1 00:00:00,000 --> 00:00:01,940 ANNOUNCER: The following content is provided under a 2 00:00:01,940 --> 00:00:03,690 Creative Commons license. 3 00:00:03,690 --> 00:00:06,640 Your support will help MIT OpenCourseWare continue to 4 00:00:06,640 --> 00:00:09,990 offer high quality educational resources for free. 5 00:00:09,990 --> 00:00:12,830 To make a donation or to view additional materials from 6 00:00:12,830 --> 00:00:16,760 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:16,760 --> 00:00:18,010 ocw.mit.edu. 8 00:00:32,619 --> 00:00:33,710 PROFESSOR: Hi. 9 00:00:33,710 --> 00:00:38,740 Our lecture for today I've entitled 'Logarithms Without 10 00:00:38,740 --> 00:00:39,800 Exponents'. 11 00:00:39,800 --> 00:00:43,160 And before I try to clarify that, let me make a couple of 12 00:00:43,160 --> 00:00:44,390 general remarks. 13 00:00:44,390 --> 00:00:48,720 First of all, we have now completed the rudiments, the 14 00:00:48,720 --> 00:00:51,240 basics so to speak, of both differential 15 00:00:51,240 --> 00:00:52,550 and integral calculus. 16 00:00:52,550 --> 00:00:56,340 Consequently, what our next ambition will be is to take 17 00:00:56,340 --> 00:01:00,020 these principles and apply them to special functions 18 00:01:00,020 --> 00:01:02,260 which are worthy of investigation. 19 00:01:02,260 --> 00:01:04,890 And you see, in connection with this, the first function 20 00:01:04,890 --> 00:01:08,670 that I've chosen to talk about is called the logarithm. 21 00:01:08,670 --> 00:01:11,630 Now the other interesting thing is that we are used to 22 00:01:11,630 --> 00:01:14,570 logarithms from high school where they were viewed as 23 00:01:14,570 --> 00:01:16,980 being a different kind of notation for 24 00:01:16,980 --> 00:01:18,840 talking about exponents. 25 00:01:18,840 --> 00:01:21,500 Now, in the same way when we treated the circular 26 00:01:21,500 --> 00:01:25,030 functions, we mentioned that one did not have to invent 27 00:01:25,030 --> 00:01:27,670 triangles to talk about trigonometry. 28 00:01:27,670 --> 00:01:30,490 The interesting point is that one does not have to talk 29 00:01:30,490 --> 00:01:33,960 about exponents to invent the logarithm function. 30 00:01:33,960 --> 00:01:37,130 This is why I have entitled the lesson, as I say, 31 00:01:37,130 --> 00:01:40,010 'Logarithms without Exponents'. 32 00:01:40,010 --> 00:01:43,100 You see, what I would like to do is try to mimic, especially 33 00:01:43,100 --> 00:01:45,930 from an engineering point of view, how many of these 34 00:01:45,930 --> 00:01:48,610 mathematical topics originated. 35 00:01:48,610 --> 00:01:51,790 Let's look at a rather straightforward physical 36 00:01:51,790 --> 00:01:56,050 principle, a principle that I'm sure we believe permeates 37 00:01:56,050 --> 00:01:58,340 many real life situations. 38 00:01:58,340 --> 00:02:01,990 It's what I call the rule of compound interest. 39 00:02:01,990 --> 00:02:06,000 Namely, let's suppose we have a physical situation in which 40 00:02:06,000 --> 00:02:07,490 the quantity that we're measuring, 41 00:02:07,490 --> 00:02:08,590 which we'll call 'm'. 42 00:02:08,590 --> 00:02:12,310 In other words, the rate of change of the quantity is 43 00:02:12,310 --> 00:02:18,520 proportional to the amount present, 'dm/ dt' equals 'km'. 44 00:02:18,520 --> 00:02:22,750 Now notice that this is a pretty harmless statement. 45 00:02:22,750 --> 00:02:24,750 The rate of change is proportional 46 00:02:24,750 --> 00:02:25,920 to the amount present. 47 00:02:25,920 --> 00:02:29,520 We certainly would expect that many experiments would run 48 00:02:29,520 --> 00:02:32,820 this way, and there seems to be nothing at all supernatural 49 00:02:32,820 --> 00:02:34,270 about this kind of an assumption. 50 00:02:34,270 --> 00:02:38,090 At any rate, let's apply our previous principles, and see 51 00:02:38,090 --> 00:02:41,130 if we can't, from this differential equation, figure 52 00:02:41,130 --> 00:02:43,360 out what 'm' has to be. 53 00:02:43,360 --> 00:02:46,990 Separating the variables, we get that 'dm' over 'm' is 54 00:02:46,990 --> 00:02:48,860 equal to 'kdt'. 55 00:02:48,860 --> 00:02:53,130 And now integrating, meaning taking the inverse derivative, 56 00:02:53,130 --> 00:02:54,130 we have what? 57 00:02:54,130 --> 00:02:57,310 That the integral of 'dm' over 'm' is equal 58 00:02:57,310 --> 00:02:59,830 to 'kt' plus a constant. 59 00:02:59,830 --> 00:03:03,300 Now you see, the interesting point is at this stage of the 60 00:03:03,300 --> 00:03:08,400 game, we do not know explicitly how to find a 61 00:03:08,400 --> 00:03:12,970 function whose derivative with respect to 'm' is '1 over m'. 62 00:03:12,970 --> 00:03:14,710 In fact, that's the problem that 63 00:03:14,710 --> 00:03:16,700 underlies today's lecture. 64 00:03:16,700 --> 00:03:20,130 And we will try to answer this question, showing how once we 65 00:03:20,130 --> 00:03:23,760 tackle this from a philosophic point of view, the rest of the 66 00:03:23,760 --> 00:03:27,530 details follow from material that we've already learned. 67 00:03:27,530 --> 00:03:30,590 At any rate, focusing our attention on the problem, the 68 00:03:30,590 --> 00:03:34,140 question is this: to determine a function, which I'll call 69 00:03:34,140 --> 00:03:36,240 capital 'L of x'-- 70 00:03:36,240 --> 00:03:40,820 the 'L' sort of to forewarn us of the fact that we will 71 00:03:40,820 --> 00:03:43,220 somehow get a logarithm out of this-- 72 00:03:43,220 --> 00:03:47,050 such that 'L prime of x' is '1 over x'. 73 00:03:47,050 --> 00:03:50,690 And you see, what I want you to notice is that if you write 74 00:03:50,690 --> 00:03:53,750 '1 over x' in exponential form, in other words, if you 75 00:03:53,750 --> 00:03:57,340 try to write '1 over x' as 'x to the n', notice that to 76 00:03:57,340 --> 00:04:00,460 solve this problem, 'L of x' would be what? 77 00:04:00,460 --> 00:04:04,570 The integral of 'x' to the 'minus 1 dx'. 78 00:04:04,570 --> 00:04:07,930 In other words, this is the case of integral ''x to the n' 79 00:04:07,930 --> 00:04:11,290 dx', with 'n' equal to minus 1. 80 00:04:11,290 --> 00:04:14,350 And you'll recall that when we studied the recipe-- 81 00:04:14,350 --> 00:04:16,430 let me just write that over here-- when we studied the 82 00:04:16,430 --> 00:04:20,740 recipe of how you integrate 'x to the n', we saw that that 83 00:04:20,740 --> 00:04:27,240 was 'x to the 'n + 1'' over 'n + 1' plus a constant. 84 00:04:27,240 --> 00:04:30,810 And we then observed that this doesn't even make sense when 85 00:04:30,810 --> 00:04:33,910 'n' is equal to minus 1, because we have a 0 86 00:04:33,910 --> 00:04:35,060 denominator. 