Transcript No Slide Title

Higher
Maths:
Unit
3.3
Higher
Maths:
Unit
3.3
Higher
Maths:
Unit
3.3
Higher
Maths:
Unit
3.3
Higher
Maths:
Unit
3.3
Higher
Maths:
Unit
3.3
Higher
Maths:
Unit
3.3
Higher
Maths:
Unit
3.3
Higher
Maths:
Unit
Higher Maths: Unit3.3
3.3
The Exponential &
Logarithmic Functions
Exponential Growth & Decay
Worked Example:
Joan puts £2500 into a savings account earning 13%
interest per annum. How much money will she have if
she leaves it there for 15 years?
nnn
nn
Let £A(n) be the amount in her account after n years,
then:
A(0)  2500
A(1)  2500  0.13 2500  2500 1.131
A(2)  A(1)  0.13  A(1)  2500 1.132
A(3)  A(2)  0.13 A(2)  2500 1.133
A(n)  A(n 1)  0.13 A(n 1)  2500 1.13 n
A(15)  2500 1.1315  15635.68
n
Example 1:
The population of an urban district is decreasing at the
rate of 2% per year.
(a) Taking P0 as the initial population, find a formula for
the population, Pn, after n years.
(b) How long will it take for the population to drop from
900 000 to 800 000?
(a)
?
P(1)  P0  0.02P0  0.98P0
P(2)  P1  0.02P1  0.98P1  0.98  0.98P0  0.982 P0
P(3)  P2  0.02P2  0.98P2  0.98  0.982 P0  0.983 P0
Suggests that
Pn  0.98 P0
n
Using our formula
Pn  0.98 P0
n
And setting up the graphing calculator
In the fifth year the population drops
below 800 000
Example 2:
The rabbit population on an island increases by 15%
each year. How many years will it take for the
population to at least double?
Let
P0 be the initial population
P(1)  P0  0.15P0  1.15P0
2
P(2)  P1  0.15P1  1.15P1  1.15 P0
P(3)  P2  0.15P2  1.15P2  1.153 P0
P(n)  1.15n P0
Set
P0  1
After 4 years the
population
doubles
A Special Exponential Function – the “Number” e
The letter e represents the value 2.718….. (a
never ending decimal). This number occurs often
in nature
f(x) = 2.718..x = ex
is called the exponential
function to the base e.
Although we now think of logarithms as the exponents to which one must raise the base
to get the required number, this is a modern way of thinking. In 1624 e almost made it
into the mathematical literature, but not quite. In that year Briggs gave a numerical
approximation to the base 10 logarithm of e but did not mention e itself in his work.
Certainly by 1661 Huygens understood the relation between the rectangular hyperbola
and the logarithm. He examined explicitly the relation between the area under the
rectangular hyperbola yx = 1 and the logarithm. Of course, the number e is such that the
area under the rectangular hyperbola from 1 to e is equal to 1. This is the property that
makes e the base of natural logarithms, but this was not understood by mathematicians at
this time, although they were slowly approaching such an understanding
Example 3:
The mass of a fixed quantity of radioactive substance
decays according to the formula m = 50e-0.02t,
where m is the mass in grams
and t is the time in years.
What is the mass after 12 years?
Linking the Exponential Function and the
Logarithmic Function
In chapter 2.2 we
x
f ( x)  2
y
found that the
f 1 ( x) log 2 x exponential function
has an inverse
(0,1)
function, called the
logarithmic function.
(1, 0)
x
log a 1  0
log a a  1
y  a x  x  log a y
The log function is the inverse of the exponential
function, so it ‘undoes’ the exponential function:
f(x) = 2x
ask yourself:
1
2
21 = 2
so
log22 = 1
“2 to what power gives 2?”
2
4
22 = 4
so
log24 =
“2 to what power gives 4?”
3
8
23 = 8
so
log28 =
4
16
24 = 16
so
log216 =
2
3
“2 to what power gives 8?”
4
“2 to what power gives 16?”
f(x) = log2x
Example 4:
(a)log381 =
4
“….
3 to what power gives 81
….?”
(b)log42 =
1
2
“….
4 to what power gives 2….?”
(c)log3  1  = -3
 27 
 1 
“….
3 to what power gives  ….?”

