WEBVTT 00:00:00.000 --> 00:00:01.940 ANNOUNCER: The following content is provided under a 00:00:01.940 --> 00:00:03.690 Creative Commons license. 00:00:03.690 --> 00:00:06.640 Your support will help MIT OpenCourseWare continue to 00:00:06.640 --> 00:00:09.990 offer high quality educational resources for free. 00:00:09.990 --> 00:00:12.830 To make a donation or to view additional materials from 00:00:12.830 --> 00:00:16.760 hundreds of MIT courses, visit MIT OpenCourseWare at 00:00:16.760 --> 00:00:18.010 ocw.mit.edu. 00:00:32.619 --> 00:00:33.710 PROFESSOR: Hi. 00:00:33.710 --> 00:00:38.740 Our lecture for today I've entitled 'Logarithms Without 00:00:38.740 --> 00:00:39.800 Exponents'. 00:00:39.800 --> 00:00:43.160 And before I try to clarify that, let me make a couple of 00:00:43.160 --> 00:00:44.390 general remarks. 00:00:44.390 --> 00:00:48.720 First of all, we have now completed the rudiments, the 00:00:48.720 --> 00:00:51.240 basics so to speak, of both differential 00:00:51.240 --> 00:00:52.550 and integral calculus. 00:00:52.550 --> 00:00:56.340 Consequently, what our next ambition will be is to take 00:00:56.340 --> 00:01:00.020 these principles and apply them to special functions 00:01:00.020 --> 00:01:02.260 which are worthy of investigation. 00:01:02.260 --> 00:01:04.890 And you see, in connection with this, the first function 00:01:04.890 --> 00:01:08.670 that I've chosen to talk about is called the logarithm. 00:01:08.670 --> 00:01:11.630 Now the other interesting thing is that we are used to 00:01:11.630 --> 00:01:14.570 logarithms from high school where they were viewed as 00:01:14.570 --> 00:01:16.980 being a different kind of notation for 00:01:16.980 --> 00:01:18.840 talking about exponents. 00:01:18.840 --> 00:01:21.500 Now, in the same way when we treated the circular 00:01:21.500 --> 00:01:25.030 functions, we mentioned that one did not have to invent 00:01:25.030 --> 00:01:27.670 triangles to talk about trigonometry. 00:01:27.670 --> 00:01:30.490 The interesting point is that one does not have to talk 00:01:30.490 --> 00:01:33.960 about exponents to invent the logarithm function. 00:01:33.960 --> 00:01:37.130 This is why I have entitled the lesson, as I say, 00:01:37.130 --> 00:01:40.010 'Logarithms without Exponents'. 00:01:40.010 --> 00:01:43.100 You see, what I would like to do is try to mimic, especially 00:01:43.100 --> 00:01:45.930 from an engineering point of view, how many of these 00:01:45.930 --> 00:01:48.610 mathematical topics originated. 00:01:48.610 --> 00:01:51.790 Let's look at a rather straightforward physical 00:01:51.790 --> 00:01:56.050 principle, a principle that I'm sure we believe permeates 00:01:56.050 --> 00:01:58.340 many real life situations. 00:01:58.340 --> 00:02:01.990 It's what I call the rule of compound interest. 00:02:01.990 --> 00:02:06.000 Namely, let's suppose we have a physical situation in which 00:02:06.000 --> 00:02:07.490 the quantity that we're measuring, 00:02:07.490 --> 00:02:08.590 which we'll call 'm'. 00:02:08.590 --> 00:02:12.310 In other words, the rate of change of the quantity is 00:02:12.310 --> 00:02:18.520 proportional to the amount present, 'dm/ dt' equals 'km'. 00:02:18.520 --> 00:02:22.750 Now notice that this is a pretty harmless statement. 00:02:22.750 --> 00:02:24.750 The rate of change is proportional 00:02:24.750 --> 00:02:25.920 to the amount present. 00:02:25.920 --> 00:02:29.520 We certainly would expect that many experiments would run 00:02:29.520 --> 00:02:32.820 this way, and there seems to be nothing at all supernatural 00:02:32.820 --> 00:02:34.270 about this kind of an assumption. 00:02:34.270 --> 00:02:38.090 At any rate, let's apply our previous principles, and see 00:02:38.090 --> 00:02:41.130 if we can't, from this differential equation, figure 00:02:41.130 --> 00:02:43.360 out what 'm' has to be. 00:02:43.360 --> 00:02:46.990 Separating the variables, we get that 'dm' over 'm' is 00:02:46.990 --> 00:02:48.860 equal to 'kdt'. 00:02:48.860 --> 00:02:53.130 And now integrating, meaning taking the inverse derivative, 00:02:53.130 --> 00:02:54.130 we have what? 00:02:54.130 --> 00:02:57.310 That the integral of 'dm' over 'm' is equal 00:02:57.310 --> 00:02:59.830 to 'kt' plus a constant. 00:02:59.830 --> 00:03:03.300 Now you see, the interesting point is at this stage of the 00:03:03.300 --> 00:03:08.400 game, we do not know explicitly how to find a 00:03:08.400 --> 00:03:12.970 function whose derivative with respect to 'm' is '1 over m'. 00:03:12.970 --> 00:03:14.710 In fact, that's the problem that 00:03:14.710 --> 00:03:16.700 underlies today's lecture. 00:03:16.700 --> 00:03:20.130 And we will try to answer this question, showing how once we 00:03:20.130 --> 00:03:23.760 tackle this from a philosophic point of view, the rest of the 00:03:23.760 --> 00:03:27.530 details follow from material that we've already learned. 00:03:27.530 --> 00:03:30.590 At any rate, focusing our attention on the problem, the 00:03:30.590 --> 00:03:34.140 question is this: to determine a function, which I'll call 00:03:34.140 --> 00:03:36.240 capital 'L of x'-- 00:03:36.240 --> 00:03:40.820 the 'L' sort of to forewarn us of the fact that we will 00:03:40.820 --> 00:03:43.220 somehow get a logarithm out of this-- 00:03:43.220 --> 00:03:47.050 such that 'L prime of x' is '1 over x'. 00:03:47.050 --> 00:03:50.690 And you see, what I want you to notice is that if you write 00:03:50.690 --> 00:03:53.750 '1 over x' in exponential form, in other words, if you 00:03:53.750 --> 00:03:57.340 try to write '1 over x' as 'x to the n', notice that to 00:03:57.340 --> 00:04:00.460 solve this problem, 'L of x' would be what? 00:04:00.460 --> 00:04:04.570 The integral of 'x' to the 'minus 1 dx'. 00:04:04.570 --> 00:04:07.930 In other words, this is the case of integral ''x to the n' 00:04:07.930 --> 00:04:11.290 dx', with 'n' equal to minus 1. 00:04:11.290 --> 00:04:14.350 And you'll recall that when we studied the recipe-- 00:04:14.350 --> 00:04:16.430 let me just write that over here-- when we studied the 00:04:16.430 --> 00:04:20.740 recipe of how you integrate 'x to the n', we saw that that 00:04:20.740 --> 00:04:27.240 was 'x to the 'n + 1'' over 'n + 1' plus a constant. 