87 00:04:35,060 --> 00:04:37,140 From the other point of view, the way to look at 88 00:04:37,140 --> 00:04:38,700 this is we said what? 89 00:04:38,700 --> 00:04:42,490 That when you differentiate, you lower the exponent by 1. 90 00:04:42,490 --> 00:04:47,160 Notice that to wind up with an exponent of minus 1, we would 91 00:04:47,160 --> 00:04:51,340 have had to start with an exponent of 0. 92 00:04:51,340 --> 00:04:53,950 But when you differentiate 'x' to the 0, that being a 93 00:04:53,950 --> 00:05:00,010 constant, the derivative of 'x' to the 0 is 0. 94 00:05:00,010 --> 00:05:03,870 It's not '1 over x'. 95 00:05:03,870 --> 00:05:07,620 You see, in other words, this particular recipe that we're 96 00:05:07,620 --> 00:05:10,590 talking about, we don't have working for us. 97 00:05:10,590 --> 00:05:13,180 In other words, this is an interesting point again. 98 00:05:13,180 --> 00:05:15,770 You say gee whiz, if the recipe for the integral ''x to 99 00:05:15,770 --> 00:05:18,650 the n' dx' works for everything except 'n' equals 100 00:05:18,650 --> 00:05:20,730 minus 1, nothing's perfect. 101 00:05:20,730 --> 00:05:22,180 Let's be content. 102 00:05:22,180 --> 00:05:23,700 Since it works for every number except 103 00:05:23,700 --> 00:05:25,710 that one, why worry? 104 00:05:25,710 --> 00:05:28,040 This is analogous to when we talked about taking 105 00:05:28,040 --> 00:05:30,220 derivatives and saw that we get a 0 over 106 00:05:30,220 --> 00:05:31,850 0 form every time. 107 00:05:31,850 --> 00:05:34,260 You see, if you pick a function at random, the 108 00:05:34,260 --> 00:05:36,890 likelihood that the limit of 'f of x' as 'x' approaches 109 00:05:36,890 --> 00:05:40,820 'a', the likelihood of that being 0 over 0 is very small, 110 00:05:40,820 --> 00:05:43,760 except when you form the derivative, you 111 00:05:43,760 --> 00:05:45,860 always get 0 over 0. 112 00:05:45,860 --> 00:05:49,560 And in a similar way, granted that integral ''x' to the 113 00:05:49,560 --> 00:05:52,810 minus 1 dx' is one very special case. 114 00:05:52,810 --> 00:05:56,990 What we have seen is that this comes up every single time 115 00:05:56,990 --> 00:06:00,620 that we want to solve a problem of the form 'dm/ dt' 116 00:06:00,620 --> 00:06:02,040 equals 'km'. 117 00:06:02,040 --> 00:06:05,330 In other words then, what it really boils down to is that 118 00:06:05,330 --> 00:06:09,350 unless we can come up with a function 'L of x' whose 119 00:06:09,350 --> 00:06:13,400 derivative is '1 over x', we cannot solve the problem that 120 00:06:13,400 --> 00:06:15,230 we started off today's lesson with. 121 00:06:15,230 --> 00:06:18,610 In other words, we must leave it in the form integral 'dm 122 00:06:18,610 --> 00:06:21,760 over m' equals 'kt' plus a constant. 123 00:06:21,760 --> 00:06:25,900 Well at any rate, let's see what such a function capital 124 00:06:25,900 --> 00:06:27,640 'L of x' must look like. 125 00:06:27,640 --> 00:06:31,050 I'm going to utilize both the differential calculus approach 126 00:06:31,050 --> 00:06:34,450 and the integral calculus approach. 127 00:06:34,450 --> 00:06:37,540 By differential calculus, assuming that 'y' equals 'L of 128 00:06:37,540 --> 00:06:40,510 x', what do we know about the function 'y'? 129 00:06:40,510 --> 00:06:44,110 By definition, we've defined it to be that function whose 130 00:06:44,110 --> 00:06:46,330 derivative is '1 over x'. 131 00:06:46,330 --> 00:06:48,850 So we know its derivative is '1 over x'. 132 00:06:48,850 --> 00:06:53,150 The next point is that knowing that 'y prime' is '1 over x', 133 00:06:53,150 --> 00:06:55,490 and even though we can't integrate '1 over x', we can 134 00:06:55,490 --> 00:06:57,390 certainly differentiate '1 over x'. 135 00:06:57,390 --> 00:06:59,640 We can now differentiate '1 over x'. 136 00:06:59,640 --> 00:07:02,760 The derivative of '1 over x' is minus '1 over 'x squared'', 137 00:07:02,760 --> 00:07:04,870 and we find that 'y double prime' is 138 00:07:04,870 --> 00:07:06,840 minus '1 over 'x squared''. 139 00:07:06,840 --> 00:07:11,100 By the way, we should observe that we must shy away from 'x' 140 00:07:11,100 --> 00:07:14,400 equaling 0, because you see, we would have a 0 denominator 141 00:07:14,400 --> 00:07:15,360 in that case. 142 00:07:15,360 --> 00:07:19,780 If we think of most physical situations, notice that we 143 00:07:19,780 --> 00:07:21,240 talk about the amount-- 144 00:07:21,240 --> 00:07:23,160 the rate of change is proportional 145 00:07:23,160 --> 00:07:24,380 to the amount present. 146 00:07:24,380 --> 00:07:27,760 From a physical point of view, the amount present can't be 147 00:07:27,760 --> 00:07:31,110 negative, so let's put the restriction on here that the 148 00:07:31,110 --> 00:07:36,790 domain of 'L' will be all positive numbers. 149 00:07:39,330 --> 00:07:41,060 Now here's the interesting thing. 150 00:07:41,060 --> 00:07:46,660 Intuitively, I can now visualize how the function 'L 151 00:07:46,660 --> 00:07:47,950 of x' must look. 152 00:07:47,950 --> 00:07:51,100 Namely, since its first derivative is '1 over x' and 153 00:07:51,100 --> 00:07:53,620 'x' is greater than 0, that tells me that the curve is 154 00:07:53,620 --> 00:07:55,050 always rising. 155 00:07:55,050 --> 00:07:58,750 And secondly, since '1 over 'x squared'' is always positive, 156 00:07:58,750 --> 00:08:01,780 minus '1 over 'x squared'' is always negative, that tells me 157 00:08:01,780 --> 00:08:04,650 that my curve is always spilling water. 158 00:08:04,650 --> 00:08:07,140 In essence then, what must the curve do? 159 00:08:07,140 --> 00:08:10,110 The curve must be rising but spilling water. 160 00:08:10,110 --> 00:08:13,100 In other words, the curve 'L of x' belongs to this 161 00:08:13,100 --> 00:08:14,410 particular family. 162 00:08:14,410 --> 00:08:17,940 Notice that once we have one curve which we call 'y' equals 163 00:08:17,940 --> 00:08:21,800 'L of x', by any displacement parallel-- 164 00:08:21,800 --> 00:08:24,430 vertical displacement, we get what? 165 00:08:24,430 --> 00:08:28,820 A member called 'y' equals 'L of x' plus 'c'. 166 00:08:28,820 --> 00:08:31,360 All that changes is what? 167 00:08:31,360 --> 00:08:34,640 The point at which the curve crosses the x-axis, but 168 00:08:34,640 --> 00:08:37,140 whichever member of the family we pick, what 169 00:08:37,140 --> 00:08:38,789 typifies 'L of x'? 170 00:08:38,789 --> 00:08:40,870 Its derivative is '1 over x'. 