49: A Practical 49: A Practical Application of Log LawsApplication of Log Laws

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 1: AS Core Vol. 1: AS Core ModulesModules

More Laws of Logs

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Module C2

MEI/OCR

More Laws of Logs

If, from an experiment, we have a set of values of x and y that we think may be related we often plot them on a graph.

If the relationship can be approximated by a straight line, a line of best fit can easily be drawn through the data.

However, it is not easy to draw a curve through data.

naxy xaby or If we think that a relationship of the form

fits the data, where a and b are constants, we can use logs to obtain a straight line.

More Laws of Logs

Method: Suppose we believe a relationship of the form exists between x and y. Then,

naxy

Take logs:

)log(log naxy

naxy

Simplify:nxay logloglog ( Law 1 )

This equation now represents a straight line

where

yY log

ylog ( Law 3 ) alog n xlog

cY m X

More Laws of Logs

Method: Suppose we believe a relationship of the form exists between x and y. Then,

naxy

Take logs:

)log(log naxy

naxy

Simplify:nxay logloglog ( Law 1 )

This equation now represents a straight line

where

yY log

ylog ( Law 3 ) alog n xlog

cY m X

More Laws of Logs

X

Method: Suppose we believe a relationship of the form exists between x and y. Then,

naxy

Take logs:

)log(log naxy

naxy

Simplify:nxay logloglog ( Law 1 )

This equation now represents a straight line

where

yY log

ylog ( Law 3 ) alog n xlog

cY m

xX log

More Laws of Logs

Method: Suppose we believe a relationship of the form exists between x and y. Then,

naxy

Take logs:

)log(log naxy

naxy

Simplify:nxay logloglog ( Law 1 )

This equation now represents a straight line

where

yY log

ylog ( Law 3 ) alog n xlog

cY m XxX log

ac logand

More Laws of Logs

Method: Suppose we believe a relationship of the form exists between x and y. Then,

naxy

Take logs:

)log(log naxy

naxy

Simplify:nxay logloglog ( Law 1 )

This equation now represents a straight line

where

yY log

ylog ( Law 3 ) alog n xlog

cY m XxX log

ac log nm and

More Laws of Logs

e.g. 1 It is believed that the following data may represent a relationship between x and y of the form . Draw a suitable straight line graph to confirm this and estimate the values of a and n.

naxy

x 2.5 3.1 4.3 5 5.9 7.1 8.1

y 9 15 26 35 43 57 69

cY m X

naxy ylog alog n xlog

We have seen thatWe have seen that

so, to get the straight line we need to plot a graph of against .ylog xlog

More Laws of Logs

x 2.5 3.1 4.3 5 5.9 7.1 8.1

y 9 15 26 35 43 57 69

log x 0.4 0.49 0.63 0.7 0.77 0.85 0.91

log y 0.95 1.18 1.41 1.54 1.63 1.76 1.84

Using logs to base 10 we get

so the graph is as follows:

More Laws of Logs

ylog alog n xlogcY m X

7150

850

XcY 71

The constant, c, cannot be read off the graph because the intercept on the y-axis is not shown.

From the graph, the gradient, m

Instead, we substitute the coordinates of any point on the graph ( not from the table ). e.g.

)1,40( )40(711 c 30c

More Laws of Logs

We now have

xnay logloglog

XY 7130 and

So, 71n

and 30log a

We finally need to find a so we must get rid of the log. This is called anti-logging.