00:04:27.240 --> 00:04:30.810 And we then observed that this doesn't even make sense when 00:04:30.810 --> 00:04:33.910 'n' is equal to minus 1, because we have a 0 00:04:33.910 --> 00:04:35.060 denominator. 00:04:35.060 --> 00:04:37.140 From the other point of view, the way to look at 00:04:37.140 --> 00:04:38.700 this is we said what? 00:04:38.700 --> 00:04:42.490 That when you differentiate, you lower the exponent by 1. 00:04:42.490 --> 00:04:47.160 Notice that to wind up with an exponent of minus 1, we would 00:04:47.160 --> 00:04:51.340 have had to start with an exponent of 0. 00:04:51.340 --> 00:04:53.950 But when you differentiate 'x' to the 0, that being a 00:04:53.950 --> 00:05:00.010 constant, the derivative of 'x' to the 0 is 0. 00:05:00.010 --> 00:05:03.870 It's not '1 over x'. 00:05:03.870 --> 00:05:07.620 You see, in other words, this particular recipe that we're 00:05:07.620 --> 00:05:10.590 talking about, we don't have working for us. 00:05:10.590 --> 00:05:13.180 In other words, this is an interesting point again. 00:05:13.180 --> 00:05:15.770 You say gee whiz, if the recipe for the integral ''x to 00:05:15.770 --> 00:05:18.650 the n' dx' works for everything except 'n' equals 00:05:18.650 --> 00:05:20.730 minus 1, nothing's perfect. 00:05:20.730 --> 00:05:22.180 Let's be content. 00:05:22.180 --> 00:05:23.700 Since it works for every number except 00:05:23.700 --> 00:05:25.710 that one, why worry? 00:05:25.710 --> 00:05:28.040 This is analogous to when we talked about taking 00:05:28.040 --> 00:05:30.220 derivatives and saw that we get a 0 over 00:05:30.220 --> 00:05:31.850 0 form every time. 00:05:31.850 --> 00:05:34.260 You see, if you pick a function at random, the 00:05:34.260 --> 00:05:36.890 likelihood that the limit of 'f of x' as 'x' approaches 00:05:36.890 --> 00:05:40.820 'a', the likelihood of that being 0 over 0 is very small, 00:05:40.820 --> 00:05:43.760 except when you form the derivative, you 00:05:43.760 --> 00:05:45.860 always get 0 over 0. 00:05:45.860 --> 00:05:49.560 And in a similar way, granted that integral ''x' to the 00:05:49.560 --> 00:05:52.810 minus 1 dx' is one very special case. 00:05:52.810 --> 00:05:56.990 What we have seen is that this comes up every single time 00:05:56.990 --> 00:06:00.620 that we want to solve a problem of the form 'dm/ dt' 00:06:00.620 --> 00:06:02.040 equals 'km'. 00:06:02.040 --> 00:06:05.330 In other words then, what it really boils down to is that 00:06:05.330 --> 00:06:09.350 unless we can come up with a function 'L of x' whose 00:06:09.350 --> 00:06:13.400 derivative is '1 over x', we cannot solve the problem that 00:06:13.400 --> 00:06:15.230 we started off today's lesson with. 00:06:15.230 --> 00:06:18.610 In other words, we must leave it in the form integral 'dm 00:06:18.610 --> 00:06:21.760 over m' equals 'kt' plus a constant. 00:06:21.760 --> 00:06:25.900 Well at any rate, let's see what such a function capital 00:06:25.900 --> 00:06:27.640 'L of x' must look like. 00:06:27.640 --> 00:06:31.050 I'm going to utilize both the differential calculus approach 00:06:31.050 --> 00:06:34.450 and the integral calculus approach. 00:06:34.450 --> 00:06:37.540 By differential calculus, assuming that 'y' equals 'L of 00:06:37.540 --> 00:06:40.510 x', what do we know about the function 'y'? 00:06:40.510 --> 00:06:44.110 By definition, we've defined it to be that function whose 00:06:44.110 --> 00:06:46.330 derivative is '1 over x'. 00:06:46.330 --> 00:06:48.850 So we know its derivative is '1 over x'. 00:06:48.850 --> 00:06:53.150 The next point is that knowing that 'y prime' is '1 over x', 00:06:53.150 --> 00:06:55.490 and even though we can't integrate '1 over x', we can 00:06:55.490 --> 00:06:57.390 certainly differentiate '1 over x'. 00:06:57.390 --> 00:06:59.640 We can now differentiate '1 over x'. 00:06:59.640 --> 00:07:02.760 The derivative of '1 over x' is minus '1 over 'x squared'', 00:07:02.760 --> 00:07:04.870 and we find that 'y double prime' is 00:07:04.870 --> 00:07:06.840 minus '1 over 'x squared''. 00:07:06.840 --> 00:07:11.100 By the way, we should observe that we must shy away from 'x' 00:07:11.100 --> 00:07:14.400 equaling 0, because you see, we would have a 0 denominator 00:07:14.400 --> 00:07:15.360 in that case. 00:07:15.360 --> 00:07:19.780 If we think of most physical situations, notice that we 00:07:19.780 --> 00:07:21.240 talk about the amount-- 00:07:21.240 --> 00:07:23.160 the rate of change is proportional 00:07:23.160 --> 00:07:24.380 to the amount present. 00:07:24.380 --> 00:07:27.760 From a physical point of view, the amount present can't be 00:07:27.760 --> 00:07:31.110 negative, so let's put the restriction on here that the 00:07:31.110 --> 00:07:36.790 domain of 'L' will be all positive numbers. 00:07:39.330 --> 00:07:41.060 Now here's the interesting thing. 00:07:41.060 --> 00:07:46.660 Intuitively, I can now visualize how the function 'L 00:07:46.660 --> 00:07:47.950 of x' must look. 00:07:47.950 --> 00:07:51.100 Namely, since its first derivative is '1 over x' and 00:07:51.100 --> 00:07:53.620 'x' is greater than 0, that tells me that the curve is 00:07:53.620 --> 00:07:55.050 always rising. 00:07:55.050 --> 00:07:58.750 And secondly, since '1 over 'x squared'' is always positive, 00:07:58.750 --> 00:08:01.780 minus '1 over 'x squared'' is always negative, that tells me 00:08:01.780 --> 00:08:04.650 that my curve is always spilling water. 00:08:04.650 --> 00:08:07.140 In essence then, what must the curve do? 00:08:07.140 --> 00:08:10.110 The curve must be rising but spilling water. 00:08:10.110 --> 00:08:13.100 In other words, the curve 'L of x' belongs to this 00:08:13.100 --> 00:08:14.410 particular family. 00:08:14.410 --> 00:08:17.940 Notice that once we have one curve which we call 'y' equals 00:08:17.940 --> 00:08:21.800 'L of x', by any displacement parallel-- 00:08:21.800 --> 00:08:24.430 vertical displacement, we get what? 00:08:24.430 --> 00:08:28.820 A member called 'y' equals 'L of x' plus 'c'. 00:08:28.820 --> 00:08:31.360 All that changes is what? 00:08:31.360 --> 00:08:34.640 The point at which the curve crosses the x-axis, but 00:08:34.640 --> 00:08:37.140 whichever member of the family we pick, what 00:08:37.140 --> 00:08:38.789 typifies 'L of x'? 