27
 
Rules of Logarithms
Rules for indices:
Rules for Logs:
a a  a
m n
loga xy  loga x  loga y
a a  a
mn
x
log a    log a x  log a y
 y
m
n
m
n
a 
m n
a
mn
loga x  p loga x
p
Example 5:
Simplify:
a)
Since
103  1000
log102 + log10500
 log10 (2  500)
 log10 1000
3
b) log363 – log37
 63 
 log 3  
 7 
 log3 9
2
Since
32  9
c)
1
1
log 2 16  log 2 8
2
3
1 1 1 1
1 1 1 1
1
1 2 2 2 2 2 12 13 3 3 3 13
log 16  log 8
2
3
3
 log2 16  log2 8
 log2 4  log2 2
 2 1
Since
 1
2
Since
log2 4  2
2
log2 2  1
Using your Calculator
You have 2 logarithm buttons on your calculator:
10 x
log which stands for log10 and its inverse log
ex
ln
which stands for loge and its inverse ln
Try finding log10100on your calculator
2
Solving Exponential Equations
Solve
5  11
x
51 = 5 and 52 = 25 so we can see that x lies
between ……and…………..
1
2
Taking logs of both sides and applying the rules
log10 5  log10 11
x x x x x x
x
x log10 5  log10 11
log10 11
x
 1.489
log10 5
x
For the formula P(t) = 50e-2t:
a) evaluate P(0)
b) for what value of t is P(t) = ½P(0)?
20
P(0)  50e
 50
1
1
b)
P(0)   50  25
2
2
2t
25  50e
Could we have
known this?
1
2 t
e
2
1
2t
ln    ln  e 
2
a)
1
2t
ln    ln  e 
2
0.693  2t ln  e
0.693  2t 1
0.693
t
2
0.346  t
Example
The formula A = A0e-kt gives the amount of a
radioactive substance after time t minutes. After 4
minutes 50g is reduced to 45g.
(a) Find the value of k to two significant figures.
(b) How long does it take for the substance to
reduce to half it original weight?
(a)
t4
A(0)  50
A(4)  45
45  50e

Take logs of
4 k

both sides  ln(45)  ln  50e
4 k
 ln(45)  ln 50  ln e
4 k
 ln 45  ln 50  ln e
4 k
4 k
ln 45  ln 50  ln e
 45 
 ln    4k ln e
 50 

0.1054  4k

0.1054
k
4

0.0263  k
remember
ln e  1
(b) How long does it take for the substance to
reduce to half it original weight?




1
0.0263t
A(0)  A(0)e
2
1
0.0263t
e
2
1
0.0263t
ln    ln  e

2
0.693  0.0263t ln  e
0.693  0.0263t  t  26.35
Experiment and Theory
When conducting an
experiment scientists may
analyse the data to find if a
formula connecting the
variables exists.
Data from an experiment
may result in a graph of the
form shown in the diagram,
indicating exponential
growth. A graph such as
this implies a formula of the
type y = kxn
y
x
We can find this formula by using logarithms:
log y
If
y  kx
Then
log y  log kx
n
n
log x
So
n
n
log
k
log y  log kx  log x
logkxk  n log x
log y  log
Y  mX  c
n
Compare this to
log y  Y
So
mn
c  log k
logkxk  n log x
log y  log
n
Is the equation of a
straight line
From
log y
y  kx
n
We see by taking
logs that we can
reduce this problem
to a straight line
problem where:
log x
So
And
logkxk  n log x
log y  log
log y  Y
n
mn
c  log k
Worked Example:
The following data was collected during an experiment:
x
50.1
194.9
501.2
707.9
y
20.9
46.8
83.2
102.3
a) Show that y and x are related by the formula y = kxn
.
b) Find the values of k and n and state the formula that
connects x and y.
a)
Taking logs of x and y gives:
logx
logy
1.69
1.32
2.28
1.67
2.70
1.92
2.84
2.00
Plotting these points
We get a
straight line
and hence the
formula
connecting X
and Y is of the
form
log10 Y
3
2.5
2
1.5
1
y  mx  c
0.5
0.5
1
1.5
2
2.5
3
log10 X
b)Since the points lie on a straight line, we can say
that:
If
y  kx
Then
log y  log kx
So
n
n
n
n
log
k
log y  log kx  log x
logkxk  n log x
log y  log
Y  mX  c
n
Compare this to
By selecting points on the graph and substituting into this
equation we get using
1.69,1.32  2.84, 2.00
1.32  1.69m  c
2.00  2.84m  c
Subtract
0.68  1.15m
0.68
m
1.15
0.6  m
So
1.32  1.69  0.6  c
0.3  c
So we have
Y  0.6 X  0.3
Compare this to
logkxk  n log x
log y  log
n
n  0.6
so
log k  0.3
and
log k  0.3
solving
k  10
0.3
k  1.99
so
y  kx  y  1.99x
n
0.6
You can
always
check this
on your
graphics
calculator