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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