04 - Solving Logarithmic Equations - Part 1 - Equations with Log(x) - By Math and Science
Transcript
| 00:00 | Hello . Welcome back to this lesson in dealing with | |
| 00:03 | logarithms . The title of this lesson is called solving | |
| 00:06 | logarithmic equations . If you go up to a random | |
| 00:09 | person and you say you know what we're going to | |
| 00:10 | talk about solving logarithmic equations . Most people are gonna | |
| 00:14 | think you're crazy or they're gonna think that whatever you're | |
| 00:16 | talking about is extremely complicated . What we're gonna do | |
| 00:19 | is we're gonna learn how to solve equations that contain | |
| 00:21 | logarithms . Another way to say that is logarithmic equations | |
| 00:24 | and by the end of it , once you get | |
| 00:26 | practice with it I hope you'll understand that . It's | |
| 00:28 | not really very much more involved than solving regular equations | |
| 00:31 | in algebra . The things that you have to keep | |
| 00:33 | in the back of your mind in order to solve | |
| 00:35 | any equation that contains the algorithm is mostly that you | |
| 00:38 | have to understand the laws of algorithms . Now actually | |
| 00:41 | I did several lessons on the laws of algorithms and | |
| 00:44 | just in the last couple of lessons . So if | |
| 00:45 | you haven't watched those you really have to understand that | |
| 00:48 | in order to get this , I went ahead and | |
| 00:50 | erase them from the board because we need to commit | |
| 00:52 | them to memory anyway . And so ultimately you need | |
| 00:55 | to have them in your mind . The number one | |
| 00:57 | thing you need to know is that whenever you have | |
| 00:58 | a long algorithm , you can kill that law algorithm | |
| 01:01 | or annihilated by raising that law algorithm to an exponent | |
| 01:04 | with the same base because the exponential function and the | |
| 01:08 | law algorithmic function are inverse functions of each other . | |
| 01:10 | So whenever you do the opposite function , then you | |
| 01:14 | annihilate and it disappears . And that's one way that | |
| 01:17 | we're gonna use to solve these equations . Other than | |
| 01:19 | that , you have to understand the laws of algorithms | |
| 01:22 | . What happens when you multiply something together ? It | |
| 01:24 | becomes addition of law algorithms . What happens when you | |
| 01:27 | divide something ? It becomes a subtraction of logarithms and | |
| 01:30 | so on and so forth . So we have to | |
| 01:32 | just go ahead and do it . Uh And that's | |
| 01:34 | what we're gonna do here . Let's take a look | |
| 01:35 | at our first problem . Log rhythm base three of | |
| 01:39 | some number X plus log rhythm Base three of some | |
| 01:43 | number X -6 Is equal to three . Now this | |
| 01:47 | is what we call a log arrhythmic equation . It | |
| 01:49 | has an equal sign . So it's a it's an | |
| 01:51 | equation and it contains algorithms Now in this case the | |
| 01:55 | unknown quantity is X . But that X appears in | |
| 01:57 | two locations . So what we have to do for | |
| 02:00 | these kinds of equations is very similar to regular equations | |
| 02:03 | . We want to isolate and get X on one | |
| 02:05 | side of the equal sign all by itself . That's | |
| 02:07 | all we want to do . And we're going to | |
| 02:09 | have to use the laws of logarithms to do that | |
| 02:12 | . So we see immediately , we have a term | |
| 02:14 | over here that's a log arrhythmic term and a term | |
| 02:16 | over here , it's the same exact base . And | |
| 02:19 | we know that when we're ADL algorithms together , we | |
| 02:21 | can combine them to basically one single term as follows | |
| 02:26 | . You can combine it like this , it has | |
| 02:27 | to be a long algorithm base three of X , | |
| 02:31 | which is this term times X -6 , which is | |
| 02:36 | this term . And that's going to be equal to | |
| 02:38 | three . If you do not understand this step , | |
| 02:40 | then it just means you never watch the lessons on | |
| 02:43 | the laws of algorithms . You can think of it | |
| 02:45 | going backwards . We have things multiply two things multiply | |
| 02:48 | this term times this term inside of the algorithms . | |
| 02:50 | So we can break it out into addition of those | |
| 02:52 | algorithm in terms that is the law of logarithms we've | |
| 02:54 | already learned in the past . So now we have | |
| 02:57 | to continue to multiply through to get X by itself | |
| 03:00 | log base three . This is going to be x | |
| 03:02 | squared minus six X on the inside . Just multiplying | |
| 03:05 | the X through And then we get equal to three | |
| 03:09 | . All right . So as often happens in math | |
