[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.76,0:00:04.95,Default,,0000,0000,0000,,In this video, we're going to be\Nlooking at logarithms, but
Dialogue: 0,0:00:04.95,0:00:09.14,Default,,0000,0000,0000,,before we can deal with\Nlogarithms, what we need to have
Dialogue: 0,0:00:09.14,0:00:11.43,Default,,0000,0000,0000,,a look at our indices 16.
Dialogue: 0,0:00:12.46,0:00:19.16,Default,,0000,0000,0000,,We know that we can write 16 as\Na power of two, so this is 2.
Dialogue: 0,0:00:19.80,0:00:22.66,Default,,0000,0000,0000,,Raised to the power 4.
Dialogue: 0,0:00:23.24,0:00:27.35,Default,,0000,0000,0000,,Now need to give some names to\Nsome of these numbers.
Dialogue: 0,0:00:28.66,0:00:31.13,Default,,0000,0000,0000,,So this number here 4.
Dialogue: 0,0:00:32.81,0:00:37.24,Default,,0000,0000,0000,,Gets called a number of\Ndifferent things. I just called
Dialogue: 0,0:00:37.24,0:00:38.57,Default,,0000,0000,0000,,it a power.
Dialogue: 0,0:00:40.01,0:00:43.19,Default,,0000,0000,0000,,Sometimes it gets called.
Dialogue: 0,0:00:43.94,0:00:45.20,Default,,0000,0000,0000,,Exponent
Dialogue: 0,0:00:46.29,0:00:52.79,Default,,0000,0000,0000,,Sometimes. Index.\NSo that's the four. What about
Dialogue: 0,0:00:52.79,0:00:56.82,Default,,0000,0000,0000,,this number? This number doesn't\Nusually get a name, but it does
Dialogue: 0,0:00:56.82,0:01:02.86,Default,,0000,0000,0000,,have one. It's referred to\Nas the base.
Dialogue: 0,0:01:03.63,0:01:06.97,Default,,0000,0000,0000,,So another example might be
Dialogue: 0,0:01:06.97,0:01:12.86,Default,,0000,0000,0000,,64. 64 is 8 to\Nthe power 2.
Dialogue: 0,0:01:13.89,0:01:20.62,Default,,0000,0000,0000,,So in this case two is\Nthe power, the exponent or
Dialogue: 0,0:01:20.62,0:01:25.52,Default,,0000,0000,0000,,the index, and here 8 is\Nthe base.
Dialogue: 0,0:01:26.83,0:01:33.19,Default,,0000,0000,0000,,So having got the language clear\Nfor indices or powers, what
Dialogue: 0,0:01:33.19,0:01:35.60,Default,,0000,0000,0000,,about logarithms? Well.
Dialogue: 0,0:01:36.22,0:01:41.84,Default,,0000,0000,0000,,Let's see what the motivation\Nmight be for wanting to use
Dialogue: 0,0:01:41.84,0:01:42.35,Default,,0000,0000,0000,,logarithms.
Dialogue: 0,0:01:43.08,0:01:49.49,Default,,0000,0000,0000,,No.\NAs we've just
Dialogue: 0,0:01:49.49,0:01:53.59,Default,,0000,0000,0000,,seen, 16 is 2\Nraised to the
Dialogue: 0,0:01:53.59,0:01:55.35,Default,,0000,0000,0000,,power of 4.
Dialogue: 0,0:01:56.95,0:02:03.88,Default,,0000,0000,0000,,8. Similarly, is 2 raised\Nto the power of 3?
Dialogue: 0,0:02:05.36,0:02:12.34,Default,,0000,0000,0000,,Now, supposing we want it to\Nmultiply 16 by 8, well, one way
Dialogue: 0,0:02:12.34,0:02:15.56,Default,,0000,0000,0000,,of doing it would be long
Dialogue: 0,0:02:15.56,0:02:20.64,Default,,0000,0000,0000,,multiplication. We can do that\Nand the answer is 128.
Dialogue: 0,0:02:21.97,0:02:26.01,Default,,0000,0000,0000,,That is another way rather than\Nhaving to go through this
Dialogue: 0,0:02:26.01,0:02:29.68,Default,,0000,0000,0000,,multiplication. Some which if\Nthe numbers were other than 16
Dialogue: 0,0:02:29.68,0:02:34.82,Default,,0000,0000,0000,,and eight would be very long.\NCan we not make use of the way
Dialogue: 0,0:02:34.82,0:02:35.92,Default,,0000,0000,0000,,we've expressed these?
Dialogue: 0,0:02:37.00,0:02:40.38,Default,,0000,0000,0000,,16 times by 8.
Dialogue: 0,0:02:41.11,0:02:47.61,Default,,0000,0000,0000,,Is 2 to the power four times by\N2 to the power 3.
Dialogue: 0,0:02:48.11,0:02:53.31,Default,,0000,0000,0000,,From laws of indices, we know\Nthat when we do this kind of
Dialogue: 0,0:02:53.31,0:02:57.71,Default,,0000,0000,0000,,calculation, what we do is we\Nsimply add the indices together.
Dialogue: 0,0:02:58.96,0:03:02.95,Default,,0000,0000,0000,,So what was a\Nmultiplication? Some we've
Dialogue: 0,0:03:02.95,0:03:05.80,Default,,0000,0000,0000,,reduced to an addition\Nsome.
Dialogue: 0,0:03:07.30,0:03:11.72,Default,,0000,0000,0000,,Similarly, we could do this\Nsort of thing with division,
Dialogue: 0,0:03:11.72,0:03:19.23,Default,,0000,0000,0000,,so if we have 16 / 8, that\Nwould be 2 to the power, 4 / 2
Dialogue: 0,0:03:19.23,0:03:24.54,Default,,0000,0000,0000,,to the power three, and that\Nwould just be too, because we
Dialogue: 0,0:03:24.54,0:03:26.31,Default,,0000,0000,0000,,would subtract the indices.
Dialogue: 0,0:03:27.35,0:03:31.22,Default,,0000,0000,0000,,Now, if we had a table of
Dialogue: 0,0:03:31.22,0:03:38.16,Default,,0000,0000,0000,,numbers. Where we listed these\Npowers then all we would need to
Dialogue: 0,0:03:38.16,0:03:40.75,Default,,0000,0000,0000,,do is look them up.
Dialogue: 0,0:03:41.38,0:03:47.91,Default,,0000,0000,0000,,Do addition. And then look back\Nagain at what this 2 to the
Dialogue: 0,0:03:47.91,0:03:50.60,Default,,0000,0000,0000,,power 7 actually meant that it
Dialogue: 0,0:03:50.60,0:03:57.74,Default,,0000,0000,0000,,means 128. Now this idea\Nis the whole basis of
Dialogue: 0,0:03:57.74,0:04:03.34,Default,,0000,0000,0000,,logarithms. And they were\Ndevised in the late 16th
Dialogue: 0,0:04:03.34,0:04:07.59,Default,,0000,0000,0000,,century by two mathematicians\Nworking in dependently John
Dialogue: 0,0:04:07.59,0:04:09.72,Default,,0000,0000,0000,,Napier and Henry Briggs.
