Solve the system: (Hint: Let A = ln w, B = ln x, C = ln y, and...

Question

Solve the system: (Hint: Let A = ln w, B = ln x, C = ln y, and D = ln z. Solve the system for A, B, C, and D. Then use the logarithmic equations to find w, x, y, and z.)

$\begin{cases}2 \ln w + \ln x + 3 \ln y - 2 \ln z = -6 \4 \ln w + 3 \ln x + \ln y - \ln z = -2 \\ln w + \ln x + \ln y + \ln z = -5 \\ln w + \ln x - \ln y - \ln z = 5\end{cases}$

Accepted Answer

everyone Welcome back in this problem. We are asked to solve the system of equations given below. Okay, we're giving a system of equations. Four variables. And we have four equations. You could choose any method you want to solve this, we could use elimination substitution. Any number of those methods. But when you have four equations with four variables like this, it gets very very messy very quickly. Okay, If you're doing all of those substitution and you're keeping track of all of those terms can get really messy. So when you're given a system like this, sometimes the easiest way to deal with this is to use an augmented matrix. Okay? When we do that, it keeps everything nice and neat. So let's go ahead and write out this system in matrix form. Okay, what we're gonna do is we're going to write the matrix. It's gonna be a four by four matrix. Ok. It's gonna be multiplied by the parameters we're looking for. So the variables we're looking for, which in this case is lawn W. Lawn X. One of why? One of Z. And then we have this equal to that right hand side. 36 negative six negative three. How do we fill in this matrix as four x four matrix? We're gonna take the coefficients corresponding to each variable. Okay, so the first row is going to correspond to the first equation and we're just gonna take the coefficients corresponding to each of these variables W X, Y. Z. So we get one -2 -1. 3. Okay, for the second equation we get 2 - -1.2. For the third equation we get to negative 32 negative six. And for the last equation we get three negative negative five. Okay. And so this is what that matrix vector product looks like. And I've written it out like this. Instead of going right to augmented matrix form to remind you that in this case we don't just have W. X. Y. Z. We have lawn W. Lawn lawn Y. Lawn Z. So when we get the results of this augmented matrix, there's gonna be a little bit more work to do with those lawns. All right? So if we write this as an augmented matrix, we get the following 1223 in the first column negative to negative four negative three negative three in the second column negative one negative 1 to 2 in the third column and three to negative six negative five in the fourth column. Okay? And then we draw the line And the result on the right hand side. 3 6 -6 -3. Okay. Alright. So this is our augmented matrix. Now in order to use this to solve our system of equations. What we want to do is we want to reduce this Okay, to an upper triangular matrix. You okay? We reduce this to an upper triangular matrix and it's going to be much easier for us to solve. Okay, we can work kind of 11 term out of time for our variables and you'll see that when we get there. Okay. So let's start trying to reduce this matrix. The first thing we're gonna do is we're gonna take and let me write the steps in blue. We're gonna take row two. Okay? And we're gonna replace it by rho two minus two times row one. Okay. And so what we're doing is we're getting rid of we're trying to get rid of this first column. So we have a zero there. Okay. Upper triangular, we're only gonna have one number In the first column. Every other row is going to have a zero. Alright, so if we do that well the first row is gonna stay the same 1, -1, 3. With three on the other side. The second row is going to be replaced by the second row minus two times row one. Okay, this is gonna be two minus two, it's gonna give us zero minus four minus negative two times two, negative two times negative two. Sorry. So we're gonna minus four plus four. This is also going to be zero. Okay, we're gonna get negative one minus two times negative one. That's going to give us positive one And then we have -2 times three. This is going to give us -4. On the right hand side, we have 6 -2 times three. That's going to be zero. The remaining rows stay the same. So two negative negative six, negative six and three negative 32 -5 -3. Okay. Alright, so now we're gonna go to the next matrix OK And in this case we are going to do the same thing but we want to get this to that's in row three, column one. We want to get that to be zero. How do we do that? Well, we're going to take row three, similarly to what we did with row two. We're going to replace row three with row three minus two times row one. Okay? And we're gonna get the following matrix. The first row is going to be left alone one negative two negative 13. On the right hand side three 001, On the right hand side zero. Okay, now row three we're subtracting two times 01. So we get two minus two. That gives us the zero here that we wanted. We get negative 3 -2 times negative two. That's gonna give us one. Okay? We have to -2 times negative one. That's going to give us four. We have negative six minus two times three. This is going to give us negative 12 and then we have negative six minus two times three. Again. This gives us negative 12. Again in the final roll, we leave alone three negative 32 negative five, negative three. Okay. All right. Next thing we're gonna do the same thing one more time to get a zero in this first column. Row four. Okay. So in this case we're going to take row four and we're gonna replace it with a row four and now this time because we have a three there we need to subtract three times row one. Okay. Alright. And this is gonna give us draw a little arrow here. This is gonna give us the following matrix. First row stays the same, 2nd, real stays the same. 3rd row stays the same. Okay? And this final row, what do you get? Three minus three times one? Zero negative three minus three times negative two. This is going to give us three, Then we get to -3 times negative one gives us five. We get negative five minus three times three gives us negative 14 And finally we get negative 3 -3 times three. We get negative 12. Okay. Alright, so we've gotten rid of these terms in this first column. We're left with the first row having a one, the rest of them having a zero. Okay, so that's great there. Now we need to keep going and get this into a full upper triangular matrix. Ok? Alright, so let's now try to get a zero In this position of row four, row four, column two. Let's try to get a zero in that position, move up and give ourselves some more space to work. Okay now in order to do this and I'm just going to draw an arrow here. Hopefully