Transcript Welcome to Year 11 Mathematics B

Welcome to Year 11
Mathematics B Term 3
Topic 5: Exponential &
Logarithmic Functions I
Index Laws
Solutions of equations involving indicies
Definitions of ax and logax for a > 1
Logarithmic laws
Graphs of, and the relationships
between, y = ax , y = logax for different
values of a (done via an assignment)
Applications of exponential and
logarithmic functions
Index Laws
An expression in the form an is called a
power.
a is the base
n is the index
There are a number of rules involving
powers
1. am  an = am+n
Consider
100  10 000 = 1 000 000
+
=
102  104 = 106
2. am  an = am-n
Consider
100000
 100
1000
=
105  103 = 102
3. (am)n = amn
Consider
(102)3 = 102  102  102 = 106
4. (ab)m = am.bm
Consider
(3t)4
= 3t  3t  3t  3t
= (3  3  3  3)  (t  t  t  t)
= 34  t4
Exercise 9.1, page 335, No. 5 & 8
5. a0 = 1 (a  0)
103 = 1 000
102 = 100
101 = 10
100 = 1
 10
 10
 10
6.
a
m
1
 m
a
103 = 1 000
102 = 100
101 = 10
100 = 1
1
10-1= 0.1 =101
3
5  ?
1
5  3
5
3
Exercise 9.1 Page 335
No. 1, 2, 5 & 8
7.
1
n
a  n a1
Use your calculator to evaluate …
9
1
2
25
1
2
1
2
81
27
1
3
1
n
7.
a  n a1
Use your calculator to evaluate ...
1
2
9  3  2 91
1
2
25  5  2 251
1
2
2
1
3
3
81  9  81
1
27  3  27
1
Fractional indices produce
surds (roots)
m
n
8.
a  a
n
m
Use your calculator to evaluate these.
3
2
2
5
2
2
9  27  9
3
4  32  4
2
3
5
27  9  27
3
2
Consider
3
4
16  16
4
3
Do you raise to the power of 3 first, or
do you take the 4th root first?
Do it both ways and check for yourself
4
4
16  4096  8
3
16 
3
4
 16   2
4
3
3
8
Therefore, it can be done both
ways!
Exercise 9.1 Page 335
No. 3, 4, 6 & 7
Summary of Index Laws
For all real values of a, b, m and n
 =
am  an = am-n
(am)n = amn
(ab)m = am.bm
am
an
am+n
a0 = 1 (a  0)
1
-n
a = a
n
a  a
1
m
m
1
a n  n am
m
1. Set 18.2 worksheet
2. Additional Exercises 1.2
3. Worksheet 7.1
Numbers 1 – 5
4. Q-Set Exponential 1
Equations With Indices
am = an, then we have powers with the
same base being equal, therefore their
indices must also be equal.
 Solving the most simple index equations
involves this concept, e.g.
 If
Solve for x if 2x = 32
2x = 32
2x = 25
x=5
Try these
3x = 81
5x = 78 125
49x = 7
Index Equations
If the bases are not the same, we must
attempt to express both sides of the
equation as powers of the same base.
Consider this.
9x = 243
We see that 243 is not an integer power
of 9, so we must find a common base
that we can express both 9 and 243 in.
9 = 32
243 = 35
If
9x = 243
Then (32)x = 35
32x = 35
2x = 5
x = 2.5
Now try these.
2x = 43
9x = 2 187
8x = 47.5
Graphic Calculator Solutions
What can we do when the numbers in the
equation cannot be written as powers
with a common base?
Consider 3x = 12
Method 1. A Graphical Approach
1.
2.
3.
4.
5.
Enter y = 3x into Y1 in the Y= editor
Enter y = 12 into Y2 in the Y= editor
Set window range, x = -1 to 4, y = 0 to
14
Graph
Choose intersect from the Calculate
menu and follow the prompts
1.
2.
Method 2. A Calculator Approach
Choose the Solve option from the
menu (Menu 3, 1).
Enter the equation, be sure to have it
rearranged to equal zero.
solve(3^(x)-12=0,x)
Attempt the following
Use both methods
1.
2.
3.
4.
5.
6.
2x = 9
2x = 15
1 – 5x = 25
3x + 7 = 21
10x = 25
7 – 10x = -49
Additional Exercises 10.2
Numbers 2 – 10
Worksheet 7.1
Numbers 6 - 10
Exercise 9.2 Page 345
No. 1 & 2
Additional Exercises 10.2
Number 1
Quadratic Form
1.
2.
3.
22x – 5(2x) + 4 = 0
This equation has three features that
make it a quadratic equation.
The bases are the same (base 2).
One index is twice the other.
There are three terms
A simple piece of substitution will enable
us to factorise this as a normal
quadratic.
22x – 5(2x) + 4 = 0
A simple piece of substitution will enable
us to factorise this as a normal
quadratic.
22x – 5(2x) + 4 = 0
1. Let A = 2x
A simple piece of substitution will enable
us to factorise this as a normal
quadratic.
22x – 5(2x) + 4 = 0
1. Let A = 2x  22x = (2x)2 = A2
A simple piece of substitution will enable
us to factorise this as a normal
quadratic.
22x – 5(2x) + 4 = 0
1. Let A = 2x  22x = (2x)2 = A2
2.  Equation becomes A2 – 5A + 4 = 0
3. Factorise:
(A – 4) (A – 1) = 0
4. Solve: A = 4
2x = 4
x=2
or
A=1
2x = 1
x=0
Logarithms
A logarithm is an index.