Dialogue: 0,0:04:11.16,0:04:14.59,Default,,0000,0000,0000,,So what exactly is
Dialogue: 0,0:04:14.59,0:04:19.29,Default,,0000,0000,0000,,a logarithm? Let's\Nstart with this 16
Dialogue: 0,0:04:19.29,0:04:22.87,Default,,0000,0000,0000,,is 2 to the power 4.
Dialogue: 0,0:04:24.28,0:04:29.26,Default,,0000,0000,0000,,This is the power index or the\Nexponent, and this is the base.
Dialogue: 0,0:04:30.61,0:04:38.34,Default,,0000,0000,0000,,If we take the logarithm, which\Nwe usually write as log to
Dialogue: 0,0:04:38.34,0:04:45.57,Default,,0000,0000,0000,,base 2. Of this number\N16, then the logarithm is
Dialogue: 0,0:04:45.57,0:04:50.51,Default,,0000,0000,0000,,4. This power index\Nor exponent becomes
Dialogue: 0,0:04:50.51,0:04:56.09,Default,,0000,0000,0000,,the logarithm, so we\Ntake logs to a base.
Dialogue: 0,0:04:58.28,0:05:04.82,Default,,0000,0000,0000,,And it's the base that gets\Nraised to a particular power in
Dialogue: 0,0:05:04.82,0:05:10.82,Default,,0000,0000,0000,,indices, and that power becomes\Nthe logarithm. So these two are
Dialogue: 0,0:05:10.82,0:05:15.18,Default,,0000,0000,0000,,equivalent statements. If we\Nwrite one, we automatically
Dialogue: 0,0:05:15.18,0:05:16.81,Default,,0000,0000,0000,,imply the other.
Dialogue: 0,0:05:17.48,0:05:24.68,Default,,0000,0000,0000,,So if I say that 64 is\N8 to the power, two, that is
Dialogue: 0,0:05:24.68,0:05:29.82,Default,,0000,0000,0000,,exactly the same as saying that\Nthe log of 64.
Dialogue: 0,0:05:30.38,0:05:36.03,Default,,0000,0000,0000,,Two base eight is 2. These two\Nstatements are exactly the same.
Dialogue: 0,0:05:36.03,0:05:42.16,Default,,0000,0000,0000,,So if I write a statement down\Nthis side, is the exactly the
Dialogue: 0,0:05:42.16,0:05:46.86,Default,,0000,0000,0000,,same statement written down\Nhere? They both mean the same
Dialogue: 0,0:05:46.86,0:05:53.93,Default,,0000,0000,0000,,thing, so this side I say the\Nlog to base three of 27 is 3.
Dialogue: 0,0:05:53.93,0:06:00.100,Default,,0000,0000,0000,,what I am saying is that 3 to\Nthe power three is equal to 27.
Dialogue: 0,0:06:02.59,0:06:07.64,Default,,0000,0000,0000,,So these statements down the\Nleft here or exactly the same as
Dialogue: 0,0:06:07.64,0:06:09.75,Default,,0000,0000,0000,,these statements down the right.
Dialogue: 0,0:06:10.83,0:06:16.34,Default,,0000,0000,0000,,So. We can write\Nthis down as a general
Dialogue: 0,0:06:16.34,0:06:19.35,Default,,0000,0000,0000,,statement. If we\Nhave a number X.
Dialogue: 0,0:06:20.46,0:06:24.44,Default,,0000,0000,0000,,And we can write X as a
Dialogue: 0,0:06:24.44,0:06:29.68,Default,,0000,0000,0000,,number A. Raised to the power\Nand then the equivalent
Dialogue: 0,0:06:29.68,0:06:33.56,Default,,0000,0000,0000,,statement in logarithms is to\Nsay that the log.
Dialogue: 0,0:06:34.30,0:06:41.04,Default,,0000,0000,0000,,To the base a of\NX is equal to N.
Dialogue: 0,0:06:41.04,0:06:43.74,Default,,0000,0000,0000,,These two are equivalent
Dialogue: 0,0:06:43.74,0:06:51.13,Default,,0000,0000,0000,,statements. Let's just develop a\Nlittle bit of that, supposing X
Dialogue: 0,0:06:51.13,0:06:53.55,Default,,0000,0000,0000,,were equal to 10.
Dialogue: 0,0:06:55.10,0:07:00.72,Default,,0000,0000,0000,,Then we can write 10 as 10\Nto the power one.
Dialogue: 0,0:07:01.86,0:07:08.43,Default,,0000,0000,0000,,So if I now take the\Nlog to base 9:50.
Dialogue: 0,0:07:09.28,0:07:14.32,Default,,0000,0000,0000,,Then be'cause 10 can be written\Nas 10 to the power one.
Dialogue: 0,0:07:14.85,0:07:20.91,Default,,0000,0000,0000,,Then it's logarithm to the base.\N10 must be one. Similarly, if I
Dialogue: 0,0:07:20.91,0:07:23.70,Default,,0000,0000,0000,,had, X is equal to 2.
Dialogue: 0,0:07:24.29,0:07:27.96,Default,,0000,0000,0000,,Then two is just 2\Nto the power one.
Dialogue: 0,0:07:29.14,0:07:31.25,Default,,0000,0000,0000,,And so if I take the log.
Dialogue: 0,0:07:31.90,0:07:37.34,Default,,0000,0000,0000,,To base two off to again,\Nthat's just one.
Dialogue: 0,0:07:38.38,0:07:43.62,Default,,0000,0000,0000,,So let's make that general. If X\Nwere equal to a, then we know
Dialogue: 0,0:07:43.62,0:07:47.73,Default,,0000,0000,0000,,that we can write that as A to\Nthe power one.
Dialogue: 0,0:07:48.53,0:07:56.09,Default,,0000,0000,0000,,And so that the log to base\Na of a is there for one.
Dialogue: 0,0:07:56.93,0:08:03.82,Default,,0000,0000,0000,,And this gives us a general rule\Nthat works for any of our base
Dialogue: 0,0:08:03.82,0:08:09.66,Default,,0000,0000,0000,,numbers a. Let's see if we can\Ngenerate some laws of
Dialogue: 0,0:08:09.66,0:08:15.26,Default,,0000,0000,0000,,logarithms. We know that there\Nare laws of indices. Can we have
Dialogue: 0,0:08:15.26,0:08:20.38,Default,,0000,0000,0000,,similar laws of logarithms, and\Nin one respect we've already had
Dialogue: 0,0:08:20.38,0:08:21.78,Default,,0000,0000,0000,,the first law?
Dialogue: 0,0:08:21.78,0:08:28.83,Default,,0000,0000,0000,,So. Let's have a look. Let's\Ntake 2 numbers X&Y and what we
Dialogue: 0,0:08:28.83,0:08:34.82,Default,,0000,0000,0000,,want to be able to do is\Nmultiply X&Y together. As a
Dialogue: 0,0:08:34.82,0:08:40.80,Default,,0000,0000,0000,,general thing, so we say X is A\Nto the power N.