171 00:08:40,870 --> 00:08:43,990 In other words, somehow or other, we visualize that 'L of 172 00:08:43,990 --> 00:08:48,850 x' exists, and its graph is something like this. 173 00:08:48,850 --> 00:08:51,860 Now suppose we want a more tangible form, namely how do 174 00:08:51,860 --> 00:08:54,760 you compute 'L of x' for a given 'x'? 175 00:08:54,760 --> 00:08:57,690 I thought what might be a nice review now is to see how we 176 00:08:57,690 --> 00:09:01,300 can use our integral calculus approach, and in particular 177 00:09:01,300 --> 00:09:03,730 the second fundamental theorem of interval calculus. 178 00:09:06,920 --> 00:09:10,270 By the integral calculus approach, we do is is we pick 179 00:09:10,270 --> 00:09:14,000 any positive number 'a', and once picked, we fix it. 180 00:09:14,000 --> 00:09:18,650 You see, we have a great deal of freedom in how we choose 181 00:09:18,650 --> 00:09:19,990 the positive number a. 182 00:09:19,990 --> 00:09:22,020 But let's pick one and leave it here. 183 00:09:22,020 --> 00:09:25,480 Now what we'll do is in the 'y-t' plane, we'll draw the 184 00:09:25,480 --> 00:09:29,140 curve 'y' equals '1 over t'. 185 00:09:29,140 --> 00:09:32,550 And what we'll do now, we'll study the area of the region 186 00:09:32,550 --> 00:09:36,910 'R' where 'R' is bounded above by 'y' equals '1 over t', on 187 00:09:36,910 --> 00:09:41,310 the left by 't' equals 'a', on the right by 't' equals 'x', 188 00:09:41,310 --> 00:09:44,420 and below by the t-axis. 189 00:09:44,420 --> 00:09:48,080 Now we've already seen that the function that we get by 190 00:09:48,080 --> 00:09:52,750 taking the definite integral from 'a' 'to x', 'dt over t', 191 00:09:52,750 --> 00:09:56,590 has the property that its derivative is '1 over x'. 192 00:09:56,590 --> 00:09:59,310 In other words, recall from the fundamental theorem, this 193 00:09:59,310 --> 00:10:03,120 is just a generalization of the fact that if 'f of t' is a 194 00:10:03,120 --> 00:10:07,480 continuous function, then the integral from 'a' to 'x', ''f 195 00:10:07,480 --> 00:10:11,120 of t' dt', is a function of 'x'. 196 00:10:11,120 --> 00:10:13,730 And its derivative with respect to 'x' 197 00:10:13,730 --> 00:10:16,600 is just 'f of x'. 198 00:10:16,600 --> 00:10:20,490 You see, in other words, I can now view capital 'L of x' as 199 00:10:20,490 --> 00:10:24,920 being an area under the curve 'y' equals '1 over t'. 200 00:10:24,920 --> 00:10:28,610 Notice that I have a degree of freedom here, namely what that 201 00:10:28,610 --> 00:10:32,000 area is depends on where I choose 'a'. 202 00:10:32,000 --> 00:10:36,490 Notice that if I choose a differently, changing 'a'-- 203 00:10:36,490 --> 00:10:39,610 let's call this 'a prime', say-- 204 00:10:39,610 --> 00:10:43,880 changing 'a' changes the area under the curve, but notice it 205 00:10:43,880 --> 00:10:46,150 changes it by a fixed amount. 206 00:10:46,150 --> 00:10:50,160 In other words, notice that shifting 'a' just changes 'L 207 00:10:50,160 --> 00:10:52,440 of x' by a constant, just like the 208 00:10:52,440 --> 00:10:54,500 ordinary in definite integral. 209 00:10:54,500 --> 00:10:57,540 At any rate, these are the two approaches that we have. 210 00:10:57,540 --> 00:11:00,960 And if we superimpose them, all we're saying is what? 211 00:11:00,960 --> 00:11:02,910 If you start with the differential calculus 212 00:11:02,910 --> 00:11:06,970 approach, 'y' equals capital 'L of x' must be a member of 213 00:11:06,970 --> 00:11:11,820 this family, crossing the axis at some point (a,0) . 214 00:11:11,820 --> 00:11:14,770 And from the integral calculus point of view, if you take the 215 00:11:14,770 --> 00:11:18,010 curve 'y' equals '1 over t' and compute the area under 216 00:11:18,010 --> 00:11:22,420 that curve from 'a' to 'x', that area 217 00:11:22,420 --> 00:11:25,690 function is 'L of x'. 218 00:11:25,690 --> 00:11:30,290 Now we'll let that rest for the time being, and now have a 219 00:11:30,290 --> 00:11:32,180 brief digression. 220 00:11:32,180 --> 00:11:34,560 You see, sooner or later, I've got to get to what a logarithm 221 00:11:34,560 --> 00:11:37,060 is, and we might just as well do that now. 222 00:11:37,060 --> 00:11:40,620 So let's now take a look and see what we mean by a 223 00:11:40,620 --> 00:11:42,280 logarithmic function. 224 00:11:42,280 --> 00:11:46,010 You see, quite frequently in mathematics, what one does is 225 00:11:46,010 --> 00:11:49,950 one deals with a particular special case in which one is 226 00:11:49,950 --> 00:11:51,220 interested. 227 00:11:51,220 --> 00:11:54,980 And having deduced very nice results, he looks at this 228 00:11:54,980 --> 00:11:58,320 thing and says, you know, all of these results came from a 229 00:11:58,320 --> 00:12:01,530 particular recipe, a particular structure that this 230 00:12:01,530 --> 00:12:03,130 system obeyed. 231 00:12:03,130 --> 00:12:06,860 What he then does is he takes this recipe or structure, 232 00:12:06,860 --> 00:12:11,280 gives it a general name, and now is able to extend this 233 00:12:11,280 --> 00:12:14,870 property to a larger class of objects. 234 00:12:14,870 --> 00:12:16,830 For example, let's look at this way. 235 00:12:16,830 --> 00:12:19,990 What is the nice thing about logarithms in the traditional 236 00:12:19,990 --> 00:12:21,270 sense of the word? 237 00:12:21,270 --> 00:12:24,030 Remember when we learned logarithms in high school, by 238 00:12:24,030 --> 00:12:27,800 use of logarithms, we were able to replace multiplication 239 00:12:27,800 --> 00:12:30,420 problems by addition problems, et cetera. 240 00:12:30,420 --> 00:12:33,680 In other words, remember the key structural property was 241 00:12:33,680 --> 00:12:36,610 that the logarithm of a product was the sum of the 242 00:12:36,610 --> 00:12:37,490 logarithms. 243 00:12:37,490 --> 00:12:40,620 That was the structural thing that we used over and over and 244 00:12:40,620 --> 00:12:41,650 over again. 245 00:12:41,650 --> 00:12:44,200 In fact, it's rather interesting to point out that 246 00:12:44,200 --> 00:12:47,050 in many cases where we used the properties of logarithms, 247 00:12:47,050 --> 00:12:50,370 we never really had to know what a logarithm was. 248 00:12:50,370 --> 00:12:52,080 All we had to know was what? 249 00:12:52,080 --> 00:12:55,190 What properties did it obey, and did we have a book of 250 00:12:55,190 --> 00:12:58,610 tables so we could look up the logarithm when we had to. 251 00:12:58,610 --> 00:13:02,810 Well at any rate, using this as motivation, we now define a 252 00:13:02,810 --> 00:13:06,250 logarithmic function, specifically a function f is 253 00:13:06,250 --> 00:13:11,630 called logarithmic if for all 'x1', 'x2' in the domain of 254 00:13:11,630 --> 00:13:16,140 'f', 'f' of the product of 'x1' and 'x2' is 'f of x1' 255 00:13:16,140 --> 00:13:17,720 plus 'f of x2'. 