18 - Properties of Logarithms (Log x) - Part 1 - Laws of Logs - Calculate Logs & Simplify - Free Educational videos for Students in K-12

18 - Properties of Logarithms (Log x) - Part 1 - Laws of Logs - Calculate Logs & Simplify - By Math and Science

Transcript


00:00	Hello . Welcome back . The title of this lesson
00:02	is called the laws of logarithms . So in this
00:06	lesson , what we're gonna do is really dive into
00:08	the properties of logarithms and some of the most important
00:11	core laws of logarithms that you're going to be using
00:14	over and over and over again . So this material
00:17	is every bit as important as things that you might
00:19	have learned in the past , like laws of exponents
00:21	or laws of radicals . The laws of logarithms are
00:24	things that you would use through the rest of algebra
00:26	and pre calculus and trig and calculus in 12 and
00:30	three and on and on and on because algorithms are
00:32	not gonna go away . So what I want to
00:34	do in the beginning is I want to just have
00:35	them on the board . I don't want to hit
00:37	you over the head with them , but I want
00:38	to talk to you about what they really mean .
00:40	And then we're gonna solve some problems in the next
00:43	few lessons to use these laws of logarithms . So
00:45	you know how to use them . And I have
00:46	a lesson about two or three lessons from now ,
00:48	where I'm going to derive all of these . So
00:50	if you want to know where these laws come from
00:51	, I want you to kind of stick with me
00:53	and we'll do that a little bit later . I
00:54	don't want to bog the discussion down at the beginning
00:57	with deriving where they come from until we get some
00:59	practice . But if you stick with me , we
01:01	will talk about exactly where these laws come from and
01:03	it's very , very understandable . So here are the
01:05	laws of logarithms . I have five of them on
01:08	on the board here . Typically in most books ,
01:11	you're only gonna see three laws , you'll probably see
01:13	these 334 and five , you'll see them boxed in
01:17	on your paper number one and number two on my
01:19	board I think are so important . So that's why
01:21	I want to have them up at the top .
01:23	Because really and truly when we learn logarithms , we
01:26	learn them for a reason . We use all of
01:28	these laws , but we use number one and number
01:30	two all the time in dealing with equations and in
01:32	calculus and trig and other places down the road .
01:35	So they're so important . I want to put them
01:36	here . So even if your book says there's only
01:38	three laws of logarithms , I want you to now
01:40	know there's really five at least what I consider to
01:42	be five laws of log robes . The first to
01:45	come from . The fact that allow algorithm is an
01:49	inverse function from the exponential function . If you don't
01:51	remember that , then you need to go back to
01:53	learn to my basic lessons in what algorithm really is
01:56	. But if you remember what we have , we
01:58	have inverse functions is we have a function number one
02:01	and function number two . In this case an exponential
02:03	function and the law algorithm , which they're in verses
02:05	of when each other of one another . If we
02:08	stick an input into the first function , calculator result
02:12	, stick that thing into the second function , the
02:14	inverse function , what it does is it kind of
02:16	un does or it reverses the calculation of the first
02:19	function , that's what an inverse does . So when
02:22	you put a function with its inverse and put them
02:24	in a chain like that , basically , it kind
02:27	of does the first function and what you get back
02:29	is exactly what you started with , that's what an
02:31	inverse function is . And that's what these first two
02:33	laws are telling you what we're saying is the base
02:35	be law algorithm . You have to replace B with
02:37	a number , you know , every all these algorithms
02:39	have numbers . Base two logarithms , base 10 logarithms
02:42	, whatever . So in this case we're keeping in
02:44	general were saying the bases be if you take a
02:46	base be law algorithm of something that is itself an
02:50	exponential be raised to the power of some number K
02:54	. And then what happens is the law algorithm kind
02:56	of UN does or it annihilates the exponential . They
02:59	kind of cancel each other and all you get back
03:01	is K . That's extremely important for you to remember
03:04	. And in my opinion , most books don't emphasize
03:06	it enough . It is one of the most important
03:08	, most useful , most critical things that you don't
03:10	understand about logarithms . They can undo or annihilate or
03:15	if you want to use a rough word , they
03:16	can kill an exponential . So here you have this
03:19	exponential B to the power of K . Could be
03:21	two to the power of three or three to the
03:23	power of five or whatever . If I take a
03:25	long algorithm that has the same base of that whole
03:28	thing then the log rhythm kills the exponential , they
03:31	disappear and all you have left is K . So
03:33	you can see right away how it's useful for solving
03:35	equations because a lot of times I want to solve
03:38	for something but I may need to kill that exponential
03:41	to get rid of it . So that's how I
03:42	do it with a log rhythm . Now the law
03:44	number two is exactly the same thing in reverse .