Dialogue: 0,0:08:41.32,0:08:45.10,Default,,0000,0000,0000,,And why is A to the power
Dialogue: 0,0:08:45.10,0:08:51.78,Default,,0000,0000,0000,,N? Now this statement X is A\Nto the power, N is the
Dialogue: 0,0:08:51.78,0:08:58.12,Default,,0000,0000,0000,,equivalent of saying the log of\NX to base a is equal to N.
Dialogue: 0,0:08:58.62,0:09:04.85,Default,,0000,0000,0000,,This statement is the equivalent\Nof saying the log of Y also to
Dialogue: 0,0:09:04.85,0:09:07.72,Default,,0000,0000,0000,,base a is equal to M.
Dialogue: 0,0:09:08.73,0:09:15.88,Default,,0000,0000,0000,,We want to multiply these two\Ntogether so X times by Y is
Dialogue: 0,0:09:15.88,0:09:22.48,Default,,0000,0000,0000,,A to the power N times by\NA to the power M.
Dialogue: 0,0:09:23.73,0:09:30.29,Default,,0000,0000,0000,,And what do we get here? A to\Nthe power N times by 8 to the
Dialogue: 0,0:09:30.29,0:09:35.62,Default,,0000,0000,0000,,power N means that we add the\Nindices together. So that's A to
Dialogue: 0,0:09:35.62,0:09:37.26,Default,,0000,0000,0000,,the N Plus M.
Dialogue: 0,0:09:38.35,0:09:41.52,Default,,0000,0000,0000,,So what is the logarithm?
Dialogue: 0,0:09:42.15,0:09:45.25,Default,,0000,0000,0000,,Two base a of XY.
Dialogue: 0,0:09:45.92,0:09:49.26,Default,,0000,0000,0000,,While quite clearly from the\Ndefinition we've had.
Dialogue: 0,0:09:49.82,0:09:57.32,Default,,0000,0000,0000,,X times by Y is A to the power\NN plus M, so the logarithm is N
Dialogue: 0,0:09:57.32,0:10:05.10,Default,,0000,0000,0000,,plus N. But N we know is\Nthe log of X to base a log
Dialogue: 0,0:10:05.10,0:10:12.97,Default,,0000,0000,0000,,of X to base A and similarly M\Nis the log of Y to base A.
Dialogue: 0,0:10:13.02,0:10:20.98,Default,,0000,0000,0000,,And so there we\Nhave our first law
Dialogue: 0,0:10:20.98,0:10:26.64,Default,,0000,0000,0000,,of logarithms. That if we\Nwant to multiply 2 numbers
Dialogue: 0,0:10:26.64,0:10:31.70,Default,,0000,0000,0000,,together, we get the log of the\Nproduct by adding the logs of
Dialogue: 0,0:10:31.70,0:10:33.26,Default,,0000,0000,0000,,the two individual numbers.
Dialogue: 0,0:10:34.18,0:10:40.28,Default,,0000,0000,0000,,That's our first law. What about\Nour second law again?
Dialogue: 0,0:10:40.79,0:10:47.22,Default,,0000,0000,0000,,Let's start with X can be\Nwritten as A to the power N.
Dialogue: 0,0:10:47.75,0:10:50.65,Default,,0000,0000,0000,,Now, what if we take X?
Dialogue: 0,0:10:51.34,0:10:54.48,Default,,0000,0000,0000,,And we raise it to the power M.
Dialogue: 0,0:10:55.69,0:10:56.96,Default,,0000,0000,0000,,That would be.
Dialogue: 0,0:10:58.23,0:11:03.13,Default,,0000,0000,0000,,A to the N all raised to\Nthe power M.
Dialogue: 0,0:11:04.19,0:11:09.59,Default,,0000,0000,0000,,And by our laws of indices, we\Nknow that in order to do that,
Dialogue: 0,0:11:09.59,0:11:14.61,Default,,0000,0000,0000,,we multiply the N and the M\Ntogether, giving us NN as the
Dialogue: 0,0:11:14.61,0:11:15.77,Default,,0000,0000,0000,,power of A.
Dialogue: 0,0:11:16.51,0:11:22.92,Default,,0000,0000,0000,,So what's the logarithm here?\NLog of X to the power M to
Dialogue: 0,0:11:22.92,0:11:24.40,Default,,0000,0000,0000,,the base A.
Dialogue: 0,0:11:25.74,0:11:32.12,Default,,0000,0000,0000,,Equals well, by definition, that\Nmust be this number here. End
Dialogue: 0,0:11:32.12,0:11:33.86,Default,,0000,0000,0000,,times by N.
Dialogue: 0,0:11:34.77,0:11:36.100,Default,,0000,0000,0000,,An times by N.
Dialogue: 0,0:11:37.58,0:11:44.57,Default,,0000,0000,0000,,Equals.\NWell, this statement here it's
Dialogue: 0,0:11:44.57,0:11:50.22,Default,,0000,0000,0000,,equivalent statement. Is that\Nthe log of X?
Dialogue: 0,0:11:50.79,0:11:57.33,Default,,0000,0000,0000,,To base a is equal to N, so\Ninstead of NI can write this.
Dialogue: 0,0:11:58.33,0:12:05.18,Default,,0000,0000,0000,,And I must multiply it by NM\Ntimes the log of X to base a.
Dialogue: 0,0:12:05.18,0:12:11.13,Default,,0000,0000,0000,,There we have our second law of\Nlogarithms that if we want to
Dialogue: 0,0:12:11.13,0:12:14.32,Default,,0000,0000,0000,,raise a number to a given power.
Dialogue: 0,0:12:15.70,0:12:20.46,Default,,0000,0000,0000,,Then we can do that by taking\Nthe log, multiplying that log by
Dialogue: 0,0:12:20.46,0:12:25.58,Default,,0000,0000,0000,,M. That will give us the log of\Nthe number to the given power,
Dialogue: 0,0:12:25.58,0:12:30.71,Default,,0000,0000,0000,,and then we can look it up in\Nreverse. So our second law of
Dialogue: 0,0:12:30.71,0:12:35.97,Default,,0000,0000,0000,,logarithms. Our third law, well,\Nin a way we've already met with
Dialogue: 0,0:12:35.97,0:12:39.95,Default,,0000,0000,0000,,third law because now we've done\Nmultiplication of two different
Dialogue: 0,0:12:39.95,0:12:43.53,Default,,0000,0000,0000,,numbers. Repeated multiplication\Nof the same number. So what
Dialogue: 0,0:12:43.53,0:12:46.32,Default,,0000,0000,0000,,we're going to look at now is
Dialogue: 0,0:12:46.32,0:12:52.73,Default,,0000,0000,0000,,division. So again, will define\NX to be A to the power N.
Dialogue: 0,0:12:53.60,0:12:57.41,Default,,0000,0000,0000,,And why to be A to the power M?
Dialogue: 0,0:12:57.96,0:13:05.03,Default,,0000,0000,0000,,Will write down what that means\Nin terms of logarithms. Log of X
Dialogue: 0,0:13:05.03,0:13:08.30,Default,,0000,0000,0000,,to the base A is N.
Dialogue: 0,0:13:08.86,0:13:15.47,Default,,0000,0000,0000,,The log of Y to the\Nbase A is N.