256 00:13:17,720 --> 00:13:20,440 In other words, since we know what nice properties a 257 00:13:20,440 --> 00:13:23,670 logarithm has, let's define any function 'f' to be 258 00:13:23,670 --> 00:13:27,460 logarithmic if 'f' of a product is equal to 259 00:13:27,460 --> 00:13:31,380 the sum of the 'f's. 260 00:13:31,380 --> 00:13:34,400 Now you see, this may sound like a pretty simple 261 00:13:34,400 --> 00:13:37,460 statement, but it's quite demanding. 262 00:13:37,460 --> 00:13:39,000 In other words, notice-- 263 00:13:39,000 --> 00:13:41,220 by the way, somebody says, I wonder if there are any 264 00:13:41,220 --> 00:13:42,610 logarithmic functions? 265 00:13:42,610 --> 00:13:45,430 You see, notice that by the way we began, there must be at 266 00:13:45,430 --> 00:13:47,260 least one logarithmic function, 267 00:13:47,260 --> 00:13:50,230 namely the usual logarithm. 268 00:13:50,230 --> 00:13:52,580 We know there's at least one function which has these 269 00:13:52,580 --> 00:13:53,880 properties. 270 00:13:53,880 --> 00:13:56,550 So the set of all logarithmic functions is certainly 271 00:13:56,550 --> 00:13:57,920 not the empty set. 272 00:13:57,920 --> 00:14:01,180 At any rate, let's see what we can deduce about any 273 00:14:01,180 --> 00:14:02,820 logarithmic function. 274 00:14:02,820 --> 00:14:06,690 Well for example, I claim that if 'f' is logarithmic, 'f of 275 00:14:06,690 --> 00:14:08,020 1' must be 0. 276 00:14:08,020 --> 00:14:09,150 Why is that? 277 00:14:09,150 --> 00:14:13,110 Well, notice that 1 can be written as 1 times 1. 278 00:14:13,110 --> 00:14:16,530 And since 'f of 1' times 1 is 'f of 1' plus 'f 279 00:14:16,530 --> 00:14:18,610 of 1', we have what? 280 00:14:18,610 --> 00:14:23,040 That 'f of 1' is equal to 'f of 1' plus 'f of 1', therefore 281 00:14:23,040 --> 00:14:27,110 twice 'f of 1' is 'f of 1', and therefore by algebraic 282 00:14:27,110 --> 00:14:29,780 manipulation, 'f of 1' is 0. 283 00:14:29,780 --> 00:14:33,330 Which, by the way, does check out with the usual logarithm, 284 00:14:33,330 --> 00:14:37,570 the logarithm of 1 to any base 'b' is 0. 285 00:14:37,570 --> 00:14:41,900 Secondly, if 'f of '1 over x'' is defined, you see in other 286 00:14:41,900 --> 00:14:45,170 words, I'm not sure whether '1 over x' is in the domain of 287 00:14:45,170 --> 00:14:48,620 'f' just because 'x' is, but suppose '1 over x' is in the 288 00:14:48,620 --> 00:14:49,920 domain of 'f'. 289 00:14:49,920 --> 00:14:55,345 Then 'f of '1 over x'' is equal to minus 'f of x'. 290 00:14:55,345 --> 00:15:00,050 In other words, 'f' of a number is minus 'f' of the 291 00:15:00,050 --> 00:15:01,890 reciprocal of that number. 292 00:15:01,890 --> 00:15:05,570 Again, a familiar logarithmic property, why does it follow 293 00:15:05,570 --> 00:15:07,050 from our basic definition? 294 00:15:07,050 --> 00:15:11,250 Well notice that we already know that 'f of 1' is 0. 295 00:15:11,250 --> 00:15:16,120 We also know that 1 is equal to 'x' times '1 over x', 296 00:15:16,120 --> 00:15:17,990 provided 'x' is not 0. 297 00:15:17,990 --> 00:15:22,280 By the way, if 'x' were 0, 1 over 0 wouldn't be defined, so 298 00:15:22,280 --> 00:15:23,930 we wouldn't be worrying about that anyway. 299 00:15:23,930 --> 00:15:26,140 But notice now we have what? 300 00:15:26,140 --> 00:15:31,550 If 'x' is not 0, 'f of 'x times '1 over x'' is 0, but by 301 00:15:31,550 --> 00:15:35,370 the logarithmic property, that's also 'f of x' plus 'f 302 00:15:35,370 --> 00:15:36,750 of '1 over x''. 303 00:15:36,750 --> 00:15:41,880 Since these two together add up to 0, 'f of '1 over x' must 304 00:15:41,880 --> 00:15:44,280 just be minus 'f of x'. 305 00:15:44,280 --> 00:15:46,780 And we'll just take a few more properties like this just to 306 00:15:46,780 --> 00:15:49,130 make sure that you see what properties logarithmic 307 00:15:49,130 --> 00:15:50,650 functions have. 308 00:15:50,650 --> 00:15:53,490 Remember again the ordinary logarithm, the logarithm of a 309 00:15:53,490 --> 00:15:56,270 quotient was the difference of the logarithms. 310 00:15:56,270 --> 00:15:59,400 Again, 'f of 'x over y'' is equal to 'f of 311 00:15:59,400 --> 00:16:01,130 x' minus 'f of y'. 312 00:16:01,130 --> 00:16:02,530 Why is that the case? 313 00:16:02,530 --> 00:16:04,710 Well, look at 'f of 'x over y''. 314 00:16:04,710 --> 00:16:08,560 That says 'f of x' times '1 over y'. 315 00:16:08,560 --> 00:16:12,040 But again, by logarithmic properties, 'f' of a product 316 00:16:12,040 --> 00:16:13,440 is the sum of the 'f's. 317 00:16:13,440 --> 00:16:18,310 Therefore 'f of 'x times '1 over y''' is 'f of x' plus 'f 318 00:16:18,310 --> 00:16:19,550 of '1 over y''. 319 00:16:19,550 --> 00:16:24,020 But we just saw that 'f of '1 over y' is minus 'f of y'. 320 00:16:24,020 --> 00:16:27,240 So we have this familiar result now. 321 00:16:27,240 --> 00:16:30,640 And finally, by mathematical induction, if 'n' is any 322 00:16:30,640 --> 00:16:35,300 positive integer, 'f of x' to the n-th power is just 'n' 323 00:16:35,300 --> 00:16:36,810 times 'f of x'. 324 00:16:36,810 --> 00:16:38,630 And the proof is rather clear. 325 00:16:38,630 --> 00:16:42,070 Namely, 'f of x' to the n-th power when 'n' is a positive 326 00:16:42,070 --> 00:16:46,160 integer means 'f of 'x' times 'x' times 'x' 327 00:16:46,160 --> 00:16:48,440 times 'x'', 'n' times. 328 00:16:48,440 --> 00:16:52,140 And since the logarithmic property says that 'f' of a 329 00:16:52,140 --> 00:16:56,120 product is a sum of the 'f's, that's equal to 'f of x' plus 330 00:16:56,120 --> 00:17:00,360 'f of x' plus et cetera plus 'f of x', 'n' times, and that 331 00:17:00,360 --> 00:17:03,290 precisely is just 'n' times 'f of x'. 332 00:17:03,290 --> 00:17:05,730 In other words, notice that our definition of a 333 00:17:05,730 --> 00:17:10,740 logarithmic function, simply that 'f of 'x1 times x2'' is 334 00:17:10,740 --> 00:17:15,930 equal to 'f of x1' plus 'f of x2' allows us to deduce all of 335 00:17:15,930 --> 00:17:19,970 the familiar properties that were known to us in terms of 336 00:17:19,970 --> 00:17:23,369 the traditional meaning of logarithm. 337 00:17:23,369 --> 00:17:26,420 And now we come to the most important question of today's 338 00:17:26,420 --> 00:17:30,200 lecture, and that is what does all of this discussion have to 339 00:17:30,200 --> 00:17:34,220 do with the function that we've called capital 'L of x'? 