| 03:12 | , there's almost always multiple ways to proceed . When | |
| 03:15 | you saw the equations , there's lots of different paths | |
| 03:17 | you can take . So when you get to la | |
| 03:19 | algorithms , there's actually even more paths you can take | |
| 03:22 | and they all should get to the same answer . | |
| 03:24 | So if I'm doing something here and you're thinking I'd | |
| 03:27 | rather do it a different way . That's fine . | |
| 03:28 | Do it your way . As long as you get | |
| 03:30 | to the same answer , I'm fine . Now when | |
| 03:31 | some of these problems , I'm gonna go ahead and | |
| 03:33 | work the problem two completely different ways because I want | |
| 03:36 | you to understand that no matter what you pick , | |
| 03:38 | as long as you're following the correct rules , you're | |
| 03:41 | always going to get to the same exact correct answer | |
| 03:44 | . All right . So , there's two ways to | |
| 03:45 | proceed . I'm gonna call this 1st 1 method . | |
| 03:50 | It's not really a method , but it's it's kind | |
| 03:52 | of the way I'm doing it number one . And | |
| 03:53 | then over here I'll do number two . So , | |
| 03:56 | what do I have ? This is a long algorithm | |
| 03:57 | , has a base three so I can write three | |
| 03:59 | which is the base to the power of three , | |
| 04:01 | which is what's on the right hand side is equal | |
| 04:03 | to what I'm taking the law algorithm of X squared | |
| 04:05 | minus six eggs . You see in this way , | |
| 04:08 | I've completely eliminated the law algorithm from the equation . | |
| 04:11 | It's gone . So that law algorithm where the variable | |
| 04:14 | is wrapped up inside of that law algorithm , it's | |
| 04:15 | now gone . I don't have to worry about that | |
| 04:17 | because I used the definition of the law algorithm , | |
| 04:19 | The base to the power of whatever the law algorithm | |
| 04:22 | returned is equal to this . And that is just | |
| 04:24 | simply a polynomial now . So what you have is | |
| 04:28 | you have X squared minus six X . I'm gonna | |
| 04:30 | move this to the other side . So to be | |
| 04:32 | a minus sign . And three times three is nine | |
| 04:35 | . And so what you're going to actually not three | |
| 04:36 | times three is nine . It's three , not three | |
| 04:38 | squared is three cubed . So you're gonna have 27 | |
| 04:41 | . Sorry about that . So we subtract it , | |
| 04:43 | bring it over here , it's gonna be a negative | |
| 04:44 | sign 27 that's equal to zero . So now your | |
| 04:47 | log arrhythmic equation just now reduced to uh to a | |
| 04:52 | polynomial that we've solved many many times . So let's | |
| 04:56 | try to factor this thing open parentheses equal to zero | |
| 05:00 | X times X . We can pick whatever you want | |
| 05:03 | here . But ultimately nine times three is going to | |
| 05:05 | work . We need a minus sign and we need | |
| 05:06 | a minus sign here . So the way that's going | |
| 05:08 | to work is minus and plus double check yourself X | |
| 05:12 | times X is x squared . This times this is | |
| 05:14 | negative 27 . Here's a negative nine X . Here's | |
| 05:17 | a positive three X . You add that you get | |
| 05:19 | a negative six X . So it appears that we | |
| 05:23 | have two solutions . We have x minus nine is | |
| 05:26 | equal to zero , so X can be equal to | |
| 05:28 | nine and we have X minus three is equal to | |
| 05:31 | zero and that means X is equal to three . | |
| 05:34 | Right ? Actually there's a plus sign here . Sorry | |
| 05:36 | about that . So this could be a negative right | |
| 05:38 | there . So it appears that we have two solutions | |
| 05:40 | X minus +90 X plus three zero . So X | |
| 05:43 | can be nine or x can be negative three . | |
| 05:45 | So if you were to circle those answers , you | |
| 05:49 | would actually be wrong or at least partially wrong . | |
| 05:51 | Can you spot what the problem is ? It's kind | |
| 05:53 | of tricky . But the problem is this negative sign | |
| 05:56 | here , Right ? Remember what you're doing ? You're | |
| 06:00 | trying to find the values of X that satisfy the | |
| 06:03 | original equation that you have , which is this one | |
| 06:05 | , X appears here and X appears here . But | |
| 06:08 | when you look at the plot of the law algorithm | |
| 06:10 | function , a logarithms is never ever ever defined . | |
| 06:14 | When you stick a negative value into that algorithm , | |
| 06:17 | it doesn't exist or at least it doesn't exist in | |
| 06:19 | terms of real numbers . Why is that if you | |
| 06:21 | remember back to the plot of algorithm , it's the | |
