Linear Law & Graphical Logarithms | Class #08 | A2 Mathematics 9709

Generated automatically from the lecture captions, so spoken maths may be transcribed imperfectly.

Okay. So, the main This is a very simple short topic. And there's not much over here that they can ask you. But, the main purpose of this linear law topic, this is related to trigonometry. Oh, sorry. This is related to logs and explanation exponential equation. Logs and exponential equations. And the main purpose of this thing is to linearize a non-linear equation. Non-linear equation. For example, we know that a straight line Not an example right now. We know a straight line has an equation y is equal to m x + c. Right? And your on your x-axis, you have the uh your horizontal axis is your x. And your vertical axis is your y.

And the values on the x-axis are replaced over here in the place of x and the values on the y-axis represent these y values. And then when you when you uh sketch this, you get a straight line. Always. And a straight line will always have y is equal to mx + c as the general equation of a straight line. This is the background. So, now if you have an equation which is non-linear, for example, if you have y is equal to a b power x. Now, you can definitely see this is not in the linear form. It is not in the y is equal to mx + c form. So, the graph of this thing will be also not linear. It would be something else.

It would be a curve something. But, now what we will do, we will use the rules of logs, the rules of exponentials, indices, and we would change our x's a little bit, and we will linearize it. We will convert into a straight line. So, what do we do? First of all, we use the logs, the rules of logs to change it into this format. And how? What do we do? We use logs on both sides. We take ln We can take ln on both sides. So, this would become ln y is equals to ln a b x. Further simplifying, this would become ln y is equals to ln a plus ln b power x. And further simplifying, this would become ln y is equals to

x ln b plus ln a. Now, this is the equation that you have formed. You have the this format. And the linear format is what? The linear is y is equals to m x plus c. So, after comparison over here, we can see that it is somewhat in the same what do you say, format as the y equals to mx plus c. Look how. You can see that now your y is now equals to Yes, ln y. Your y is now ln y. Your x is still x. But the gradient in this case, the gradient in this can be compared with ln b. Yes, your gradient is ln b. And your c, the y-intercept is ln a. >> ln a. >> But still, if you would sketch this line, ln y is equals to x ln b plus ln a

on the on the Cartesian coordinate system where your vertical axis is the y and the horizontal axis is the x, you will still get a curve. Now, to make it linear, to linearize it, what do we do? We change the the x is as well. Now, how would we change the x is 8 minutes? We would change the x is like this. Now, on your y axis after conversion, we can see that your y has now become ln y. So, this means on your y axis it would not be y. It would be now ln y. And your x is still x. So, on your x axis it will be x. And now, if you would draw the sketch the points, this would be the y intercept which

would be ln a. Now, it would become a straight line because on the y axis you have ln y, on the x axis you have x, so the gradient will be ln b and the y intercept will be ln a. This is how you linearize. First of all, you convert it, write it in that format. You separate the terms, you get it in the three terms like y and mx and then c. So, you can have something else as well. It won't be ln y always. You can have something else as well. But then, after converting into this thing, you compare and you then you see what your y has changed into and what your x has changed into. And then, you change the

x is as well accordingly. And then, if you would plot them, this this would this would make a straight line. And any point on this line as well will be equal to not x {comma} y. It will be now x {comma} ln y. Not y. It would be ln y because on your y axis it is ln y now. That's why. So, this is how your this is what everything will be revolved around. Your equation will not be linear. This is not linear. This is one of the examples of the equations that they can give you. There can be many other types of equation that they can give you which will not be linear, which will not be y is equal to mx plus c. But you

would try to make it into that format by using logs or exponential rules. This is and that would make you make you the straight line that you wanted. You linearized a non non non-linear equation. Now there are not many um combinations or different ways that the questions come can come in. We'll do an example for each of them and you will see that this is not a very difficult one. And uh you get very repetitive kind of questions over here as well. So This is a very I directly move on to the past papers you know. And relate the information the knowledge which I give you right now with the question. How will we apply it

over here? I'll do this one and then you can copy down both of them with the example as well. So in this example the equation the actual equation it is what? It is y is equals to capital A which is unknown e minus k x squared. And they have linearized it look this is the graph is given for the same equation but on your y axis you can see it is ln y. And on your x axis it is not x it is x squared. So this is x axis. So and then they have given you two points on this graph as well. So the points also represent not x comma y it will be x squared comma ln y. And then they're asking you to get the values of the unknowns the A and K.

