Click to Start Higher Maths Unit 3 Chapter 3 Logarithms Experiment & Theory.

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Transcript of Click to Start Higher Maths Unit 3 Chapter 3 Logarithms Experiment & Theory.

Page 1: Click to Start Higher Maths Unit 3 Chapter 3 Logarithms Experiment & Theory.

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Page 2: Click to Start Higher Maths Unit 3 Chapter 3 Logarithms Experiment & Theory.

Higher Maths

Unit 3 Chapter 3

Logarithms

Experiment & Theory

Page 3: Click to Start 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

ny ax xy ab

Polynomial function Exponential function

Page 4: Click to Start Higher Maths Unit 3 Chapter 3 Logarithms Experiment & Theory.

ny ax

Polynomial function

xy ab

Exponential function

Often it is difficult to know which model to choose

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A useful way is to take logarithms

(i) For ny ax

log log ny ax

log log log ny a x

log log logy a n x

This is like logY a nX

or rearranging logY nX a Y mX c

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A useful way is to take logarithms

(ii) For xy ab

log log xy ab

log log log xy a b

log log logy a x b

This is like log logY a x b

or rearranging log logY b x a Y mx c

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ny ax

logY nX a

log log logy a n x

Plotting log y against log x gives us a straight lineIn the case ofa simple polynomial

xy ablog log logy a x b

log logY b x a

In the case ofAn exponential Plotting log y against x gives us a straight line

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By drawing the straight line graph

The constants for

the gradient m

the y-intercept c can be found

log log logy a n x ny ax

xy ab log log logy a x b

c m

c m

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1.0 1.62.0

2.3

x

y

2.1

2.2

1.1 1.2 1.3 1.4 1.5

Example The table shows the result of an experiment

x 1.1 1.2 1.3 1.4 1.5 1.6

y 2.06 2.11 2.16 2.21 2.26 2.30

How are x and y related ?

A quick sketch

x

x

x

x

x

x

suggests

ny ax

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Now take logarithms of x and y

0.04 0.08 0.11 0.15 0.18 0.20

0.31 0.32 0.33 0.34 0.35 0.36

We can now draw the best fitting straight line

10log x

10log y

Take 2 points on the line

(0.04, 0.31) and (0.18, 0.35)

0.35 0.310.29

0.18 0.04m

To find c, use: y mx c

0.35 0.29 0.18 c

0.30c

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0.29m 0.30c

Recall our initial suggestionny ax

Taking logs of both sides log log ny ax

log log log ny a x

log log logy a n x

0.30c 0.29m

logY nX a

n = 0.29

10log 0.3a 0.310 1.995...a

a = 2 (1 dp)

0.32y xOur model is approximately

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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

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Qu. 1 Assumeny ax Express equation in logarithmic form

log log ny ax

log log log ny a x

log log logy a n x

From the graph:

m = 0.70.6 0.2

0.6666....0.6 0

m

c = 0.2

0.71.6y xRelation between x and y is:

Find the relation between x and y

ny ax

n = 0.7 10log 0.2a 0.210a 1.584...a

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Qu. 2 Assumeny ax Express equation in logarithmic form

log log ny ax

log log log ny a x

log log logy a n x

From the graph:

m = -0.70.4 0

0.6666....0 0.6

m

c = 0.4

0.72.5y xRelation between x and y is:

Find the relation between x and y

ny ax

n = -0.7 10log 0.4a 0.410a 2.511...a

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Qu. 3

ny ax

When log10 y is plotted against log10 x, a best fitting straight line

log log ny ax

log log log ny a x

log log logy a n x

Given

m = 2 c = -0.8

20.2y xRelation between x and y is:

has gradient 2 and passes through the point (0.6, 0.4)

ny ax

n = 2 10log 0.8a 0.810a 0.158...a

Fit this data to the model

Using

y mx c 0.4 2 0.6 c

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Qu. 4

ny ax

When log10 y is plotted against log10 x, a best fitting straight line

log log ny ax

log log log ny a x

log log logy a n x

Given

m = -1 c = 1.1

112.6y xRelation between x and y is:

has gradient -1 and passes through the point (0.9, 0.2)

ny ax

n = -1 10log 1.1a 1.110a 12.589...a

Fit this data to the model

Using

y mx c 0.2 1 0.9 c

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Qu. 5 Assumexy ab Express equation in logarithmic form

log log xy ab

log log log xy a b

log log logy a x b

From the graph:

m = 0.00250.3 0.15

0.002560 0

m

c = 0.15

1.4 1.01x

y Relation between x and y is:

Find the relation between x and y

xy ab

10log 0.15a 1.412...a 10log 0.0025b 1.005...b

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Qu. 6 Assumexy ab Express equation in logarithmic form

log log xy ab

log log log xy a b

log log logy a x b

From the graph:

m = -0.0150.6 0

0.0150 40

m

c = 0.6

4.0 0.97x

y Relation between x and y is:

Find the relation between x and y

xy ab

10log 0.6a 3.981...a 10log 0.015b 0.966...b

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2002 Paper I

11. The graph illustrates the law ny kx

If the straight line passes through A(0.5, 0) and B(0, 1). Find the values of k and n. ( 4 )

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logeP p

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 logeQ q

b) Show that p and q satisfy a relationshipof the form bp aq

stating the values of a and b. ( 4 )

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