00:08:38.789 --> 00:08:40.870 Its derivative is '1 over x'. 00:08:40.870 --> 00:08:43.990 In other words, somehow or other, we visualize that 'L of 00:08:43.990 --> 00:08:48.850 x' exists, and its graph is something like this. 00:08:48.850 --> 00:08:51.860 Now suppose we want a more tangible form, namely how do 00:08:51.860 --> 00:08:54.760 you compute 'L of x' for a given 'x'? 00:08:54.760 --> 00:08:57.690 I thought what might be a nice review now is to see how we 00:08:57.690 --> 00:09:01.300 can use our integral calculus approach, and in particular 00:09:01.300 --> 00:09:03.730 the second fundamental theorem of interval calculus. 00:09:06.920 --> 00:09:10.270 By the integral calculus approach, we do is is we pick 00:09:10.270 --> 00:09:14.000 any positive number 'a', and once picked, we fix it. 00:09:14.000 --> 00:09:18.650 You see, we have a great deal of freedom in how we choose 00:09:18.650 --> 00:09:19.990 the positive number a. 00:09:19.990 --> 00:09:22.020 But let's pick one and leave it here. 00:09:22.020 --> 00:09:25.480 Now what we'll do is in the 'y-t' plane, we'll draw the 00:09:25.480 --> 00:09:29.140 curve 'y' equals '1 over t'. 00:09:29.140 --> 00:09:32.550 And what we'll do now, we'll study the area of the region 00:09:32.550 --> 00:09:36.910 'R' where 'R' is bounded above by 'y' equals '1 over t', on 00:09:36.910 --> 00:09:41.310 the left by 't' equals 'a', on the right by 't' equals 'x', 00:09:41.310 --> 00:09:44.420 and below by the t-axis. 00:09:44.420 --> 00:09:48.080 Now we've already seen that the function that we get by 00:09:48.080 --> 00:09:52.750 taking the definite integral from 'a' 'to x', 'dt over t', 00:09:52.750 --> 00:09:56.590 has the property that its derivative is '1 over x'. 00:09:56.590 --> 00:09:59.310 In other words, recall from the fundamental theorem, this 00:09:59.310 --> 00:10:03.120 is just a generalization of the fact that if 'f of t' is a 00:10:03.120 --> 00:10:07.480 continuous function, then the integral from 'a' to 'x', ''f 00:10:07.480 --> 00:10:11.120 of t' dt', is a function of 'x'. 00:10:11.120 --> 00:10:13.730 And its derivative with respect to 'x' 00:10:13.730 --> 00:10:16.600 is just 'f of x'. 00:10:16.600 --> 00:10:20.490 You see, in other words, I can now view capital 'L of x' as 00:10:20.490 --> 00:10:24.920 being an area under the curve 'y' equals '1 over t'. 00:10:24.920 --> 00:10:28.610 Notice that I have a degree of freedom here, namely what that 00:10:28.610 --> 00:10:32.000 area is depends on where I choose 'a'. 00:10:32.000 --> 00:10:36.490 Notice that if I choose a differently, changing 'a'-- 00:10:36.490 --> 00:10:39.610 let's call this 'a prime', say-- 00:10:39.610 --> 00:10:43.880 changing 'a' changes the area under the curve, but notice it 00:10:43.880 --> 00:10:46.150 changes it by a fixed amount. 00:10:46.150 --> 00:10:50.160 In other words, notice that shifting 'a' just changes 'L 00:10:50.160 --> 00:10:52.440 of x' by a constant, just like the 00:10:52.440 --> 00:10:54.500 ordinary in definite integral. 00:10:54.500 --> 00:10:57.540 At any rate, these are the two approaches that we have. 00:10:57.540 --> 00:11:00.960 And if we superimpose them, all we're saying is what? 00:11:00.960 --> 00:11:02.910 If you start with the differential calculus 00:11:02.910 --> 00:11:06.970 approach, 'y' equals capital 'L of x' must be a member of 00:11:06.970 --> 00:11:11.820 this family, crossing the axis at some point (a,0) . 00:11:11.820 --> 00:11:14.770 And from the integral calculus point of view, if you take the 00:11:14.770 --> 00:11:18.010 curve 'y' equals '1 over t' and compute the area under 00:11:18.010 --> 00:11:22.420 that curve from 'a' to 'x', that area 00:11:22.420 --> 00:11:25.690 function is 'L of x'. 00:11:25.690 --> 00:11:30.290 Now we'll let that rest for the time being, and now have a 00:11:30.290 --> 00:11:32.180 brief digression. 00:11:32.180 --> 00:11:34.560 You see, sooner or later, I've got to get to what a logarithm 00:11:34.560 --> 00:11:37.060 is, and we might just as well do that now. 00:11:37.060 --> 00:11:40.620 So let's now take a look and see what we mean by a 00:11:40.620 --> 00:11:42.280 logarithmic function. 00:11:42.280 --> 00:11:46.010 You see, quite frequently in mathematics, what one does is 00:11:46.010 --> 00:11:49.950 one deals with a particular special case in which one is 00:11:49.950 --> 00:11:51.220 interested. 00:11:51.220 --> 00:11:54.980 And having deduced very nice results, he looks at this 00:11:54.980 --> 00:11:58.320 thing and says, you know, all of these results came from a 00:11:58.320 --> 00:12:01.530 particular recipe, a particular structure that this 00:12:01.530 --> 00:12:03.130 system obeyed. 00:12:03.130 --> 00:12:06.860 What he then does is he takes this recipe or structure, 00:12:06.860 --> 00:12:11.280 gives it a general name, and now is able to extend this 00:12:11.280 --> 00:12:14.870 property to a larger class of objects. 00:12:14.870 --> 00:12:16.830 For example, let's look at this way. 00:12:16.830 --> 00:12:19.990 What is the nice thing about logarithms in the traditional 00:12:19.990 --> 00:12:21.270 sense of the word? 00:12:21.270 --> 00:12:24.030 Remember when we learned logarithms in high school, by 00:12:24.030 --> 00:12:27.800 use of logarithms, we were able to replace multiplication 00:12:27.800 --> 00:12:30.420 problems by addition problems, et cetera. 00:12:30.420 --> 00:12:33.680 In other words, remember the key structural property was 00:12:33.680 --> 00:12:36.610 that the logarithm of a product was the sum of the 00:12:36.610 --> 00:12:37.490 logarithms. 00:12:37.490 --> 00:12:40.620 That was the structural thing that we used over and over and 00:12:40.620 --> 00:12:41.650 over again. 00:12:41.650 --> 00:12:44.200 In fact, it's rather interesting to point out that 00:12:44.200 --> 00:12:47.050 in many cases where we used the properties of logarithms, 00:12:47.050 --> 00:12:50.370 we never really had to know what a logarithm was. 00:12:50.370 --> 00:12:52.080 All we had to know was what? 00:12:52.080 --> 00:12:55.190 What properties did it obey, and did we have a book of 00:12:55.190 --> 00:12:58.610 tables so we could look up the logarithm when we had to. 00:12:58.610 --> 00:13:02.810 Well at any rate, using this as motivation, we now define a 00:13:02.810 --> 00:13:06.250 logarithmic function, specifically a function f is 00:13:06.250 --> 00:13:11.630 called logarithmic if for all 'x1', 'x2' in the domain of 00:13:11.630 --> 00:13:16.140 'f', 'f' of the product of 'x1' and 'x2' is 'f of x1' 00:13:16.140 --> 00:13:17.720 plus 'f of x2'. 