Find >> [snorts] >> A and K. So what can we do now? First of all we convert into a form which is linearized and so that we can uh understand what is the gradient, what is the Y intercept. So, what do we do? We use the ln on both sides. So, ln is equals to ln A E minus K X squared. ln Y is still ln Y. This would become ln A plus ln E minus K X squared. ln Y is still ln Y. ln A and minus K X squared comes down and becomes ln E. ln E we know it is equal to 1. When the base and the inside is same, ln has the log base E. And inside is also E. So, it becomes zero, sorry, one. And so, this becomes ln Y

is equal to minus K X squared plus ln A. And right now, also you can see on the Y axis we have ln Y. On the X axis we have X X squared. And so, your gradient and C. So, you can see that the gradient will be minus K. So, the gradient on this of the two points that they have given you, if you find the gradient, that will be equal to minus A. And the Y intercept will be C. So, this means if you find the gradient, gradient base Y2 minus Y1 X2 minus X1. So, Y2 minus Y1 upon X2 minus X1. This will be equal to minus K because after conversion that's what we've got. So, use your calculator. 0.32 minus 0.76 upon 1.69 minus 0.64.

So, your K is coming out as Give your answer correct to two decimal places. So, minus 0.44. Minus K is equal to minus 0.42, sorry. So, this your K is coming out as minus 0.42. And now they're asking you to get the value of A as well. Now, how can you do that? You can use one of the points, any one of the points as well. So, you can substitute your coordinate, one of the coordinates also that they have given you. Look, it is 0.7 0.64, {comma} 0.76. This one This one. So, the value of ln y is 0.76. So, complete thing 0.76. ln y is equal to 0.76. And minus K, it is 0.419 something which we found over here.

41907 047 something like that. times x squared. The value of x squared for that point is 0.64. plus ln A. Now, you'll move things around and find the values of A. Value of A. So, 0.76 minus 0.419 times 0.64. So, ln A is 0.491 84 is equal to ln A. And to find the value of A, you would you would change it into exponential form. So, this would become A is equal to e power 0.49184. Or taking e on both side. So, e power 0.49184. So, capital A comes out as 1.6 4 correct to two decimal places. So, this is your answer. This is how your questions will be. So, you changed a non-linear into a linear format, and

then you compared, and then you get the gradient, and you compared and you found that your gradient is equal to minus K and your y intercept is ln A. And then you found the values your values of A and K. This is a very typical question. If you have any questions, ask. This is something very unique. I also did not ever think that this could be done as well. But, um this comes in very, very handy sometimes. Because working with a straight line is much easier than working with a curve, right? So, the actual if you would draw the curve, you can't do it without your uh proper curve and formulation and everything.

Okay, let's do this one. Similar to the last thing. So, your actual equation, it is Y is equals to capital A X power M. And then, on your X axis, you can see ln X, and on your Y axis, you can see ln Y. So, somehow they have used logs and after comparison with the linear equation, they have got your They have changed the Y with ln Y and the X with ln X. So, let's see. So, we would apply logs over here ln on both sides. So, ln Y ln K X M ln Y ln K plus ln X M ln Y is equals to M ln X plus ln K. So, you can see Y is equals to M X plus C. So, your Y axis is now ln Y, your X axis is the ln X. And your gradient M is still M, and your

Y intercept is C, which is ln K. Okay. So, if you find the gradient, the Y intercept you we see 0 {comma} 2. So, this means the ln K directly we know that ln K is equal to two, right? The Y intercept. Right now, the Y intercept for ln Y against ln X. You would write ln This is the graph again of ln Y against ln X. Your Y axis is your ln Y. Vertical axis is your ln Y. Horizontal is your ln X. So, you can see Y intercept is two directly. You can compare. So, ln K is two. So, your K will come out as e power two. If you compare convert into exponential format from log form. So, this is your e. And then your

