49: a practical application of log laws © christine crisp “teach a level maths” vol. 1: as core...
TRANSCRIPT
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
"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"
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
More Laws of Logs
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