Dialogue: 0,0:13:16.50,0:13:22.36,Default,,0000,0000,0000,,And we're going to have a look\Nat X divided by Y.
Dialogue: 0,0:13:23.64,0:13:31.62,Default,,0000,0000,0000,,So that's A to the N divided by\NA to the N and our laws of
Dialogue: 0,0:13:31.62,0:13:37.11,Default,,0000,0000,0000,,indices tell us that we do this\Ncalculation by subtracting these
Dialogue: 0,0:13:37.11,0:13:40.61,Default,,0000,0000,0000,,indices, so that's A to the N
Dialogue: 0,0:13:40.61,0:13:47.76,Default,,0000,0000,0000,,minus M. So what's the\Nlog of this quantity log to
Dialogue: 0,0:13:47.76,0:13:54.70,Default,,0000,0000,0000,,base a of X divided by\NY equals or by definition it
Dialogue: 0,0:13:54.70,0:13:57.59,Default,,0000,0000,0000,,must be this index here.
Dialogue: 0,0:13:58.29,0:14:01.30,Default,,0000,0000,0000,,An minus M.
Dialogue: 0,0:14:02.38,0:14:08.31,Default,,0000,0000,0000,,And N. Is this quantity the log\Nto base a of X?
Dialogue: 0,0:14:08.88,0:14:12.82,Default,,0000,0000,0000,,Log to base A 4X.
Dialogue: 0,0:14:13.62,0:14:16.78,Default,,0000,0000,0000,,And this quantity M is up here.
Dialogue: 0,0:14:17.51,0:14:24.50,Default,,0000,0000,0000,,The log. Of Y\Ntwo base A and. So there
Dialogue: 0,0:14:24.50,0:14:27.94,Default,,0000,0000,0000,,is our third law of logarithms.
Dialogue: 0,0:14:29.05,0:14:32.34,Default,,0000,0000,0000,,That if we want the\Nlog of a quotient.
Dialogue: 0,0:14:33.48,0:14:38.31,Default,,0000,0000,0000,,The log of a division some. Then\Nwe subtract the logs of the two
Dialogue: 0,0:14:38.31,0:14:40.04,Default,,0000,0000,0000,,numbers that were working with.
Dialogue: 0,0:14:41.35,0:14:44.08,Default,,0000,0000,0000,,1 final
Dialogue: 0,0:14:44.08,0:14:47.89,Default,,0000,0000,0000,,point. Final general point.
Dialogue: 0,0:14:49.26,0:14:55.62,Default,,0000,0000,0000,,What happens if we have A\Nto the power 0?
Dialogue: 0,0:14:56.62,0:15:01.63,Default,,0000,0000,0000,,We know that anything\Nraised to the power zero
Dialogue: 0,0:15:01.63,0:15:03.86,Default,,0000,0000,0000,,from indices is one.
Dialogue: 0,0:15:06.29,0:15:09.60,Default,,0000,0000,0000,,Well, what does that mean?
Dialogue: 0,0:15:10.24,0:15:11.95,Default,,0000,0000,0000,,About the log.
Dialogue: 0,0:15:12.77,0:15:16.19,Default,,0000,0000,0000,,To base A of
Dialogue: 0,0:15:16.19,0:15:23.33,Default,,0000,0000,0000,,one. Well, since we can write\N1 as A to the power zero, that
Dialogue: 0,0:15:23.33,0:15:28.42,Default,,0000,0000,0000,,means that the log of one to\Nbase a must be 0.
Dialogue: 0,0:15:29.20,0:15:33.28,Default,,0000,0000,0000,,And so the log of one in any\Nbase is 0.
Dialogue: 0,0:15:34.02,0:15:39.79,Default,,0000,0000,0000,,Now that does make sense. Let's\Njust think about it. If you were
Dialogue: 0,0:15:39.79,0:15:45.34,Default,,0000,0000,0000,,doing multiplication. One times\Nby 6 is 6 * 1 doesn't affect the
Dialogue: 0,0:15:45.34,0:15:51.37,Default,,0000,0000,0000,,six, so if we were to do the\Nsame in logs, we want the log of
Dialogue: 0,0:15:51.37,0:15:55.89,Default,,0000,0000,0000,,one plus the log of six.\NWouldn't want to change the log
Dialogue: 0,0:15:55.89,0:16:00.79,Default,,0000,0000,0000,,of six simply because we knew we\Nwere multiplying by one, so it
Dialogue: 0,0:16:00.79,0:16:04.56,Default,,0000,0000,0000,,makes sense for the log of one\Nto be 0.
Dialogue: 0,0:16:05.71,0:16:11.76,Default,,0000,0000,0000,,Let's have a look now at some\Nmore examples of calculating
Dialogue: 0,0:16:11.76,0:16:17.26,Default,,0000,0000,0000,,logarithms or various numbers to\Nparticular basis. So let's take
Dialogue: 0,0:16:17.26,0:16:21.11,Default,,0000,0000,0000,,the log of 512 to base 2.
Dialogue: 0,0:16:21.97,0:16:29.89,Default,,0000,0000,0000,,Well, this is the same as asking\Nwhat's 512 as a power of two
Dialogue: 0,0:16:29.89,0:16:36.12,Default,,0000,0000,0000,,512 equals 2 to what power?\NWhat's that power up there?
Dialogue: 0,0:16:36.12,0:16:41.78,Default,,0000,0000,0000,,Well, 512 is in fact 2 to\Nthe power 9.
Dialogue: 0,0:16:42.58,0:16:44.52,Default,,0000,0000,0000,,And so by definition.
Dialogue: 0,0:16:45.11,0:16:48.88,Default,,0000,0000,0000,,The log of 512 to base 2.
Dialogue: 0,0:16:49.40,0:16:50.99,Default,,0000,0000,0000,,Is 9.
Dialogue: 0,0:16:52.78,0:16:54.61,Default,,0000,0000,0000,,What about the log?
Dialogue: 0,0:16:55.45,0:17:01.06,Default,,0000,0000,0000,,To base eight of\None over 64.
Dialogue: 0,0:17:01.74,0:17:09.43,Default,,0000,0000,0000,,This is the same as asking\Nwhat is one over 64 as
Dialogue: 0,0:17:09.43,0:17:13.98,Default,,0000,0000,0000,,a power. Of\Neight.
Dialogue: 0,0:17:15.91,0:17:20.74,Default,,0000,0000,0000,,Well, what is that? We know that\None over 64.
Dialogue: 0,0:17:21.29,0:17:23.61,Default,,0000,0000,0000,,Is 64.
Dialogue: 0,0:17:25.26,0:17:28.52,Default,,0000,0000,0000,,To the minus one.
Dialogue: 0,0:17:30.72,0:17:38.15,Default,,0000,0000,0000,,We also know that 64 is 8\Nsquared, 8 times by 8 and so
Dialogue: 0,0:17:38.15,0:17:45.06,Default,,0000,0000,0000,,that tells us that one over 64\Nis 8 tool, the minus 2.
Dialogue: 0,0:17:45.59,0:17:52.22,Default,,0000,0000,0000,,And so the logarithm of one over\N64 to base eight is minus.