03:47	Just like a log can kill an exponential . An
03:50	exponential can kill algorithm . See here we have a
03:53	long algorithm , a base beale algorithm of some number
03:56	but we take that whole thing and we raise it
03:58	B to the power of that . So we have
04:00	an exponential base B . But what we're taking the
04:03	exponential of YsL algorithm again . Same base B .
04:06	So the exponential cancels or annihilates the law algorithm giving
04:10	you the number back . Now , if you remember
04:12	back to in verses and discussions of inverse is we
04:15	did a lot of that recently . Again you take
04:17	an input into the first function , then run that
04:20	through the inverse and what you get back is exactly
04:22	what you put in . That's what an inverse function
04:24	is . So here if you think about it ,
04:27	if the input to this whole process is K right
04:31	then the first function you run it through is B
04:33	to the power of K . And then you take
04:35	that result and you run it through its inverse ,
04:37	which is a long algorithm , log rhythm based be
04:39	So you take an input of K . Run it
04:41	through the function which is an exponential , then do
04:44	its inverse , which is the logarithms . So they
04:46	all kill each other and all you get back out
04:48	is what you start with K . Here , you're
04:51	taking the law algorithm of a number in right ?
04:53	And so you're running it through the first function ,
04:55	which is a long algorithm and then you get the
04:57	result of that and you do its inverse . You
04:59	take B to raise to that power , which is
05:01	the inverse of the algorithm . So they kill each
05:02	other again and you get what you start with C
05:04	. In both cases you get when you started with
05:06	that's what an inverse function is . Okay now the
05:09	rest of these guys were going to use them as
05:11	we kind of encounter problems here in just a minute
05:13	. But I'm going to derive all of these in
05:15	a future lesson as I said . But basically when
05:18	you have two things multiplied together and you take there'll
05:20	algorithm then it becomes like addition of logarithms . So
05:24	multiplication of things that functions or numbers . When you
05:29	take the law algorithm becomes addition of logarithms . So
05:31	multiplication becomes addition of logarithms . Okay , likewise division
05:36	of numbers once you take the algorithm becomes like subtraction
05:40	. So it's kind of like it's like a transformation
05:42	rule like multiplication kind of becomes addition of logarithms and
05:46	division kind of becomes like subtraction . That kind of
05:49	it is subtraction of logarithms . It's kind of a
05:52	transformation . If you don't like working with multiplication .
05:54	No problem . You can make it into addition .
05:56	You gotta take algorithms first though to make it happen
05:58	. But you can do that right If you don't
06:00	like division . Maybe it's really hard to do take
06:03	the algorithm then your division goes away and you only
06:05	have subtraction . So some problems are really hard .
06:07	Maybe you have a very complicated division of two .
06:10	Very complicated things . No problem . Take the algorithm
06:12	and then division goes away , it becomes subtraction .
06:15	But then you have the added baggage . If you
06:16	have to take algorithms , right ? And then there
06:19	last one here is the odd man out , I
06:22	guess these kind of go together and these kind of
06:24	go together . This is kind of the odd man
06:25	out . And so basically if you're taking the law
06:27	algorithm of something raised to a power um where uh
06:32	yeah something raised to a party notice the base is
06:34	not the same , like it was up here ,
06:36	like it could be different . Right ? So anyway
06:38	, what you can do is take this exponents and
06:40	bring it to the outside of the law algorithm .
06:43	So basically everything else stays the same longer than based
06:46	B . Of M . That's here , we just
06:48	take the exponent and we go backwards and we can
06:50	pull it out in front K times the log of
06:52	that . So that's very useful as well when solving
06:55	some equations . So again , multiplication of things becomes
06:58	addition . When you're dealing with the algorithms , division
07:00	of things again becomes subtraction . When you're dealing with
07:03	the algorithms , exponents becomes where I can just take
07:06	the exponent out and bring it onto the front ,
07:08	times the log rhythm of the basic expression and then
07:11	the other two at the top just basically mean that
07:13	exponential can annihilate logarithms and logarithms can annihilate exponentials .
