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Higher Maths Unit 3 Chapter 3 Logarithms Experiment & Theory

In experimental work Introduction data can often be modelled by equations of the form: we often want to create a mathematical model Polynomial function Exponential function

Polynomial function Exponential function Often it is difficult to know which model to choose

A useful way is to take logarithms (i)For This is like or rearranging

A useful way is to take logarithms (ii)For This is like or rearranging

Plotting log y against log x gives us a straight line In the case of a simple polynomial In the case of An exponential Plotting log y against x gives us a straight line

By drawing the straight line graph The constants for the gradient m the y-intercept c can be found cm c m

x y Example The table shows the result of an experiment x y How are x and y related ? A quick sketch x x x x x x suggests

Now take logarithms of x and y We can now draw the best fitting straight line Take 2 points on the line (0.04, 0.31) and (0.18, 0.35) To find c, use:

Recall our initial suggestion Taking logs of both sides n = 0.29 a = 2 (1 dp) Our model is approximately

Putting it into practice 1.From the Graph find the gradient 2.From the Graph find or calculate the y-intercept Make sure the graph shows the origin, if reading it directly 3.Take logs of both sides of suggested function 4.Arrange into form of a straight line 5.Compare gradients and y-intercept

Qu. 1 AssumeExpress equation in logarithmic form From the graph: m = 0.7 c = 0.2 Relation between x and y is: Find the relation between x and y n = 0.7

Qu. 2 AssumeExpress equation in logarithmic form From the graph: m = -0.7 c = 0.4 Relation between x and y is: Find the relation between x and y n = -0.7

Qu. 3 When log 10 y is plotted against log 10 x, a best fitting straight line Given m = 2 c = -0.8 Relation between x and y is: has gradient 2 and passes through the point (0.6, 0.4) n = 2 Fit this data to the model Using

Qu. 4 When log 10 y is plotted against log 10 x, a best fitting straight line Given m = -1 c = 1.1 Relation between x and y is: has gradient -1 and passes through the point (0.9, 0.2) n = -1 Fit this data to the model Using

Qu. 5 AssumeExpress equation in logarithmic form From the graph: m = c = 0.15 Relation between x and y is: Find the relation between x and y

Qu. 6 AssumeExpress equation in logarithmic form From the graph: m = c = 0.6 Relation between x and y is: Find the relation between x and y

2002  Paper I 11.The graph illustrates the law If the straight line passes through A(0.5, 0) and B(0, 1). Find the values of k and n. ( 4 )

2000  Paper II B11. The results of an experiment give rise to the graph shown. a) Write down the equation of the line in terms of P and Q. ( 2 ) It is given that and b)Show that p and q satisfy a relationship of the form stating the values of a and b. ( 4 )

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