Introduction to Exponents and Logarithms

1 Mathematics Learning CentreIntroduction toExponents and LogarithmsChristopher Thomasc 1998 University of SydneyAcknowledgementsParts of section 1 of this booklet rely a great deal on the presentation given in the bookletof the same name, written by Peggy Adamson for the Mathematics Learning Centre in1987. The remainder is Nicholas, Sue Gordon and Trudy Weibel read pieces of earlier drafts of this like to thank them for their extremely helpful comments on the contents andlayout. In addition, Duncan Turpie performed the laborious task of final proof ThomasDecember 1991 This booklet was revised in 1998 by Jackie Introduction .. Exponents with the Same Exponents with Different Scientific Notation.

2 Summary .. 102 Exponential The Functionsy=2xandy=2 The functionsy=bxandy=b The Functionsy=exandy=e Summary .. Introduction .. Logarithms to Base 10 (Common Logarithms ).. Logarithms to Base b .. Logarithms to Basee(Natural Logarithms ) .. Exponential Functions Summary .. 234 Solutions to Solutions to Exercises from Section 1 .. Solutions to Exercises from Section 2 .. Solutions to Exercises from Section 3.. 29 Mathematics Learning Centre, University of IntroductionWhenever we use expressions like 73or 25weare using symbol 25means 2 2 2 2 2 symbol is spoken as two raised to thepowerfive , two to the power five or simply two to the five.

3 The expression 25is just ashorthand way of writing multiply 2 by itself 5 times . The number 2 is called thebase,and 5 , ifbis any real number thenb3stands forb b the base, and 3the a whole number,bnstands forb b b written inexponential form,and we callbthe base andnthe exponent , power or names are used when the exponent is 2 or 3. The expressionb2is usuallyspoken as bsquared , and the expressionb3as bcubed . Thus two cubed means23=2 2 2= Exponents with the Same BaseWewill begin with a very simple definition. Ifbis any real number andnis a positiveinteger thenbnmeansbmultiplied by itselfntimes. The rules for the behaviour ofexponents follow naturally from this , let s try multiplying two numbers in exponential form.

4 For example23 24=(2 2 2) (2 2 2 2)=2 2 2 2 2 2 2 7factors=27=23+ like this suggest the following general 1:bn bm=bn+ is, tomultiplytwonumbers in exponential form (with the same base), s look at what happens when we divide two numbers in exponential form. For example,3634=3 3 3 3 3 33 3 3 3=3 3 3 3 3 33 3 3 3=3 3=32=36 Learning Centre, University of Sydney2 This leads us to another general 2:bnbm=bn words, todividetwonumbers in exponential form (with the same base) , wesubtracttheir not yet given any meaning to negative Exponents , sonmust be greater thanmfor this rule to make sense. In a moment we will see what happens ifnis not greater look at what happens when a number in exponential form is raised to some ,(22)3=(2 2) (2 2) (2 2)=26=22 suggest another general 3:(bm)n=bmnThat is, to raise a number in exponential form to a power, wemultiplythe 54=52+4=56= +3 1= ( 5 3)3=( 5 3)3=( 2)3= 2 3= 6 +43m+1 simplifies to 3n m+ the following expressions using a calculator where ( )35.

5 (3+ )4 Simplify these, or at least change them around a + +3y9.(3x) +2z3z4 Mathematics Learning Centre, University of Sydney3 Until now we have only considered Exponents which are positive integers, such as 7 or189. Our intention is to extend this notation to cover Exponents which are not necessarilypositive integers, for example 5, or11331,ornumbers such as Just as wecan make sense of expressions like 5189,wewant to be able to make sense of expressionssuch as more than this, we want to make sense of these expressions in such awaythat rules 1, 2 and 3 remain valid. It is not at all obvious how we should interpretan expression really make sense to think of it as 5 multiplied by plan is this: if we want rules 1, 2 and 3 to hold for general Exponents then we willtry defining expressions like 511331to be whatever they must be in order that rules 1, 2 and3remain valid.