07:18	Okay inwards that is going to save you so much
07:21	time because in most textbooks you flip the page and
07:23	flip the page , you don't even know what they're
07:24	trying to say . This is all you need to
07:26	know now . What we need to do is start
07:28	solving some problems simplifying and solving some simple simplifications using
07:32	these laws of log room . So what we're gonna
07:34	do is get started with that right now let's take
07:38	our first problem . What if I give you ,
07:41	I want you to simplify the following law algorithm base
07:44	two of the number two to the fourth power .
07:48	What is that equal to now ? In the past
07:51	the way you would do it would be a different
07:52	a different way we'll talk about in just a second
07:54	. But now that you know that law algorithms can
07:58	annihilate exponential is just giving you back what's in the
08:00	exponential then you look back here and you say that's
08:03	exactly what I'm doing . I have an exponential two
08:05	to the power of something the bases to but I'm
08:07	taking the law algorithm of that thing with exactly the
08:10	same base . So the law algorithm completely annihilates the
08:13	exponential and all you have is the number four that
08:15	comes out , the four goes in , goes through
08:18	the function , then it gets run through its inverse
08:20	with the same base . And so what you get
08:22	back is what you started with , which is the
08:23	number four . Now the old way that we would
08:26	do that . So you can just write this down
08:27	without any math and you know , the annihilate ,
08:29	you just write it down . But the way we
08:30	do it in the past , as we would say
08:32	, the base to the power of something , it's
08:35	got a equal to what we're taking , the long
08:37	rhythm of which is two to the fourth power .
08:39	And you can then see since the basis of the
08:40	same X has to be equal to four . And
08:43	so this longer than reduces to four . So ,
08:45	you see these laws of logarithms really are not that
08:47	different from what you've been doing before , but a
08:49	lot of times , especially with very large equations ,
08:51	it's gonna be much easier to understand these rules and
08:55	use them rather than some of these other techniques that
08:57	we have learned . All right , let's take a
09:00	look at the next problem . Let's say we have
09:03	la algorithm base five of the number five to the
09:08	power of six . Again , you don't have to
09:11	calculate anything . You have a base five to the
09:14	power of six . This is an exponential , but
09:16	you're taking a long algorithm of that thing . So
09:18	the log of the exponential , same base means they
09:21	annihilate each other and all I have left is a
09:23	number six back . Because I take that six ,
09:26	I run it through this exponential function , then I
09:28	take the result and I run it through its inverse
09:30	the law algorithm . And so everything annihilates and I
09:32	get what I start with again , it would be
09:34	the same thing as we had . The way we
09:36	have done it before . You would say base five
09:39	to the power of something is equal to what I'm
09:41	taking the log five to the power of six .
09:43	And you didn't know that X is equal to six
09:45	. So you could do it that way if you
09:46	want . I'm just trying to show you how to
09:49	use the laws of logarithms here . All right now
09:54	let's switch gears a little bit . Instead of taking
09:56	a log of an exponential , let's take the exponential
09:59	of the law algorithm . Let's say we have to
10:02	to the power of law algorithm based too . Uh
10:06	of the # seven . So this is all in
10:09	the exponent of the two . So it's exactly reversed
10:12	instead of doing the exponential first . And then taking
10:15	the law algorithm here , we're going to take the
10:17	law algorithm first and then we're gonna take the answer
10:19	and raise that to to the power of two .
10:22	So notice how you have a base to in a
10:24	base to here . So this exponential completely cancels with
10:27	the logger in them and you don't even have to
10:29	do any math , you just say the answer is
10:30	seven , Right ? You just say that the answer
10:32	is seven . And that comes back to this rule
10:35	right here . It's a little harder to read without
10:37	numbers in my opinion , but that's what it's saying
10:39	. Base be raised to a law algorithm with a
10:41	base be of some number . The logarithms completely annihilate
10:44	the exponential . And so you just left with what
10:46	you start with because you've taken the seven . You
10:49	ran it through algorithm based too . You've got the
10:52	answer then you ran it through its inverse , which
10:54	is an exponential with the same base . And so
10:56	you get back what you start with . That's what
10:57	an exponential is or that's what an inverse function is
11:02	. All right . What if we have 11 Raised
11:06	to the power of the law algorithm , base 11
11:09	of the number three ? Well , you can see
11:10	right away you have a longer than base 11 .