Dialogue: 0,0:17:52.78,0:17:53.36,Default,,0000,0000,0000,,2.
Dialogue: 0,0:17:55.58,0:17:58.73,Default,,0000,0000,0000,,What about? The log
Dialogue: 0,0:17:58.73,0:18:05.53,Default,,0000,0000,0000,,of. 25\NTool base 5.
Dialogue: 0,0:18:06.95,0:18:14.03,Default,,0000,0000,0000,,Well, this is the same as\Nasking what is 25 written as
Dialogue: 0,0:18:14.03,0:18:16.39,Default,,0000,0000,0000,,5 to the power.
Dialogue: 0,0:18:16.91,0:18:20.85,Default,,0000,0000,0000,,What's that power that index?\NWhat goes there?
Dialogue: 0,0:18:21.39,0:18:28.40,Default,,0000,0000,0000,,Well, what we do know is that\Nit's 5 squared and so the log of
Dialogue: 0,0:18:28.40,0:18:31.66,Default,,0000,0000,0000,,25 two base five is just two.
Dialogue: 0,0:18:32.87,0:18:37.89,Default,,0000,0000,0000,,I want to do another calculation\Nwhich is going to look very
Dialogue: 0,0:18:37.89,0:18:42.90,Default,,0000,0000,0000,,similar to this one and I'll\Nneed to compare it with this
Dialogue: 0,0:18:42.90,0:18:47.08,Default,,0000,0000,0000,,calculation, so I'm going to\Nrepeat this statement over the
Dialogue: 0,0:18:47.08,0:18:50.66,Default,,0000,0000,0000,,page. So we've got log.
Dialogue: 0,0:18:51.26,0:18:57.72,Default,,0000,0000,0000,,Of 25 to base\Nfive, we know that
Dialogue: 0,0:18:57.72,0:19:03.97,Default,,0000,0000,0000,,that's two. Now what if\NI interchange the number and the
Dialogue: 0,0:19:03.97,0:19:11.27,Default,,0000,0000,0000,,base? So that I'm asking\Nthe question, what's the log of
Dialogue: 0,0:19:11.27,0:19:13.75,Default,,0000,0000,0000,,five to base 25?
Dialogue: 0,0:19:14.63,0:19:16.59,Default,,0000,0000,0000,,Now that's the same as asking.
Dialogue: 0,0:19:17.98,0:19:19.62,Default,,0000,0000,0000,,If I have 5.
Dialogue: 0,0:19:20.17,0:19:26.51,Default,,0000,0000,0000,,How can I write it as a\Npower of 25 Watts that
Dialogue: 0,0:19:26.51,0:19:27.56,Default,,0000,0000,0000,,index there?
Dialogue: 0,0:19:28.73,0:19:35.00,Default,,0000,0000,0000,,Well, one of the things I do\Nknow is that the square root of
Dialogue: 0,0:19:35.00,0:19:40.83,Default,,0000,0000,0000,,25 is 5 and I can write a\Nsquare root as a power.
Dialogue: 0,0:19:41.66,0:19:42.66,Default,,0000,0000,0000,,1/2
Dialogue: 0,0:19:43.73,0:19:48.28,Default,,0000,0000,0000,,So what that tells me is\Nthat the log of five to
Dialogue: 0,0:19:48.28,0:19:49.79,Default,,0000,0000,0000,,base 25 is 1/2.
Dialogue: 0,0:19:51.00,0:19:56.13,Default,,0000,0000,0000,,And what seems to have happened\Nhere is that by interchanging
Dialogue: 0,0:19:56.13,0:19:58.46,Default,,0000,0000,0000,,the number with the base.
Dialogue: 0,0:19:59.56,0:20:02.68,Default,,0000,0000,0000,,We got one over the logarithm.
Dialogue: 0,0:20:04.23,0:20:07.12,Default,,0000,0000,0000,,Let's check that by looking\Nat another example.
Dialogue: 0,0:20:09.05,0:20:14.97,Default,,0000,0000,0000,,We know that eight can be\Nwritten as 2 to the power
Dialogue: 0,0:20:14.97,0:20:19.90,Default,,0000,0000,0000,,three, and of course that\Nmeans that the log of
Dialogue: 0,0:20:19.90,0:20:22.85,Default,,0000,0000,0000,,eight to base two is 3.
Dialogue: 0,0:20:24.04,0:20:30.77,Default,,0000,0000,0000,,But what if we interchange the\Nnumber with the base? So we ask
Dialogue: 0,0:20:30.77,0:20:35.95,Default,,0000,0000,0000,,ourselves what's the log of two\Nto the base 8?
Dialogue: 0,0:20:36.80,0:20:41.35,Default,,0000,0000,0000,,And that's the equivalent of\Nsaying to ourselves. How can I
Dialogue: 0,0:20:41.35,0:20:46.32,Default,,0000,0000,0000,,write 2 as a power of eight?\NWhat goes there? What's that
Dialogue: 0,0:20:46.32,0:20:53.79,Default,,0000,0000,0000,,index? Well, two is\Nthe cube root of
Dialogue: 0,0:20:53.79,0:20:59.96,Default,,0000,0000,0000,,8. Which we write like that. But\Nanother way of writing the cube
Dialogue: 0,0:20:59.96,0:21:03.74,Default,,0000,0000,0000,,root is to write it as 8 to the
Dialogue: 0,0:21:03.74,0:21:10.77,Default,,0000,0000,0000,,power 1/3. That tells us that\Nthe log of two to base eight
Dialogue: 0,0:21:10.77,0:21:16.84,Default,,0000,0000,0000,,is 1/3. And so we see again,\Nthat by interchanging the number
Dialogue: 0,0:21:16.84,0:21:18.09,Default,,0000,0000,0000,,and the base.
Dialogue: 0,0:21:18.67,0:21:21.97,Default,,0000,0000,0000,,We get one over the original
Dialogue: 0,0:21:21.97,0:21:27.40,Default,,0000,0000,0000,,logarithm. And that's true in\Ngeneral. I haven't proved it by
Dialogue: 0,0:21:27.40,0:21:31.89,Default,,0000,0000,0000,,showing you those two examples,\Nbut I have been able to
Dialogue: 0,0:21:31.89,0:21:37.19,Default,,0000,0000,0000,,demonstrate it and that is true\Nin general that if we have the
Dialogue: 0,0:21:37.19,0:21:39.64,Default,,0000,0000,0000,,log of B to base A.
Dialogue: 0,0:21:40.28,0:21:46.01,Default,,0000,0000,0000,,Then that is equal\Nto one over the log
Dialogue: 0,0:21:46.01,0:21:49.20,Default,,0000,0000,0000,,of A to base be.
Dialogue: 0,0:21:52.62,0:21:55.57,Default,,0000,0000,0000,,Now. We've done a lot of work
Dialogue: 0,0:21:55.57,0:22:00.71,Default,,0000,0000,0000,,with different bases. So the\Nquestion that we might ask is,
Dialogue: 0,0:22:00.71,0:22:04.77,Default,,0000,0000,0000,,are there any standard basis?\NAre logarhythms calculated using
Dialogue: 0,0:22:04.77,0:22:08.83,Default,,0000,0000,0000,,particular basis? And of course\Nthe answer is yes.
