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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:
y  ax
n
Polynomial function
y  ab
x
Exponential function
y  ax
n
Polynomial function
y  ab
x
Exponential function
Often it is difficult to know which model to choose
A useful way is to take logarithms
(i)
For
y  ax
n
log y  log ax
n
log y  log a  log xn
log y  log a  n log x
This is like
or rearranging
Y  log a  nX
Y  nX  log a
Y  mX  c
A useful way is to take logarithms
(ii)
For
y  ab
x
log y  log ab
x
log y  log a  log bx
log y  log a  x log b
This is like
Y  log a  x log b
or rearranging
Y   log b  x  log a
Y  mx  c
y  ax
n
log y  log a  n log x
Y  nX  log a
In the case of
a simple polynomial
y  ab
Plotting log y against log x gives us a straight line
x
log y  log a  x log b
Y   log b  x  log a
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
y  axn
log y  log a  n log x
c
y  ab x
can be found
m
log y  log a  x log b
c
m
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 ?
y
x
2.3
x
A quick sketch
x
2.2
x
suggests
y  axn
x
2.1
x
2.0
1.0
1.1
1.2
1.3
1.4
1.5
1.6
x
Now take logarithms of x and y
log10 x
0.04
0.08
0.11
0.15
0.18
0.20
log10 y
0.31
0.32
0.33
0.34
0.35
0.36
We can now draw the best fitting straight line
Take 2 points on the line
(0.04, 0.31) and
(0.18, 0.35)
0.35  0.31
m
 0.29
0.18  0.04
To find c, use:
y  mx  c
0.35  0.29  0.18  c
c  0.30
m  0.29
c  0.30
Recall our initial suggestion
Taking logs of both sides
y  ax
n
log y  log ax
n
log y  log a  log x
n = 0.29
log10 a  0.3
n
log y  log a  n log x
Y  nX  log a
a  100.3  1.995...
a = 2 (1 dp)
Our model is approximately
m  0.29
y  2x
0.3
c  0.30
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
y  axn
Express equation in logarithmic form
Find the relation between x and y
y  axn
log y  log axn
log y  log a  log xn
log y  log a  n log x
From the graph:
m
n = 0.7
0.6  0.2
 0.6666....
0.6  0
log10 a  0.2
m = 0.7
 a  100.2
Relation between x and y is:
c = 0.2
 a  1.584...
y  1.6x
0.7
Qu. 2
Assume
y  axn
Express equation in logarithmic form
Find the relation between x and y
y  axn
log y  log axn
log y  log a  log xn
log y  log a  n log x
From the graph:
m
0.4  0
  0.6666....
0  0.6
n = -0.7
log10 a  0.4
m = -0.7
 a  100.4
Relation between x and y is:
c = 0.4
 a  2.511...
y  2.5x
0.7
Qu. 3
When log10 y is plotted against log10 x, a best fitting straight line
has gradient 2 and passes through the point (0.6, 0.4)
Fit this data to the model
y  axn
y  axn
log y  log axn
log y  log a  log xn
log y  log a  n log x
Given
m= 2
n=2
Using
y  mx  c
0.4  2  0.6  c
log10 a  0.8  a  100.8
Relation between x and y is:
c = -0.8
 a  0.158...
y  0.2 x
2
Qu. 4
When log10 y is plotted against log10 x, a best fitting straight line
has gradient -1 and passes through the point (0.9, 0.2)
Fit this data to the model
y  axn
y  axn
log y  log axn
log y  log a  log xn
log y  log a  n log x
Given
m = -1
n = -1
Using
y  mx  c
log10 a  1.1
0.2   1 0.9  c
 a  101.1
Relation between x and y is:
c = 1.1
 a  12.589...
y  12.6 x
1
Qu. 5
Assume
y  ab x
Express equation in logarithmic form
Find the relation between x and y
y  ab x
log y  log abx
log y  log a  log bx
log y  log a  x log b
From the graph:
m
0.3  0.15
 0.0025
60  0
log10 a  0.15
m = 0.0025
 a  1.412...
Relation between x and y is:
c = 0.15
log10 b  0.0025  b  1.005...
y  1.4 1.01
x
Qu. 6
Assume
y  ab x
Express equation in logarithmic form
Find the relation between x and y
y  ab x
log y  log abx
log y  log a  log bx
log y  log a  x log b
From the graph:
m
0.6  0
  0.015
0  40
log10 a  0.6
m = -0.015
 a  3.981...
Relation between x and y is:
c = 0.6
log10 b  0.015
y  4.0  0.97 
 b  0.966...
x
2002  Paper I
11.
The graph illustrates the law
y  kx n
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 rise to the graph shown.
a) Write down the equation of the line in terms of P and Q.
(2)
It is given that
P  log e p and
Q  log e q
b) Show that p and q satisfy a relationship
of the form
b
p  aq
stating the values of a and b.
(4)
THE END