Transcript 3)_C2_Exponentials_and_Logarithms

Exponentials and Logarithms
• This chapter is focused on functions
which are exponential
• These functions change at an
increasing/decreasing rate
• Logarithms are used to solve problems
involving exponential functions
Exponentials and Logarithms
y
Graphs of Exponential Functions
8
7
You need to be familiar with the
function;
ya
x
6
5
a0
where
4
3
For example, y = 2x, y = 5x and so
on…
1) Draw the graph of y = 2x
-3
-2
-1
0
1
2
3
y
1/
8
1/
4
1/
2
1
2
4
8
Remember:
2 3
1
-3
x
1
23
2
-2
Any graph of
basic shape
2
3
x
-1
1
y  ax
will be the same
It always passes through (0,1) as
anything to the power 0 is equal to 1
3A
Exponentials and Logarithms
Graphs of Exponential Functions
All pass through
(0,1)
Here are a few more examples of
graphs where y  a x
y = 3x
30
Notice that either
side of (0,1), the
biggest/smallest
values switch
25
20
y
They never go
below 0
15
y = 2x
10
5
y = 1.5x
0
-3
-2
-1
0
x
1
2
Above (0,1), y = 3x
is the biggest
value, below (0,1),
it is the smallest…
3
3A
Exponentials and Logarithms
Graphs of Exponential Functions
The graph y = (1/2)x is a
reflection of y = 2x
Here are a few more examples of
graphs where y  a x
9
y = 2x
8
1
y 
2
7
6
x
y
5
4
y  2
3

1 x
y = (1/2)x
2
1
0
-3
-2
-1
0
1
2
3
y  2 x
x
3A
Exponentials and Logarithms
Writing expressions as Logarithms
log a n  x
means that a x  n
‘a’ is known as the ‘base’ of the
logarithm…
1) Write 25 = 32 as a logarithm…
25  32
log 2 32  5
Effectively, the 2
stays as the ‘first’
number…
The 32 and the 5
‘switch positions’
2) Write as a logarithm:
a) 103 = 1000
103  1000
log10 1000  3
b) 54 = 625
54  625
log5 625  4
c) 210 = 1024
210  1024
log 2 1024  10
3B
Exponentials and Logarithms
Writing expressions as Logarithms
log a n  x
means that a x  n
Find the value of:
a)
log3 81
What power do I
raise 3 to, to get
81?
log3 81  4
b)
log 4 0.25
What power do I
raise 4 to, to get
0.25?
0.25 is 1/4
log 4 0.25  1
Remember, 4 1 
1
4
3B
Exponentials and Logarithms
Writing expressions as Logarithms
log a n  x
means that a x  n
Find the value of:
c)
log0.5 4
What power do I
raise 0.5 to, to get
4?
0.5 = 1/2
0.52 = 1/4
d) log a ( a 5 )
What power do I
raise ‘a’ to, to get
a5?
log a (a 5 )  5
0.5-2 = 4
log0.5 4  2
3B
Exponentials and Logarithms
Calculating logarithms on a Calculator
On your calculator, you can calculate a logarithm.
 Using the log button on the calculator automatically chooses base 10,
ie) log20 will work out what power you must raise 10 to, to get 20
 To work out log20, all you do is type log20 into the calculator!
 log20 = 1.301029996….
 1.30 to 3sf
3C
Exponentials and Logarithms
Proof of the first rule:
Laws of logarithms
You do not need to know proofs of
these rules, but you will need to learn
and use them:
log a xy  log a x  log a y
x
log a    log a x  log a y
 y
log a ( x)  k log a x
k
1
log a     log a x
 x
(The Multiplication
law)
(The Division law)
(The Power law)
Suppose that;
log a x  b and
log a y  c
ab  x
ac  y
xy  a  a
b
xy  a
c
bc
log a xy  b  c
‘a must be raised to the power (b+c)
to get xy’
3D
Exponentials and Logarithms
Laws of logarithms
log a xy  log a x  log a y
log a ( x) k  k log a x
Write each of these as a single
logarithm:
x
log a    log a x  log a y
 y
1
log a     log a x
 x
1) log3 6  log3 7
2) log 2 15  log 2 3
log3 (6  7)
log 2 (15  3)
log 5 32  log 5 23
log3 42
log 2 5
log5 9  log5 8
3)
2log5 3  3log5 2
log5 (9  8)
log5 72
3D
Exponentials and Logarithms
Laws of logarithms
log a xy  log a x  log a y
log a ( x) k  k log a x
Write each of these as a single
logarithm:
x
log a    log a x  log a y
 y
1
log a     log a x
 x
4)
1
log10 3  4 log10  
2
1
log10 3  log10  
2
4
1
log10 3  log10  
 16 
1

log10  3  
 16 
log10 48
Alternatively,
using rule 4
log10 3  log10 16
log10 (3 16)
log10 48
3D
Exponentials and Logarithms
Laws of logarithms
log a xy  log a x  log a y
log a ( x) k  k log a x
Write in terms of logax, logay and logaz
x
log a    log a x  log a y
 y
1
log a     log a x
 x
1)
2)
log a ( x 2 yz 3 )
 x 
log a  3 
y 
log a ( x 2 )  log a y  log a ( z 3 )
log a x  log a ( y )3
2log a x  log a y  3log a z
log a x  3log a y
3D
Exponentials and Logarithms
Laws of logarithms
log a xy  log a x  log a y
log a ( x) k  k log a x
Write in terms of logax, logay and logaz
x
log a    log a x  log a y
 y
1
log a     log a x
 x
x y
3) log a 
 z