| 06:25 | inverse of the exponential function . right ? So if | |
| 06:28 | this is X and this is fx and were plotting | |
| 06:30 | a log rhythm here , then what's gonna basically happen | |
| 06:34 | is it goes through the number one , That's this | |
| 06:37 | spot right here , It starts very close to the | |
| 06:38 | axis , it goes through this point and then it | |
| 06:41 | bends over something like this . So this is an | |
| 06:43 | ascent . Oh it never really gets to the axis | |
| 06:46 | right here . So these are negative values of X | |
| 06:48 | negative one , negative two , negative three . And | |
| 06:51 | this is positive one , positive to positive three and | |
| 06:53 | so on . So the law algorithm has values for | |
| 06:56 | 1234567 And all also very , very close to 0.5 | |
| 07:00 | point 3.1 point oh one . But it does not | |
| 07:03 | exist over here . You cannot put a negative value | |
| 07:06 | into a log a rhythm . Um And this comes | |
| 07:08 | about because this law algorithm is a reflection of the | |
| 07:12 | exponential function . It's a reflection of an exponential function | |
| 07:15 | across this diagonal line , Y is equal to X | |
| 07:18 | . So if you can imagine an exponential function doing | |
| 07:21 | something like this , then reflecting it gives the inverse | |
| 07:24 | , and that's why it's down here . So just | |
| 07:27 | like an exponential function doesn't exist down here . A | |
| 07:29 | log rhythmic function does not exist over here because it's | |
| 07:32 | a reflection . So what you need to know is | |
| 07:35 | that negative values are not allowed to be stuck into | |
| 07:38 | law algorithms , and so you cannot have a negative | |
| 07:41 | value like this . So what you say is not | |
| 07:45 | a solution , if you take a negative three and | |
| 07:48 | try to stick it in a calculator , it's going | |
| 07:50 | to give you an error because it's not defined , | |
| 07:52 | it doesn't exist . It's like saying , give me | |
| 07:53 | a square rectangle , square rectangle , I'm sorry , | |
| 07:56 | a square triangle or give me a square circle . | |
| 07:58 | It's not gonna work . There's no it's not defined | |
| 08:01 | . So we throw away this and we say this | |
| 08:03 | is our solution , we have one solution of this | |
| 08:06 | equation . Alright , one solution to this equation . | |
| 08:10 | If you stick to number nine in here and calculate | |
| 08:12 | this log and here and subtract and calculate this log | |
| 08:14 | and add it , then you're gonna get the number | |
| 08:16 | three . Now that was method number one . Using | |
| 08:18 | the definition of the large rhythm , there's always different | |
| 08:20 | ways to do a problem . So I want to | |
| 08:23 | talk briefly about what I'm going to call Method number | |
| 08:26 | two because there's always more than one way to do | |
| 08:31 | something . So let's go around here and try to | |
| 08:34 | keep things as tidy as I can . Okay , | |
| 08:37 | so we have this log arrhythmic equation here , we | |
| 08:39 | combine the logarithms , we get down to this point | |
| 08:42 | , this is the point where we branch off and | |
| 08:43 | we say there's two methods . What else can we | |
| 08:45 | do at this step ? Right here ? Remember we | |
| 08:48 | talked about the fact that a log a rhythm is | |
| 08:51 | the inverse function of an exponential . They are opposites | |
| 08:54 | of one another , they annihilate each other , they | |
| 08:56 | kill each other , they undo each other . Whatever | |
| 08:59 | word you want to do , the opposite of algorithm | |
| 09:01 | is an exponential . The opposite of an exponential is | |
| 09:04 | a logarithms . We've done problems like this in the | |
| 09:07 | past but this is the first time where we're going | |
| 09:09 | to solve more complicated equation using this . So if | |
| 09:12 | I don't want to do this but I want to | |
| 09:13 | start at this point then what I can do is | |
| 09:15 | I can sense this is a logarithms that I want | |
| 09:18 | to get rid of . I can raise both sides | |
| 09:20 | With a base of three . I can say 3 | |
| 09:23 | to the power of lager and um base three X | |
| 09:26 | squared minus six X . Is equal to three to | |
| 09:30 | the power of three . See what I've done . | |
| 09:32 | I can do anything . I want to both sides | |
| 09:34 | of this equation . As long as I'm doing it | |
| 09:35 | to both sides I choose to raise the left hand | |
| 09:37 | and put it in the exponent of the three and | |
| 09:40 | the right hand side . I'm gonna also put it | |
| 09:41 | in the exponent of the three . So I'm basically | |
| 09:44 | the equation is still balanced . I've raised both sides | |
