f() =log b for various values of b. We have required the base of our logarithms, b, tobegreater than. In fact we can take logarithms to any base b provided b>0 and b.itismore usual though to use b>, and in this booklet we will always use a base b>. Figure 7 shows graphs of the functions y = log b for various values of b. As you can see from these graphs, the logarithm functions behave in a similar fashion for different bases b, providing b>. All of what we said earlier remains true for log b if 0 is replaced by b. Inparticular the five rules of logarithms remain true. Let us restate these to be applicable to log b.
Mathematics Learning Centre, University of Sydney 2 For areal number b>: Rule : Forany real number, log b b = Rule 2: Forany real number >0, b log b = Rule 3: Forany real numbers >0 and y>0, log b y = log b + log b y Rule 4: Forany real numbers >0 and y>0, log = log b y b log b y Rule 5: For real numbers and n, with >0, log b n = n log b Eercises 4. Simplify these epressions. a. log 2 2 2 +y b. 5 log 5 +y 3 c. log 7 49 uv d. 3 log 9 u 3w 5. Rewrite the following epressions so that they involve only one logarithm. a. 2 log 3 ( + y) 3 log 3 (y)+log 3 2 b. log 6 y 4 log 6 ( + y) c. 4 log 7 y 2 + log 7 ( 2 + y 2 ) 2.5 log 7 Now that we have shown how to define logarithms to any base b>, let us see how these logarithms are related to each other. We will consider logarithms to two bases a> and b>. By rule 2, = a log a. Taking logarithms to base b of both sides of this equation yields log b = log b (a log a ) = log a log b a (by rule 5). This, our sith rule of logarithms, tells us how logarithms to different bases are related. Rule 6: For numbers >0, a> and b>, log b = log b a log a. From this rule we see that log b a log a b = log b b =,and so log b a = log a b. This fact enables us to calculate the logarithm of a number to any base from a calculator which calculates logarithms to one base only. Eample: If your calculator only has logarithms to base 0 on it, how can you find log 7 9? Solution: By rule 6, log 7 9 = log 7 0 log 0 9 = log 0 7 log 0 9 and the last epression can be evaluated by any calculator which can evaluate logarithms to base 0. Eercises 6. Using a calculator, find the following logarithms. a. log 3 7 b. log 5 2 c. log 22 4 d. log 4 8
Mathematics Learning Centre, University of Sydney 22 3.4 Logarithms to Base e (Natural Logarithms) Logarithms to the base 0 are commonly used, because we use a decimal number system and not a base 8 system, or a base 2 system. If humans were born with 3 toes (or if sloths could count) then logarithms to base 3 might be in common use. Apart from the fact that we use a decimal number system, there is no reason for us to prefer logarithms to base 0 over logarithms to any other base. Indeed, we mentioned in section 2 that the function y = e is a very important function in mathematics, and it is therefore reasonable to epect that logarithms to base e will also assume special importance. They do, and are given the name Natural Logarithms or Napierian Logarithms. They are even given a special symbol, ln, so that ln = log e. One of the graphs in Figure 7 is a plot of the function y = log e =ln. Notice that the function y =ln behaves in a similar fashion to the function y = log 0. This comes as no surprise to us since we have already seen in section 2 that the functions e and 0 are very similar to each other. 