WEBVTT 00:00:00.000 --> 00:00:08.980 CHRISTINE BREINER: Welcome back to recitation. 00:00:08.980 --> 00:00:11.350 Today we're going to talk about some rules of logarithms 00:00:11.350 --> 00:00:12.950 that you're going to need to remember. 00:00:12.950 --> 00:00:15.880 We're going to prove why one of them is true, and then I'm 00:00:15.880 --> 00:00:18.760 going to ask you to use these rules to take a derivative of 00:00:18.760 --> 00:00:20.220 a function. 00:00:20.220 --> 00:00:24.130 So let's just look at these rules first. So I want to 00:00:24.130 --> 00:00:26.380 point out as I'm talking about these rules, the first three 00:00:26.380 --> 00:00:28.410 are written with natural log. 00:00:28.410 --> 00:00:32.640 But one can also write them in any base as long as the base 00:00:32.640 --> 00:00:34.520 is the same all the way across. 00:00:34.520 --> 00:00:37.380 So in any legitimate base that one is allowed to use, so with 00:00:37.380 --> 00:00:41.470 a positive base, one can use it all the way across instead 00:00:41.470 --> 00:00:43.100 of the natural log. 00:00:43.100 --> 00:00:46.310 So the first one says that the natural log of a product is 00:00:46.310 --> 00:00:48.800 equal to the sum of the natural logs. 00:00:48.800 --> 00:00:51.570 So natural log M times N is equal to the natural log of M 00:00:51.570 --> 00:00:53.755 plus natural log of N. 00:00:53.755 --> 00:00:57.080 The second one says the natural log of a quotient is 00:00:57.080 --> 00:00:59.770 equal to the difference of the natural logs. 00:00:59.770 --> 00:01:03.760 So natural log of M divided by N is equal to natural log of M 00:01:03.760 --> 00:01:06.130 minus natural log of N. 00:01:06.130 --> 00:01:09.530 This third one says that the natural log of something 00:01:09.530 --> 00:01:13.700 raised to a power is that power as a coefficient times 00:01:13.700 --> 00:01:14.990 the natural log of the something. 00:01:14.990 --> 00:01:18.460 So natural log of M to the k is equal to k times the 00:01:18.460 --> 00:01:20.240 natural log of M. 00:01:20.240 --> 00:01:22.270 And what I want to point out is that there's a distinct 00:01:22.270 --> 00:01:24.140 difference where the power is. 00:01:24.140 --> 00:01:27.840 So if the power is inside the argument then this rule holds, 00:01:27.840 --> 00:01:30.160 but if the power is outside the argument-- 00:01:30.160 --> 00:01:31.470 so if it's natural log of M, the whole 00:01:31.470 --> 00:01:33.130 thing raised to a power-- 00:01:33.130 --> 00:01:34.910 this does not work. 00:01:34.910 --> 00:01:38.110 This is not equal to what's written above. 00:01:38.110 --> 00:01:39.880 And then the third-- the fourth one-- sorry. 00:01:39.880 --> 00:01:42.310 The fourth one is a change of base formula. 00:01:42.310 --> 00:01:47.140 So if I have, if I have log base something b, that maybe I 00:01:47.140 --> 00:01:51.430 want to change the base of M, I can rewrite 00:01:51.430 --> 00:01:53.820 that in the base e. 00:01:53.820 --> 00:01:56.450 I can write that as natural log of M divided by 00:01:56.450 --> 00:01:58.500 natural log of b. 00:01:58.500 --> 00:02:00.390 And I want to point out a common mistake people make is 00:02:00.390 --> 00:02:02.450 sometimes they confuse the second and the fourth because 00:02:02.450 --> 00:02:04.240 they both have quotients. 