Dialogue: 0,0:22:08.84,0:22:13.78,Default,,0000,0000,0000,,What are those bases? Well, one\Nof the common. The two common
Dialogue: 0,0:22:13.78,0:22:19.14,Default,,0000,0000,0000,,ones is 10. If you look on your\NCalculator, you will see a
Dialogue: 0,0:22:19.14,0:22:20.79,Default,,0000,0000,0000,,button that's labeled log.
Dialogue: 0,0:22:21.46,0:22:28.79,Default,,0000,0000,0000,,Just log. That button that's\Nlabeled log gives you logs to
Dialogue: 0,0:22:28.79,0:22:33.88,Default,,0000,0000,0000,,base 10. So for instance, if you\Nput in 100.
Dialogue: 0,0:22:34.39,0:22:39.58,Default,,0000,0000,0000,,And press the log key. It will\Ngive you the answer to because
Dialogue: 0,0:22:39.58,0:22:42.37,Default,,0000,0000,0000,,the log of 100 to base 10.
Dialogue: 0,0:22:43.24,0:22:48.51,Default,,0000,0000,0000,,That's the log of 10\Nsquared to base. 10
Dialogue: 0,0:22:48.51,0:22:50.86,Default,,0000,0000,0000,,clearly gives us 2.
Dialogue: 0,0:22:52.05,0:22:55.69,Default,,0000,0000,0000,,What's the other common base?\NWell, the other common bases
Dialogue: 0,0:22:55.69,0:23:01.13,Default,,0000,0000,0000,,base E. To letter, not a\Nnumber you might say, but
Dialogue: 0,0:23:01.13,0:23:06.09,Default,,0000,0000,0000,,remember pie is also a letter,\Na Greek letter, and the number
Dialogue: 0,0:23:06.09,0:23:10.65,Default,,0000,0000,0000,,and E and π share something in\Ncommon. They both have
Dialogue: 0,0:23:10.65,0:23:14.79,Default,,0000,0000,0000,,infinite decimal expansions,\Nso E is a very special number.
Dialogue: 0,0:23:16.37,0:23:22.69,Default,,0000,0000,0000,,How big is it? Well to three\Ndecimal places? It's 2.718, but
Dialogue: 0,0:23:22.69,0:23:27.96,Default,,0000,0000,0000,,that's the three decimal places.\NRemember, it's got an infinite
Dialogue: 0,0:23:27.96,0:23:32.77,Default,,0000,0000,0000,,decimal expansion. So if you\Nlook on your Calculator, you
Dialogue: 0,0:23:32.77,0:23:35.44,Default,,0000,0000,0000,,will see a button that's got Ln
Dialogue: 0,0:23:35.44,0:23:43.06,Default,,0000,0000,0000,,on it. And that button Ln stands\Nfor logs to base E. So if you
Dialogue: 0,0:23:43.06,0:23:50.55,Default,,0000,0000,0000,,put the number 3 in it and press\NLn, it will give you the log to
Dialogue: 0,0:23:50.55,0:23:52.42,Default,,0000,0000,0000,,base E of three.
Dialogue: 0,0:23:53.77,0:23:56.66,Default,,0000,0000,0000,,These logs have a name.
Dialogue: 0,0:23:57.16,0:24:03.22,Default,,0000,0000,0000,,Sometimes they're called Napier\NIan Logarhythms After John
Dialogue: 0,0:24:03.22,0:24:08.53,Default,,0000,0000,0000,,Napier, and sometimes they're\Ncalled natural logarithms.
Dialogue: 0,0:24:10.97,0:24:16.08,Default,,0000,0000,0000,,Logs to base E of the logs that\Nget used in calculus. So it is
Dialogue: 0,0:24:16.08,0:24:19.84,Default,,0000,0000,0000,,important to know about logs\Nbecause they're going to come up
Dialogue: 0,0:24:19.84,0:24:23.59,Default,,0000,0000,0000,,as a regular part of the\Ncalculus, and in particular when
Dialogue: 0,0:24:23.59,0:24:24.61,Default,,0000,0000,0000,,we're solving differential
Dialogue: 0,0:24:24.61,0:24:30.84,Default,,0000,0000,0000,,equations. So let's see if we\Ncan develop a way of using
Dialogue: 0,0:24:30.84,0:24:33.73,Default,,0000,0000,0000,,logarithms and see a use for
Dialogue: 0,0:24:33.73,0:24:38.55,Default,,0000,0000,0000,,them. This isn't going to be a\Nusing terms of calculus, but in
Dialogue: 0,0:24:38.55,0:24:40.10,Default,,0000,0000,0000,,terms of solving a particular
Dialogue: 0,0:24:40.10,0:24:46.81,Default,,0000,0000,0000,,kind of equation. So supposing\Nwe've got 3 to the power
Dialogue: 0,0:24:46.81,0:24:48.70,Default,,0000,0000,0000,,X equals 5.
Dialogue: 0,0:24:49.78,0:24:54.02,Default,,0000,0000,0000,,How would we solve that\Nequation? The unknown is up
Dialogue: 0,0:24:54.02,0:24:56.99,Default,,0000,0000,0000,,here, it's an index, it's\Nan exponent.
Dialogue: 0,0:24:58.53,0:25:03.70,Default,,0000,0000,0000,,We need to get it down out\Nof being an exponent down to
Dialogue: 0,0:25:03.70,0:25:08.08,Default,,0000,0000,0000,,being an ordinary number and\None of the ways of doing
Dialogue: 0,0:25:08.08,0:25:12.86,Default,,0000,0000,0000,,that is to take logarithms,\Nso I'm going to take the log
Dialogue: 0,0:25:12.86,0:25:15.25,Default,,0000,0000,0000,,of both sides to base 10.
Dialogue: 0,0:25:16.59,0:25:18.72,Default,,0000,0000,0000,,So I'm going to have the log.
Dialogue: 0,0:25:19.72,0:25:26.29,Default,,0000,0000,0000,,Of three to the power, X is\Nequal to the log of five. Now
Dialogue: 0,0:25:26.29,0:25:30.98,Default,,0000,0000,0000,,notice I haven't written the 10\Nin on the base.
Dialogue: 0,0:25:31.71,0:25:36.16,Default,,0000,0000,0000,,And that's because I'm using the\Nconvention that I've just
Dialogue: 0,0:25:36.16,0:25:40.16,Default,,0000,0000,0000,,expressed that log now means log\Nto base 10.
Dialogue: 0,0:25:41.45,0:25:46.78,Default,,0000,0000,0000,,Well, I've got the log of three\Nraised to the power X and of
Dialogue: 0,0:25:46.78,0:25:52.50,Default,,0000,0000,0000,,course one of the things that I\Ndo know from my laws of logs is
Dialogue: 0,0:25:52.50,0:25:57.83,Default,,0000,0000,0000,,that I'm raising fun raising to\Nthe power X. Then what I need to
Dialogue: 0,0:25:57.83,0:26:00.12,Default,,0000,0000,0000,,do is multiply the logarithm by
Dialogue: 0,0:26:00.12,0:26:07.07,Default,,0000,0000,0000,,X. Well now log of three\Nand log of five aren't just
Dialogue: 0,0:26:07.07,0:26:08.65,Default,,0000,0000,0000,,numbers, nothing more.