3.5 Eponential Functions Revisited In section 2 we saw how much the eponential functions resemble each other. If b> then the eponential function b looks very much like any of the other eponential functions with base greater than, and if b< then b looks a lot like any of the eponential functions with base less than one. We will now be able to see more clearly what is going on here. Consider the function y =2.Now2=e log e 2,sowecan write 2 =(e log e 2 ) = e log e 2. We have been able to write the function 2 as a function involving the base e, though the eponent is now not simply, but multiplied by some fied number, namely log e 2. Similarly, we could write 5 = e log e 5 9 = e log e 9 7 = 4 log 4 7 e = 3 log 3 e We can write all eponential functions in the form y = e k, where k is some constant which may be negative. Eercises 7. Write each of the following functions in the form y = e k for a suitable constant k. a. y =0 b. y =7.5 c. y =4 d. y =( 4 )
Mathematics Learning Centre, University of Sydney 23 3.6 Summary Forany real number b> and any >0, log b is equal to that number to which b must be raised to obtain the number. One can think of log b as the answer to the question b? =. The number log b is called the logarithm to base b of. The function log b satisfies the following rules: Rule : Forany real number, log b b = Rule 2: Forany real number >0, b log b = Rule 3: Forany real numbers >0 and y>0, log b y = log b + log b y Rule 4: Forany real numbers >0 and y>0, log = log b y b log b y Rule 5: For real numbers and n, with >0, log b n = n log b Rule 6: For numbers >0, a> and b>, log b = log b a log a. Logarithms to base 0 are in common use and for this reason they are called Common Logarithms. Logarithms to base e are of special importance. They are often called natural logarithms or Napierian logarithms, and the symbol ln is often used for them. Thus ln = log e. Any eponential function may be written in the form e k, where the constant k may be negative. 3.7 Eercises Without using a calculator, find the following numbers. 8. log 0 0 9 9. log e e 5 e 0. log 2 6. log 7 7 3 7 2. ln e2 e 2 3. ln e 7 log 2 4. 5 log 5 32.7 5. e ln 9 2 6. e ln 3 27 Rewrite the following epressions using the rules of logarithms, and simplify where possible. 00 7. log 2 0 8. ln y 3 4 9. log.3 z 7 9y e.37 4 2 y 3 20. log 3 3 y 2 27z 2 2. ln(e 2.4 6 ) 22. log 5 25 3 0.2y 2 Using the rules of logarithms, rewrite the following epressions so that just one logarithm appears in each. 23. 3 log 2 + log 2 30 + log 2 y log 2 w 24. 2ln ln y + a ln w 25. 2(ln +lny) 26. log 3 e ln 8 + log 3 5 log 5 w 27. log 7 0 log 0 2 log 7 49 28. log 0 0. log 6 2 log 6 y + log 6 4 log 4 e Given that log e 5.6094, and log e 7.9459, find the following numbers without using a calculator ecept to perform multiplication or division. 29. log 5 e 30. log 5 7 3. log 5 7 2 32. log 49 5 33. log 49 25 34. log e 25