gradient You can find the gradient. Gradient will be equal to m. Which is y2 - y1. So, 10.2 minus two upon 6 minus 0. So, your m comes out as So, easiest five marks, I think. This was even shorter than the previous one. Minus two upon six. So, 41 upon three and they're asking you to write it in correct to two decimal places. So, 1.37. And e power two. 7.4. 7.4 No, 7.39. So, yeah. Any questions? Anything? Okay. >> The variables X and Y satisfy this equation. 5 Y + 1 is equals to 2 3 X. By taking logs, show that the graph of Y against X is a straight line. So, they're telling us that your Y against X, this means

after comparison, you will get your Y on the vertical axis and you will get your X on the vertical axis somehow. They're telling us, let's see. So, then after comparison, we will get the value of the gradient and the intercept, and let's see. So, ln on both sides. So, ln 5 Y + 1 ln 2 3 X. Y + 1 would come down and multiply. And 3 X can come down and multiply. And then ln Y ln 5 + ln 5 is equals to 3 X ln 2. Move everything on one side. So, Y ln 5 is equals to 3 X ln 2 - ln 5. So, Y will be 3 ln 2 upon ln 5 X minus ln 5 upon ln 5. Isn't it? You're dividing ln 5 on this side. So, ln 5 ln 5 cancel

So, this means after simple comparison >> [snorts] >> your gradient comes out as 3 ln 2 upon ln 5. And your C comes out as 1 minus 1. Y intercept is minus 1. And then you can see that it is it is in the form Y is equals to MX + C. And if you make your graph of Y against X, you will get a straight line because it is in that form. Y is equals to MX + C. You don't have to change the X's. So, that's why Y against X will be a straight line. Then, find the exact value of the gradient of this line. So, we have found it. Exact value of the gradient of the line. It is this one. And state the coordinates of the point

at which the line cuts the Y axis. Here it is. It is cutting the Y axis at the Already, everything we have found. 0 {comma} minus 1 is the point where it is crossing the Y axis, which is the Y intercept. And the gradient is this. So, you have done the complete question. Okay, let's do question number one. And then prepare. You will be You people will do the question, the next one. After this one. The variables X and Y satisfy the equation X power N Y is equal to C, capital C. X is equal to 1.1, Y is equal to 5.2, X is equal to 3.2, Y is equal to 1.05. Sometimes, you won't need anything as well. You just uh

You can do substitution and then solve it simultaneously. You can do that as well. For example, in this one, look. X is equal to 1.1, so 1.1 power N and Y is 5.2. 5.2 is equal to C. You form this as one of the equation. The other one would be 3.2 power N into 1.05 is equal to C again. And now you can solve this simultaneously. The C and the C can be substituted. So, 1.1 N 5.2 is equal to 3.2n. And 1.05. And then and then and then what can you do? 1.1 power n divided by 3.2 power n is equals to 1.05 divided by 5.2. And now you can combine them as 1.1 divided by 3.2 using rules of indices. Exponentials.

And 1.0 5 divided by 5.2. And now you can convert them into logs. Convert into logarithmic form so that you can get the answer. So log of the base base is 1.1 upon 3.2. And the inside will be 1.05 divided by 5.2. It is equals to n. You can directly get the answer for the value of n. And then you can substitute it to get the value of c as well. So log this is 1.1 divided by 3.2. And this is 1.05 upon 5.2. So 1.4 9 8 2 is your value of n. And now you want you would want to find the value of c. So 3.2 power this 1.4 9 8 2 multiplied by 1.05 is equals to c. So this answer whatever that you've gotten

3.2 power of this answer multiplied by 1. 05. So 5.998. These are the two answers direct. No no need of doing that that thing as well. So you in the second part explain why the graph of ln y against ln x is a straight line. This is the same thing you would linearize it. You would take ln on both sides and then you will see that y will be replaced with ln y and x will be replaced with ln x. So x power n y is equal to c x power n y is equal to c. So ln on both sides so ln x n y is equal to ln c. ln x n plus ln y is equal to ln c. n comes down ln x ln y ln c. Move the make ln y the subject so ln y

will be equal to minus n ln x plus ln c. So you can see y is equal to mx plus c. So your it is now linearized. So ln y against ln x will be now a straight line. Easy? Any questions?