11:12	You're taking that thing and you're saying the base 11
11:15	race to that . So the exponential annihilates what the
11:17	law algorithm and you can say that the answer is
11:19	three without any further work , because it falls directly
11:22	under the laws of algorithms . I'm doing these kind
11:26	of simple problems with numbers because I want you to
11:28	understand what the laws are really saying . But you're
11:30	gonna have to trust me here . When you get
11:31	to more advanced math , you might have an entire
11:34	equation which is an exponential and you want to kill
11:36	that thing . So you're gonna take a log rhythm
11:38	of both sides in order to kill whatever it is
11:41	. You're trying to isolate or to simplify something and
11:44	so knowing these laws that will save you a ton
11:46	of time when doing real problems . All right ,
11:49	let's do something a little more challenging . Let's say
11:52	we have eight to the power of two plus X
11:56	. Now this two plus X is all in the
11:57	exponent of eight of the number eight is equal to
12:00	two . And I want to solve this equation .
12:02	Now actually this equation , we've solved it before because
12:05	remember we learned about the exponential function and then we
12:08	learn how to solve exponential equations and the way we
12:11	solved it before , as we said , okay I
12:12	can write eight as two to the power of three
12:16	. So I can say this is two to the
12:17	power three and that's raised to this . And then
12:20	when I do the multiplication and set the exponents equal
12:22	, I can solve for X . We've done that
12:23	in the past . I'm gonna show you an alternative
12:25	way now that you understand the laws of logarithms notice
12:30	that when I take a log a rhythm of an
12:32	exponential with the same base it kills it . And
12:35	I just get back what I start with here .
12:37	So that's what I have . I have an exponential
12:39	with a base raised to the power of K .
12:41	A base raised to the power of quote unquote K
12:44	. So if I want to get rid of this
12:46	exponential right ? I can take the law algorithm of
12:48	both sides . So I'll take the law algorithm and
12:51	I'm gonna take the base eight algorithm of the quantity
12:54	eight to the power of two plus X . Taking
12:58	the log rhythm of the whole thing . And if
13:00	I do it to the left I have to do
13:01	it to the right because that's how equations work .
13:03	So you see I'm allowed to do what I want
13:05	to both sides . I'm gonna take the base eight
13:07	law algorithm . Why am I choosing base eight ?
13:09	Because this exponential has a base of eight . So
13:11	I know from the laws of algorithms if I take
13:13	a log base eight of an exponential that also has
13:16	a base eight , the exponential completely annihilates with the
13:20	law algorithm . And all I have left is what
13:21	I started within the exponent two plus X . It's
13:23	gone . It's like I don't want to strike through
13:25	but it's like you can just strike through it ,
13:27	it's gone . They cancel each other . It's like
13:29	adding two and then subtracting to , you don't you
13:32	don't have any change . It's like multiplying by five
13:34	and then immediately divided by five . Running something through
13:37	an exponential and then immediately threw a log a rhythm
13:40	of the same base . Just one does the whole
13:41	thing and you get back what you started with in
13:43	this case is what the exponential is , where you
13:45	started . It goes through the exponential through the law
13:47	. That's what you get at the end on the
13:49	right hand side . You still have to deal with
13:51	this log Base eight of the # two . That's
13:54	true , that's there . But we know how to
13:55	do logs . So let's go ahead and solve this
13:59	. Let's go off here to solve that log base
14:03	eight of the number two . How do we solve
14:05	that ? It's the base eight to the power of
14:07	something is equal to two . But you know that
14:09	eight can be written as two to the power of
14:11	three . That's all I did there . So you
14:14	multiply the exponents three times . Access three eggs and
14:18	I can equate these exponents three . X . Is
14:20	one , so then X . Is one third .
14:22	I did all this on the side just to figure
14:24	out that the right hand side was equal to one
14:26	third . So I have two plus X is one
14:30	third . So all I have to do to solve
14:33	this equation is say that X is 1/3 two right
14:39	now . A lot of you can do this kind
14:41	of thing in your mind in your head . That's
14:42	no problem . But for those of us who can't
14:44	this is what you would do , you have to
14:45	get a common denominator . So you have one third
14:48	minus 2/1 . That's what you're subtracting . But then
14:52	I'm gonna multiply this fraction by 3/3 so I can
14:54	get a common denominator of three , so I'm gonna
14:57	have one third minus two times three is 66 3rd
15:01	. And so what I'm gonna have is X is
15:03	going to be equal to the common denominators , so
15:05	1 -6 is negative five thirds . What were you
15:10	trying to do ? You were trying to solve for
15:11	X . And that's what we did we solve for
15:13	X . Now it's a little confusing because when I
15:15	came over here , I also introduced an X .