340 00:17:34,220 --> 00:17:35,830 And that's what will tackle next. 341 00:17:35,830 --> 00:17:39,670 Let's take a look and see whether capital 'L of x' is 342 00:17:39,670 --> 00:17:43,220 indeed a logarithmic function. 343 00:17:43,220 --> 00:17:48,540 Well, if 'L' is to be a logarithmic function, if 'b' 344 00:17:48,540 --> 00:17:53,220 is any constant, in particular 'L of 'b times x'' had better 345 00:17:53,220 --> 00:17:55,830 be equal to 'L of b' plus 'L of x'. 346 00:17:55,830 --> 00:17:58,550 That's the definition of logarithmic. 347 00:17:58,550 --> 00:18:00,210 Now we don't know if that's true. 348 00:18:00,210 --> 00:18:04,390 What do we have at our disposal to be able to see 349 00:18:04,390 --> 00:18:06,190 whether we can solve this problem or not? 350 00:18:06,190 --> 00:18:09,260 How do we, using calculus, check to see whether two 351 00:18:09,260 --> 00:18:10,750 functions are equal? 352 00:18:10,750 --> 00:18:12,860 Remember the standard approach is what? 353 00:18:12,860 --> 00:18:15,730 Take the derivative of both sides. 354 00:18:15,730 --> 00:18:17,970 If the derivatives are equal, then the 355 00:18:17,970 --> 00:18:19,930 functions differ by a constant. 356 00:18:19,930 --> 00:18:22,530 And if we can show that that constant is 0, then the two 357 00:18:22,530 --> 00:18:24,380 functions are equal. 358 00:18:24,380 --> 00:18:26,990 So let's take a look and see how that works over here. 359 00:18:26,990 --> 00:18:31,230 First of all, let's see what the derivative of 'L of bx' is 360 00:18:31,230 --> 00:18:32,650 with respect to 'x'. 361 00:18:32,650 --> 00:18:35,430 And by the way, here again is the beauty of what we mean 362 00:18:35,430 --> 00:18:39,280 when we say that all of the study that we're making allows 363 00:18:39,280 --> 00:18:42,340 us to utilize every fundamental result that we've 364 00:18:42,340 --> 00:18:43,450 learned before. 365 00:18:43,450 --> 00:18:46,630 We've learned that the basic definition of 'L' is what? 366 00:18:46,630 --> 00:18:49,980 That if you differentiate 'L' with respect to the given 367 00:18:49,980 --> 00:18:52,960 variable, you get 1 over that variable. 368 00:18:52,960 --> 00:18:56,940 We want the derivative of 'L of bx' with respect to 'x'. 369 00:18:56,940 --> 00:19:00,140 Now, you see the idea is the derivative of 'L of u' with 370 00:19:00,140 --> 00:19:02,590 respect to 'u' is '1 over u'. 371 00:19:02,590 --> 00:19:06,330 You see by the chain rule, if we were differentiating 'L of 372 00:19:06,330 --> 00:19:12,050 bx' with respect to 'bx', that would be '1 over bx'. 373 00:19:12,050 --> 00:19:13,290 I shouldn't say by the chain rule yet. 374 00:19:13,290 --> 00:19:16,200 What I'm saying is if we differentiate 'L of bx' with 375 00:19:16,200 --> 00:19:19,680 respect to 'bx', we would have '1 over bx'. 376 00:19:19,680 --> 00:19:22,150 But we're not differentiating with respect to 'bx', we're 377 00:19:22,150 --> 00:19:24,410 differentiating with respect to 'x'. 378 00:19:24,410 --> 00:19:26,410 And this is where the chain rule comes in. 379 00:19:26,410 --> 00:19:31,660 Namely, we rewrite the derivative of 'L of bx' with 380 00:19:31,660 --> 00:19:35,730 respect to 'x' as the derivative 'L of bx' with 381 00:19:35,730 --> 00:19:39,020 respect to 'bx' times the derivative of 'bx' with 382 00:19:39,020 --> 00:19:40,170 respect to 'b'. 383 00:19:40,170 --> 00:19:41,700 And in this way we get what? 384 00:19:41,700 --> 00:19:43,860 The first factor is '1 over bx'. 385 00:19:43,860 --> 00:19:45,990 The second factor is 'b' itself-- 386 00:19:45,990 --> 00:19:47,310 remember 'b' is a constant-- 387 00:19:47,310 --> 00:19:52,200 and therefore the derivative of 'L of bx' is '1 over x'. 388 00:19:52,200 --> 00:19:54,860 On the other hand, what is the derivative of 'L of 389 00:19:54,860 --> 00:19:56,440 b' plus 'L of x'? 390 00:19:56,440 --> 00:20:00,460 Since 'b' is a constant, the derivative 'L of b' is 0. 391 00:20:00,460 --> 00:20:03,250 And by definition, the derivative of 'L of x' with 392 00:20:03,250 --> 00:20:05,440 respect to 'x' is '1 over x'. 393 00:20:05,440 --> 00:20:09,710 And now you see, comparing these two results, we see that 394 00:20:09,710 --> 00:20:14,430 'L of bx' and 'L of b' plus 'L of x have the same derivative, 395 00:20:14,430 --> 00:20:18,220 hence they differ by a constant which I'll call 'c'. 396 00:20:18,220 --> 00:20:21,830 In other words, notice that capital 'L' is what I call 397 00:20:21,830 --> 00:20:23,300 almost logarithmic. 398 00:20:23,300 --> 00:20:26,670 If it weren't for this factor constant 'c' in here, it would 399 00:20:26,670 --> 00:20:28,810 be a logarithmic function. 400 00:20:28,810 --> 00:20:30,820 Well at any rate, what do we have? 401 00:20:30,820 --> 00:20:34,730 We know that capital 'L of bx' is equal to capital 'L of b' 402 00:20:34,730 --> 00:20:38,450 plus capital 'L of x' plus some constant 'c'. 403 00:20:38,450 --> 00:20:42,190 To evaluate the constant, we only have to evaluate it at 404 00:20:42,190 --> 00:20:46,030 one element in the domain, let's pick 'x' equal to 1. 405 00:20:46,030 --> 00:20:49,100 The motif being that if 'x' is 1, notice that on the left 406 00:20:49,100 --> 00:20:52,070 hand side you have an 'L of b' term, on the right hand side 407 00:20:52,070 --> 00:20:54,590 you have an 'L of b' term, and they will cancel. 408 00:20:54,590 --> 00:20:58,570 In other words, if 'x' is 1, this equation becomes 'L of b' 409 00:20:58,570 --> 00:21:02,500 equals 'L of b' plus 'L of 1' plus 'c', from which it 410 00:21:02,500 --> 00:21:06,180 follows that 'c' is minus 'L of 1'. 411 00:21:06,180 --> 00:21:08,060 'c' is minus 'L of 1'. 412 00:21:08,060 --> 00:21:13,010 Now what do we want to 'c' to be if 'c' were equal to 0? 413 00:21:13,010 --> 00:21:17,510 If 'c' where equal to 0, this would be logarithmic, and all 414 00:21:17,510 --> 00:21:24,350 we need to make 'c' equal to 0 is to set 'L of 1' equal to 0. 415 00:21:24,350 --> 00:21:28,105 In other words, summarizing this result, capital 'L of x' 416 00:21:28,105 --> 00:21:31,830 is logarithmic if capital 'L of 1' is 0. 417 00:21:31,830 --> 00:21:35,530 Now remember, we have a whole family of 'L's that work. 418 00:21:35,530 --> 00:21:39,180 What we're saying now is let's pick the member of the family 419 00:21:39,180 --> 00:21:42,660 of 'L's that passes through the point (1,0) . 