this is clear that we're coming around and we're working with the matrix that we left off with their Okay, Alright. We're gonna take row four And we're gonna replace it with row four minus three times row three. Okay. In order to get rid of that three to get a zero in that term, so our matrix is going to become first row stays the same one, negative two negative 13, three, second row stays the same. 001 negative 40, 3rd row stays the same. 01 or -12 -12. And the final row we have 0. We have 3 -30. We get 5 -3 times negative one negative seven. We get -14 -3 times negative 12. This is going to give us 22 and I think I misspoke when I did this negative seven, that should have been 5 -3 times four. Okay. Alright. In this final number negative 12 -3 times negative 12, that's going to give us 24. Okay? Alright. So now what we see we have this first row that has four terms that are non zero. We have this third row has zero and then three terms that are not. Okay, so let's go ahead and move that up. So let's swap row two and row three. Okay? And then we can leave that row alone. So let's take row two and swap it with row three. That's this operation here. Remember we're just doing row operations to try to get this matrix this augmented matrix into upper triangular so that we can solve the system of equations that it represents. Okay. So we're swapping the rows, so row two now is 014, negative 12, negative 12. And row three is 001, negative four zero. The last row stays the same. 00, negative 7, 22 24. Alright, so we're really close. The last thing we want to do is we want to get a zero in this fourth call Earth. Sorry, fourth row third column. We want to get that to be a zero. So in order for that to be zero. Well let's take row four and replace it with row four plus seven times row three. Okay. We don't want to add it to either row 102 because then we're going to introduce ones or numbers back in where we have these first two zeros. So let's add it to row three and we're going to have to add seven times row three in order to cancel out that negative seven. Okay? And this is going to give the following resulting matrix. The first row was the same one, negative two negative 13, 3, 2nd row. The same -12, -12. Third row 001, negative 4000. Okay, now we're taking negative seven and we're adding seven. That's gonna give us zero. And then we're going to do 22 we're going to add seven times negative four. That's going to give us negative six. And then we have 24 plus seven times zero. This is going to be 24. Okay. Alright. And what you'll see is that we now have an upper triangular matrix. Okay? So this matrix forms a triangle of non zeros on the top portion and we have zeros on the bottom. This is exactly what we wanted. Okay. Because this is going to allow us to solve one term at a time. Okay? So when you have an upper triangular matrix like this, what you want to do is you want to start with the bottom? What does the bottom represent? Well this tells us remember the first column relates to lawn W. The second column lawn X. The third column lawn Y. And the fourth column Lindsay. Okay, so this tells us that negative six times lawn Z is equal to 24. Okay? So that's easy. We just have lawns E. And Z. That we have to solve for everything else is numbers. We can go ahead and do that. So we're going to get that lawn Z Is equal to -4. Okay? If we go E to the exponent. Okay? In order to remove the lawn, we're gonna get Z. Is equal to E. To the negative four. And remember if we have a negative exponent we can write this as one over E. To the positive four. Move down, give us some more room. I'll leave our upper triangular matrix so we can see it. Okay, so now we found a Z value. So that's great. We have this value. Now we're gonna move up one row in our upper triangular form of our matrix. And what is that going to tell us? Well, again, the third column represents why? So we have long y minus for lawn, Z is equal to zero. Okay, well now we have two variables or Y. Sorry and Z. But we know what Z is. So we're going to plug in what we know about Z. We're going to have long y minus four lawn Z. Which was negative for equals zero. Well this tells us that long, Y is equal to negative and so Y. Is equal to E. To the negative 16. And again we can write this as one over E. To the 16. Okay? When we have a negative exponent we can write it as positive and put it in the denominator. Alright, so we've done we've done why we have two variables left. Okay, now moving up to the next row. Okay, you'll notice because we have upper triangular. Each row above has one extra variable and that's why you put it into upper triangular. Okay, so we start at the bottom, we sold for one variable. Then we move up to two variables, but we know one of them. Okay now we're moving up to three variables but we're going to know too. So we get long necks Plus 4 1, Y -12. Lawns E. Is equal to -12. Yeah. So long max is what we're trying to find. We know on Y. We just found it as negative 16. We have four times negative 16 -12 times lawns E. Which was negative for a is equal to negative 12. So we get lawn X - Plus 48 equals -12. We get lawn X. Is equal to four. Okay? Which tells us that X. Is equal to E. To the exponent four. Alright. We found X. Or sorry? Yeah X. We found why we found Z. All we have left to do is W. And that's going to come from that top row of our matrix. Okay so looking at the coefficients of the top row, we are going to have lawn W -2. Lawn X minus one. Y plus three. Lawns Is equal to three. Okay, I'm gonna scroll down to give us some more room to work this out. And we are gonna get lawn W minus two times Lawn X. Which was four minus long Y. Which was negative 16 Plus three times lawns which was -4 is equal to three. Okay, so we get lawn W minus eight plus 16 minus 12 is equal to three. Moving all of the numbers to the right hand side. Lawn W. Is equal to seven and so W. Is equal to E. To the seven. Okay. Alright. So that was a long process. But what we did, let me go back to the top here was we took our system of equations, four equations for variables. That would have been really messy to solve a substitution or any other method. We put it into an augmented matrix, we reduced that matrix upper triangular form and then we could solve for each of those variables one at a time. Okay, What we found is that our solution was the point W. X. Y. Z. Where W. Is E. To the seven, X. Is E to the four, Y is one over E to the 16 and Z is one over E. To the four. So that's going to be answer. B. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Logarithmic Properties and Transformations

Solving Systems of Linear Equations

Exponentiation to Reverse Logarithms

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