Logarithmic Functions and the Log Laws

Mathematics Learning Centre Logarithmic Functions and the Log Laws Christopher Thomas c 998 University of Sydney

Mathematics Learning Centre, University of Sydney Logarithms. 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. 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.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..2 Logarithms to base 0 (Common Logarithms) We will begin by considering the function y =0, graphed in Figure. y 20 5 0 5-2 -.5 - -0.5 0.5.5 2 Figure : Graph of f() =0 We know that, 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 towhich 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 2 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 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. 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.

Mathematics Learning Centre, University of Sydney 3 The graph of y = log 0 is shown in Figure 2. y 0.5 2 4 6 8 0 2 4 6-0.5 - Figure 2: Graph of f() =log 0 You should pay attention to several important features of this graph. The graph intercepts the -ais at =.Inother words, log 0 =0.Weshould epect this because we know that 0 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 of0then 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 4 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 to which 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 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 and 2. We can use the rules of eponents 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). 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.

Mathematics Learning Centre, University of Sydney 5 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 2.3 Logarithms to Base b ( 2 z 3 ) 4 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

Mathematics Learning Centre, University of Sydney 6 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>. y 2 y = log 2 y = log e 0 y = log 0 2 3 4 5 6 7 8 - -2 Figure 3: Graph of f() =log b for various values of b. Figure 3 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. For a real 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 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.

Mathematics Learning Centre, University of Sydney 7 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..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, 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 3 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 that the functions e and 0 are very similar to each other..5 Eponential functions revisited In a previous lecture, 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.

Mathematics Learning Centre, University of Sydney 8 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..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..7 Eercises. 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 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

Mathematics Learning Centre, University of Sydney 9 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 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 6. Using a calculator, find the following logarithms. a. log 3 7 b. log 5 2 c. log 22 4 d. log 4 8 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 ) 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 0.8 Solutions to eercises. 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 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 2 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 0.9459 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