15:17	This this over here was just for the purpose of
15:19	finding out what this law algorithm was . I should
15:21	have used probably a different variable here . I could
15:23	use Y or or a or B . Or anything
15:25	else . Just to use a dummy variable to calculate
15:28	this . But you see what we're doing here ,
15:30	You might argue with me and say , well taking
15:32	the law algorithm of both sides is way harder than
15:34	the other way we did it the other way we
15:36	did it , we just wrote eight as two to
15:38	the power of three , raised to this power and
15:41	we go through it . We've done that before .
15:42	It's actually easier to solve it using the other techniques
15:45	. But that's not the point . The point is
15:47	I'm trying to teach you how to do logarithms ,
15:49	how to use logarithms because I promise you there are
15:52	many , many problems where it's much , much faster
15:55	to use algorithm . So we have to kind of
15:57	crawl before we can walk a little more work here
15:59	. But we get the same answer here . If
16:00	you go back to your notes , we already did
16:02	this problem before you get exactly the same thing .
16:04	So the first couple of problems were mostly geared in
16:09	trying to get you to understand how to solve these
16:11	equations . The next few problems are going to be
16:14	uh well it's a mixture , but we're going to
16:16	be trying to use a little bit more of the
16:18	last three laws of logarithms . Remember multiplication . When
16:21	you have , dealing with the algorithms becomes addition division
16:25	. When you're dealing with logarithms become subtraction and raising
16:28	to an exponent just pulls the exponent out in front
16:30	and it's a log of the whatever it is you
16:32	were doing to begin with . So let's slide over
16:36	and try to tackle a few more to give us
16:38	some more practice with that . What if I have
16:40	log rhythm base nine of the number X is negative
16:46	one half . Now there's I guess I lied to
16:50	you a little bit . We're probably going to be
16:52	using a little bit more of the the first couple
16:55	of laws here . I forgot about this problem .
16:56	But how would you solve this ? You want to
16:58	figure out what this is . There's lots of ways
17:00	to do it . We've solved this problem before .
17:01	You already know you can solve this by saying nine
17:04	to the power of this is equal to this .
17:06	We've solved that exact problem into the power of that
17:09	is equal to X . And boom , you solve
17:10	for X . That's not the point here . The
17:12	point is for me to teach you how to use
17:14	these laws of logarithms to solve problems . We've already
17:16	done . So how would we kind of get rid
17:19	of the law algorithm ? We would have to do
17:22	the exponential to both sides . We have to use
17:24	the exponential , but it has to be the same
17:25	base . So I can write this as nine to
17:27	the power of log base nine of X . And
17:31	then if I do that on the left , I
17:32	have to make it nine to the power of one
17:34	half on the right . I take and make this
17:36	an exponential with a base nine . This exponential the
17:39	base nine . I know I have the same basis
17:41	. Exponential cancels with the log . So I'm left
17:43	with X . Nine to the negative one half .
17:46	Notice this is exactly what you get . If you
17:48	say nine to the power of this is equal to
17:51	this , Nine to the power of this is equal
17:53	to this . Using the laws of logarithms and using
17:56	the techniques we've learned before , yield exactly the same
17:58	thing . So , you know , you're on the
17:59	right track . So what we have here is 1/9
18:02	to the one half . And so you get X
18:04	is one third because of the square root of nine
18:07	is three there ? Same answer as before this problem
18:11	. And the last one we've already solved those problems
18:14	. All right . So now we want to switch
18:15	our gears to using Law # three . Law #
18:19	four . Law # five . So what we want
18:21	to do here is right . In terms of log
18:30	Base two of em and log Base two of end
18:34	. You'll see what I mean when I give you
18:36	the first problem , that's how we're simplifying these .