Mathematics Learning Centre, University of Sydney 24 4 Solutions to Eercises 4. Solutions to Eercises from Section. 3 4 =3 3 3 3=8 2. 4 2 3 3 =4 4 3 3 3=6 27 = 432 3..5 7 =.5 7 4 =3.375.5 4 4. (2.7 2 ) 3 =2.7 2 3 = 387.420489 5. (3 + 2.2 3 ) 4 =3.648 4 = 34695.732 6. 2 2+y 2 =2 2+y =2 +y =2 2 y 7. 7 9.5.2 = 7+9.5.2 = 5.3 8. 4 +3y =4 4 3y 9. (3 ) z =3 z 0. 3+2z 3 z 4 =3 +2z z 4 =3 + 7z 4. 5 = 5 2. 3 2 2 3 = 9 8 =72 3. 5 6 9 =5 3 = 5 3 = 25 4. 2.2 4 2 3 =2.2 2 =.20666 2.2 2 5. (6 2 ) 2 =6 2 2 = =.00077604938276 6 4 6. 7. ( 2 + y 2 )( 2 y 2 ) = 2 2 2 y 2 + y 2 2 y 2 y 2 = y ( 3 3 3 3 4 y 4 )( 4 + 2 y 4 + 4 y 2 + y 4 ) = 4 4 + 4 2 y 4 + 4 4 y 2 + 4 y 4 y 3 3 4 4 y 4 2 y 4 y 4 4 y 2 y 4 y 4 = + 3 3 4 y 4 + 2 y 2 + 4 y 4 3 3 4 y 4 2 y 2 4 y 4 y = y 8. 3 2 + y 2 2 + y 2 = ( 2 + y 2 ) 2 + y 2 =
Mathematics Learning Centre, University of Sydney 25 9. 2 2 ( y) ( 2 y 2 )( 2 + y 2 ) 2 2 ( y) = 2 2 + 2 y 2 2 y 2 y 2 y 2 = 2 2 ( y) y = ( ) 2 2 = 20. ( 2 ) 2 = 2 2 = 2. 25 3 2 = (25 2 ) 3 =5 3 = 25 22. ( 8) 5 4 is not defined. 23. 8 5 4 = (8 4 ) 5 =3 5 = 243 24. ( 27) 3 2 is not defined. 25. ( 27) 2 3 =(( 27) 3 ) 2 =( 3) 2 =9 26. ( 4 y) 2 = 4 2 y 2 = 2 y 2 27. 6 3 36 3 =(6 36) 3 =6 28. ( y 3 4 ) 4 = y 3 4 = y 3 4 29. ( 8 6 ) 4 = 8 4 6 4 = 3 2 30. (a 3 b ) 3 = a 3 3 b 3 = a 9 b 3 3. 0.0049 = 4.9 0 3 32. 3. 0 2 4.2 0 8 =3. 4.2 0 2 8 =3.02 0 6 =.302 0 5 33. 2.7 0 2 +4.3 0 = 270 + 0.43 = 270.43 =2.7043 0 2 34. 8327 = 8.327 0 3 35. 2. 0 2 8.7 0 3 =2. 8.7 0 2 3 =8.27 0 =.827 36. 9 2 = 9=3 37. 6 3 4 = (6 4 ) 3 =2 3 =8 38. ( 5 ) = 5 =5 39. (3 ) 2 =3 2 = 3 2 = 9 40. ( 5 2 ) 2 =( 2 5 )2 = 4 25
Mathematics Learning Centre, University of Sydney 26 4. ( 8) 3 2 is not defined. 42. ( 27 8 ) 2 3 =(( 27 8 ) 3 ) 2 =( 3 2 )2 = 9 4 43. 5 27 5 24 =5 27 24 =5 3 = 25 44. 8 2 2 2 =(8 2) 2 =6 2 =4 45. ( 25) 2 3 =(( 25) 3 ) 2 =( 5) 2 =25 46. 3n+2 3 n 2 =3 n+2 (n 2) =3 4 =8 47. ( 6 6 )=( 6 6 ) 2 = 6 2 6 2 = 4 3 48. (a 2 + b 2 ) 2 =(a 2 ) 2 +2a 2 b 2 +(b 2 ) 2 = a +2a 2 b 2 + b 49. ( 2 + y 2 ) 2 2 ( 2 + y 2 ) 2 = ( 2 + y 2 ) 2 2 ( 2 + y 2 ) 2 = (2 + y 2 ) 2 ( 2 + y 2 ) 2 2 ( 2 + y 2 ) 2 = 2 + y 2 2 ( 2 + y 2 ) 2 = y 2 ( 2 + y 2 ) 2 50. 2 + 2 5. = 2 2 + 2 =+ 2 (u 2 2 2 2 2 2 3 v 3 )(u 3 +(uv) 3 + v 3 ) = u 3 u 3 + u 3 (uv) 3 + u 3 v 3 v 3 u 3 v 3 (uv) 3 v 3 v 3 = u v 52. 0.0037 = 3.7 0 3 53. 2.5 0 7 3.54 0 =2.5 3.54 0 7 =7.6 0 6 54. 3.47 0 7 7.4 0 3 =3.47 7.4 0 7 3 =25.678 0 4 =2.5687 0 5 55. (2.7 0 65 ) 3 =(0.27 0 66 ) 3 =(0.27) 3 (0 66 ) 3 0.646 0 66 3 =6.46 0 22 56. 5.98 0 6 3.7 0 5 = 5980000 370000 = 560000 = 5.6 0 6 57. (3.8 0 27 ) (2.45 0 8 ) = 3.8 2.45 027 ( 8) =.55 0 35 58. (.7) 8 is undefined. 59. ( 3) 2 9 is defined. 60. ( 4.8) 6 4 =( 4.8) 3 7 is defined. 6. (π) 2 is defined 62. ( π) 8 4 =( π) 4 7 is defined.
