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 --> 00:00:15.450
We're going to prove
why one of them is true,

00:00:15.450 --> 00:00:18.310
and then I'm going to ask you
to use these rules to take

00:00:18.310 --> 00:00:20.220
a derivative of a function.

00:00:20.220 --> 00:00:23.430
So let's just look
at these rules first.

00:00:23.430 --> 00:00:25.820
So I want to point out, as
I'm talking about these rules,

00:00:25.820 --> 00:00:28.410
the first three are
written with natural log.

00:00:28.410 --> 00:00:30.800
But one can also
write them in any base

00:00:30.800 --> 00:00:34.520
as long as the base is the
same all the way across.

00:00:34.520 --> 00:00:36.400
So in any legitimate
base that one

00:00:36.400 --> 00:00:38.890
is allowed to use, so
with a positive base,

00:00:38.890 --> 00:00:41.470
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 --> 00:00:46.130
So the first one says that
the natural log of a product

00:00:46.130 --> 00:00:48.769
is equal to the sum
of the natural logs.

00:00:48.769 --> 00:00:51.060
So natural log of M times N
is equal to the natural log

00:00:51.060 --> 00:00:53.755
of M plus natural log of N.

00:00:53.755 --> 00:00:56.940
The second one says the
natural log of a quotient

00:00:56.940 --> 00:00:59.770
is equal to the difference
of the natural logs.

00:00:59.770 --> 00:01:02.200
So natural log of
M divided by N is

00:01:02.200 --> 00:01:06.130
equal to natural log of
M minus natural log of N.

00:01:06.130 --> 00:01:09.760
This third one says that the
natural log of something raised

00:01:09.760 --> 00:01:13.366
to a power is that
power, as a coefficient,

00:01:13.366 --> 00:01:14.990
times the natural
log of the something.

00:01:14.990 --> 00:01:18.350
So natural log of M to
the k is equal to k times

00:01:18.350 --> 00:01:19.977
the natural log of M.

00:01:19.977 --> 00:01:21.810
And what I want to point
out is that there's

00:01:21.810 --> 00:01:24.140
a distinct 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 --> 00:01:33.130
so if it's natural log of M, the
whole thing raised to a power--

00:01:33.130 --> 00:01:34.910
this does not work.

00:01:34.910 --> 00:01:38.089
This is not equal to
what's written above.

00:01:38.089 --> 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 --> 00:01:46.160
So if I have, if I have
log base something,

00:01:46.160 --> 00:01:50.760
b-- maybe I want to
change the base-- of M,

00:01:50.760 --> 00:01:53.820
I can rewrite that
in the base e.

00:01:53.820 --> 00:01:56.020
I can write that
as natural log of M

00:01:56.020 --> 00:01:58.071
divided by natural log of b.

00:01:58.071 --> 00:02:00.070
And I want to point out,
a common mistake people

00:02:00.070 --> 00:02:02.403
make is sometimes they confuse
the second and the fourth

00:02:02.403 --> 00:02:04.240
because they both
have quotients.

00:02:04.240 --> 00:02:06.320
But notice that
the second one is

00:02:06.320 --> 00:02:08.890
the natural log of a
quotient, and the fourth one

00:02:08.890 --> 00:02:11.330
is about the quotient
of natural logs.

00:02:11.330 --> 00:02:14.030
So that's a distinct difference,
and hopefully then you

00:02:14.030 --> 00:02:17.480
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 --> 00:02:21.670
So now what I'd like
to do is using what

00:02:21.670 --> 00:02:24.030
we know about 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 --> 00:02:31.970
Well, in order to make
this top line make sense

00:02:31.970 --> 00:02:34.550
we know that M and N
have to be positive.

00:02:34.550 --> 00:02:36.347
And so I can find--
actually, let

00:02:36.347 --> 00:02:37.680
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 --> 00:02:50.422
that e to the a equals M and
e to the b is equal to N.

00:02:50.422 --> 00:02:52.969
And let me just write
out also what that means.