| 09:47 | to the power of three . But here I have | |
| 09:49 | a power base three and then in the expanded to | |
| 09:51 | have a log rhythm based three . So the exponential | |
| 09:55 | base three cancels or annihilates or kills the law algorithm | |
| 09:58 | based three . And all I have left down here | |
| 10:00 | is X squared minus six X poof ! This it's | |
| 10:03 | like you could draw a line through the log and | |
| 10:05 | through the base and this is what I have left | |
| 10:07 | equals nine . Keep one to say nine . This | |
| 10:10 | is three cubed is 27 . Look at what I've | |
| 10:12 | written down here , X squared minus six X . | |
| 10:15 | Is equal to 27 which is exactly what I have | |
| 10:17 | here . X squared minus six , X minus 27 | |
| 10:21 | is equal to zero . It's exactly the same polynomial | |
| 10:23 | we wrote down right here . So whatever way you | |
| 10:26 | want to do , I want you to do that | |
| 10:28 | . However I do want you to start thinking about | |
| 10:31 | oh I want to kill that logarithms . Gonna raise | |
| 10:33 | that thing to an exponential . Oh I want to | |
| 10:34 | kill that exponential . I'm gonna I'm gonna take a | |
| 10:36 | log rhythm on that boy because then I can get | |
| 10:38 | rid of it right ? I know that . It | |
| 10:40 | doesn't really seem like it's something you want to do | |
| 10:43 | in these kinds of problems here . It seems like | |
| 10:45 | maybe it's easier to do it this way and whatever | |
| 10:47 | you want to do . But I'm telling you that | |
| 10:49 | as you get some more complicated problems , you're going | |
| 10:51 | to want to annihilate and eliminate both sides of an | |
| 10:53 | equation by doing the opposite or the inverse operation . | |
| 10:56 | And that means if you have a log then you | |
| 10:58 | take the exponential of both sides to kill that log | |
| 11:01 | and so on and you get exactly the same solution | |
| 11:03 | either way , we're not going to do that every | |
| 11:08 | single time . I'm not going to solve the problem | |
| 11:09 | twice every single time , but a few times along | |
| 11:12 | the way , I will do that problem . Number | |
| 11:14 | two , let's say we have log rhythm and here's | |
| 11:18 | something weird . The base is actually not a number | |
| 11:20 | . It's a letter . Call it a . I'll | |
| 11:22 | explain in a second Of the variable X . is | |
| 11:25 | equal to two times algorithm . Base A three Plus | |
| 11:30 | logger in them . Base A . of five . | |
| 11:34 | All right . So you might look at that and | |
| 11:35 | say I've never seen an equation like that makes no | |
| 11:37 | sense at all . Well , it's because you haven't | |
| 11:39 | seen it before so once you see it the first | |
| 11:41 | time it's not so scary . What's going on here | |
| 11:43 | is you know , normally we have basis of logarithms | |
| 11:46 | . Base two logarithms based three logarithms . Base 10 | |
| 11:48 | logarithms whatever here . I'm just saying I don't care | |
| 11:51 | what the bases its base a some numbers here . | |
| 11:54 | I don't know what it is but it's something and | |
| 11:56 | whatever that is . It's the same base for this | |
| 11:57 | algorithm and it's the same base for this algorithm Is | |
| 12:01 | not what I'm trying to solve for . Though . | |
| 12:02 | I'm trying to solve for the variable X . X | |
| 12:05 | is my unknown . It just so happens that in | |
| 12:07 | this equation base a base a base , it doesn't | |
| 12:10 | actually matter . You're going to find out as we | |
| 12:12 | do the solution that the base the particular base of | |
| 12:14 | this algorithm . It doesn't matter if I have a | |
| 12:17 | base to in all three of those locations are a | |
| 12:19 | base 10 and all three of those locations , the | |
| 12:20 | answer will be the same . So don't let that | |
| 12:23 | stress you out . Just treat it as a base | |
| 12:26 | and follow the rules of logarithms . What do I | |
| 12:29 | want to do again ? There's a couple of different | |
| 12:31 | ways I'm gonna do this , but let's start working | |
| 12:33 | on the right hand side , let's say on the | |
| 12:35 | left log a of the unknown variable X . Here | |
| 12:38 | I have a two times log . So I'm gonna | |
| 12:40 | bring that too and put it in the exponents . | |
| 12:42 | That's one of the rules of logarithms that we learned | |
| 12:44 | in the past . That I can take an outside | |
| 12:47 | multiplier number and move it into the exponents on the | |
| 12:50 | inside of a log rhythm like this . And then | |
| 12:52 | I have logged again base A . Of the number | |
| 12:55 | five . So on the right hand side I have | |
| 12:57 | a log rhythm plus they log rhythm . And I | |