420 00:21:42,660 --> 00:21:45,320 And because that's logarithmic, let's give that a 421 00:21:45,320 --> 00:21:46,320 special name. 422 00:21:46,320 --> 00:21:50,160 In essence, we will define this symbol. 423 00:21:50,160 --> 00:21:51,750 It's written 'ln of x'. 424 00:21:51,750 --> 00:21:55,200 It will later be called the natural logarithm of 'x'. 425 00:21:55,200 --> 00:21:58,120 I'm trying to avoid the word logarithm here as much as 426 00:21:58,120 --> 00:22:01,400 possible, because I want you to see that we haven't had to 427 00:22:01,400 --> 00:22:04,880 use exponents at all in making this kind of a definition. 428 00:22:04,880 --> 00:22:09,340 But let's define 'ln of x' to be the member of the family 429 00:22:09,340 --> 00:22:12,080 capital 'L of x' plus 'c' which passes 430 00:22:12,080 --> 00:22:14,320 through the point (1,0). 431 00:22:14,320 --> 00:22:17,950 In essence then, what is 'ln of x'? 432 00:22:17,950 --> 00:22:20,800 I'll call it natural log, it's easier for me to say. 433 00:22:20,800 --> 00:22:24,390 The natural 'log of x' is that function whose derivative with 434 00:22:24,390 --> 00:22:28,470 respect to 'x' is '1 over x', and characterized by 435 00:22:28,470 --> 00:22:30,980 the fact that what? 436 00:22:30,980 --> 00:22:33,620 If the input is 1, the output is 0. 437 00:22:33,620 --> 00:22:35,260 In other words, the graph passes through 438 00:22:35,260 --> 00:22:37,660 the point (1 , 0). 439 00:22:37,660 --> 00:22:41,650 Again, notice that in terms of what we said earlier in 440 00:22:41,650 --> 00:22:46,190 today's lesson, we talked about the functions 'L of x', 441 00:22:46,190 --> 00:22:48,880 and talked about, in the differential calculus 442 00:22:48,880 --> 00:22:52,290 approach, the curve could go through any 443 00:22:52,290 --> 00:22:53,470 point on the x-axis. 444 00:22:53,470 --> 00:22:56,440 What we're saying is now if the point on the x-axis that 445 00:22:56,440 --> 00:23:00,860 the curve crosses at is (1 , 0), that's what 'lnx' will 446 00:23:00,860 --> 00:23:03,140 mean, the natural log function. 447 00:23:03,140 --> 00:23:06,910 In terms of the integral calculus approach, if the 448 00:23:06,910 --> 00:23:11,010 particular 'a' value that we choose, the fixed value that 449 00:23:11,010 --> 00:23:14,310 we're going to study the area under, notice that all we're 450 00:23:14,310 --> 00:23:16,730 saying is pick 'a' to be 1. 451 00:23:16,730 --> 00:23:19,190 And by the way, as a quick check, all 452 00:23:19,190 --> 00:23:19,940 we're saying is what? 453 00:23:19,940 --> 00:23:23,090 The 'natural log of x' is defined to be the area under 454 00:23:23,090 --> 00:23:25,200 this curve as a function of 'x'. 455 00:23:25,200 --> 00:23:28,090 In terms of the definite integral, that's the integral 456 00:23:28,090 --> 00:23:30,760 from '1 to x,' 'dt over t'. 457 00:23:30,760 --> 00:23:33,880 And notice that if you pick 'x' to be 1, the natural log 458 00:23:33,880 --> 00:23:39,330 of 1 is the integral from 1 to 1 'dt over t', and that's 0, 459 00:23:39,330 --> 00:23:42,030 just as it should be. 460 00:23:42,030 --> 00:23:44,340 So at any rate now, that tells me how to 461 00:23:44,340 --> 00:23:46,170 define the natural log. 462 00:23:46,170 --> 00:23:48,260 Am I doing this in terms of exponents? 463 00:23:48,260 --> 00:23:48,790 No. 464 00:23:48,790 --> 00:23:50,270 I am trying to do what? 465 00:23:50,270 --> 00:23:53,830 Find a function whose derivative with respect to 'x' 466 00:23:53,830 --> 00:23:57,430 is '1 over x', and I'd also like that function, since it's 467 00:23:57,430 --> 00:24:01,010 that close to being a logarithmic function, to be a 468 00:24:01,010 --> 00:24:02,970 logarithmic function. 469 00:24:02,970 --> 00:24:07,250 In other words, this is how I invented the function 'ln of 470 00:24:07,250 --> 00:24:09,190 x', the 'natural log of x'. 471 00:24:09,190 --> 00:24:12,310 It's derivative with respect to 'x' is '1 over x', and the 472 00:24:12,310 --> 00:24:16,480 natural log of a product is the sum of the natural logs. 473 00:24:16,480 --> 00:24:20,050 And by the way, that's exactly how we use this material. 474 00:24:20,050 --> 00:24:22,970 Let me just take a few minutes and go over something that we 475 00:24:22,970 --> 00:24:25,380 already had an answer for, just to show 476 00:24:25,380 --> 00:24:26,680 you how this works. 477 00:24:26,680 --> 00:24:30,650 Let me try to rederive the product rule using logarithms. 478 00:24:30,650 --> 00:24:33,150 Suppose 'u' and 'v' are differentiable functions of 479 00:24:33,150 --> 00:24:37,360 'x', and I want to find the derivative of 'u' times 'v' 480 00:24:37,360 --> 00:24:38,440 with respect to 'x'. 481 00:24:38,440 --> 00:24:41,290 In other words, I'd like to find 'dy dx'. 482 00:24:41,290 --> 00:24:44,480 OK, I take the natural log of both sides. 483 00:24:44,480 --> 00:24:47,770 In other words, I say if 'y' equals 'u' times 'v', the 484 00:24:47,770 --> 00:24:53,780 natural log 'y' equals natural log 'u times v'. 485 00:24:53,780 --> 00:24:56,550 Now what is the property that the natural log function has? 486 00:24:56,550 --> 00:24:59,840 I deliberately chose it to be that member of the l family 487 00:24:59,840 --> 00:25:01,180 that was logarithmic. 488 00:25:01,180 --> 00:25:05,480 The natural log of 'u times v' is 'natural log u' plus 489 00:25:05,480 --> 00:25:07,590 'natural log v'. 490 00:25:07,590 --> 00:25:11,700 Now, can I differentiate this implicitly 491 00:25:11,700 --> 00:25:12,760 with respect to 'x'? 492 00:25:12,760 --> 00:25:15,500 You see how all of our old stuff keeps 493 00:25:15,500 --> 00:25:17,040 coming up in new context. 494 00:25:17,040 --> 00:25:20,110 I take this equation and I differentiate it implicitly 495 00:25:20,110 --> 00:25:21,690 with respect to 'x'. 496 00:25:21,690 --> 00:25:23,800 The derivative of the left hand side is '1 497 00:25:23,800 --> 00:25:26,160 over y', 'dy dx'. 498 00:25:26,160 --> 00:25:29,020 The derivative of the right hand side is what? 499 00:25:29,020 --> 00:25:31,380 Well, the derivative of 'log u' with respect to 'u' is '1 500 00:25:31,380 --> 00:25:33,260 over u', but I'm differentiating it with 501 00:25:33,260 --> 00:25:36,740 respect to 'x', so I must use the correction factor by the 502 00:25:36,740 --> 00:25:38,530 chain rule 'du dx'. 503 00:25:38,530 --> 00:25:41,670 And similarly, the derivative of 'log v' with respect to 'x' 504 00:25:41,670 --> 00:25:44,020 is '1 over v', 'dv dx'. 505 00:25:44,020 --> 00:25:47,790 And now multiplying through by 'y', I have obtained that 'dy 506 00:25:47,790 --> 00:25:53,170 dx' is ''y over u' du dx' plus ''y over v' dv dx'. 