Dialogue: 0,0:26:09.40,0:26:15.99,Default,,0000,0000,0000,,So in the same way that I\Nmight solve an equation such as
Dialogue: 0,0:26:15.99,0:26:21.06,Default,,0000,0000,0000,,three X equals 12 by dividing\Nboth sides by three.
Dialogue: 0,0:26:21.07,0:26:28.25,Default,,0000,0000,0000,,Then I'm going to do exactly the\Nsame with this, and I'm going to
Dialogue: 0,0:26:28.25,0:26:31.84,Default,,0000,0000,0000,,divide both sides by the log of
Dialogue: 0,0:26:31.84,0:26:37.26,Default,,0000,0000,0000,,three. And that calculation can\Nbe finished off using a
Dialogue: 0,0:26:37.26,0:26:42.73,Default,,0000,0000,0000,,Calculator. We take the log of 5\Ndivided by the log of three
Dialogue: 0,0:26:42.73,0:26:45.87,Default,,0000,0000,0000,,using our Calculator and get the\Nanswer for X.
Dialogue: 0,0:26:46.71,0:26:48.85,Default,,0000,0000,0000,,Let's just have a look at one
Dialogue: 0,0:26:48.85,0:26:55.71,Default,,0000,0000,0000,,more example. If we have three\Nto the X is equal to 5 to the
Dialogue: 0,0:26:55.71,0:27:01.02,Default,,0000,0000,0000,,X minus two. What then again,\Nthe unknown is X and again it's
Dialogue: 0,0:27:01.02,0:27:06.32,Default,,0000,0000,0000,,upstairs, so to speak. It's in\Nthe index in the power and we
Dialogue: 0,0:27:06.32,0:27:11.22,Default,,0000,0000,0000,,need to bring it back downstairs\Nto the normal level. So again
Dialogue: 0,0:27:11.22,0:27:16.93,Default,,0000,0000,0000,,will take logs of both sides,\Nlog of three to the power X is
Dialogue: 0,0:27:16.93,0:27:20.19,Default,,0000,0000,0000,,equal to the log of 5 to the
Dialogue: 0,0:27:20.19,0:27:25.00,Default,,0000,0000,0000,,power. Minus two in the same\Nway, if we're raising to the
Dialogue: 0,0:27:25.00,0:27:29.44,Default,,0000,0000,0000,,power, then in order to get the\Nlogarithm we need to multiply.
Dialogue: 0,0:27:30.19,0:27:37.39,Default,,0000,0000,0000,,The log of three by X\Nand the log of five by
Dialogue: 0,0:27:37.39,0:27:39.19,Default,,0000,0000,0000,,X minus 2.
Dialogue: 0,0:27:40.36,0:27:44.53,Default,,0000,0000,0000,,Well, this is now just an\Nordinary equation log of three
Dialogue: 0,0:27:44.53,0:27:49.08,Default,,0000,0000,0000,,and log of five and nothing more\Nthan numbers. So let's multiply
Dialogue: 0,0:27:49.08,0:27:50.97,Default,,0000,0000,0000,,out this bracket this side.
Dialogue: 0,0:27:51.68,0:27:59.48,Default,,0000,0000,0000,,So we're going to have\NX log 5 - 2
Dialogue: 0,0:27:59.48,0:28:05.90,Default,,0000,0000,0000,,log 5. Let's gather together\Nsome terms in X and move this as
Dialogue: 0,0:28:05.90,0:28:10.97,Default,,0000,0000,0000,,a number over to this site so\Nit's positive. So we're going to
Dialogue: 0,0:28:10.97,0:28:15.65,Default,,0000,0000,0000,,add this to both sides, which\Nwill give Me 2 log 5.
Dialogue: 0,0:28:16.68,0:28:24.65,Default,,0000,0000,0000,,And I'm going to take this away\Nfrom both sides, so X log 5
Dialogue: 0,0:28:24.65,0:28:26.92,Default,,0000,0000,0000,,minus X log 3.
Dialogue: 0,0:28:27.99,0:28:33.32,Default,,0000,0000,0000,,I've got all my ex is together\Nhere, so I'm going to take X out
Dialogue: 0,0:28:33.32,0:28:35.09,Default,,0000,0000,0000,,as a common factor 2.
Dialogue: 0,0:28:35.78,0:28:43.28,Default,,0000,0000,0000,,Log 5 is X times log 5 minus\Nlog 3 and close the bracket. So
Dialogue: 0,0:28:43.28,0:28:50.78,Default,,0000,0000,0000,,now to get X all I need to\Ndo is to divide both sides by
Dialogue: 0,0:28:50.78,0:28:57.78,Default,,0000,0000,0000,,this number here log 5 minus log\N3 because that's all it is. It's
Dialogue: 0,0:28:57.78,0:29:04.28,Default,,0000,0000,0000,,just a number, so let's do that\Nright. The line down again to
Dialogue: 0,0:29:04.28,0:29:05.78,Default,,0000,0000,0000,,log 5 is.
Dialogue: 0,0:29:05.79,0:29:12.88,Default,,0000,0000,0000,,Equal to X times log\N5 minus log 3.
Dialogue: 0,0:29:14.92,0:29:21.86,Default,,0000,0000,0000,,And I'm going to divide both\Nsides by this number 2 log
Dialogue: 0,0:29:21.86,0:29:28.79,Default,,0000,0000,0000,,5 divided by log 5 minus\Nlog of three is equal to
Dialogue: 0,0:29:28.79,0:29:35.27,Default,,0000,0000,0000,,X. OK, I've got X. You can\Nput this lump of numbers here
Dialogue: 0,0:29:35.27,0:29:40.40,Default,,0000,0000,0000,,into a Calculator and work them\Nout. One way of making that
Dialogue: 0,0:29:40.40,0:29:44.68,Default,,0000,0000,0000,,calculation alittle bit simpler\Nthough, is to notice that here
Dialogue: 0,0:29:44.68,0:29:49.39,Default,,0000,0000,0000,,we are subtracting two logs and\Nif we're subtracting two logs,
Dialogue: 0,0:29:49.39,0:29:54.53,Default,,0000,0000,0000,,that means we are in fact\Ndividing five by three in terms
Dialogue: 0,0:29:54.53,0:30:00.09,Default,,0000,0000,0000,,of the numbers themselves. So we\Ncan write this as two log 5.
Dialogue: 0,0:30:00.14,0:30:07.13,Default,,0000,0000,0000,,Over log of\N5 / 3.
Dialogue: 0,0:30:07.42,0:30:11.29,Default,,0000,0000,0000,,And that gives us perhaps\Na slightly better form to
Dialogue: 0,0:30:11.29,0:30:12.84,Default,,0000,0000,0000,,do the calculations in.