| 12:59 | can now combine those algorithm base A . Of the | |
| 13:03 | number X . Log rhythm , Base A . of | |
| 13:06 | three times 3 is nine times the five . So | |
| 13:10 | basically multiplying inside because multiplying um uh and the inside | |
| 13:16 | of algorithm becomes edition of algorithms . So when I | |
| 13:19 | continue , I have log Base . a . 45 | |
| 13:25 | . And now you see exactly what the solution of | |
| 13:27 | the equation is . I have a base a log | |
| 13:30 | rhythm on the left and a base a log rhythm | |
| 13:32 | on the right . It's the same exact algorithm , | |
| 13:33 | same exact base . So the only difference between the | |
| 13:35 | two is what exes and it has to be equal | |
| 13:37 | to 45 . And that's how you write the answer | |
| 13:40 | . Now I got into a position where everything else | |
| 13:43 | was the same so that X had to be equal | |
| 13:45 | to what I can read off on the right hand | |
| 13:47 | side of the equation . Right ? And now you | |
| 13:49 | can see that it doesn't matter what base you put | |
| 13:51 | in here . If I chose this to be base | |
| 13:54 | to base two in base to all that would happen | |
| 13:57 | is I would get base to here . In base | |
| 13:59 | to hear . X would still be 45 . If | |
| 14:01 | I choose to base 10 , 10 and 10 . | |
| 14:03 | All that's gonna happen is it will be based 10 | |
| 14:05 | in base 10 X will still be equal to 45 | |
| 14:08 | . Don't get so wrapped up when you see that | |
| 14:10 | you don't think you can get to the end from | |
| 14:13 | the beginning . You have to just start working and | |
| 14:15 | start carrying things through logically . Okay , now there's | |
| 14:19 | always one more than one way to do something . | |
| 14:21 | And I want to show you that way . Let's | |
| 14:23 | write the equation over again . Log Base A of | |
| 14:27 | X two times log Base A of three plus log | |
| 14:34 | Base A of five . All right . So , | |
| 14:38 | I told you , I said , you know , | |
| 14:39 | ultimately I want these this X variable over here . | |
| 14:42 | So forget about the rest of this thing . I | |
| 14:43 | want X by itself , but X is wrapped up | |
| 14:46 | inside of algorithm . A bass ale algorithm . How | |
| 14:49 | do I get X by itself ? It's wrapped up | |
| 14:50 | inside of a package . How do I get rid | |
| 14:52 | of that algorithm ? I have to do the inverse | |
| 14:54 | . The opposite function . The opposite of algorithm is | |
| 14:57 | an exponential function . So if I take the left | |
| 15:00 | hand side and I say base A . And I | |
| 15:03 | raise it and put this thing in the exponent base | |
| 15:06 | A . Of X . I have taken what I | |
| 15:09 | have and I've raised it to the exponential of the | |
| 15:11 | base A exponential . So I have an exponential and | |
| 15:14 | log rhythm up here , they cancel each other . | |
| 15:17 | So that's how X . Is gonna drop out on | |
| 15:19 | the left . But I have to do the same | |
| 15:20 | operation on the right hand side . So it'll be | |
| 15:22 | a to the power of two times log A three | |
| 15:27 | plus log a . They say five . I know | |
| 15:32 | it looks crazy . All I'm doing is I'm raising | |
| 15:34 | the left hand side of the equation to the power | |
| 15:36 | A to the power of that and then a to | |
| 15:38 | the power of this . But the nice thing about | |
| 15:40 | it is because log is exactly the inverse of A | |
| 15:43 | . Then on the left hand side the only thing | |
| 15:45 | left is exit drops out . It's like it cancels | |
| 15:47 | with it just like 4/4 . They cancel 5/5 , | |
| 15:51 | they cancel 10/10 . They cancel A . To the | |
| 15:53 | power of log . They canceled . So the X | |
| 15:55 | drops out on the right hand side . I have | |
| 15:57 | all of this stuff . But notice that in the | |
| 16:00 | exponent here I have a plus side . So I | |
| 16:02 | can write this experiment is a to the power of | |
| 16:05 | to log Base A . three multiplied by a log | |
| 16:13 | base A five . How can I write it as | |
| 16:15 | a multiplication ? Because these exponents are added . So | |
| 16:18 | because the base is the same , I can add | |
| 16:20 | these exponents which is exactly what I would have here | |
| 16:23 | . So I'm just breaking up going in reverse . | |
| 16:25 | I have an addition of exponents . I'm gonna treat | |
| 16:26 | as multiplication and I can add these experts and get | |
| 16:29 | what I have uh there . Right ? So now | |
| 16:33 | notice what's going on over here . I have two | |
| 16:35 | times log base A of this . So I need | |
| 16:37 | to take care of that . So it'll be a | |
| 16:40 | log base A three squared . This floats in front | |