00:02:04.240 --> 00:02:07.290 But notice that the second one is the natural log of a 00:02:07.290 --> 00:02:09.940 quotient, and the fourth one is about the quotient of 00:02:09.940 --> 00:02:11.330 natural logs. 00:02:11.330 --> 00:02:13.910 So that's a distinct difference, and hopefully then 00:02:13.910 --> 00:02:17.480 you see that they are not, these two statements are not, 00:02:17.480 --> 00:02:19.770 in fact, the same statement. 00:02:19.770 --> 00:02:22.250 So now what I'd like to do is, using what we know about 00:02:22.250 --> 00:02:24.030 exponential and log functions-- 00:02:24.030 --> 00:02:26.380 I want to prove number one. 00:02:26.380 --> 00:02:28.630 So let's set out to do that. 00:02:28.630 --> 00:02:32.250 Well, in order to make this top line make sense we know 00:02:32.250 --> 00:02:34.550 that M and N have to be positive. 00:02:34.550 --> 00:02:36.100 And so I can find-- 00:02:36.100 --> 00:02:37.680 actually, let me write first what we're doing. 00:02:37.680 --> 00:02:40.940 We're going to prove one. 00:02:40.940 --> 00:02:45.000 So with M and N both positive I can find values a and b such 00:02:45.000 --> 00:02:50.860 that e to the a equals M and e to the b is equal to N. And 00:02:50.860 --> 00:02:53.320 let me just write out also what that means, because 00:02:53.320 --> 00:02:56.380 exponential and log functions are inverses of one another. 00:02:56.380 --> 00:03:01.510 This means that a is equal to natural log of M and b is 00:03:01.510 --> 00:03:04.570 equal to natural log of N. So these are equivalent 00:03:04.570 --> 00:03:06.350 statements. 00:03:06.350 --> 00:03:08.400 This statement and this statement are equivalent. 00:03:08.400 --> 00:03:10.970 This statement and this statement are equivalent. 00:03:10.970 --> 00:03:13.300 So now let's use that information to try and solve 00:03:13.300 --> 00:03:14.860 the problem. 00:03:14.860 --> 00:03:17.360 To try and prove number one. 00:03:17.360 --> 00:03:21.440 So the natural a log of M times N, well, what is that? 00:03:21.440 --> 00:03:24.320 M is e to the a, N is e to the b. 00:03:24.320 --> 00:03:29.150 So I can write this as natural log of e to the a 00:03:29.150 --> 00:03:31.490 times e to the b. 00:03:31.490 --> 00:03:33.120 What's e to the a times e to the b? 00:03:33.120 --> 00:03:35.770 This is where we use our rules of exponents. 00:03:35.770 --> 00:03:40.270 e to the a times e to the b is e to the a plus b. 00:03:40.270 --> 00:03:43.700 So this is natural log of e to the a plus b. 00:03:43.700 --> 00:03:45.150 And now, what's the point? 00:03:45.150 --> 00:03:48.040 The point is that natural log in exponential functions are 00:03:48.040 --> 00:03:50.450 inverses of one another, or natural log of 00:03:50.450 --> 00:03:52.450 e to the x is x. 00:03:52.450 --> 00:03:58.040 So natural log of e to the a plus b is just a plus b. 00:03:58.040 --> 00:04:00.490 And I've already recorded for you what those are-- 00:04:00.490 --> 00:04:05.250 it's natural log of M plus natural log of N. So notice 00:04:05.250 --> 00:04:07.230 we've done when we set out to do-- 00:04:07.230 --> 00:04:10.650 natural log of the quantity M times N is equal to natural 00:04:10.650 --> 00:04:14.260 log of M plus natural log of N. 00:04:14.260 --> 00:04:16.750 And in a similar flavor one could immediately do number 00:04:16.750 --> 00:04:22.400 two, and number three follows quite similarly, as well. 00:04:22.400 --> 00:04:25.030 It uses, obviously, these are going to use different rules 00:04:25.030 --> 00:04:28.740 for exponents besides the product of two exponential 00:04:28.740 --> 00:04:32.200 functions is equal to the sum of the powers. 