00:13:17.720 --> 00:13:20.440 In other words, since we know what nice properties a 00:13:20.440 --> 00:13:23.670 logarithm has, let's define any function 'f' to be 00:13:23.670 --> 00:13:27.460 logarithmic if 'f' of a product is equal to 00:13:27.460 --> 00:13:31.380 the sum of the 'f's. 00:13:31.380 --> 00:13:34.400 Now you see, this may sound like a pretty simple 00:13:34.400 --> 00:13:37.460 statement, but it's quite demanding. 00:13:37.460 --> 00:13:39.000 In other words, notice-- 00:13:39.000 --> 00:13:41.220 by the way, somebody says, I wonder if there are any 00:13:41.220 --> 00:13:42.610 logarithmic functions? 00:13:42.610 --> 00:13:45.430 You see, notice that by the way we began, there must be at 00:13:45.430 --> 00:13:47.260 least one logarithmic function, 00:13:47.260 --> 00:13:50.230 namely the usual logarithm. 00:13:50.230 --> 00:13:52.580 We know there's at least one function which has these 00:13:52.580 --> 00:13:53.880 properties. 00:13:53.880 --> 00:13:56.550 So the set of all logarithmic functions is certainly 00:13:56.550 --> 00:13:57.920 not the empty set. 00:13:57.920 --> 00:14:01.180 At any rate, let's see what we can deduce about any 00:14:01.180 --> 00:14:02.820 logarithmic function. 00:14:02.820 --> 00:14:06.690 Well for example, I claim that if 'f' is logarithmic, 'f of 00:14:06.690 --> 00:14:08.020 1' must be 0. 00:14:08.020 --> 00:14:09.150 Why is that? 00:14:09.150 --> 00:14:13.110 Well, notice that 1 can be written as 1 times 1. 00:14:13.110 --> 00:14:16.530 And since 'f of 1' times 1 is 'f of 1' plus 'f 00:14:16.530 --> 00:14:18.610 of 1', we have what? 00:14:18.610 --> 00:14:23.040 That 'f of 1' is equal to 'f of 1' plus 'f of 1', therefore 00:14:23.040 --> 00:14:27.110 twice 'f of 1' is 'f of 1', and therefore by algebraic 00:14:27.110 --> 00:14:29.780 manipulation, 'f of 1' is 0. 00:14:29.780 --> 00:14:33.330 Which, by the way, does check out with the usual logarithm, 00:14:33.330 --> 00:14:37.570 the logarithm of 1 to any base 'b' is 0. 00:14:37.570 --> 00:14:41.900 Secondly, if 'f of '1 over x'' is defined, you see in other 00:14:41.900 --> 00:14:45.170 words, I'm not sure whether '1 over x' is in the domain of 00:14:45.170 --> 00:14:48.620 'f' just because 'x' is, but suppose '1 over x' is in the 00:14:48.620 --> 00:14:49.920 domain of 'f'. 00:14:49.920 --> 00:14:55.345 Then 'f of '1 over x'' is equal to minus 'f of x'. 00:14:55.345 --> 00:15:00.050 In other words, 'f' of a number is minus 'f' of the 00:15:00.050 --> 00:15:01.890 reciprocal of that number. 00:15:01.890 --> 00:15:05.570 Again, a familiar logarithmic property, why does it follow 00:15:05.570 --> 00:15:07.050 from our basic definition? 00:15:07.050 --> 00:15:11.250 Well notice that we already know that 'f of 1' is 0. 00:15:11.250 --> 00:15:16.120 We also know that 1 is equal to 'x' times '1 over x', 00:15:16.120 --> 00:15:17.990 provided 'x' is not 0. 00:15:17.990 --> 00:15:22.280 By the way, if 'x' were 0, 1 over 0 wouldn't be defined, so 00:15:22.280 --> 00:15:23.930 we wouldn't be worrying about that anyway. 00:15:23.930 --> 00:15:26.140 But notice now we have what? 00:15:26.140 --> 00:15:31.550 If 'x' is not 0, 'f of 'x times '1 over x'' is 0, but by 00:15:31.550 --> 00:15:35.370 the logarithmic property, that's also 'f of x' plus 'f 00:15:35.370 --> 00:15:36.750 of '1 over x''. 00:15:36.750 --> 00:15:41.880 Since these two together add up to 0, 'f of '1 over x' must 00:15:41.880 --> 00:15:44.280 just be minus 'f of x'. 00:15:44.280 --> 00:15:46.780 And we'll just take a few more properties like this just to 00:15:46.780 --> 00:15:49.130 make sure that you see what properties logarithmic 00:15:49.130 --> 00:15:50.650 functions have. 00:15:50.650 --> 00:15:53.490 Remember again the ordinary logarithm, the logarithm of a 00:15:53.490 --> 00:15:56.270 quotient was the difference of the logarithms. 00:15:56.270 --> 00:15:59.400 Again, 'f of 'x over y'' is equal to 'f of 00:15:59.400 --> 00:16:01.130 x' minus 'f of y'. 00:16:01.130 --> 00:16:02.530 Why is that the case? 00:16:02.530 --> 00:16:04.710 Well, look at 'f of 'x over y''. 00:16:04.710 --> 00:16:08.560 That says 'f of x' times '1 over y'. 00:16:08.560 --> 00:16:12.040 But again, by logarithmic properties, 'f' of a product 00:16:12.040 --> 00:16:13.440 is the sum of the 'f's. 00:16:13.440 --> 00:16:18.310 Therefore 'f of 'x times '1 over y''' is 'f of x' plus 'f 00:16:18.310 --> 00:16:19.550 of '1 over y''. 00:16:19.550 --> 00:16:24.020 But we just saw that 'f of '1 over y' is minus 'f of y'. 00:16:24.020 --> 00:16:27.240 So we have this familiar result now. 00:16:27.240 --> 00:16:30.640 And finally, by mathematical induction, if 'n' is any 00:16:30.640 --> 00:16:35.300 positive integer, 'f of x' to the n-th power is just 'n' 00:16:35.300 --> 00:16:36.810 times 'f of x'. 00:16:36.810 --> 00:16:38.630 And the proof is rather clear. 00:16:38.630 --> 00:16:42.070 Namely, 'f of x' to the n-th power when 'n' is a positive 00:16:42.070 --> 00:16:46.160 integer means 'f of 'x' times 'x' times 'x' 00:16:46.160 --> 00:16:48.440 times 'x'', 'n' times. 00:16:48.440 --> 00:16:52.140 And since the logarithmic property says that 'f' of a 00:16:52.140 --> 00:16:56.120 product is a sum of the 'f's, that's equal to 'f of x' plus 00:16:56.120 --> 00:17:00.360 'f of x' plus et cetera plus 'f of x', 'n' times, and that 00:17:00.360 --> 00:17:03.290 precisely is just 'n' times 'f of x'. 00:17:03.290 --> 00:17:05.730 In other words, notice that our definition of a 00:17:05.730 --> 00:17:10.740 logarithmic function, simply that 'f of 'x1 times x2'' is 00:17:10.740 --> 00:17:15.930 equal to 'f of x1' plus 'f of x2' allows us to deduce all of 00:17:15.930 --> 00:17:19.970 the familiar properties that were known to us in terms of 00:17:19.970 --> 00:17:23.369 the traditional meaning of logarithm. 