| 16:45 | and makes goes into the square here because it's sitting | |
| 16:48 | outside of the law algorithm . That's another one of | |
| 16:50 | our rules of logarithms dealing with Xbox , multiplied by | |
| 16:54 | a raise to the power of log A . To | |
| 16:58 | the fifth . I'm not to the fifth log of | |
| 17:01 | the number five . So here we're very close , | |
| 17:03 | we have X on the left . What do we | |
| 17:05 | have on the right ? The a raise to the | |
| 17:07 | power of log base A . The logs in the | |
| 17:09 | egg completely canceled . All I have is three square | |
| 17:13 | . This cancels cancels . I have the number comes | |
| 17:15 | and drops down the A . And the log base | |
| 17:17 | a cancel here . Leaving me with five now I | |
| 17:20 | have nine times five X . Is equal to 45 | |
| 17:23 | . It's exactly the same answer we got here . | |
| 17:25 | Now you might look at this and say , oh | |
| 17:26 | , there's more steps . I don't like that . | |
| 17:28 | Oh , there's fewer steps . I like this . | |
| 17:29 | I'm trying to show you both ways mostly . So | |
| 17:31 | you understand how it's done because later on , some | |
| 17:35 | equations will be really hard to solve this way , | |
| 17:37 | especially in advanced calculus . So I'm trying to teach | |
| 17:39 | you , you know the path to get from here | |
| 17:42 | to there . All right . You get the same | |
| 17:44 | answer either way , but you have complete freedom on | |
| 17:47 | how to proceed . As long as you're following the | |
| 17:49 | correct rules . All right . Let's go back to | |
| 17:52 | a little more of an easier problem . Let's say | |
| 17:54 | we have low algorithm base B of X plus three | |
| 17:59 | . I don't know what the base is . I | |
| 18:00 | don't care what some bit . Number B equals log | |
| 18:04 | base B of the number one erased us . Of | |
| 18:08 | the number eight minus log base B of the number | |
| 18:12 | two . Yeah . All right . So , there's | |
| 18:14 | lots of different ways to do this . I'm not | |
| 18:16 | going to solve every problem . Two ways . Right | |
| 18:18 | . But you by now know that there's two ways | |
| 18:20 | to do it . I can use the laws of | |
| 18:22 | law algorithms , which is how most textbooks will probably | |
| 18:24 | show you . Or I can because because I want | |
| 18:28 | this variable X to pop out . I can raise | |
| 18:30 | both sides . Base be raised to the to this | |
| 18:33 | and then base be raised to that and I'll get | |
| 18:35 | the same exact answer either way . Just like I | |
| 18:37 | did in the last problem . So you have to | |
| 18:39 | pick away . There is no correct way . I | |
| 18:41 | will do some problems one way and some problems . | |
| 18:43 | Another way . This way let's just use the laws | |
| 18:45 | of logarithms , log base B X plus three on | |
| 18:50 | the right hand side . What do we have A | |
| 18:51 | subtraction of logs . So I can write this as | |
| 18:54 | log base B . Of 8/2 because this minus this | |
| 18:58 | becomes division 8/2 . So what I'll have is log | |
| 19:02 | base B . X plus three of equals log base | |
| 19:09 | B . Of what ? For now you see what | |
| 19:11 | you have log on the left , log on the | |
| 19:13 | right . Base be based B . Everything is the | |
| 19:15 | same . So that means this thing inside of the | |
| 19:17 | log has to be equal to four . X plus | |
| 19:20 | three is equal to four . So then X . | |
| 19:22 | When you subtract the three you just get a one | |
| 19:24 | . This is the final answer . I'm not going | |
| 19:27 | to solve every problem every way . But if you | |
| 19:29 | were to do this the other way , if you | |
| 19:31 | would raise the left hand side , B to the | |
| 19:33 | power of this and the right hand side B to | |
| 19:36 | the power of this , it would look really really | |
| 19:38 | similar to what we did here . It's almost the | |
| 19:40 | same exact problem . Everything would fall out so that | |
| 19:43 | you would still get X is equal to one . | |
| 19:44 | I'm just from now on , I'm just gonna pick | |
| 19:47 | a path . I'm not gonna particularly choose one specific | |
| 19:50 | way to do things . Okay ? Um let me | |
| 19:54 | slide to the next board to do the next problem | |
| 19:57 | because it's a little longer and then our last problem | |
| 19:59 | is actually going to be fairly short . What if | |
| 20:02 | we have log base A . Of the variable x | |
| 20:07 | minus log Base AX -5 is equal to log base | |
| 20:15 | A . Of the number six . Again , I | |
| 20:17 | can do what I want . I can use the | |
| 20:19 | laws of logarithms . I have subtraction of two logarithms | |