On the calculator, the anti-log usually shares a button with the log. For base 10 it is marked .

x10

02a30log a ( 2 s.f. )

More Laws of Logs

so, yY log ( as before )

For a relationship of the form we work in a similar way.

xaby

xaby Take logs: )log(log xaby

xbay logloglog ylog alog x blog

mY c X

More Laws of Logs

so, yY log ( as before )

but xX

For a relationship of the form we work in a similar way.

xaby

xaby Take logs: )log(log xaby

xbay logloglog ylog alog x blog

mY c X

More Laws of Logs

so, yY log ( as before )

but xX

For a relationship of the form we work in a similar way.

xaby

xaby Take logs: )log(log xaby

xbay logloglog

blogThe gradient, m =

ylog alog x blog mY c X

More Laws of Logs

so, yY log ( as before )

but xX

For a relationship of the form we work in a similar way.

xaby

xaby Take logs: )log(log xaby

xbay logloglog

blogThe gradient, m =

and c = alog

ylog alog x blog mY c X

More Laws of Logs

so, yY log ( as before )

but xX

We plot against x.

ylog

For a relationship of the form we work in a similar way.

xaby

xaby Take logs: )log(log xaby

xbay logloglog

blogThe gradient, m =

and c = alog

ylog alog x blog mY c X

More Laws of Logs

SUMMARY

naxy xaby The relationships and can both be reduced to straight lines by taking logs.• For ,

naxy )log(log naxy nxay logloglog

)log(log xaby xbay logloglog

• For ,xaby

ylog alog n xlogcY m X

ylog alog x blog

cY mX

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Exercise1. Use a suitable straight line graph to show

that the data fit a relationship of the formxaby

x 5 15 25 30 35 40

y 2.4 6.3 16.3 26.2 42.2 67.9

Estimate the values of a and b to 2 s.f.

2. Explain how you would use a graph to estimate the values of a and n for a set of x and y data thought to fit a relationship of the form naxy

More Laws of Logs

Solutions1.

xaby

Plot against x.ylog

x 5 15 25 30 35 40

log y 0.38 0.80 1.21 1.42 1.63 1.83

040log bm

20log ac

11 b(2 s.f.)

61 a(2 s.f.)

bxay logloglog

More Laws of Logs

2.

naxy • Convert the equation , by taking logs, to get .logloglog xnay

cmXY

• Read off the value where the line meets the Y-axis to find c OR substitute a pair of ( X, Y ) values into .

• Calculate values of and .xlog ylog• Plot a graph of against .)(log Xx )(log Yy

• Measure the gradient, m, of the graph to obtain n.

• Draw the line of best fit.

• Use and antilog to find a. ca log

Solutions

More Laws of Logs

More Laws of Logs

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

More Laws of Logs

SUMMARY naxy xaby The relationships and can both be reduced to straight lines by taking logs.• For ,

naxy )log(log naxy nxay logloglog

)log(log xaby xbay logloglog

• For ,xaby

ylog alog n xlogcY m X

ylog alog x blog

cY mX

Plot againstylog xlog

Plot againstylog x

More Laws of Logs

e.g. 1 It is believed that the following data may represent a relationship between x and y of the form . Draw a suitable straight line graph to confirm this and estimate the values of a and n.

naxy

6957433526159y

8.17.15.954.33.12.5x

cY m X

naxy ylog alog n xlog

We have seen thatWe have seen that

so, to get the straight line we need to plot a graph of against .ylog xlog

More Laws of Logs

6957433526159y

8.17.15.954.33.12.5x

1.841.761.631.541.411.180.95log y

0.910.850.770.70.630.490.4log x

Using logs to base 10 we get

so the graph is as follows:

More Laws of Logs

ylog alog n xlogcY m X

7150

850

XcY 71

The constant, c, cannot be read off the graph because the intercept on the y-axis is not shown.

From the graph, the gradient, m

Instead, we substitute the coordinates of any point on the graph. e.g. )1,40(

)40(711 c 30c

More Laws of Logs

e.g. Use a suitable straight line graph to show that the data fit a relationship of the formxaby

67.942.226.216.36.32.4y

40353025155x

Estimate the values of a and b to 2 s.f. xaby

Plot against x.ylog

1.831.631.421.210.800.38log y

40353025155x

bxay logloglog

More Laws of Logs

040log bm

20log ac

11 b(2 s.f.)

61 a(2 s.f.)

bxay logloglog