18:39	So what if I give you log base to of
18:43	the quantity em to the six power times in to
18:47	the third power . Now you have a little bit
18:48	of a combination of things happening here . You have
18:51	two things multiplied together , but you also have exponents
18:54	in there . So remember when you have two things
18:57	multiplied together and you're taking the longer than you can
18:59	just break them apart and make it the law algorithm
19:01	of both of those things . So do that part
19:04	first . Right . So what's this going to be
19:06	long algorithm based . Two of em to the six
19:10	plus log rhythm base two of end to the third
19:13	because these are multiplied together just like it was in
19:15	the law . So this thing is the law algorithm
19:18	of that . This thing is a log rhythm of
19:19	that . We add them together . So the multiplication
19:21	is now gone , it becomes edition . But then
19:23	we remember our last law of logarithms is any time
19:26	we're taking the log of something with an experiment ,
19:28	we can take the exponents out in front and just
19:31	take the log of what is left over and again
19:33	, we're gonna prove all of these later . We
19:35	haven't done that yet but we have exactly that situation
19:37	here . The six can come in front of the
19:39	log making six times log Base two of em plus
19:44	sign stays here . The three can come out in
19:46	front log base to and then six law to base
19:51	two of em three log base two of em .
19:53	This is the final answer . So we had to
19:56	write these in terms of log base two log event
19:58	and based to log event . That's exactly what we've
20:00	done . All right , put a little dot here
20:02	. If you want to denote the multiplication . All
20:05	right . How many more problems do I have ?
20:07	I think I have enough . I think I want
20:10	to do one more problem on this board . And
20:13	yeah , I think I want to do one more
20:15	on this board and then I have plenty of space
20:17	and the other . What if I have log base
20:21	to of the number m times the square root of
20:25	N . So I have two things multiplied together of
20:28	course . And so when I have two things multiplied
20:30	together , what do I do ? I break this
20:33	guy up into its own log rhythm of each thing
20:35	and I add them together . Multiplication becomes addition when
20:39	you're dealing with the algorithm . So what do I
20:40	have here ? This is going to then become log
20:45	Base two of em plus log base to Of what
20:51	is this square root event ? It's just another number
20:53	. Square root of N . Now if you circle
20:55	this on your paper , I probably would give you
20:57	most credit because at this point , you know how
20:59	to apply the law of logarithms . You recognize you're
21:02	multiplying something so you break it apart and make an
21:04	addition of algorithms . That's fine . But there's a
21:07	little bit more that you can do here because if
21:10	you think about it , what this is really equal
21:13	to is log base two of them plus log Base
21:18	to this radical can be written as into the 1/2
21:21	power . So this exponent here on the second term
21:25	, we can use this one here to pull the
21:27	exponent in front and that's going to basically make it
21:30	a little bit uh simpler because notice we want to
21:33	write it in terms of this and this , but
21:35	this is a square into the one half , this
21:38	is not log of end , this is into the
21:39	one half , so we have to do something to
21:41	get it there . Log base two of em plus
21:45	the exponent comes out one half times log base to
21:51	end . This is the final answer . Log base
21:53	two of them plus one half log base two of
21:55	end . Now we have written it in terms of
21:57	this and this . The previous step , even though
22:00	it kind of was halfway , there really wasn't because
22:02	this is the log base two of the square root
22:04	event . That's not exactly what we were trying to
22:07	accomplish . All right , we have two more just
22:11	to get kind of our feet wet and continue working
22:13	on this . Um and what if we have log
22:20	Mhm base to again , I'm using base to a
22:23	lot , you know , based do here for this
22:25	problem . But just keep in mind these laws of
22:27	logarithms apply and are used for any base of m
22:32	to the fourth power over , in To the 3rd
22:36	power . So it's a base two logarithms of the
22:38	quotient . And if you remember any time you're dividing
22:42	uh in something inside of al algorithm , it becomes
22:45	subtraction of the logarithms . So that's what you have
22:48	to really do their and so then what we're gonna
22:50	have then is log Base tube of em to the
22:54	4th power minus log base to end to the third
22:58	power . So you've effectively split this up in the
23:00	subtraction . But then you realize you have these exponents
23:02	here . So we use the other law of logarithms
23:04	, pull the exponent in front four times log base
23:08	two of them minus the three can come down three
23:11	times log base to event . And this will be
23:15	the final answer . Four times based to log of
23:17	m minus three times log base two . Event ,
23:21	that's the correct answer . All right , we have
23:25	one more . This was a little bit longer ,
23:26	but it's not that difficult . You just have to
23:28	take it one step at a time . What if
23:30	you have log base to of em over and cubed
23:37	. But this whole thing has a square root around
23:39	it . That's what you're taking the log rhythm of
23:41	. So it looks like a lot of stuff to
23:43	unpack and it is so let's go all over here
23:45	and do it one step at a time . This
23:47	whole thing has a square root around the whole thing
23:49	. So first we need to get rid of that
23:51	and make it a fraction Log Base two of mm
23:55	over into the third power . But the square root
23:58	now becomes a power of one half . You always
24:00	want to do that pretty much all the time .
