exponentials and logarithms this chapter is focused on functions which are exponential these...
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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
Graphs of Exponential Functions
You need to be familiar with the function;
For example, y = 2x, y = 5x and so on…
1) Draw the graph of y = 2x
3A
xy a 0a where
84211/21/4
1/8y
3210-1-2-3x
Remember:
323
1
2
x
y
1
2
3
4
5
7
6
8
1 2 3-1-2-3
Any graph of will be the same basic shape
It always passes through (0,1) as anything to the power 0 is equal to 1
xy a
Exponentials and Logarithms
Graphs of Exponential Functions
Here are a few more examples of graphs where
3A
xy a
0
5
10
15
20
25
30
-3 -2 -1 0 1 2 3
x
y
y = 3x
y = 2x
y = 1.5x
All pass through (0,1)
They never go below 0
Notice that either side of (0,1), the biggest/smallest
values switch
Above (0,1), y = 3x is the biggest
value, below (0,1), it is the
smallest…
Exponentials and Logarithms
Graphs of Exponential Functions
Here are a few more examples of graphs where
3A
xy a
0
1
2
3
4
5
6
7
8
9
-3 -2 -1 0 1 2 3
x
y
y = 2x
y = (1/2)x
The graph y = (1/2)x is
a reflection of y = 2x
1
2
x
y
12x
y
2 xy
Exponentials and Logarithms
Writing expressions as Logarithms
‘a’ is known as the ‘base’ of the logarithm…
1) Write 25 = 32 as a logarithm …
3B
loga n x xa nmeans that
52 32
2log 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
b) 54 = 625
c) 210 = 1024
310 1000
10log 1000 3
45 625
5log 625 4
102 1024
2log 1024 10
Exponentials and Logarithms
Writing expressions as Logarithms
3B
loga n x xa nmeans that
Find the value of:
a) 3log 81
What power do I raise 3 to, to get
81?
3log 81 4
b) 4log 0.25
What power do I raise 4 to, to get
0.25?
4log 0.25 1 0.25 is 1/4
Remember, 14 1
4
Exponentials and Logarithms
Writing expressions as Logarithms
3B
loga n x xa nmeans that
Find the value of:
c) 0.5log 4
What power do I raise 0.5 to, to
get 4?
0.5log 4 2
d) 5log ( )a a
What power do I raise ‘a’ to, to get
a5?
5log ( ) 5a a 0.5 = 1/2
0.52 = 1/4
0.5-2 = 4
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
Laws of logarithms
You do not need to know proofs of these rules, but you will need to learn and use them:
3D
log log loga a axy x y
log log loga a a
xx y
y
log ( ) logka ax k x
1log loga a xx
(The Multiplication law)
(The Division law)
(The Power law)
Proof of the first rule:
Suppose that;loga x b loga y can
d
ba x ca y
xy b ca a
xy b ca
loga xy b c ‘a must be raised to the power
(b+c) to get xy’
Exponentials and Logarithms
Laws of logarithms
Write each of these as a single logarithm:
3D
log log loga a axy x y
log log loga a a
xx y
y
log ( ) logka ax k x
1log loga a xx
1) 3 3log 6 log 7
3log (6 7)
3log 42
2) 2 2log 15 log 3
2log (15 3)
2log 5
3) 5 52log 3 3log 2
2 35 5log 3 log 2
5 5log 9 log 8
5log (9 8)
5log 72
Exponentials and Logarithms
Laws of logarithms
Write each of these as a single logarithm:
3D
log log loga a axy x y
log log loga a a
xx y
y
log ( ) logka ax k x
1log loga a xx
4) 10 10
1log 3 4log
2
4
10 10
1log 3 log
2
10 10
1log 3 log
16
10
1log 3
16
10log 48
Alternatively, using rule 4
10 10log 3 log 16
10log (3 16)
10log 48
Exponentials and Logarithms
Laws of logarithms
Write in terms of logax, logay and logaz
3D
log log loga a axy x y
log log loga a a
xx y
y
log ( ) logka ax k x
1log loga a xx
1) 2 3log ( )a x yz
2 3log ( ) log log ( )a a ax y z
2log log 3loga a ax y z
2) 3loga
x
y
3log log ( )a ax y
log 3loga ax y
Exponentials and Logarithms
Laws of logarithms
Write in terms of logax, logay and logaz
3D
log log loga a axy x y
log log loga a a
xx y
y
log ( ) logka ax k x
1log loga a xx
3) logax y
z
4)
log log loga a ax y z
1
2log log ( ) loga a ax y z
1log log log
2a a ax y z
4loga
x
a
4log log ( )a ax a
log 4loga ax a
log 4a x = 1
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.
3 20x
10 10log (3 ) log 20x
10 10log 3 log 20x
10
10
log 20
log 3x
1.3010...
0.4771...x
2.73x
‘Take logs’
You can bring the power
down…
Divide by log103
Make sure you use the exact
answers to avoid rounding errors..
3E
(3sf)
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…
1 27 3x x ‘Take logs’
3E
1 2log(7 ) log(3 )x x
( 1) log 7 ( 2) log3x x
log 7 log 7 log3 2log3x x
log 7 log3 2log3 log 7x x
(log 7 log3) 2log3 log 7x
2log3 log 7
(log 7 log3)x
0.297x
Bring the powers down
Multiply out the brackets
Rearrange to get ‘x’s together
Factorise to isolate the x
termDivide by (log7-log3)
Be careful when typing it
all in! (3dp)
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
When you raise a number to a power, the answer cannot be negative…
25 7(5 ) 30 0x x Sub in ‘y = 5x’
3E
2 7 30 0y y
( 10)( 3) 0y y
10y 3y or
5 3x log5 log3x
log5 log3x log3
log5x
0.68x
y2 = 5x x 5xy2 = 52x
Factorise
You have 2 possible answers
‘Take logs’
Bring the power down
Divide by log5
Make sure it is accurate…
(2dp)
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
3F
loga x m
ma x
log ( )mb a log ( )b x
logbm a logb x
m log
logb
b
x
a
loga x log
logb
b
x
a
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)
Exponentials and Logarithms
Changing the base
3F
loga x log
logb
b
x
a
4log 9 10
10
log 9
log 4
4log 9 1.58 (2dp)
Special case
logab x log x
b
a
25log 10 10
5
log 2
43log 10 10
3
log 4
Exponentials and Logarithms
Changing the base
Find the value of log811 to 3.s.f
3F
loga x log
logb
b
x
a
8log 11 10
10
log 11
log 8
8log 11 1.15 (3sf)
Alternatively…8log 11
8 11x
10 10log (8 ) log 11x
10 10log 8 log 11x
10
10
log 11
log 8x
‘Take logs’
Power law
Divide by log108
Exponentials and Logarithms
Changing the base
Solve the equation:
log5x + 6logx5 = 5
3F
loga x log
logb
b
x
a
logab x log x
b
a
5log 6log 5 5xx
5log x5
6
log x
y 6
y5
2y 6 5y
5
2 5 6 0y y
( 2)( 3) 0y y
2 or 3y y
5log 2x
5log 3x
25x
125x
25 x
35 x
Use the ‘special case’ rule
Let log5x = y
Multiply by y
Rearrange like a quadratic
Factorise
Solve for y
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
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