ECM1110: Engineering mathematics
Term 1:
Lecture 14: Exponential and logarithmic
functions
Aileen MacGregor
A.M.Macgregor@exeter.ac.uk
Harrison 271
College of Engineering, Mathematics
& Physical Sciences
Algebra
17)Exponential and logarithmic functions
Log-linear graphs
Log-log graphs
15) Exponential functions
An exponential function is one where a number is raised to a power which
is itself a function of x .
x
x
For example y  2 , y  3 , or y  e x
We can see the graphs of exponential functions
e is a constant = 2.718… and is very
important.
It turns up throughout many scientific
and natural processes.
Notice
 that all graphs go through the point (0,1)
 ex > 0 for all x
 The graph of ex is steeper than 2x but not
as steep as 3x
x
The function y  e has the property that for all values of x
the slope of the graph at a point is equal to the value of the function y.
Discharge of a capacitor
Here, before the switch is closed, the
capacitor has a voltage V across it.
C
R
Suppose the switch is closed at time t = 0.
A current then flows in the circuit and the
voltage v across the capacitor decays with
time.
The voltage across the capacitor is given by the piecewise function
V
v  t
Ve RC
t0
t0
Logarithmic functions
The logarithmic and exponential functions are related – one is the inverse
of the other.
We talk of logs to a certain base and write log2 x for the log of x to base 2.
This means ‘what power does 2 have to be raised to give x’.
If
y  log2 x
then
2 y  2log 2 x  x
So 2 raised to the power
log 2 of any number x returns the result x.
In general a raised to the power loga cancel each other out.
Bases which are frequently used are 10 and e.
Write
And
logex as ln x
log x is usually used for
log10 x
Rules of logs
There are standard rules for logs to any base which can be derived
from their relationship with the exponential functions.
loga x  loga y  loga xy
x
log a x  log a y  log a
y
log a x p
 p log a x
And remember
that
loga a
loga 1
1
0
Examples
1) Express in log notation
a)
b)
4  64
3
2) a) Find x if
3) Simplify
4) Express
5) Simplify
log 2 32  x
2
27 3
9
b) Find y if
y  log 3 729
log a 4  2 log a 3  log a 6
x3
log a 2 in terms of
y z
log a x, log a y, log a z
log 2 8  log16 2
6) In a vibrational system, the logarithmic decrement, δ, is defined as
 ket1 
  ln t 
2
 ke

where ζ is the damping ratio, ω is the angular frequency, t1 ,t2 represent
time and k is a constant.
Show that
  t2  t1 
ln x
2
1.5
1
0.5
0
-0.5 0
1
2
3
4
5
-1
-1.5
-2
-2.5
Notice that:
the graph goes through the point (1,0)
ln(x) only exists for x > 0
y = lnx is a reflection of y = ex in the line y = x
John Napier
Scotland
1550 – 1617
Log-linear graphs
Suppose we have the following data where we believe the values of x and
x
y are related by a law of the form y  ab where a and b are constants.
x
y
0.6
4.5
1.3
7.4
1.9
11.2
2.4
15.8
3.7
39.0
4.5
68.0
6.5
271.5
Verify that the law relating y and x is of the form y  ab x and determine the
approximate values of a and b.
y  ab x
 
log y  log ab x
log y  x log b  log a
compare with
y = mx + c
straight line of
gradient log b
intercept log a
Making a table of values of x, y and log y, gives
x
y
log y
0.6
4.5
0.65
1.3
7.4
0.87
1.9
11.2
1.05
2.4
15.8
1.20
3.7
39.0
1.59
4.5
68.0
1.83
6.5
271.5
2.43
log y vs x
3
2.5
log y
2
1.5
1
0.5
0
-1
0
1
2
Thus the law is verified,
3
x
y  ab x
4
5
6
7
is the relationship between x and y.
Now to find a, intercept is approximately at 0.47
Thus
log a  0.47
 a  100.47
a = 2.95
To find b, choose two points on the graph and calculate the gradient
gradient  log b 
2.0  0.75
 0.3125
5 1
 b  100.3125
b = 2.05
law relating x and y is
approximately
y  32 
x
Or log-linear graph paper.
Linear scale
Logarithmic scale
one-cycle paper.
The number of times
the pattern of markings
is repeated signifies
the number of cycles,
the distance each
cycle occupies on a
logarithmic scale being
the same.
one cycle can be used
to signify values from
0.1 to 1,
or from 1 to 10,
or from 10 to 100 etc.
To depict a set of numbers
from 0.3 to 189 say, would
require 4 cycles
(0.1 to 1,
1 to 10,
10 to 100,
and 100 to 1000).
first cycle
second cycle
3 cycle log-linear paper
third cycle
Example
The quantities x and y are believed to be related by a law of the form y  ab x
The values of x and corresponding values of y are shown
x
y
−0.9
2.5
0.25
6.0
0.9
10.0
2.1
25.0
2.8
42.5
3.7
85.0
4.8
198.0
Plot the graph on log-linear graph paper, and determine the values of a and b.
y  ab x
log y  log a  x log b
gradient  log b
log102  log10 1


3.9  0.9
3
 b  10

1
3
y  5 2.15 x
x = 0 , a = 5.0
= 2.15

Similarly, use log-log graph paper to graph functions of the form
Using rules of logs
y  ax n
 
log y  log ax n
log y  log a  n log x
Compare with Y = mX+ c
So using log-log paper to plot y  ax n
gives a straight line with gradient n, and intercept at a.
third cycle
second cycle
first cycle
1 x 3 log-log graph paper
one cycle
Example
The power dissipated by a resistor was measured for various values of current
flowing in the resistor and the results are shown below:
Current, I, amperes
1.3
2.4
3.7
4.9
5.8
Power, P, watts
37
127
301
528
740
Prove that the law relating current and power is of the form P  RI n
where R and n are constant, and determine the law.
P  RI n
log P  log R  n log I
,
gradient = n

log 300  log 50
 1.98
log 3.7  log1.5
y  22.5 I 2
x = 0 , R = 22.5