24:02	Make your radicals into exponents like this right now you're
24:07	taking the log rhythm of something . This whole something
24:10	here raised to the one half . A lot of
24:12	students will look at this and start trying to apply
24:16	this rule to subtract the logarithms . But you gotta
24:18	look bigger picture than that . The thing that you're
24:19	doing on the outside of that is this whole thing
24:22	is to the one half power . So the proper
24:23	thing order of operations wise is to take the one
24:26	half out in front times long , Base two of
24:31	em over into the third power like this . So
24:35	you're taking that expanded out now you're taking the base
24:38	to log of a straight quotient like this , so
24:42	then what you'll have is now this one half is
24:44	multiplied times the log of the final results . So
24:47	you have to really should open up a bracket or
24:49	parenthesis because this is going to expand , it'll algorithm
24:52	based two of em minus because it's division logger them
24:57	, Base two of end to the third power because
25:00	that is what is on the bottom . Now this
25:02	whole thing is in a bracket because it's one half
25:04	times this this happens to be this , so there's
25:07	a bunch of different ways you can do it if
25:09	you want . Let's just do it like this ,
25:11	let's keep the one half on the outside , log
25:15	base two of em minus now this is a log
25:17	but you have a power here , so this power
25:19	can drop down also three times a log base to
25:23	event and then you have one half times this whole
25:27	thing . So then when you multiply the one half
25:29	in , you have one half times log base two
25:33	of minus . When you take the one half times
25:35	the three , you're going to get three halves log
25:39	based to event . Let me double check myself one
25:42	half , log base two of them minus three halves
25:44	, log base two of em . This is the
25:46	final answer . This is the most simplified form ,
25:48	remember we're trying to write them in terms of log
25:50	base two of them and log base two event .
25:52	So these problems are specifically crafted to get you to
25:58	force you to use the laws of logarithms and you
26:00	can see that rarely are you doing just one or
26:03	the other ? Like in this problem we had to
26:05	deal with the exponent bringing it down , then we
26:08	had to do the subtraction because it was a quota
26:10	, then we had another expanded to bring down .
26:11	Then we had to multiply the one half back in
26:13	. So we were doing a lot of different things
26:15	. And so you can see these laws of logarithms
26:17	, they work together . So again , to recap
26:20	, there are five primary laws of logarithms that I
26:22	want you to really remember and burn in your mind
26:25	when you have the log base B of an exponential
26:28	with the same base , they kill each other and
26:30	you just get back the exponents . When you have
26:32	an exponential of algorithms , same base , they kill
26:35	each other and you just get back what you were
26:36	doing the algorithm of the next minute there . When
26:39	you multiply things and you take the law algorithm ,
26:41	it becomes addition of logarithms . When you divide things
26:44	and take the algorithm it becomes subtraction of logarithms .
26:47	And when you have an exponent of something raised to
26:50	something else powered , the exponent can always come out
26:52	in front of the law algorithm multiplied by law algorithm
26:55	of what's left over same base obviously . Where do
26:59	these laws come from ? We talked about extensively where
27:01	the first to come from the last three . I
27:03	haven't given you any proof of where they come from
27:05	but I promise you they're not hard to understand .
27:07	What I want to do is make sure you understand
27:09	all of these problems , solve them yourself , make
27:12	sure you grab a paper and solve them yourself .
27:15	Go and do the next lesson or two with me
27:17	, Get some more practice with using these laws and
27:19	then I have a lesson on the books that I'm
27:21	gonna show you and derive exactly where they come from
27:23	. So you'll understand where everything is . So solve
27:25	these yourself , follow me on to the next lesson
27:27	and continue to conquer and practice the concept of the
27:30	laws of logarithms .

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