507 00:25:53,170 --> 00:25:56,470 But remembering that 'y' is equal to 'u' times 'v', this 508 00:25:56,470 --> 00:26:02,060 becomes 'v 'du dx'' plus 'u 'dv dx', and we have arrived 509 00:26:02,060 --> 00:26:05,870 at the familiar product rule using logarithmic 510 00:26:05,870 --> 00:26:07,120 differentiation. 511 00:26:07,120 --> 00:26:09,290 You see the point that's really important to stress 512 00:26:09,290 --> 00:26:12,190 here it that it wasn't important whether the natural 513 00:26:12,190 --> 00:26:13,950 log was an exponent or not. 514 00:26:13,950 --> 00:26:16,440 The important thing that we used about logarithms, whether 515 00:26:16,440 --> 00:26:18,700 it was our high school course, whether it's going to be in 516 00:26:18,700 --> 00:26:22,140 our college calculus course, the important thing was what? 517 00:26:22,140 --> 00:26:25,580 That when we wanted to write the log of a product, it was 518 00:26:25,580 --> 00:26:27,580 the sum of the logs. 519 00:26:27,580 --> 00:26:29,720 And now the only additional fact that we have from the 520 00:26:29,720 --> 00:26:32,820 calculus approach is that the derivative of 'log x' with 521 00:26:32,820 --> 00:26:36,800 respect to 'x' was defined to be '1 over x'. 522 00:26:36,800 --> 00:26:41,140 By the way, since there is a tendency to think of 523 00:26:41,140 --> 00:26:44,970 traditional logarithms when one uses the word logarithm, 524 00:26:44,970 --> 00:26:46,780 if we so wanted-- 525 00:26:46,780 --> 00:26:48,850 and there's no reason why we have to do this-- 526 00:26:48,850 --> 00:26:51,940 if we wanted to associate a base with 527 00:26:51,940 --> 00:26:53,540 the natural log system-- 528 00:26:53,540 --> 00:26:56,650 this is just a little aside, and we'll talk about this more 529 00:26:56,650 --> 00:26:59,720 perhaps in the learning exercises or in the 530 00:26:59,720 --> 00:27:01,780 supplementary notes as need be-- 531 00:27:01,780 --> 00:27:03,090 but the idea is this. 532 00:27:03,090 --> 00:27:05,970 Notice that in a traditional logarithm system, if the base 533 00:27:05,970 --> 00:27:09,380 is 'b', the base is characterized by the fact that 534 00:27:09,380 --> 00:27:12,400 the 'log of 'b to the base b'' is 1. 535 00:27:12,400 --> 00:27:18,100 Thus, if we use 'e' to denote the base for the natural log, 536 00:27:18,100 --> 00:27:21,820 whatever 'e' is, it must be characterized by the fact that 537 00:27:21,820 --> 00:27:24,050 the 'natural log of e' is 1. 538 00:27:24,050 --> 00:27:27,430 And the reason that I put the word base in quotation marks 539 00:27:27,430 --> 00:27:31,150 here is that I would like you to observe that again, if I 540 00:27:31,150 --> 00:27:34,640 have never heard of the word exponent, it still makes sense 541 00:27:34,640 --> 00:27:38,650 to say find the number e such that the 'natural 542 00:27:38,650 --> 00:27:40,220 log of e' is 1. 543 00:27:40,220 --> 00:27:43,290 In fact, I can give you two interpretations for that 544 00:27:43,290 --> 00:27:47,510 number 'e', and as a byproduct, even show you why 545 00:27:47,510 --> 00:27:50,430 the second fundamental theorem of integral calculus is as 546 00:27:50,430 --> 00:27:53,346 powerful as it really is. 547 00:27:53,346 --> 00:27:56,160 See, the idea is this. 548 00:27:56,160 --> 00:27:59,300 To find what 'e' is, all we're saying is take the curve 'y' 549 00:27:59,300 --> 00:28:02,620 equals 'natural log x'. 550 00:28:02,620 --> 00:28:05,000 Look to see where the y-coordinate is 1. 551 00:28:05,000 --> 00:28:08,130 In other words, draw the line 'y' equals 1. 552 00:28:08,130 --> 00:28:12,020 Where that line intercepts the curve 'y' equals 'log x', that 553 00:28:12,020 --> 00:28:13,160 x-coordinate is 'e'. 554 00:28:13,160 --> 00:28:16,700 In other words, the 'natural log of e' must be 1, so this 555 00:28:16,700 --> 00:28:19,650 is geometrically how I would locate 'e' using 556 00:28:19,650 --> 00:28:21,130 differential calculus. 557 00:28:21,130 --> 00:28:26,250 If I wanted to use integral calculus, notice that the 'log 558 00:28:26,250 --> 00:28:28,440 of e' by definition is what? 559 00:28:28,440 --> 00:28:31,960 The integral from 1 to 'e', 'dt over t'. 560 00:28:31,960 --> 00:28:34,750 That's the area of this region 'R'. 561 00:28:34,750 --> 00:28:36,600 Now what do I want 'e' to be? 562 00:28:36,600 --> 00:28:41,420 I want 'e' to be that number that makes this area 1. 563 00:28:41,420 --> 00:28:45,650 In other words, I want the 'natural log of e' to be 1. 564 00:28:45,650 --> 00:28:50,170 So again, in the same way that one can think of pi as being 565 00:28:50,170 --> 00:28:53,490 geometrically constructed, notice I can construct 'e' 566 00:28:53,490 --> 00:28:55,970 geometrically, namely again what? 567 00:28:55,970 --> 00:29:00,450 I take the curve 'y' equals '1 over t' from 't' equals 1, 568 00:29:00,450 --> 00:29:04,120 bounded below by the t-axis, and I keep shifting over to 569 00:29:04,120 --> 00:29:09,890 the right until the area of this region 'R' is exactly 1. 570 00:29:09,890 --> 00:29:14,410 The 't' value that makes this area 1 is called 'e'. 571 00:29:14,410 --> 00:29:17,880 And by the way, notice again the power of what I mean by 572 00:29:17,880 --> 00:29:19,200 the area approach. 573 00:29:19,200 --> 00:29:22,430 Notice I can start now to get estimates on what 574 00:29:22,430 --> 00:29:23,270 'e' must look like. 575 00:29:23,270 --> 00:29:25,530 Let me show you what I'm driving at over here. 576 00:29:25,530 --> 00:29:28,840 Let's suppose we take the curve 'y' equals '1 over t' 577 00:29:28,840 --> 00:29:30,760 from 1 to 2. 578 00:29:30,760 --> 00:29:33,480 Now what do we mean by natural log 2? 579 00:29:33,480 --> 00:29:38,280 Natural log 2 is by definition the area of this region here. 580 00:29:38,280 --> 00:29:40,980 Now notice that the smallest height of this region, since 581 00:29:40,980 --> 00:29:43,880 the curve is 'y' equals '1 over t', the smallest height 582 00:29:43,880 --> 00:29:48,080 of this region is 1/2, and the tallest height, the highest 583 00:29:48,080 --> 00:29:50,020 height in this region is 1. 584 00:29:50,020 --> 00:29:54,390 Consequently, whatever the area of this region is, it's 585 00:29:54,390 --> 00:29:57,280 less than the area of the inscribed rectangle-- 586 00:29:57,280 --> 00:30:01,460 it's greater than the area of the inscribed rectangle, and 587 00:30:01,460 --> 00:30:05,330 less than the area of the circumscribed rectangle. 588 00:30:05,330 --> 00:30:08,620 Now notice that both of these rectangles have base 1, and 589 00:30:08,620 --> 00:30:10,350 we're going from 1 to 2. 