00:17:23.369 --> 00:17:26.420 And now we come to the most important question of today's 00:17:26.420 --> 00:17:30.200 lecture, and that is what does all of this discussion have to 00:17:30.200 --> 00:17:34.220 do with the function that we've called capital 'L of x'? 00:17:34.220 --> 00:17:35.830 And that's what will tackle next. 00:17:35.830 --> 00:17:39.670 Let's take a look and see whether capital 'L of x' is 00:17:39.670 --> 00:17:43.220 indeed a logarithmic function. 00:17:43.220 --> 00:17:48.540 Well, if 'L' is to be a logarithmic function, if 'b' 00:17:48.540 --> 00:17:53.220 is any constant, in particular 'L of 'b times x'' had better 00:17:53.220 --> 00:17:55.830 be equal to 'L of b' plus 'L of x'. 00:17:55.830 --> 00:17:58.550 That's the definition of logarithmic. 00:17:58.550 --> 00:18:00.210 Now we don't know if that's true. 00:18:00.210 --> 00:18:04.390 What do we have at our disposal to be able to see 00:18:04.390 --> 00:18:06.190 whether we can solve this problem or not? 00:18:06.190 --> 00:18:09.260 How do we, using calculus, check to see whether two 00:18:09.260 --> 00:18:10.750 functions are equal? 00:18:10.750 --> 00:18:12.860 Remember the standard approach is what? 00:18:12.860 --> 00:18:15.730 Take the derivative of both sides. 00:18:15.730 --> 00:18:17.970 If the derivatives are equal, then the 00:18:17.970 --> 00:18:19.930 functions differ by a constant. 00:18:19.930 --> 00:18:22.530 And if we can show that that constant is 0, then the two 00:18:22.530 --> 00:18:24.380 functions are equal. 00:18:24.380 --> 00:18:26.990 So let's take a look and see how that works over here. 00:18:26.990 --> 00:18:31.230 First of all, let's see what the derivative of 'L of bx' is 00:18:31.230 --> 00:18:32.650 with respect to 'x'. 00:18:32.650 --> 00:18:35.430 And by the way, here again is the beauty of what we mean 00:18:35.430 --> 00:18:39.280 when we say that all of the study that we're making allows 00:18:39.280 --> 00:18:42.340 us to utilize every fundamental result that we've 00:18:42.340 --> 00:18:43.450 learned before. 00:18:43.450 --> 00:18:46.630 We've learned that the basic definition of 'L' is what? 00:18:46.630 --> 00:18:49.980 That if you differentiate 'L' with respect to the given 00:18:49.980 --> 00:18:52.960 variable, you get 1 over that variable. 00:18:52.960 --> 00:18:56.940 We want the derivative of 'L of bx' with respect to 'x'. 00:18:56.940 --> 00:19:00.140 Now, you see the idea is the derivative of 'L of u' with 00:19:00.140 --> 00:19:02.590 respect to 'u' is '1 over u'. 00:19:02.590 --> 00:19:06.330 You see by the chain rule, if we were differentiating 'L of 00:19:06.330 --> 00:19:12.050 bx' with respect to 'bx', that would be '1 over bx'. 00:19:12.050 --> 00:19:13.290 I shouldn't say by the chain rule yet. 00:19:13.290 --> 00:19:16.200 What I'm saying is if we differentiate 'L of bx' with 00:19:16.200 --> 00:19:19.680 respect to 'bx', we would have '1 over bx'. 00:19:19.680 --> 00:19:22.150 But we're not differentiating with respect to 'bx', we're 00:19:22.150 --> 00:19:24.410 differentiating with respect to 'x'. 00:19:24.410 --> 00:19:26.410 And this is where the chain rule comes in. 00:19:26.410 --> 00:19:31.660 Namely, we rewrite the derivative of 'L of bx' with 00:19:31.660 --> 00:19:35.730 respect to 'x' as the derivative 'L of bx' with 00:19:35.730 --> 00:19:39.020 respect to 'bx' times the derivative of 'bx' with 00:19:39.020 --> 00:19:40.170 respect to 'b'. 00:19:40.170 --> 00:19:41.700 And in this way we get what? 00:19:41.700 --> 00:19:43.860 The first factor is '1 over bx'. 00:19:43.860 --> 00:19:45.990 The second factor is 'b' itself-- 00:19:45.990 --> 00:19:47.310 remember 'b' is a constant-- 00:19:47.310 --> 00:19:52.200 and therefore the derivative of 'L of bx' is '1 over x'. 00:19:52.200 --> 00:19:54.860 On the other hand, what is the derivative of 'L of 00:19:54.860 --> 00:19:56.440 b' plus 'L of x'? 00:19:56.440 --> 00:20:00.460 Since 'b' is a constant, the derivative 'L of b' is 0. 00:20:00.460 --> 00:20:03.250 And by definition, the derivative of 'L of x' with 00:20:03.250 --> 00:20:05.440 respect to 'x' is '1 over x'. 00:20:05.440 --> 00:20:09.710 And now you see, comparing these two results, we see that 00:20:09.710 --> 00:20:14.430 'L of bx' and 'L of b' plus 'L of x have the same derivative, 00:20:14.430 --> 00:20:18.220 hence they differ by a constant which I'll call 'c'. 00:20:18.220 --> 00:20:21.830 In other words, notice that capital 'L' is what I call 00:20:21.830 --> 00:20:23.300 almost logarithmic. 00:20:23.300 --> 00:20:26.670 If it weren't for this factor constant 'c' in here, it would 00:20:26.670 --> 00:20:28.810 be a logarithmic function. 00:20:28.810 --> 00:20:30.820 Well at any rate, what do we have? 00:20:30.820 --> 00:20:34.730 We know that capital 'L of bx' is equal to capital 'L of b' 00:20:34.730 --> 00:20:38.450 plus capital 'L of x' plus some constant 'c'. 00:20:38.450 --> 00:20:42.190 To evaluate the constant, we only have to evaluate it at 00:20:42.190 --> 00:20:46.030 one element in the domain, let's pick 'x' equal to 1. 00:20:46.030 --> 00:20:49.100 The motif being that if 'x' is 1, notice that on the left 00:20:49.100 --> 00:20:52.070 hand side you have an 'L of b' term, on the right hand side 00:20:52.070 --> 00:20:54.590 you have an 'L of b' term, and they will cancel. 00:20:54.590 --> 00:20:58.570 In other words, if 'x' is 1, this equation becomes 'L of b' 00:20:58.570 --> 00:21:02.500 equals 'L of b' plus 'L of 1' plus 'c', from which it 00:21:02.500 --> 00:21:06.180 follows that 'c' is minus 'L of 1'. 00:21:06.180 --> 00:21:08.060 'c' is minus 'L of 1'. 00:21:08.060 --> 00:21:13.010 Now what do we want to 'c' to be if 'c' were equal to 0? 00:21:13.010 --> 00:21:17.510 If 'c' where equal to 0, this would be logarithmic, and all 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. 00:21:24.350 --> 00:21:28.105 In other words, summarizing this result, capital 'L of x' 00:21:28.105 --> 00:21:31.830 is logarithmic if capital 'L of 1' is 0. 00:21:31.830 --> 00:21:35.530 Now remember, we have a whole family of 'L's that work. 00:21:35.530 --> 00:21:39.180 What we're saying now is let's pick the member of the family 00:21:39.180 --> 00:21:42.660 of 'L's that passes through the point (1,0) . 