| 20:22 | . I could do that . It would probably be | |
| 20:23 | a shorter answer , just like it did in the | |
| 20:25 | last uh the last one . But I really want | |
| 20:28 | you to understand both ways of doing it . So | |
| 20:30 | instead of doing it like this where we use the | |
| 20:33 | laws of logarithms , it'll probably be pretty short to | |
| 20:35 | be honest with you . Let's do it the slightly | |
| 20:37 | longer way . Because I want to show you that | |
| 20:38 | both ways work . All right . So here have | |
| 20:41 | log base a log , base a log base a | |
| 20:45 | right . So what I wanna do is I want | |
| 20:46 | to kill all those algorithms . So on the left | |
| 20:48 | hand side I'll raise it and put all of this | |
| 20:51 | stuff in the exponent of a base A . This | |
| 20:55 | whole thing on the left hand side , AX -5 | |
| 20:59 | . This entire thing is in the exponent of a | |
| 21:01 | . Over here on a log . They say six | |
| 21:07 | . All right . But again , I have a | |
| 21:09 | subtraction in the exponents have subtraction . So I can | |
| 21:12 | write this as log in the expo A of X | |
| 21:16 | multiplied by A . To the head . I want | |
| 21:21 | to do this negative log Base AX -5 . And | |
| 21:27 | on the right hand side the A . Is gonna | |
| 21:29 | kind of cancel with the log because in verses of | |
| 21:31 | one another I'm just gonna have a six now , | |
| 21:33 | make sure you really understand the left hand side . | |
| 21:35 | All I'm saying is that I have the same base | |
| 21:37 | here , so I can add these expos if I | |
| 21:38 | add this to this in the expo and I'm gonna | |
| 21:40 | get exactly what I have before . That's why I | |
| 21:42 | had to have the negative sign there because ultimately they're | |
| 21:44 | subtracted . But so I can work in kind of | |
| 21:46 | the exponential kind of realm . If I want to | |
| 21:49 | know what do I have this log base A cancels | |
| 21:54 | with this exponential with base A . So I just | |
| 21:56 | have the X popping out from the top this A | |
| 21:59 | cancels with this . However , there's one little gotcha | |
| 22:02 | here , you have a negative sign in front before | |
| 22:04 | you can do the cancellation . This is like a | |
| 22:06 | negative one out here . So what I want you | |
| 22:08 | to do is say log base A X minus five | |
| 22:13 | to the negative one . Power this is like a | |
| 22:15 | negative one times this . So forget about everything else | |
| 22:18 | here . If you just had this this negative one | |
| 22:21 | you can float it in and put it as a | |
| 22:23 | power log . Base A X minus five to the | |
| 22:25 | negative one . Log base a X minus five to | |
| 22:27 | the negative one . I want to get rid of | |
| 22:29 | it here and bring it here . So now I | |
| 22:31 | have a very clean cancellation A canceling with log . | |
| 22:34 | Uh This is equal to six so I'm gonna have | |
| 22:36 | X multiplied by the A . It's a base cancels | |
| 22:40 | with log A . So all I have left is | |
| 22:42 | x minus five to the negative one . Let me | |
| 22:47 | just double check myself . And so now I'm home | |
| 22:49 | free . I've got rid of all the algorithms now | |
| 22:51 | it's just a basic algebra problem . Let's bring this | |
| 22:53 | to the bottom X over X -5 to the first | |
| 22:56 | power . I can drop the Prince Caesar leaving their | |
| 22:58 | whatever you want . six . So I have to | |
| 23:00 | solve for X . Mhm . So I'm gonna multiply | |
| 23:05 | the left and the right hand side by the denominator | |
| 23:07 | X -5 on the left . I'm gonna be left | |
| 23:10 | with X on the right . It'll be six times | |
| 23:12 | X -5 . I multiply the left by X -5 | |
| 23:16 | . It cancels . That's what I'm left with on | |
| 23:18 | the right and multiply by the same thing . Here's | |
| 23:19 | what I have . So I have now six X | |
| 23:22 | -30 when I distribute this in . And then I'm | |
| 23:26 | gonna move the six over here by subtraction . So | |
| 23:29 | one minus six is negative five X . And then | |
| 23:34 | I'm going to divide by negative five . I'm going | |
| 23:36 | to get a positive six X . Is equal to | |
| 23:38 | positive six X , correct . All right . And | |
| 23:42 | again , there's there's there's always more than a way | |
| 23:45 | to do the problem . But in this case I | |
| 23:47 | wanted to really show you that you can if you | |
| 23:49 | have logarithms running around your equations , just kill them | |
| 23:52 | . With exponentials , raise the left to an exponential | |
| 23:54 | , the right to an exponential , same base and | |
| 23:57 | all of those logs will get eliminated . All right | |
| 23:59 | now , just for giggles , I want to make | |