00:04:32.200 --> 00:04:34.300 It's going to use some of those other rules. 00:04:34.300 --> 00:04:36.220 And I believe that some of these other things might 00:04:36.220 --> 00:04:39.620 actually also be proven in a later lecture 00:04:39.620 --> 00:04:41.820 in the actual course. 00:04:41.820 --> 00:04:42.870 So you'll see these. 00:04:42.870 --> 00:04:45.320 But I would say, you might want to try and prove two and 00:04:45.320 --> 00:04:48.280 three, at least, on your own-- might be helpful to look at 00:04:48.280 --> 00:04:51.350 how those work using the same kind of rules here. 00:04:51.350 --> 00:04:55.380 So now what I'd like us to do is using these rules, I'd like 00:04:55.380 --> 00:04:57.150 us to take a derivative. 00:04:57.150 --> 00:05:04.040 So what I want us to look at is y equals the square root of 00:05:04.040 --> 00:05:07.440 x times x plus 4. 00:05:07.440 --> 00:05:10.580 And we'll just assume that x is bigger than 0. 00:05:10.580 --> 00:05:15.110 And I want you to find y prime. 00:05:15.110 --> 00:05:18.090 Now you could do this just brute force, cranking it out. 00:05:18.090 --> 00:05:20.560 But I'd like you to try and use the log differentiation 00:05:20.560 --> 00:05:22.690 technique in order to find this derivative. 00:05:22.690 --> 00:05:25.210 I'll give you a moment to do it and then I'll come back and 00:05:25.210 --> 00:05:26.460 I'll show you how I do it. 00:05:34.840 --> 00:05:35.100 OK. 00:05:35.100 --> 00:05:36.130 Welcome back. 00:05:36.130 --> 00:05:39.150 So I'm going to use the log differentiation and the rules 00:05:39.150 --> 00:05:42.100 I have on the side of the board there to take a 00:05:42.100 --> 00:05:44.150 derivative to find y prime. 00:05:44.150 --> 00:05:47.370 So first what we do is we take the log of both sides and then 00:05:47.370 --> 00:05:49.680 we use some of the rules of logarithms to simplify the 00:05:49.680 --> 00:05:51.660 expression on the right hand side. 00:05:51.660 --> 00:05:58.070 So I will take natural log y is equal to natural log of the 00:05:58.070 --> 00:06:03.108 square root of x times x plus 4. 00:06:03.108 --> 00:06:05.260 Now square root-- 00:06:05.260 --> 00:06:06.820 wow, sorry-- 00:06:06.820 --> 00:06:09.695 square root is the power of something raised to the 1/2. 00:06:09.695 --> 00:06:10.510 Right? 00:06:10.510 --> 00:06:11.940 That's what it means to take a square root. 00:06:11.940 --> 00:06:14.500 You can take this whole product and 00:06:14.500 --> 00:06:15.710 raise it to the 1/2. 00:06:15.710 --> 00:06:18.700 So I'm going to use rule number three and I'm going to 00:06:18.700 --> 00:06:21.990 bring that 1/2 that is a power out in front of the log. 00:06:21.990 --> 00:06:23.740 So I can rewrite this expression as 00:06:23.740 --> 00:06:28.190 1/2 log of this product. 00:06:28.190 --> 00:06:30.220 That's one too many parentheses, but that's OK. 00:06:33.920 --> 00:06:34.020 OK. 00:06:34.020 --> 00:06:38.260 So I have 1/2 the natural log of the product x and x plus 4. 00:06:38.260 --> 00:06:41.170 So now I'm going to use rule number one which changes the 00:06:41.170 --> 00:06:43.290 product, the natural log of a product into the sum of the 00:06:43.290 --> 00:06:44.570 natural logs. 00:06:44.570 --> 00:06:51.630 And I can rewrite this as 1/2 natural log x plus 1/2 natural 00:06:51.630 --> 00:06:54.330 log, its quantity x plus 4. 