Mathematics Learning Centre, University of Sydney 27 4.2 Solutions to Eercises from Section 2. The graphs of y =2.5 and y =5 appear below. In both cases the graph intercepts the y-ais at y =. In neither case does the graph intercept the -ais, though the graph does get etremely close to the -ais in both cases. 20 y 5 0 5-4 -3-2 - 2 3 4 Figure 8: Graph of y =2.5. 20 y 5 0 5-4 -3-2 - 2 3 4 Figure 9: Graph of y =5. 2. Remember that the function f() =b is increasing if b> and is decreasing if b<. a. f() =2.7 is increasing since 2.7 >. b. f() =( 2.7 ) =2.7,sothis function is also increasing. c. f() =3 =( 3 ) is decreasing since <. 3 d. f() =0.22 is decreasing since 0.22 <.
Mathematics Learning Centre, University of Sydney 28 3. Again, remember that the function f() =b is increasing if b> and is decreasing if b<. a. f() =( 5 3 ) is increasing because 5 >. 3 b. f() =( 5 3 ) =( 3 5 ) is decreasing because 3 <. 5 c. f() =( 3 5 ) =( 5 3 ) is increasing because 5 >. 3 d. f() =( 3 5 ) is decreasing because 3 <. 5 4. The graphs are drawn in Figures 2 and 22 below. Notice that the graph of f() = 2.9 is very close to the graph of f() =3, and similarly for the other pair of graphs. y 40 y = 3 y = 2.9 30 20 0-4 -3-2 - 2 3 4 Figure 0: Graphs of y =3 and y =2.9. y y = 3-40 y = 2.9-30 20 0-4 -3-2 - 2 3 4 Figure : Graphs of y =3 and y =2.9. 5. On my calculator I get e.0986 =2.999993. Now e.0986 = (e.0986 ) 3 Thus the functions 3 and e.0986 agree very closely with each other.
Mathematics Learning Centre, University of Sydney 29 4.3 Solutions to Eercises from Section 3.. a. log 0 0000 = log 0 0 4 =4 b. log = log 0 00 0 0 2 = 2 c. log 0 0.00 = log 0 0 3 = 3 d. log 0 0 2.3 =2.3 e. log 0 0 = log0 0 2 = 2 f. log 0 0 4 0 = log 0 0 0 4 = log 0 0 + 4 = 5 4 g. log 0 ( ) 3 0 = log 000 0 0 3+ 3 = 8 3 h. log = log 0 0.00 0 0 3 =3 2. a. 0 log 0 37.23 =37.23 by rule B. b. log 0 0 2y = 2 y by Rule A. c. 0 log 0 (0) =0, since log 0 (0 )= by rule A. d. log 0 0 2y = log 0 (0 2 ) 2y = log0 0 2 2y = by rule A. 4y e. 0 0log 0 =0, since 0 log 0 = by rule B. f. log 0 0 +y z = +y z by rule A. g. 0 log 0 3y z = 3y by rule B. z h. log 0 0 02 =0 2 by rule A. 3. a. log 0 3 2.5 log 0 y = log 0 3 log 0 y 2.5 = log 0 3 b. log 0 6+log 0 2 = log 0 6 2 c. 5 log 0 3 4 log 0 (y + z 2 )=log 0 (3) 5 log 0 (y + z 2 ) 4 = log 0 (3) 5 y 2.5 (y+z 2 ) 4 d. 2 log 0 y +3log 0 (z 2 y 2 )=log 0 (y) 2 + log 0 (z 2 y 2 ) 3 = log 0 (y) 2 (z 2 y 2 ) 3 e. log 0 ( + y) 3 log 0 4=log 0 ( + y) log 0 4 3 = log 0 +y 64 f. log 0 y.7 log 0 y 2 = log 0 y log 0 (y 2 ).7 = log 0 y y 3.4 = log 0 y 2.4 4. a. log 2 2 2 +y = log 2 2 ++y =2 + y b. 5 log 5 +y 3 = +y 3 c. log 7 49 uv = log 7 (7 2 ) uv = log 7 7 2uv =2uv 5. a. d. 3 log 9 u 3w =(9 2 ) log 9 u 3w =9 2 log 9 u 3w b. =9 log 9 ( u 3w ) 2 =( u 3w ) 2 2 log 3 ( + y) 3 log 3 (y)+log 3 2 = log 3 ( + y) 2 log 3 (y) 3 + log 3 2 = log 3 (( + y) 2 2 ) (y) 3 log 6 y 4 log 6 ( + y) = log 6 y log 6 ( + y) 4 = log 6 y ( + y) 4
Mathematics Learning Centre, University of Sydney 30 c. 4 log 7 y 2 + log 7 ( 2 + y 2 ) 2.5 log 7 = log 7 (y 2 ) 4 + log 7 ( 2 + y 2 ) log 7 2.5 = log 7 4 y 8 ( 2 + y 2 ) 2.5 6. a. log 3 7 = log 3 0 log 0 7 = log 0 7.2304 2.5789 log 0 3 0.477 b. log 5 2=log 5 0 log 0 2= log 0 2 0.300 0.4306 log 0 5 0.6990 c. log 22 4 = log 22 0 log 0 4 = log 0 4.46 0.8538 log 0 22.3424 d. log 4 8=log 4 0 log 0 8= log 0 8 0.903.5 (in fact log log 0 4 0.602 4 8iseactly.5 because 8=4 3 2 ) 7. a. y =0 =(e ln 0 ) (ln 0) = e b. 7.5 =(e log e 7.5 ) = e (log e 7.5) c. 4 =(e log e 4 ) = e (log e 4) d. ( 4 ) =(e log e 4 ) = e (log e 4) 8. log 0 0 9 = 9 9. log e e 5 e = log e e 6 5 = 6 5 0. log 2 6 = log 2 2 4 =4. log 7 7 3 7 = log 7 7 3 2 = 5 2 2. ln e2 e 2 =lne 2 2 = 9 3. ln e 7 log 2 = 7 log 2 = 7 2 4. 5 log 5 32.7 =32.7 5. e ln 9 2 = 9 2 6. e ln 3 27 = 3 27=3 7. log 0 00 2 9y = log 0 00 + log 0 2 log 0 9y =2+2log 0 log 0 9 log 0 y 8. ln( y 3 e.37 )=ln 3lny ln e.37 =ln 3lny.37 9. log 4 ( 4.3 z 7 )=.3+7log 2 y 3 4 z 2 log 4 3 log 4 y 20. log 3 3 y 2 27z 2 =3log 3 +2log 3 y log 3 27 2 log 3 z =3log 3 +2log 3 y 3 2 log 3 z 2. ln(e 2.4 6 )= 2.4+6ln 22. log 5 25 3 0.2y 2 = log 5 5 3 +3log 5 log 5 5 2 log 5 y =4+3log 5 2 log 5 y 23. 3 log 2 + log 2 30 + log 2 y log 2 w = log 2 30 3 y w
Mathematics Learning Centre, University of Sydney 3 24. 2ln ln y + a ln w =ln 2 ln y +lnw a =ln 2 w a y 25. 2(ln +lny) =ln(y) 2 26. log 3 e ln 8 + log 3 5 log 5 w = log 3 8 + log 3 w =4+log 3 w 27. log 7 0 log 0 2 log 7 49 = log 7 2 log 7 49 log 7 = 2+log 7 2 = 2+log 7 28. log 0 0. log 6 2 log 6 y + log 6 4 log 4 e = log 6 log 6 y 2 + log 6 e = log 6 e y 2 29. log 5 e = log e 5.6094 0.623 30. log 5 7=log 5 e log e 7= log e 7 log e 5.9459.6094.209 3. log 5 7 2 = log 5 e 2 log e 7= 2 log e 7 log e 5 2.9459.6094 2.482 32. log 49 5=log 49 e log e 5= log e 5 log e 49 = log e 5 log e 7 2.6094 2.9459 0.435 33. log 49 25 = log 49 e log e 5 2 = log e 52 log e 7 2 2.6094 2.9459 0.827 34. log e 25 = log e 5 2 2 log e 5 2.6094 3.288