Dialogue: 0,0:30:14.23,0:30:19.60,Default,,0000,0000,0000,,Now there's one further thing\Nthat I just want to have a look
Dialogue: 0,0:30:19.60,0:30:23.73,Default,,0000,0000,0000,,at and this will become quite\Nimportant when doing calculus
Dialogue: 0,0:30:23.73,0:30:27.70,Default,,0000,0000,0000,,so. Let me take the number
Dialogue: 0,0:30:27.70,0:30:33.98,Default,,0000,0000,0000,,8. And I'm going to raise\N2 to the power 8.
Dialogue: 0,0:30:35.47,0:30:38.40,Default,,0000,0000,0000,,Let me
Dialogue: 0,0:30:38.40,0:30:45.27,Default,,0000,0000,0000,,know. Take\Nlogs to base two
Dialogue: 0,0:30:45.27,0:30:48.97,Default,,0000,0000,0000,,of two to the
Dialogue: 0,0:30:48.97,0:30:55.40,Default,,0000,0000,0000,,power 8. Well, I\Nknow what that is, that's eight.
Dialogue: 0,0:30:57.11,0:31:00.61,Default,,0000,0000,0000,,I just think what's happened\Nthere. We started with a number.
Dialogue: 0,0:31:01.45,0:31:06.08,Default,,0000,0000,0000,,We took a base and we raise the\Nbase to that power.
Dialogue: 0,0:31:07.28,0:31:10.37,Default,,0000,0000,0000,,We then took logs to that bass.
Dialogue: 0,0:31:11.08,0:31:15.80,Default,,0000,0000,0000,,Of our resulting answer and we\Nended up with the number that we
Dialogue: 0,0:31:15.80,0:31:16.89,Default,,0000,0000,0000,,first started with.
Dialogue: 0,0:31:18.16,0:31:24.24,Default,,0000,0000,0000,,So any fact what we did here\Nundid what we've done there.
Dialogue: 0,0:31:25.15,0:31:30.72,Default,,0000,0000,0000,,Now when that happens, what\Nwe're saying is we have got
Dialogue: 0,0:31:30.72,0:31:31.73,Default,,0000,0000,0000,,inverse operations.
Dialogue: 0,0:31:32.75,0:31:37.73,Default,,0000,0000,0000,,That what this operation\Ndoes, raising the base to the
Dialogue: 0,0:31:37.73,0:31:42.21,Default,,0000,0000,0000,,given power is undone by what\Nthis operation does.
Dialogue: 0,0:31:43.62,0:31:47.53,Default,,0000,0000,0000,,Does it work the\Nother way around?
Dialogue: 0,0:31:49.09,0:31:52.49,Default,,0000,0000,0000,,So let's start with eight again.
Dialogue: 0,0:31:53.41,0:31:57.11,Default,,0000,0000,0000,,This time, let's take the
Dialogue: 0,0:31:57.11,0:31:59.61,Default,,0000,0000,0000,,log. To base.
Dialogue: 0,0:32:00.43,0:32:02.94,Default,,0000,0000,0000,,Two of eight.
Dialogue: 0,0:32:03.82,0:32:07.94,Default,,0000,0000,0000,,Well, I know the\Nlog.
Dialogue: 0,0:32:09.10,0:32:13.55,Default,,0000,0000,0000,,To base two of eight, well, I\Nneed to write the 18 a different
Dialogue: 0,0:32:13.55,0:32:18.32,Default,,0000,0000,0000,,way I need it is 2 to the power\Nthree and then of course gives
Dialogue: 0,0:32:18.32,0:32:21.18,Default,,0000,0000,0000,,me the answer straight away.\NThis number is 3.
Dialogue: 0,0:32:22.59,0:32:29.69,Default,,0000,0000,0000,,Now I take my base\N#2 and I raise it
Dialogue: 0,0:32:29.69,0:32:32.53,Default,,0000,0000,0000,,to this power 3.
Dialogue: 0,0:32:33.40,0:32:35.58,Default,,0000,0000,0000,,And the answers 8.
Dialogue: 0,0:32:36.99,0:32:41.97,Default,,0000,0000,0000,,So again, I've done these two\Noperations in a different order.
Dialogue: 0,0:32:42.79,0:32:47.65,Default,,0000,0000,0000,,But the number that I started\Nwith and the number that I end
Dialogue: 0,0:32:47.65,0:32:49.90,Default,,0000,0000,0000,,up with are exactly the same.
Dialogue: 0,0:32:50.86,0:32:56.87,Default,,0000,0000,0000,,So what I've got here are two\Ninverse operations. What one
Dialogue: 0,0:32:56.87,0:33:03.42,Default,,0000,0000,0000,,does the other one undoes? So in\Nterms of our standard basis,
Dialogue: 0,0:33:03.42,0:33:10.52,Default,,0000,0000,0000,,what does that mean? It means\Nthat the natural log of E to
Dialogue: 0,0:33:10.52,0:33:17.08,Default,,0000,0000,0000,,the X. Here I've taken X as\Na power, and I've raised the
Dialogue: 0,0:33:17.08,0:33:22.24,Default,,0000,0000,0000,,base E to that power X and then\NI've taken the log.
Dialogue: 0,0:33:22.77,0:33:30.41,Default,,0000,0000,0000,,The log undoes what I did\Nwith the X and leaves me
Dialogue: 0,0:33:30.41,0:33:34.24,Default,,0000,0000,0000,,with X again. Similarly, if I
Dialogue: 0,0:33:34.24,0:33:42.06,Default,,0000,0000,0000,,Hav E. Raised to the power,\Nthe log of X, this is again X
Dialogue: 0,0:33:42.06,0:33:46.86,Default,,0000,0000,0000,,because I've used these two\Nprocess is these two things.
Dialogue: 0,0:33:46.86,0:33:51.18,Default,,0000,0000,0000,,Again in combination there\Ninverse operations, so they must
Dialogue: 0,0:33:51.18,0:33:54.06,Default,,0000,0000,0000,,undo what the other one has
Dialogue: 0,0:33:54.06,0:34:00.89,Default,,0000,0000,0000,,done. Similarly.\NIf we take our base 10,
Dialogue: 0,0:34:00.89,0:34:02.51,Default,,0000,0000,0000,,then the log.
Dialogue: 0,0:34:03.75,0:34:10.97,Default,,0000,0000,0000,,Of 10 Raised to\Nthe power, X is just X.
Dialogue: 0,0:34:11.78,0:34:13.77,Default,,0000,0000,0000,,And if we take.
Dialogue: 0,0:34:14.36,0:34:21.24,Default,,0000,0000,0000,,10 And we raise it to\Nthe power. The log of X. Then
Dialogue: 0,0:34:21.24,0:34:25.58,Default,,0000,0000,0000,,again, the answer is just X\Nbecause we have undone.
Dialogue: 0,0:34:26.29,0:34:32.34,Default,,0000,0000,0000,,By raising 10 to this power, we\Nhave undone what we did by
Dialogue: 0,0:34:32.34,0:34:36.06,Default,,0000,0000,0000,,taking the log of X to base 10.
Dialogue: 0,0:34:37.94,0:34:41.59,Default,,0000,0000,0000,,This is the particularly\Nimportant one when you're doing
Dialogue: 0,0:34:41.59,0:34:46.87,Default,,0000,0000,0000,,calculus. This is the one that\Nyou will need to bear in mind.