00:06:54.330 --> 00:06:57.180 Essentially what I'm doing here is I have to distribute 00:06:57.180 --> 00:06:59.940 this 1/2 because I had one term, and then I'm going to 00:06:59.940 --> 00:07:02.540 have two terms that are added together, but the 1/2 applies 00:07:02.540 --> 00:07:04.210 to both of them. 00:07:04.210 --> 00:07:05.860 So now I have this nice setup. 00:07:05.860 --> 00:07:07.540 I have natural log of y is equal to 00:07:07.540 --> 00:07:08.990 something in terms of x. 00:07:08.990 --> 00:07:11.540 And now I can take the derivative of a both sides. 00:07:11.540 --> 00:07:14.520 Now remember, I want to find y prime, so there's some 00:07:14.520 --> 00:07:16.990 implicit differentiation going on. 00:07:16.990 --> 00:07:19.250 So let's just be careful when we do that. 00:07:19.250 --> 00:07:21.760 If I take the derivative of this side I don't just get y 00:07:21.760 --> 00:07:24.300 prime, I get y prime over y. 00:07:24.300 --> 00:07:25.710 Where does that come from? 00:07:25.710 --> 00:07:29.460 Well, d dx of this expression is the derivative of the 00:07:29.460 --> 00:07:32.460 natural log evaluated at y then times the 00:07:32.460 --> 00:07:33.310 derivative of y. 00:07:33.310 --> 00:07:35.490 You've seen this, I think, a lot by now, but just to make 00:07:35.490 --> 00:07:38.500 sure you understand where both of those come from. 00:07:38.500 --> 00:07:41.290 So when I take the derivative here I get y prime over y. 00:07:41.290 --> 00:07:43.750 When I take the derivative here with respect to x, well, 00:07:43.750 --> 00:07:46.780 derivative of natural log of x is just 1 over x. 00:07:46.780 --> 00:07:49.510 So I get 1 over 2x. 00:07:49.510 --> 00:07:52.910 And then the derivative of natural log of x plus 4, if I 00:07:52.910 --> 00:07:55.750 use the chain rule I get 1 over x plus 4 times the 00:07:55.750 --> 00:07:58.220 derivative of x plus 4, which is still just 1-- 00:07:58.220 --> 00:08:03.790 so I get 1 over 2 times x plus 4. 00:08:03.790 --> 00:08:07.050 So now I wanted us to find y prime. 00:08:07.050 --> 00:08:09.310 So to find y prime I'm going to move over 00:08:09.310 --> 00:08:11.410 a little bit more. 00:08:11.410 --> 00:08:15.240 And just notice that y prime is going to equal y 00:08:15.240 --> 00:08:16.070 times all of that. 00:08:16.070 --> 00:08:17.030 Well, I know y. 00:08:17.030 --> 00:08:18.940 So I'm going to write what y is. 00:08:18.940 --> 00:08:22.570 y is the square root of x times x plus 00:08:22.570 --> 00:08:26.450 4 times this quantity. 00:08:26.450 --> 00:08:32.680 1 over 2x plus 1 over 2 times x plus 4. 00:08:35.550 --> 00:08:39.060 So that's actually one way to write the derivative of y 00:08:39.060 --> 00:08:39.900 prime now-- 00:08:39.900 --> 00:08:42.000 or sorry-- the derivative of y. 00:08:42.000 --> 00:08:45.220 Now I could combine these two fractions into a single 00:08:45.220 --> 00:08:48.640 fraction and try and make it look a little bit nicer, or I 00:08:48.640 --> 00:08:49.530 could just leave it this way. 00:08:49.530 --> 00:08:51.420 This is technically a derivative. 00:08:51.420 --> 00:08:54.200 So if I started trying to combine things I might find 00:08:54.200 --> 00:08:55.700 out that I could have just taken the 00:08:55.700 --> 00:08:57.460 derivative the long way. 00:08:57.460 --> 00:09:01.170 So this is a nice short way to just get to a place where I 00:09:01.170 --> 00:09:03.610 can start to find out something about the 00:09:03.610 --> 00:09:05.730 derivative of y. 00:09:05.730 --> 00:09:07.930 So I guess I'll stop there.