00:21:42.660 --> 00:21:45.320 And because that's logarithmic, let's give that a 00:21:45.320 --> 00:21:46.320 special name. 00:21:46.320 --> 00:21:50.160 In essence, we will define this symbol. 00:21:50.160 --> 00:21:51.750 It's written 'ln of x'. 00:21:51.750 --> 00:21:55.200 It will later be called the natural logarithm of 'x'. 00:21:55.200 --> 00:21:58.120 I'm trying to avoid the word logarithm here as much as 00:21:58.120 --> 00:22:01.400 possible, because I want you to see that we haven't had to 00:22:01.400 --> 00:22:04.880 use exponents at all in making this kind of a definition. 00:22:04.880 --> 00:22:09.340 But let's define 'ln of x' to be the member of the family 00:22:09.340 --> 00:22:12.080 capital 'L of x' plus 'c' which passes 00:22:12.080 --> 00:22:14.320 through the point (1,0). 00:22:14.320 --> 00:22:17.950 In essence then, what is 'ln of x'? 00:22:17.950 --> 00:22:20.800 I'll call it natural log, it's easier for me to say. 00:22:20.800 --> 00:22:24.390 The natural 'log of x' is that function whose derivative with 00:22:24.390 --> 00:22:28.470 respect to 'x' is '1 over x', and characterized by 00:22:28.470 --> 00:22:30.980 the fact that what? 00:22:30.980 --> 00:22:33.620 If the input is 1, the output is 0. 00:22:33.620 --> 00:22:35.260 In other words, the graph passes through 00:22:35.260 --> 00:22:37.660 the point (1 , 0). 00:22:37.660 --> 00:22:41.650 Again, notice that in terms of what we said earlier in 00:22:41.650 --> 00:22:46.190 today's lesson, we talked about the functions 'L of x', 00:22:46.190 --> 00:22:48.880 and talked about, in the differential calculus 00:22:48.880 --> 00:22:52.290 approach, the curve could go through any 00:22:52.290 --> 00:22:53.470 point on the x-axis. 00:22:53.470 --> 00:22:56.440 What we're saying is now if the point on the x-axis that 00:22:56.440 --> 00:23:00.860 the curve crosses at is (1 , 0), that's what 'lnx' will 00:23:00.860 --> 00:23:03.140 mean, the natural log function. 00:23:03.140 --> 00:23:06.910 In terms of the integral calculus approach, if the 00:23:06.910 --> 00:23:11.010 particular 'a' value that we choose, the fixed value that 00:23:11.010 --> 00:23:14.310 we're going to study the area under, notice that all we're 00:23:14.310 --> 00:23:16.730 saying is pick 'a' to be 1. 00:23:16.730 --> 00:23:19.190 And by the way, as a quick check, all 00:23:19.190 --> 00:23:19.940 we're saying is what? 00:23:19.940 --> 00:23:23.090 The 'natural log of x' is defined to be the area under 00:23:23.090 --> 00:23:25.200 this curve as a function of 'x'. 00:23:25.200 --> 00:23:28.090 In terms of the definite integral, that's the integral 00:23:28.090 --> 00:23:30.760 from '1 to x,' 'dt over t'. 00:23:30.760 --> 00:23:33.880 And notice that if you pick 'x' to be 1, the natural log 00:23:33.880 --> 00:23:39.330 of 1 is the integral from 1 to 1 'dt over t', and that's 0, 00:23:39.330 --> 00:23:42.030 just as it should be. 00:23:42.030 --> 00:23:44.340 So at any rate now, that tells me how to 00:23:44.340 --> 00:23:46.170 define the natural log. 00:23:46.170 --> 00:23:48.260 Am I doing this in terms of exponents? 00:23:48.260 --> 00:23:48.790 No. 00:23:48.790 --> 00:23:50.270 I am trying to do what? 00:23:50.270 --> 00:23:53.830 Find a function whose derivative with respect to 'x' 00:23:53.830 --> 00:23:57.430 is '1 over x', and I'd also like that function, since it's 00:23:57.430 --> 00:24:01.010 that close to being a logarithmic function, to be a 00:24:01.010 --> 00:24:02.970 logarithmic function. 00:24:02.970 --> 00:24:07.250 In other words, this is how I invented the function 'ln of 00:24:07.250 --> 00:24:09.190 x', the 'natural log of x'. 00:24:09.190 --> 00:24:12.310 It's derivative with respect to 'x' is '1 over x', and the 00:24:12.310 --> 00:24:16.480 natural log of a product is the sum of the natural logs. 00:24:16.480 --> 00:24:20.050 And by the way, that's exactly how we use this material. 00:24:20.050 --> 00:24:22.970 Let me just take a few minutes and go over something that we 00:24:22.970 --> 00:24:25.380 already had an answer for, just to show 00:24:25.380 --> 00:24:26.680 you how this works. 00:24:26.680 --> 00:24:30.650 Let me try to rederive the product rule using logarithms. 00:24:30.650 --> 00:24:33.150 Suppose 'u' and 'v' are differentiable functions of 00:24:33.150 --> 00:24:37.360 'x', and I want to find the derivative of 'u' times 'v' 00:24:37.360 --> 00:24:38.440 with respect to 'x'. 00:24:38.440 --> 00:24:41.290 In other words, I'd like to find 'dy dx'. 00:24:41.290 --> 00:24:44.480 OK, I take the natural log of both sides. 00:24:44.480 --> 00:24:47.770 In other words, I say if 'y' equals 'u' times 'v', the 00:24:47.770 --> 00:24:53.780 natural log 'y' equals natural log 'u times v'. 00:24:53.780 --> 00:24:56.550 Now what is the property that the natural log function has? 00:24:56.550 --> 00:24:59.840 I deliberately chose it to be that member of the l family 00:24:59.840 --> 00:25:01.180 that was logarithmic. 00:25:01.180 --> 00:25:05.480 The natural log of 'u times v' is 'natural log u' plus 00:25:05.480 --> 00:25:07.590 'natural log v'. 00:25:07.590 --> 00:25:11.700 Now, can I differentiate this implicitly 00:25:11.700 --> 00:25:12.760 with respect to 'x'? 00:25:12.760 --> 00:25:15.500 You see how all of our old stuff keeps 00:25:15.500 --> 00:25:17.040 coming up in new context. 00:25:17.040 --> 00:25:20.110 I take this equation and I differentiate it implicitly 00:25:20.110 --> 00:25:21.690 with respect to 'x'. 00:25:21.690 --> 00:25:23.800 The derivative of the left hand side is '1 00:25:23.800 --> 00:25:26.160 over y', 'dy dx'. 00:25:26.160 --> 00:25:29.020 The derivative of the right hand side is what? 00:25:29.020 --> 00:25:31.380 Well, the derivative of 'log u' with respect to 'u' is '1 00:25:31.380 --> 00:25:33.260 over u', but I'm differentiating it with 00:25:33.260 --> 00:25:36.740 respect to 'x', so I must use the correction factor by the 00:25:36.740 --> 00:25:38.530 chain rule 'du dx'. 00:25:38.530 --> 00:25:41.670 And similarly, the derivative of 'log v' with respect to 'x' 00:25:41.670 --> 00:25:44.020 is '1 over v', 'dv dx'. 00:25:44.020 --> 00:25:47.790 And now multiplying through by 'y', I have obtained that 'dy 00:25:47.790 --> 00:25:53.170 dx' is ''y over u' du dx' plus ''y over v' dv dx'. 