| 24:01 | sure you understand that there's obviously an equivalent way of | |
| 24:05 | doing this here . So let's go down here and | |
| 24:07 | I'm gonna show you were going to start with the | |
| 24:09 | black line . How would we do it ? Just | |
| 24:10 | using the laws of logarithms . I have a log | |
| 24:12 | rhythm minus algorithms so I can write this as a | |
| 24:15 | division log base A . of this divided by this | |
| 24:20 | X over X -5 equals this log base A six | |
| 24:27 | . So you see just with a simple line , | |
| 24:29 | I now have basil algorithm based algorithm and the thing | |
| 24:33 | I'm taking the log rhythm of they have to be | |
| 24:34 | equal X over x minus five is equal to six | |
| 24:37 | . This is exactly where I arrived right here . | |
| 24:39 | X over x minus five is six . So from | |
| 24:42 | here the solution is the same . You will get | |
| 24:43 | the same thing . I agree with you that this | |
| 24:45 | is fewer steps . This is typically what you would | |
| 24:48 | see , you know in most textbooks , especially at | |
| 24:50 | the introductory level . But they really want you to | |
| 24:52 | understand how algorithms and exponential can annihilate each other because | |
| 24:55 | when you get farther along in math it becomes incredibly | |
| 24:58 | useful . And so I want to make sure you | |
| 25:00 | understand that . All right , we have one more | |
| 25:03 | problem . It's actually not a very long problem . | |
| 25:05 | So I'm gonna slide back here and do it underneath | |
| 25:09 | but it's a good problem . What if I have | |
| 25:12 | log rhythm based too , X squared minus nine is | |
| 25:17 | equal to four . Now , I just got through | |
| 25:20 | telling you I can kill the algorithm with an exponential | |
| 25:22 | . I can raise both sides to an exponential with | |
| 25:25 | a base to I can do that . But instead | |
| 25:27 | of doing that let's just use the definition of algorithm | |
| 25:29 | just for variety two To the power of four is | |
| 25:33 | equal to what this is inside the log rhythm . | |
| 25:36 | X squared minus nine . Now the law algorithm has | |
| 25:39 | again gone . 2 to the power of four is | |
| 25:41 | 16 . X squared minus nine . I'll move the | |
| 25:45 | nine over by additional have 25 equals X squared . | |
| 25:49 | I never like having it like this . So let's | |
| 25:50 | flip it around 25 and then we'll go over here | |
| 25:54 | . I think you can see what the answer is | |
| 25:55 | gonna be . Uh X is going to be plus | |
| 25:57 | or minus the square root of 25 Because they take | |
| 26:01 | the square root of both sides . So X can | |
| 26:03 | be five . X can be negative five . This | |
| 26:07 | is the final answer . Xs five Xs negative five | |
| 26:10 | . You have two answers you might say whoa , | |
| 26:11 | whoa , whoa , whoa , whoa . You just | |
| 26:13 | told me that I cannot have negative numbers feeding into | |
| 26:16 | a longer term . You just told me I cannot | |
| 26:18 | have negative numbers going into a long rhythm . So | |
| 26:21 | why do I have two answers here ? Well here's | |
| 26:23 | why If I put five in here it's squared . | |
| 26:27 | That's gonna be 25 -9 . That is what I'm | |
| 26:29 | taking the log rhythm of a positive number . If | |
| 26:31 | I put negative five in here I still get positive | |
| 26:34 | 25 I still get the positive number here . I'm | |
| 26:37 | still taking away algorithm of a positive number . So | |
| 26:40 | you have to check both solutions . Don't just throw | |
| 26:42 | it away if it's negative , go and see if | |
| 26:44 | the solution is requiring you to take a log rhythm | |
| 26:47 | of a negative number . Original problem , The original | |
| 26:50 | problem would have me feed a negative three directly into | |
| 26:53 | a log rhythm that's not allowed . So I could | |
| 26:54 | just kill it . But here I had to go | |
| 26:56 | and check and make sure both were valid . So | |
| 26:58 | here we have started the process of learning how to | |
| 27:00 | solve logarithmic equations . There is more work to it | |
| 27:04 | and you have to understand the rules and the laws | |
| 27:06 | of logarithms , but it's not black magic and there's | |
| 27:09 | always more than one way to do it . So | |
| 27:11 | make sure you can solve all of these yourself . | |
| 27:13 | If I were you , I'd solve both of them | |
| 27:14 | two different ways . Just to make sure you understand | |
| 27:16 | . Follow me on to the next lesson . We're | |
| 27:18 | gonna get more practice solving more complicated log arrhythmic equations | |
| 00:0-1 | . |
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