4)



loga x  log a
y  log a z
1
2
log a x  log a ( y)  log a z
1
log a x  log a y  log a z
2
 x 
log a  4 
a 
log a x  log a (a) 4
log a x  4log a a
=1
log a x  4
3D
Exponentials and Logarithms
Solving Equations using Logarithms
Logarithms allow you to solve
equations where ‘powers’ are
involved.
You need to be able to solve these
by ‘taking logs’ of each side of the
equation.
All logarithms you use on the
calculator will be in base 10.
‘Take logs’
You can bring the
power down…
Divide by log103
Make sure you
use the exact
answers to avoid
rounding errors..
3x  20
log10 (3x )  log10 20
x log10 3  log10 20
log10 20
x
log10 3
x
1.3010...
0.4771...
x  2.73
(3sf)
3E
Exponentials and Logarithms
Solving Equations using
Logarithms
The steps are essentially the
same when the power is an
expression, such as ‘x – 2’, ‘2x
+ 4’ etc…
There is more rearranging to
be done though, as well as
factorising.
Overall, you are trying to get
all the ‘x’s on one side and all
the logs on the other…
‘Take logs’
Bring the
powers down
Multiply out the
brackets
Rearrange to get
‘x’s together
Factorise to
isolate the x term
Divide by
(log7-log3)
Be careful when
typing it all in!
7 x 1  3x  2
log(7 x 1 )  log(3x  2 )
( x  1) log 7  ( x  2) log 3
x log 7  log 7  x log 3  2 log 3
x log 7  x log 3  2 log 3  log 7
x(log 7  log 3)  2 log 3  log 7
x
2 log 3  log 7
(log 7  log 3)
x  0.297
(3dp)
3E
Exponentials and Logarithms
Solving Equations using
Logarithms
You may also need to use a
substitution method with even
harder ones.
You will know to use this when
you see a logarithm that has a
similar shape to a quadratic
equation..
Let y=5x
y2 = 5x x 5x
y2 = 52x
When you raise a number to a
power, the answer cannot be
negative…
Sub in ‘y = 5x’
Factorise
You have 2
possible answers
‘Take logs’
Bring the power
down
Divide by log5
Make sure it is
accurate…
52 x  7(5x )  30  0
y 2  7 y  30  0
( y  10)( y  3)  0
y  10 or
y3
5x  3
log 5x  log 3
x log 5  log 3
log 3
x
log 5
x  0.68
(2dp)
3E
Exponentials and Logarithms
Changing the base
Your calculator will always give you
answers for log10, unless you say
otherwise.
You need to be able to change the
base if your calculator cannot do
this
You also need to be able to change
the base to solve some logarithmic
equations
Rewrite as an
equation
‘Take logs’ to a
different base
The power law –
bring the m down
Divide by logba
Sub in logax for m
(from first line)
log a x  m
am  x
log b (a m )  logb ( x)
m logb a  log b x
log b x
m 
log b a
log b x
log a x 
log b a
3F
Exponentials and Logarithms
Changing the base
Special case
log b x
log a x 
log b a
b log a x 
b
log x a
log10 9
log 4 9 
log10 4
5log 2 10 
5
log10 2
log 4 9  1.58
3
3log 4 10 
log10 4
(2dp)
3F
Exponentials and Logarithms
Changing the base
Find the value of log811 to 3.s.f
log a x 
log b x
log b a
Alternatively…
log10 11
log8 11 
log10 8
log8 11  1.15
(3sf)
‘Take logs’
Power law
Divide by log108
log8 11
8 x  11
log10 (8 x )  log10 11
x log10 8  log10 11
log10 11
x
log10 8
3F
Exponentials and Logarithms
Changing the base
Solve the equation:
log5x + 6logx5 = 5
log b x
log a x 
log b a
b log a x

b
log x a
log5 x  2
52  x
x  25
Use the ‘special
case’ rule
5 x
Rearrange like a
quadratic
6
 5
log 5 x
6
y
 5
y2  6
 5y
y
3
x  125
log 5 x 
Let log5x = y
Multiply by y
log5 x  3
log5 x  6log x 5  5

y2  5 y  6  0
Factorise
( y  2)( y  3)  0
Solve for y
y  2 or y  3
3F
Summary
• We have learnt what logarithms are
• We have learnt a number of rules which
can be used to manipulate logarithms
• We have also seen how logarithms can
help us solve equations with powers as
unknowns