00:02:52.969 --> 00:02:54.510
Because exponential
and log functions

00:02:54.510 --> 00:02:56.590
are inverses of
one another, this

00:02:56.590 --> 00:03:00.480
means that a is equal
to natural log of M

00:03:00.480 --> 00:03:03.520
and b is equal to
natural log of N.

00:03:03.520 --> 00:03:06.350
So these are
equivalent 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 --> 00:03:12.640
So now let's use
that information

00:03:12.640 --> 00:03:14.860
to try and solve 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 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 --> 00:03:29.560
So I can write this as natural
log of e to the a times e

00:03:29.560 --> 00:03:31.490
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 --> 00:03:47.890
The point is that natural
log in exponential functions

00:03:47.890 --> 00:03:52.450
are inverses of one another, or
natural log of e to the x is x.

00:03:52.450 --> 00:03:58.006
So natural log of e to the
a plus b is just a plus b.

00:03:58.006 --> 00:03:59.380
And I've already
recorded for you

00:03:59.380 --> 00:04:04.720
what those are-- it's natural
log of M plus natural log of N.

00:04:04.720 --> 00:04:07.230
So notice we've done
what we set out to do.

00:04:07.230 --> 00:04:09.780
Natural log of the
quantity M times N

00:04:09.780 --> 00:04:14.260
is equal to natural log of
M plus natural log of N.

00:04:14.260 --> 00:04:16.320
And in a similar flavor
one could immediately

00:04:16.320 --> 00:04:21.350
do number two, and number
three follows quite similarly,

00:04:21.350 --> 00:04:22.400
as well.

00:04:22.400 --> 00:04:23.900
It uses-- obviously,
these are going

00:04:23.900 --> 00:04:25.630
to use different
rules for exponents

00:04:25.630 --> 00:04:29.170
besides the product of
two exponential functions

00:04:29.170 --> 00:04:32.200
is equal to the
sum of the powers.

00:04:32.200 --> 00:04:34.155
It's going to use some
of those other rules.

00:04:34.155 --> 00:04:36.030
And I believe that some
of these other things

00:04:36.030 --> 00:04:39.620
might 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 --> 00:04:45.160
But I would say, you might
want to try and prove two

00:04:45.160 --> 00:04:46.900
and three, at
least, on your own--

00:04:46.900 --> 00:04:49.310
might be helpful to look
at how those work using

00:04:49.310 --> 00:04:51.350
the same kind of rules here.

00:04:51.350 --> 00:04:54.890
So now what I'd like us to
do is using these rules,

00:04:54.890 --> 00:04:57.150
I'd like us to
take a derivative.

00:04:57.150 --> 00:05:00.330
So what I want us
to look at is y

00:05:00.330 --> 00:05:07.440
equals the square root
of 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 --> 00:05:22.669
technique in order to
find this derivative.

00:05:22.669 --> 00:05:24.960
I'll give you a moment to do
it and then I'll come back

00:05:24.960 --> 00:05:26.210
and I'll show you how I do it.

00:05:34.600 --> 00:05:35.100
OK.

00:05:35.100 --> 00:05:36.130
Welcome back.

00:05:36.130 --> 00:05:39.220
So I'm going to use the log
differentiation and the rules I

00:05:39.220 --> 00:05:41.370
have on the side
of the board there

00:05:41.370 --> 00:05:44.150
to take a derivative
to find y prime.

00:05:44.150 --> 00:05:47.022
So first what we do is we
take the log of both sides

00:05:47.022 --> 00:05:48.980
and then we use some of
the rules of logarithms

00:05:48.980 --> 00:05:51.660
to simplify the expression
on the right-hand side.

00:05:51.660 --> 00:05:57.690
So I will take natural log
y is equal to natural log

00:05:57.690 --> 00:06:03.108
of the square root
of x times x plus 4.

00:06:03.108 --> 00:06:07.862
Now square root-- wow,
sorry-- square root

00:06:07.862 --> 00:06:09.695
is the power of something
raised to the 1/2.

00:06:09.695 --> 00:06:10.510
Right?

00:06:10.510 --> 00:06:12.301
That's what it means
to take a square root.

