VI. Exponentials and Logarithms42. Exponents42.1. The trouble with powers of negative numbers43. Logarithms44. Properties of logarithms45. Graphs of exponential functions and logarithms46. The derivative of ax and the definition of e47. Derivatives of Logarithms48. Limits involving exponentials and logarithms49. Exponential growth and decay49.1. Half time and doubling time49.2. Determining X0 and kExercisesVI. Exponentials and LogarithmsIn this chapter we first recall some facts about exponentials (xywith x > 0 and yarbitrary): they should b e familiar from algebra, or “precalculus.” What is new is perhapsthe definition of xywhen y is not a fraction: e.g., 23/4is the 4th ro ot of the third powerof 2 (4√23), but what is 2√2?Then we ask “what is the derivative of f(x) = ax?” The answer leads us to the famousnumber e ≈ 2.718 281 828 459 045 235 360 287 471 352 662 497 757 247 093 699 95 ···.Finally, we compute the derivative of f(x) = logax, and we look at things that “growexponentially.”42. ExponentsHere we go over the definition of xywhen x and y are arbitrary real numbers, withx > 0.For any real number x and any positive integer n = 1, 2, 3, . . . one definesxn=n timesz }| {x · x · ··· · xand, if x 6= 0,x−n=1xn.One defines x0= 1 for any x 6= 0.To define xp/qfor a general fractionpqone must assume that the number x is positive.One then defines(34) xp/q=q√xp.This does not tell us how to define xais the exponent a is not a fraction. One can definexafor irrational numbers a by taking limits. For example, to define 2√2, we look at thesequence of numbers you get by truncating the decimal expansion of√2, i.e.a1= 1, a2= 1.4 =1410, a3= 1.41 =141100, a4= 1.414 =14141000, ··· .Each anis a fraction, so that we know what 2anis, e.g. 2a4=1000√21414. Our definitionof 2√2then is2√2= limn→∞2an,i.e. we define 2√2as the limit of the sequence of numbers2,10√214,100√2141,1000√21414, ···(See table 2.)Here one ought to prove that this limit exists, and that its value does not depend onthe particular choice of numbers antending to a. We will not go into these details in thiscourse.It is shown in precalculus texts that the e xponential functions satisfy the followingproperties:(35) xaxb= xa+b,xaxb= xa−b,`xa´b= xab9798x 2x1.0000000000 2.0000000000001.4000000000 2.6390158215461.4100000000 2.6573716281931.4140000000 2.6647496501841.4142000000 2.6651190885321.4142100000 2.6651375617941.4142130000 2.6651431037981.4142135000 2.665144027466......Table 2. Approximating 2√2. Note that as x gets closer to√2 thequantity 2xapp e ars to converge to s ome number. This limit is ourdefinition of 2√2.provided a and b are fractions. One can show that these properties still hold if a and bare real numbers (not necessarily fractions.) Again, we won’t go through the proofs here.Now instead of considering xaas a function of x we can pick a positive number a andconsider the function f(x) = ax. This function is defined for all real numbers x (as longas the base a is positive.).42.1. The trouble with powers of negative numbersThe cube root of a negative number is well defined. For instance3√−8 = −2 because(−2)3= −8. In view of the definition (34) of xp/qwe can write this as(−8)1/3=3p(−8)1=3√−8 = −2.But there is a problem: since26=13you would think that (−8)2/6= (−8)1/3. Howeverour definition (34) tells us that(−8)2/6=6p(−8)2=6√+64 = +2.Another example:(−4)1/2=√−4 is not definedbut, even though12=24,(−4)2/4=4p(−4)2=4√+16 = 2 is defined.There are two ways out of this mess:(1) avoid taking fractional powers of negative numbers(2) when you compute xp/qfirst simplify the fraction by removing common divisorsof p and q.The safest is just not to take fractional powers of negative numbers.Given that fractional powers of negative numbers cause all these headaches it is notsurprising that we didn’t try to define xafor negative x if a is irrational. For example,(−8)πis not defined6.6There is a definition of (−8)πwhich uses complex numbers. You will see this next semester if youtake math 222.9943. LogarithmsBriefly, y = logax is the inverse function to y = ax. This means that, by definition,y = logax ⇐⇒ x = ay.In other words, logax is the answer to the question “for which number y does one havex = ay?” The number logax is called the logarithm with base a of x. In this definitionboth a and x must be positive.For instance,23= 8, 21/2=√2, 2−1=12solog28 = 3, log2`√2´=12, log212= −1.Also:log2(−3) doesn’t existbecause there is no number y for which 2y= −3 (2yis always positive) andlog−32 doesn’t exist eitherbecause y = log−32 would have to be some real number which satisfies (−3)y= 2, andwe don’t take non-integer powe rs of negative numbers.44. Properties of logarithmsIn general one haslogaax= x, and alogax= x.There is a subtle difference between these formulas: the first one holds for all real numbersx, but the second only holds for x > 0, since logax doesn’t make sense for x ≤ 0.Again, one finds the following formulas in precalculus texts:(36)logaxy = logax + logaylogaxy= logax − logaylogaxy= y logaxlogax =logbxlogbaThey follow from (35).45. Graphs of exponential functions and logarithmsFigure 27 shows the graphs of some exponential functions y = axwith different valuesof a, and figure 28 shows the graphs of y = log2x, y = log3x, log1/2x, log1/3(x) andy = log10x. Can you tell which is which? (Yes, you can.)From algebra/precalc recall:If a > 1 then f(x) = axis an increasing function.andIf 0 < a < 1 then f(x) = axis a decreasing function.In other words, for a > 1 it follows from x1< x2that ax1< ax2; if 0 < a < 1, thenx1< x2implies ax1> ax2.100-3 -2 -1 0 1 2 323Figure 27. The graphs of y = 2x, 3x, (1/2)x, (0.1)xand y = (4/5)x.1 2 3 4 5 6 7 8 9 10-3-2-10123Figure 28. Graphs of some logarithms46. The derivative of axand the definition of eTo begin, we try to differentiate the function y = 2x:d2xdx= lim∆x→02x+∆x− 2x∆x= lim∆x→02x2∆x− 2x∆x= lim∆x→02x2∆x− 1∆x= 2xlim∆x→02∆x− 1∆x.101So if we assume that the limitlim∆x→02∆x− 1∆x= Cexists then we have(37)d2xdx= C2x.On your calculator you can compute2∆x−1∆xfor smaller and smaller values of ∆x, whichleads you to suspect that the limit actually exists, and that C ≈ 0.693 147 . . .. One canin fact prove that the limit exists, but we will not do this here.Once we know (37) we can compute the