6 In other words, we will insist that rules 1, 2 and 3 remain valid for thesemore general Exponents , and hope that this requirement will tell us what the definitionsof expressions like 511331must us begin by extending the notation to include an exponent equal to 0. We want tomake sense of the expressionb0in such a way that rules 1, 2 and 3 hold. What happensto rule 2 whenn=m?Rule 2 givesbnbn=bn nor1= now we have not attached any meaning to the t make senseto talk about a number being multiplied by itself 0 times. However, if we want rule 2to continue to be valid whenn=mthen we mustdefinethe expressionb0to mean thenumber =0then we defineb0to be equal to 1. We do not attempt to give any meaning to this definition we can check that rules 1 and 3 also remain valid.

7 For example, tocheck that rule 1 still holds, ifnis a whole number andm=0then rule 1 givesbn b0=bnwhich is okay becauseb0= correct we should also check that rule 1 remains valid in the case thatm=0andn= should check that this is true and that rule 3 also remains valid underthis definition had no idea of how to extend our notation to cover a zero exponent , but ifwewish rules 1, 2 and 3 to remain valid for such an exponent then the definitionb0=1is forced on us. We have no , we have come up with a sensible definition ofb0bytakingm=nin rule 2 andseeing whatb0must be if rule 2 is to remain valid. To come up with a suitable meaningfor negative Exponents we can taken<min rule 2. For example, let s tryn=2andm= Learning Centre, University of Sydney4 Rule 2 givesb2b3=b 1or1b=b suggests that we should defineb 1to be equal definition, too, makes sensefor all values ofbexceptb= a similar way we can see that we should defineb nto mean1bn,except whenb=0,in which case it is undefined.

8 You should convince yourself of this by showing that therequirement that rule 2 remains valid forces on us the definitionsb 2=1b2andb 3= a positive integer (for examplen=17orn=178)then we defineb nto be definition makes sense for all values ofbexceptb=0,inwhich case theexpressionb nremains check that, with this definition, rules 1 and 3 also remain 3 4=30=12 1=121=1223 4+2=23 2=8 = (x 1+x 3) 1=1x 1+x 3=11x+1x3=1x2+1x3=x3x2+1 ExercisesEvaluate the following 2 315.(6 2)2 Simplify the following (x12+y12)(x12 y12)17.(x14 y14)(x34+x12y14+x14y12+y34) +y12xx12+y1219. x12(x y)(x12 y12)(x12+y12) 220.(x12)2 Mathematics Learning Centre, University of Sydney5 Pause for a moment and look at what has been achieved.

9 We have been able to giveameaning tobnfor all integer values ofn,positive, negative, and zero, and we havedone it in such a way that all three of the rules above still hold. We can give meaningto expressions like (357)13and quite a way, but there are a lot ofexponents that we cannot yet handle. For example, what meaning would we give to anexpression like 579?Our next task is to give a suitable meaning to expressions involvingfractional us start to give meaning to this expression in such a way that therules 1, 2 and 3 remain valid. If rule 2 is to hold then we must haveb12 b12=b12+12=b1= s be specific and takeb= , 412 412=4,so412is equal to a number whosesquare is 4. There are two numbers whose square is 4.

10 They are 2 and 2. We define 412to be thepositivesquare root of 4. That is, general,b12is defined to be the positive square root ofb,also written course,bmust be positive ifb12is to have any meaning for us, because if we take anyreal number and multiply itself by itself then we get a positive number. (Actually thereis a way of giving meaning to the square root of a negative number. This leads to thenotion of complex numbers, a beautiful area of mathematics which is beyond the scopeof this booklet.)That takes care of a meaning forb12ifb>0. Now 2 is to remainvalid then we must haveb13 b13 b13=b13+13+13=b1= a concrete example takeb= 813must be such that 813 813 813= just one number which when multiplied by itself 3 times gives 8.