00:06:12.301 --> 00:06:15.710
You can take this whole product
and raise it to the 1/2.

00:06:15.710 --> 00:06:18.090
So I'm going to use
rule number three

00:06:18.090 --> 00:06:21.220
and I'm going to bring that 1/2
that is a power out in front

00:06:21.220 --> 00:06:21.990
of the log.

00:06:21.990 --> 00:06:28.190
So I can rewrite this expression
as 1/2 log of this product.

00:06:28.190 --> 00:06:30.220
That's one too many
parentheses, but that's OK.

00:06:33.520 --> 00:06:34.020
OK.

00:06:34.020 --> 00:06:38.135
So I have 1/2 the natural log of
the product of x and x plus 4.

00:06:38.135 --> 00:06:40.010
So now I'm going to use
rule number one which

00:06:40.010 --> 00:06:42.380
changes the product-- the
natural log of a product

00:06:42.380 --> 00:06:44.570
into the sum of
the natural logs.

00:06:44.570 --> 00:06:48.490
And I can rewrite this
as 1/2 natural log

00:06:48.490 --> 00:06:54.330
x plus 1/2 natural log
the quantity x plus 4.

00:06:54.330 --> 00:06:56.370
Essentially what
I'm doing here is

00:06:56.370 --> 00:06:59.059
I have to distribute this
1/2 because I had one term,

00:06:59.059 --> 00:07:01.600
and then I'm going to have two
terms that are added together,

00:07:01.600 --> 00:07:04.210
but the 1/2 applies
to both of them.

00:07:04.210 --> 00:07:05.860
So now I have this nice setup.

00:07:05.860 --> 00:07:08.990
I have natural log of y is equal
to 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 --> 00:07:14.070
Now remember, I want
to find y prime,

00:07:14.070 --> 00:07:16.990
so there's some 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 --> 00:07:22.240
If I take the derivative of this
side I don't just get y prime,

00:07:22.240 --> 00:07:24.300
I get y prime over y.

00:07:24.300 --> 00:07:25.710
Where does that come from?

00:07:25.710 --> 00:07:29.280
Well, d/dx of this
expression is the derivative

00:07:29.280 --> 00:07:31.770
of the natural log
evaluated at y,

00:07:31.770 --> 00:07:33.204
then times the derivative of y.

00:07:33.204 --> 00:07:34.870
You've seen this, I
think, a lot by now,

00:07:34.870 --> 00:07:37.470
but just to make sure you
understand where both of those

00:07:37.470 --> 00:07:38.500
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 --> 00:07:52.680
And then the derivative of
natural log of x plus 4,

00:07:52.680 --> 00:07:54.705
if I use the chain
rule I get 1 over x

00:07:54.705 --> 00:07:58.220
plus 4 times the 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 --> 00:08:11.410
So to find y prime I'm going
to move over a little bit more.

00:08:11.410 --> 00:08:15.490
And just notice that y prime
is going to equal y times

00:08:15.490 --> 00:08:16.070
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 --> 00:08:26.450
y is the square root of x times
x plus 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 --> 00:08:37.860
So that's actually
one way to write

00:08:37.860 --> 00:08:42.000
the derivative of y prime now--
or sorry, the derivative of y.

00:08:42.000 --> 00:08:44.650
Now I could combine
these two fractions

00:08:44.650 --> 00:08:46.720
into a single fraction
and try and make

00:08:46.720 --> 00:08:49.530
it look a little bit nicer, or
I could just leave it this way.

00:08:49.530 --> 00:08:51.420
This is technically
a derivative.

00:08:51.420 --> 00:08:53.600
So if I started trying
to combine things

00:08:53.600 --> 00:08:55.610
I might find out that
I could have just taken

00:08:55.610 --> 00:08:57.460
the derivative the long way.

00:08:57.460 --> 00:09:00.795
So this is a nice short
way to just get to a place

00:09:00.795 --> 00:09:02.420
where I can start to
find out something

00:09:02.420 --> 00:09:05.730
about the derivative of y.

00:09:05.730 --> 00:09:07.930
So I guess I'll stop there.