00:25:53.170 --> 00:25:56.470 But remembering that 'y' is equal to 'u' times 'v', this 00:25:56.470 --> 00:26:02.060 becomes 'v 'du dx'' plus 'u 'dv dx', and we have arrived 00:26:02.060 --> 00:26:05.870 at the familiar product rule using logarithmic 00:26:05.870 --> 00:26:07.120 differentiation. 00:26:07.120 --> 00:26:09.290 You see the point that's really important to stress 00:26:09.290 --> 00:26:12.190 here it that it wasn't important whether the natural 00:26:12.190 --> 00:26:13.950 log was an exponent or not. 00:26:13.950 --> 00:26:16.440 The important thing that we used about logarithms, whether 00:26:16.440 --> 00:26:18.700 it was our high school course, whether it's going to be in 00:26:18.700 --> 00:26:22.140 our college calculus course, the important thing was what? 00:26:22.140 --> 00:26:25.580 That when we wanted to write the log of a product, it was 00:26:25.580 --> 00:26:27.580 the sum of the logs. 00:26:27.580 --> 00:26:29.720 And now the only additional fact that we have from the 00:26:29.720 --> 00:26:32.820 calculus approach is that the derivative of 'log x' with 00:26:32.820 --> 00:26:36.800 respect to 'x' was defined to be '1 over x'. 00:26:36.800 --> 00:26:41.140 By the way, since there is a tendency to think of 00:26:41.140 --> 00:26:44.970 traditional logarithms when one uses the word logarithm, 00:26:44.970 --> 00:26:46.780 if we so wanted-- 00:26:46.780 --> 00:26:48.850 and there's no reason why we have to do this-- 00:26:48.850 --> 00:26:51.940 if we wanted to associate a base with 00:26:51.940 --> 00:26:53.540 the natural log system-- 00:26:53.540 --> 00:26:56.650 this is just a little aside, and we'll talk about this more 00:26:56.650 --> 00:26:59.720 perhaps in the learning exercises or in the 00:26:59.720 --> 00:27:01.780 supplementary notes as need be-- 00:27:01.780 --> 00:27:03.090 but the idea is this. 00:27:03.090 --> 00:27:05.970 Notice that in a traditional logarithm system, if the base 00:27:05.970 --> 00:27:09.380 is 'b', the base is characterized by the fact that 00:27:09.380 --> 00:27:12.400 the 'log of 'b to the base b'' is 1. 00:27:12.400 --> 00:27:18.100 Thus, if we use 'e' to denote the base for the natural log, 00:27:18.100 --> 00:27:21.820 whatever 'e' is, it must be characterized by the fact that 00:27:21.820 --> 00:27:24.050 the 'natural log of e' is 1. 00:27:24.050 --> 00:27:27.430 And the reason that I put the word base in quotation marks 00:27:27.430 --> 00:27:31.150 here is that I would like you to observe that again, if I 00:27:31.150 --> 00:27:34.640 have never heard of the word exponent, it still makes sense 00:27:34.640 --> 00:27:38.650 to say find the number e such that the 'natural 00:27:38.650 --> 00:27:40.220 log of e' is 1. 00:27:40.220 --> 00:27:43.290 In fact, I can give you two interpretations for that 00:27:43.290 --> 00:27:47.510 number 'e', and as a byproduct, even show you why 00:27:47.510 --> 00:27:50.430 the second fundamental theorem of integral calculus is as 00:27:50.430 --> 00:27:53.346 powerful as it really is. 00:27:53.346 --> 00:27:56.160 See, the idea is this. 00:27:56.160 --> 00:27:59.300 To find what 'e' is, all we're saying is take the curve 'y' 00:27:59.300 --> 00:28:02.620 equals 'natural log x'. 00:28:02.620 --> 00:28:05.000 Look to see where the y-coordinate is 1. 00:28:05.000 --> 00:28:08.130 In other words, draw the line 'y' equals 1. 00:28:08.130 --> 00:28:12.020 Where that line intercepts the curve 'y' equals 'log x', that 00:28:12.020 --> 00:28:13.160 x-coordinate is 'e'. 00:28:13.160 --> 00:28:16.700 In other words, the 'natural log of e' must be 1, so this 00:28:16.700 --> 00:28:19.650 is geometrically how I would locate 'e' using 00:28:19.650 --> 00:28:21.130 differential calculus. 00:28:21.130 --> 00:28:26.250 If I wanted to use integral calculus, notice that the 'log 00:28:26.250 --> 00:28:28.440 of e' by definition is what? 00:28:28.440 --> 00:28:31.960 The integral from 1 to 'e', 'dt over t'. 00:28:31.960 --> 00:28:34.750 That's the area of this region 'R'. 00:28:34.750 --> 00:28:36.600 Now what do I want 'e' to be? 00:28:36.600 --> 00:28:41.420 I want 'e' to be that number that makes this area 1. 00:28:41.420 --> 00:28:45.650 In other words, I want the 'natural log of e' to be 1. 00:28:45.650 --> 00:28:50.170 So again, in the same way that one can think of pi as being 00:28:50.170 --> 00:28:53.490 geometrically constructed, notice I can construct 'e' 00:28:53.490 --> 00:28:55.970 geometrically, namely again what? 00:28:55.970 --> 00:29:00.450 I take the curve 'y' equals '1 over t' from 't' equals 1, 00:29:00.450 --> 00:29:04.120 bounded below by the t-axis, and I keep shifting over to 00:29:04.120 --> 00:29:09.890 the right until the area of this region 'R' is exactly 1. 00:29:09.890 --> 00:29:14.410 The 't' value that makes this area 1 is called 'e'. 00:29:14.410 --> 00:29:17.880 And by the way, notice again the power of what I mean by 00:29:17.880 --> 00:29:19.200 the area approach. 00:29:19.200 --> 00:29:22.430 Notice I can start now to get estimates on what 00:29:22.430 --> 00:29:23.270 'e' must look like. 00:29:23.270 --> 00:29:25.530 Let me show you what I'm driving at over here. 00:29:25.530 --> 00:29:28.840 Let's suppose we take the curve 'y' equals '1 over t' 00:29:28.840 --> 00:29:30.760 from 1 to 2. 00:29:30.760 --> 00:29:33.480 Now what do we mean by natural log 2? 00:29:33.480 --> 00:29:38.280 Natural log 2 is by definition the area of this region here. 00:29:38.280 --> 00:29:40.980 Now notice that the smallest height of this region, since 00:29:40.980 --> 00:29:43.880 the curve is 'y' equals '1 over t', the smallest height 00:29:43.880 --> 00:29:48.080 of this region is 1/2, and the tallest height, the highest 00:29:48.080 --> 00:29:50.020 height in this region is 1. 00:29:50.020 --> 00:29:54.390 Consequently, whatever the area of this region is, it's 00:29:54.390 --> 00:29:57.280 less than the area of the inscribed rectangle-- 00:29:57.280 --> 00:30:01.460 it's greater than the area of the inscribed rectangle, and 00:30:01.460 --> 00:30:05.330 less than the area of the circumscribed rectangle. 00:30:05.330 --> 00:30:08.620 Now notice that both of these rectangles have base 1, and 00:30:08.620 --> 00:30:10.350 we're going from 1 to 2. 