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This chapter focuses on utilizing logarithms to create mathematical models for experimental data. It emphasizes the significance of polynomial and exponential functions and how logarithmic transformations can simplify the analysis. By plotting logarithmic values, we can derive linear relationships, facilitating the identification of constants like the gradient and y-intercept. The chapter includes examples and exercises that guide readers through determining the relationship between variables, fitting models, and understanding the implications of these mathematical techniques in real-world applications.
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Higher Maths Unit 3 Chapter 3 Logarithms Experiment & Theory
Introduction In experimental work we often want to create a mathematical model data can often be modelled by equations of the form: Polynomial function Exponential function
Exponential function Polynomial 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
In the case of a simple polynomial Plotting log y against log x gives us a straight line 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 c m m c
y 2.3 2.2 2.1 2.0 x 1.0 1.1 1.2 1.3 1.4 1.5 1.6 Example The table shows the result of an experiment How are x and y related ? x x A quick sketch x x suggests x x
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 Assume Express equation in logarithmic form Find the relation between x and y From the graph: m = 0.7 c = 0.2 n = 0.7 Relation between x and y is:
Qu. 2 Assume Express equation in logarithmic form Find the relation between x and y From the graph: m = -0.7 c = 0.4 n = -0.7 Relation between x and y is:
When log10y is plotted against log10x, a best fitting straight line Qu. 3 has gradient 2 and passes through the point (0.6, 0.4) Fit this data to the model Given Using c = -0.8 m = 2 n = 2 Relation between x and y is:
When log10y is plotted against log10x, a best fitting straight line Qu. 4 has gradient -1 and passes through the point (0.9, 0.2) Fit this data to the model Given Using c = 1.1 m = -1 n = -1 Relation between x and y is:
Qu. 5 Assume Express equation in logarithmic form Find the relation between x and y From the graph: m = 0.0025 c = 0.15 Relation between x and y is:
Qu. 6 Assume Express equation in logarithmic form Find the relation between x and y From the graph: m = -0.015 c = 0.6 Relation between x and y is:
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 riseto 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 )
