“Teach A Level Maths”
Vol. 1: AS Core Modules
49: A Practical Application
of Log Laws
© Christine Crisp
More Laws of Logs
Module C2
MEI/OCR
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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.
If we think that a relationship of the form
y  ax n or
y  ab x
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 y  ax n
exists between x and y. Then,
Take logs:
y  ax n
log y  log(ax n )
Simplify:
log y  log a  log x

n
( Law 1 )
log y  log a  n log x ( Law 3 )
This equation now represents a straight line
where
Y  c
Y  log y
 mX
More Laws of Logs
Method:
Suppose we believe a relationship of the form y  ax n
exists between x and y. Then,
Take logs:
y  ax n
log y  log(ax n )
Simplify:
log y  log a  log x

n
( Law 1 )
log y  log a  n log x ( Law 3 )
This equation now represents a straight line
where
Y  c
Y  log y
 mX
More Laws of Logs
Method:
Suppose we believe a relationship of the form y  ax n
exists between x and y. Then,
Take logs:
y  ax n
log y  log(ax n )
Simplify:
log y  log a  log x

n
( Law 1 )
log y  log a  n log x ( Law 3 )
This equation now represents a straight line
where
Y  c
Y  log y
 mX
X  log x
More Laws of Logs
Method:
Suppose we believe a relationship of the form y  ax n
exists between x and y. Then,
Take logs:
y  ax n
log y  log(ax n )
Simplify:
log y  log a  log x

n
( Law 1 )
log y  log a  n log x ( Law 3 )
This equation now represents a straight line
where
and
Y  c
Y  log y
c  log a
 mX
X  log x
More Laws of Logs
Method:
Suppose we believe a relationship of the form y  ax n
exists between x and y. Then,
Take logs:
y  ax n
log y  log(ax n )
Simplify:
log y  log a  log x

n
( Law 1 )
log y  log a  n log x ( Law 3 )
This equation now represents a straight line
where
and
Y  c
Y  log y
c  log a
 mX
X  log x
mn
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
y  ax n . Draw a suitable straight line graph to
confirm this and estimate the values of a and n.
x
2.5
3.1
4.3
5
5.9
7.1
8.1
y
9
15
26
35
43
57
69
We have seen that

y  ax n
log y  log a  n log x
Y  c  mX
so, to get the straight line we need to plot a graph
of log y against log x .
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
Using logs to base 10 we get
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
so the graph is as follows:
More Laws of Logs
log y  log a  n log x
Y  c  mX
0  85
From the graph, the gradient, m 
 1 7
05
The constant, c, cannot be read off the graph
because the intercept on the y-axis is not shown.
Instead, we substitute the coordinates of any point
on the graph ( not from the table ). e.g. ( 0  4, 1 )
Y  c  1 7 X  1  c  1  7 ( 0  4 )  c  0 3
More Laws of Logs
We now have log y  log a  n log x
and
Y  0  3  1 7 X
So,
n  1 7
and
log a  0  3
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
x
button with the log. For base 10 it is marked 10 .
log a  0  3

a  2 0 ( 2 s.f. )
More Laws of Logs
For a relationship of the form y  ab x we work in a
similar way.
Take logs:



so,
y  ab x
log y  log(ab x )
log y  log a  log b x
log y  log a  x log b
Y  c X m
Y  log y ( as before )
More Laws of Logs
For a relationship of the form y  ab x we work in a
similar way.
Take logs:



so,
y  ab x
log y  log(ab x )
log y  log a  log b x
log y  log a  x log b
Y  c X m
Y  log y ( as before ) but
X x
More Laws of Logs
For a relationship of the form y  ab x we work in a
similar way.
Take logs:



so,
y  ab x
log y  log(ab x )
log y  log a  log b x
log y  log a  x log b
Y  c X m
Y  log y ( as before ) but
The gradient, m = log b
X x
More Laws of Logs
For a relationship of the form y  ab x we work in a
similar way.
Take logs:



so,
y  ab x
log y  log(ab x )
log y  log a  log b x
log y  log a  x log b
Y  c X m
Y  log y ( as before ) but
X x
The gradient, m = log b and c = log a
More Laws of Logs
For a relationship of the form y  ab x we work in a
similar way.
Take logs:



so,
y  ab x
log y  log(ab x )
log y  log a  log b x
log y  log a  x log b
Y  c X m
Y  log y ( as before ) but
X x
The gradient, m = log b and c = log a
We plot log y against x.
More Laws of Logs
SUMMARY
 The relationships y  ax n and y  ab x can both
be reduced to straight lines by taking logs.
•
For y  ax
n
,


•
For
y  ab x ,


log y  log(ax n )
log y  log a  log x n
log y  log a  n log x
Y  c  mX
x
log y  log(ab )
log y  log a  log b x
log y  log a  x log b
Y  c  Xm
More Laws of Logs
Exercise
1. Use a suitable straight line graph to show that the
data fit a relationship of the form y  ab x
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
y  ax n
More Laws of Logs
Solutions
1.
y  ab x 
log y  log a  x log b
Plot log y against x.
x
5
15
25
30
35
40
log y
0.38
0.80
1.21
1.42
1.63
1.83
m  log b  0  04
 b  1 1
(2 s.f.)
c  log a  0  2
 a  1 6
(2 s.f.)
More Laws of Logs
Solutions
2.
•
Convert the equation y  ax ,n by taking logs, to
get
log y  log a  n log x.
•
•
•
Calculate values of log x and log y .
Plot a graph of log y ( Y ) against log x ( X ).
Draw the line of best fit.
•
Measure the gradient, m, of the graph to
obtain n.
Read off the value where the line meets
the Y-axis to find c OR substitute a pair
of ( X, Y ) values into Y  mX  c .
Use log a  c and antilog to find a.
•
•
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
 The relationships y  ax n and y  ab x can both
be reduced to straight lines by taking logs.
•
For y  ax
n
,


•
For
Plot
y  ab x ,


Plot
log y  log(ax n )
log y  log a  log x n
log y  log a  n log x
Y  c  mX
log y against log x
log y  log(ab x )
log y  log a  log b x
log y  log a  x log b
Y  c  Xm
log y against 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
y  ax n . Draw a suitable straight line graph to
confirm this and estimate the values of a and n.
x
2.5
3.1
4.3
5
5.9
7.1
8.1
y
9
15
26
35
43
57
69
We have seen that

y  ax n
log y  log a  n log x
Y  c  mX
so, to get the straight line we need to plot a graph
of log y against log x .
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
Using logs to base 10 we get
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
so the graph is as follows:
More Laws of Logs
log y  log a  n log x
Y  c  mX
0  85
From the graph, the gradient, m 
 1 7
05
The constant, c, cannot be read off the graph
because the intercept on the y-axis is not shown.
Instead, we substitute the coordinates of any point
on the graph. e.g. ( 0  4, 1 )
Y  c  1 7 X  1  c  1  7 ( 0  4 )  c  0 3
More Laws of Logs
e.g. Use a suitable straight line graph to show that
the data fit a relationship of the form y  ab x
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.
y  ab x 
log y  log a  x log b
Plot log y against x.
x
5
15
25
30
35
40
log y
0.38
0.80
1.21
1.42
1.63
1.83
More Laws of Logs
log y  log a  x log b
m  log b  0  04  b  1 1
(2 s.f.)
c  log a  0  2  a  1 6
(2 s.f.)