590 00:30:10,350 --> 00:30:13,490 The height of the inscribed rectangle is 1/2, therefore 591 00:30:13,490 --> 00:30:16,200 the area of the small rectangle is 1/2. 592 00:30:16,200 --> 00:30:20,930 The area of the big rectangle, it's a 1 by 1 rectangle, is 1. 593 00:30:20,930 --> 00:30:25,290 And now what we have is that whatever the natural log of 2 594 00:30:25,290 --> 00:30:29,250 is, it being the area of this region, it must be what? 595 00:30:29,250 --> 00:30:32,600 Less than 1 but greater than 1/2. 596 00:30:32,600 --> 00:30:35,020 We also know that whatever 'e' is, the 'log of 597 00:30:35,020 --> 00:30:37,530 e' is equal to 1. 598 00:30:37,530 --> 00:30:39,350 Well, let's go on a little bit further. 599 00:30:39,350 --> 00:30:41,250 What about the log of 4? 600 00:30:41,250 --> 00:30:46,410 The log of 4, natural log of 4, is the log of 2 squared. 601 00:30:46,410 --> 00:30:51,290 But we've already seen that one of the properties of a 602 00:30:51,290 --> 00:30:53,960 logarithmic function is that you can bring the exponent 603 00:30:53,960 --> 00:30:56,020 down as a multiplier. 604 00:30:56,020 --> 00:30:59,340 See, 'f of 'x to the n'' is 'n 'f of x''. 605 00:30:59,340 --> 00:31:03,470 This becomes 2 log 2, and because the natural log of 2 606 00:31:03,470 --> 00:31:06,680 is greater than 1/2, twice the natural log of 2 607 00:31:06,680 --> 00:31:08,890 is more than 1. 608 00:31:08,890 --> 00:31:13,010 In other words, if log 2 is more than 1/2, twice log 2 is 609 00:31:13,010 --> 00:31:14,400 more than 1. 610 00:31:14,400 --> 00:31:17,200 In other words, if we put these three lines together, we 611 00:31:17,200 --> 00:31:20,890 have that the natural log of 2 is less than the 'natural log 612 00:31:20,890 --> 00:31:24,570 of e', which in turn is less than the natural log of 4. 613 00:31:24,570 --> 00:31:28,490 By the way, notice that since the derivative of 'log x' with 614 00:31:28,490 --> 00:31:32,660 respect to 'x' is '1 over x', and that's positive, 'log x' 615 00:31:32,660 --> 00:31:34,580 is a one to one function. 616 00:31:34,580 --> 00:31:38,250 Therefore, if the log of 2 is less than the 'log of e' is 617 00:31:38,250 --> 00:31:42,080 less than the log of 4, it follows that 2 must be less 618 00:31:42,080 --> 00:31:45,230 than 'e', which in turn must be less than 4. 619 00:31:45,230 --> 00:31:48,450 In other words, even with this crude approximation of just 620 00:31:48,450 --> 00:31:52,260 inscribing and circumscribing rectangles, I can show again 621 00:31:52,260 --> 00:31:56,410 without reference to exponents that whatever the number 'e' 622 00:31:56,410 --> 00:32:02,330 is, the number e defined by the fact that 'ln of e' is 1, 623 00:32:02,330 --> 00:32:06,400 that number 'e' is some number between 2 and 4. 624 00:32:06,400 --> 00:32:08,830 But again, we won't dwell on this too long. 625 00:32:08,830 --> 00:32:12,000 What I want to do now is to come back to a summary point, 626 00:32:12,000 --> 00:32:14,570 and at the same time, lead into the 627 00:32:14,570 --> 00:32:16,020 lecture for next time. 628 00:32:16,020 --> 00:32:20,790 Recall that we began this lecture with the problem of 629 00:32:20,790 --> 00:32:24,740 solving the situation where the rate of change was 630 00:32:24,740 --> 00:32:27,390 proportional to the amount present. 631 00:32:27,390 --> 00:32:31,180 Given 'dm/ dt' equals 'km', we separated variables and got 632 00:32:31,180 --> 00:32:34,920 down to the stage where the integral of 'dm over m' was 633 00:32:34,920 --> 00:32:36,880 'kt' plus a constant. 634 00:32:36,880 --> 00:32:40,540 And all we did in the rest of today's lecture that was at 635 00:32:40,540 --> 00:32:45,350 all different, was we invented and constructed the particular 636 00:32:45,350 --> 00:32:48,970 function whose derivative would be '1 over m', and which 637 00:32:48,970 --> 00:32:50,550 had the logarithmic property. 638 00:32:50,550 --> 00:32:52,920 In other words, we can now say that this is the 639 00:32:52,920 --> 00:32:54,340 answer to the problem. 640 00:32:54,340 --> 00:32:58,240 This problem is explicitly solved because the natural log 641 00:32:58,240 --> 00:33:00,690 function can be viewed as an area under a curve. 642 00:33:00,690 --> 00:33:02,950 We can construct it for each 'm'. 643 00:33:02,950 --> 00:33:05,400 This is then what we managed to do. 644 00:33:05,400 --> 00:33:08,800 And by the way, all I'm saying now is that since the log is a 645 00:33:08,800 --> 00:33:10,250 one to one function-- 646 00:33:10,250 --> 00:33:12,250 see, its derivative is always positive-- 647 00:33:12,250 --> 00:33:14,670 can't we talk about the inverse function? 648 00:33:14,670 --> 00:33:16,880 In other words, notice that another way of writing this is 649 00:33:16,880 --> 00:33:21,720 that id 'natural log m' is 'kt' plus 'c', 'm' itself must 650 00:33:21,720 --> 00:33:26,490 be the 'inverse log of kt' plus 'c'. 651 00:33:26,490 --> 00:33:30,450 And you see, next time what we're going to do is to 652 00:33:30,450 --> 00:33:35,610 explore what we mean by the inverse log. 653 00:33:35,610 --> 00:33:39,300 At any rate, in summarizing today's lecture, to make sure 654 00:33:39,300 --> 00:33:43,250 that we didn't fall victim to all the computational details, 655 00:33:43,250 --> 00:33:46,500 notice that all we did was physically motivated the 656 00:33:46,500 --> 00:33:50,070 necessity for inventing a function whose derivative with 657 00:33:50,070 --> 00:33:51,920 respect to 'x' was '1 over x'. 658 00:33:51,920 --> 00:33:55,900 And from that point on, everything else that we did 659 00:33:55,900 --> 00:33:59,730 followed as applications of material that came before. 660 00:33:59,730 --> 00:34:03,980 It's in this sense that the course now picks up in tempo. 661 00:34:03,980 --> 00:34:07,050 That as we go on now, we're going to be able to cover 662 00:34:07,050 --> 00:34:10,679 larger globs of material in one sitting, because we will 663 00:34:10,679 --> 00:34:14,130 find, at least for the next several lectures, that every 664 00:34:14,130 --> 00:34:18,989 new topic is basically one new idea together with all of the 665 00:34:18,989 --> 00:34:20,330 old recipes. 666 00:34:20,330 --> 00:34:22,330 At any rate, until next time, goodbye. 667 00:34:25,489 --> 00:34:28,030 ANNOUNCER: Funding for the publication of this video was 668 00:34:28,030 --> 00:34:32,750 provided by the Gabriella and Paul Rosenbaum Foundation. 669 00:34:32,750 --> 00:34:36,920 Help OCW continue to provide free and open access to MIT 670 00:34:36,920 --> 00:34:41,120 courses by making a donation at ocw.mit.edu/donate.