00:30:10.350 --> 00:30:13.490 The height of the inscribed rectangle is 1/2, therefore 00:30:13.490 --> 00:30:16.200 the area of the small rectangle is 1/2. 00:30:16.200 --> 00:30:20.930 The area of the big rectangle, it's a 1 by 1 rectangle, is 1. 00:30:20.930 --> 00:30:25.290 And now what we have is that whatever the natural log of 2 00:30:25.290 --> 00:30:29.250 is, it being the area of this region, it must be what? 00:30:29.250 --> 00:30:32.600 Less than 1 but greater than 1/2. 00:30:32.600 --> 00:30:35.020 We also know that whatever 'e' is, the 'log of 00:30:35.020 --> 00:30:37.530 e' is equal to 1. 00:30:37.530 --> 00:30:39.350 Well, let's go on a little bit further. 00:30:39.350 --> 00:30:41.250 What about the log of 4? 00:30:41.250 --> 00:30:46.410 The log of 4, natural log of 4, is the log of 2 squared. 00:30:46.410 --> 00:30:51.290 But we've already seen that one of the properties of a 00:30:51.290 --> 00:30:53.960 logarithmic function is that you can bring the exponent 00:30:53.960 --> 00:30:56.020 down as a multiplier. 00:30:56.020 --> 00:30:59.340 See, 'f of 'x to the n'' is 'n 'f of x''. 00:30:59.340 --> 00:31:03.470 This becomes 2 log 2, and because the natural log of 2 00:31:03.470 --> 00:31:06.680 is greater than 1/2, twice the natural log of 2 00:31:06.680 --> 00:31:08.890 is more than 1. 00:31:08.890 --> 00:31:13.010 In other words, if log 2 is more than 1/2, twice log 2 is 00:31:13.010 --> 00:31:14.400 more than 1. 00:31:14.400 --> 00:31:17.200 In other words, if we put these three lines together, we 00:31:17.200 --> 00:31:20.890 have that the natural log of 2 is less than the 'natural log 00:31:20.890 --> 00:31:24.570 of e', which in turn is less than the natural log of 4. 00:31:24.570 --> 00:31:28.490 By the way, notice that since the derivative of 'log x' with 00:31:28.490 --> 00:31:32.660 respect to 'x' is '1 over x', and that's positive, 'log x' 00:31:32.660 --> 00:31:34.580 is a one to one function. 00:31:34.580 --> 00:31:38.250 Therefore, if the log of 2 is less than the 'log of e' is 00:31:38.250 --> 00:31:42.080 less than the log of 4, it follows that 2 must be less 00:31:42.080 --> 00:31:45.230 than 'e', which in turn must be less than 4. 00:31:45.230 --> 00:31:48.450 In other words, even with this crude approximation of just 00:31:48.450 --> 00:31:52.260 inscribing and circumscribing rectangles, I can show again 00:31:52.260 --> 00:31:56.410 without reference to exponents that whatever the number 'e' 00:31:56.410 --> 00:32:02.330 is, the number e defined by the fact that 'ln of e' is 1, 00:32:02.330 --> 00:32:06.400 that number 'e' is some number between 2 and 4. 00:32:06.400 --> 00:32:08.830 But again, we won't dwell on this too long. 00:32:08.830 --> 00:32:12.000 What I want to do now is to come back to a summary point, 00:32:12.000 --> 00:32:14.570 and at the same time, lead into the 00:32:14.570 --> 00:32:16.020 lecture for next time. 00:32:16.020 --> 00:32:20.790 Recall that we began this lecture with the problem of 00:32:20.790 --> 00:32:24.740 solving the situation where the rate of change was 00:32:24.740 --> 00:32:27.390 proportional to the amount present. 00:32:27.390 --> 00:32:31.180 Given 'dm/ dt' equals 'km', we separated variables and got 00:32:31.180 --> 00:32:34.920 down to the stage where the integral of 'dm over m' was 00:32:34.920 --> 00:32:36.880 'kt' plus a constant. 00:32:36.880 --> 00:32:40.540 And all we did in the rest of today's lecture that was at 00:32:40.540 --> 00:32:45.350 all different, was we invented and constructed the particular 00:32:45.350 --> 00:32:48.970 function whose derivative would be '1 over m', and which 00:32:48.970 --> 00:32:50.550 had the logarithmic property. 00:32:50.550 --> 00:32:52.920 In other words, we can now say that this is the 00:32:52.920 --> 00:32:54.340 answer to the problem. 00:32:54.340 --> 00:32:58.240 This problem is explicitly solved because the natural log 00:32:58.240 --> 00:33:00.690 function can be viewed as an area under a curve. 00:33:00.690 --> 00:33:02.950 We can construct it for each 'm'. 00:33:02.950 --> 00:33:05.400 This is then what we managed to do. 00:33:05.400 --> 00:33:08.800 And by the way, all I'm saying now is that since the log is a 00:33:08.800 --> 00:33:10.250 one to one function-- 00:33:10.250 --> 00:33:12.250 see, its derivative is always positive-- 00:33:12.250 --> 00:33:14.670 can't we talk about the inverse function? 00:33:14.670 --> 00:33:16.880 In other words, notice that another way of writing this is 00:33:16.880 --> 00:33:21.720 that id 'natural log m' is 'kt' plus 'c', 'm' itself must 00:33:21.720 --> 00:33:26.490 be the 'inverse log of kt' plus 'c'. 00:33:26.490 --> 00:33:30.450 And you see, next time what we're going to do is to 00:33:30.450 --> 00:33:35.610 explore what we mean by the inverse log. 00:33:35.610 --> 00:33:39.300 At any rate, in summarizing today's lecture, to make sure 00:33:39.300 --> 00:33:43.250 that we didn't fall victim to all the computational details, 00:33:43.250 --> 00:33:46.500 notice that all we did was physically motivated the 00:33:46.500 --> 00:33:50.070 necessity for inventing a function whose derivative with 00:33:50.070 --> 00:33:51.920 respect to 'x' was '1 over x'. 00:33:51.920 --> 00:33:55.900 And from that point on, everything else that we did 00:33:55.900 --> 00:33:59.730 followed as applications of material that came before. 00:33:59.730 --> 00:34:03.980 It's in this sense that the course now picks up in tempo. 00:34:03.980 --> 00:34:07.050 That as we go on now, we're going to be able to cover 00:34:07.050 --> 00:34:10.679 larger globs of material in one sitting, because we will 00:34:10.679 --> 00:34:14.130 find, at least for the next several lectures, that every 00:34:14.130 --> 00:34:18.989 new topic is basically one new idea together with all of the 00:34:18.989 --> 00:34:20.330 old recipes. 00:34:20.330 --> 00:34:22.330 At any rate, until next time, goodbye. 00:34:25.489 --> 00:34:28.030 ANNOUNCER: Funding for the publication of this video was 00:34:28.030 --> 00:34:32.750 provided by the Gabriella and Paul Rosenbaum Foundation. 00:34:32.750 --> 00:34:36.920 Help OCW continue to provide free and open access to MIT 00:34:36.920 --> 00:34:41.120 courses by making a donation at ocw.mit.edu/donate.