Number
1
10
100
1 000
Logarithms
A logarithm is an index.
Number
1
10
100
1 000
Power
of 10
100
101
102
103
Logarithms
A logarithm is an index.
Number
1
10
100
1 000
Power
of 10
100
101
102
103
Log (Base
10)
0
1
2
3
Earthquakes, Sound & Logs
Earthquake magnitude is measured by
the amount of ground movement. This
can vary from fractions of a millimetre to
metres.
Numbers with such a wide range are
difficult to fit on a number line when
graphing.
To overcome this, the Richter scale is
based on exponents, base 10.
Richter Scale
1000000000
900000000
800000000
700000000
600000000
500000000
400000000
300000000
200000000
100000000
0
0
1
2
3
4
5
6
7
8
9
deci-Bells
Using a linear scale, the difference
successive values on a number line is
the same.
0
1
2
3
4
5
6
One unit greater
•On a logarithmic scale, the ratio
between successive values is the same.
10 time greater
0
1
2
3
4
5
6
If ax = N
Index
then x = loga N
If ax = N
Base
then x = loga N
If ax = N
Number
then x = loga N
If ax = N
then x = loga N
i.e.
ax = N  x = loga N
(N > 0, a > 0, a  1)
Why not???
Without use of a calculator, evaluate
Log10 100 000
Either…
Or…
Let x = log10 100 000
x = log10 105
x=5
Let x = log10 100 000
10x = 100 000
10x = 105
x=5
Logs (to base 10), are referred to as
common logs, and the base 10 can be
left out.
i.e. log10 100 000  log 100 000
Whereas log5 200 must include the base.
Evaluate log3 243
Either…
Or…
Let x = log3 243
x = log3 35
x =5
Let x = log3 243
 3x = 243
3x = 3 5
 x = 5
log 0.01
Let x = log 0.01
10x = 0.01
10x = 10-2
x = -2
log 10
let x  log 10
10 x  10
10 10
x
1
2
x  0.5
Additional Exercises 10.3
Properties of Common
Logarithms
1. loga MN = loga M + loga N
log10 (100  1000) = log10 100 + log10 1000
L.H.S. = log10 (100  1000)
R.H.S. = log10 100 + log10 1000
= log10 (102  103)
= log10 102 + log10 103
= log10
=5
(105)
=2+3
=5
= L.H.S.
Evaluate log3 (9  81)
= log3 9 + log3 81
= log3 32 + log3 34
=2+4
=6
i.e. 9  81 = 36
Express as a single logarithm
log3 5 + log3 7
= log3 (5  7)
= log3 35
M
2. loga
= loga M – loga N
N
100000
log 10
 log 10 100000  log 10 1000
1000
5
3
log 10 100  log 10 10  log 10 10
log 10 10  5  3
2
22
Express as a single logarithm
log2 9 – log2 5
= log2 (9  5)
= log2 1.8
Try These
log3 5 + log3 7 e) log4 3 – log4 2
b) log2 6 + log2 9 f) log6 50 + log6 2
c) log8 6 – log8 3 g) log2 6 + log2 3 – log25
d) log3 6 – log3 5 h) log5 6 + log5 10 – log5 4
a)
3. loga Np = p.loga N
Consider log10 1002 = 2.log10 100
L.H.S.
= log10 1002
= log10 (102)2
= log
4
10
10
=4
R.H.S.
= 2.log10 100
= 2.log10 102
=2  2
=4
= L.H.S.
Evaluate
log3 813
= 3.log3 81
= 3.log3 34
=34
= 12
Evaluate
log10 1
= log10 100
=0
 loga 1 = 0
Evaluate
log10 10
= log10 101
=1
 loga a = 1
Summary of Logarithm Laws
1.
If N = ax then loga N
(N > 0, a>0, a  1)
2.
loga MN = loga M + loga N
3.
M
log a
 log a M  log a N
N
4.
loga a = 1
5.
loga 1 = 0
6.
loga ax = x
7.
loga Np = p.loga N
8.
a
loga N
N
Find the value of
2 log3 6 – log3 4
= log3 62 – log3 4
= log3 36 – log3 4
= log3 (36  4)
= log3 9
= log3 32
=2
Given that log 5.24  0.7193
Find the value of log 5 240

5 240 = 5.24  1000
log (5 240) = log (5.24  1000)
= log 5.24 + log 1000
= 0.7193 + 3
= 3.7193
Use common logarithms to solve
1.8x = 15
Take logs of both sides
log 1.8x = log 15
x.log 1.8 = log 15
x = log 15  log 1.8
x ≈ 1.176  0.2553
Worksheet 7.2
x ≈ 4.607
Exercise 9.3 Page 348
No. 1, 15 to 19
Shape of Log Graphs
No. 1
y = log10 x
No. 2
y = log5 x
No. 3
y = log2 x
No. 4
y = log (x – 2)
No. 5
y = log2 (x – 1)
No. 6
y = log4 (x + 3)
No. 7
y = log (x) + 1
No. 8
y = log3 (x + 1) + 2
Application
The level of a leaking oil-storage tank
drops by 1% of the height above the
leak every hour. The leak starts at
2am and is at a height of 4.5m. The
tank is a large cylinder of height 10m
and diameter 20m. It was only ¾ full
when the leak began.
The level of a
leaking oil-storage
tank drops by 1%
of the height above
the leak every hour.
The leak starts at 2
am and is at a
height of 4.5m.
The tank is a large
cylinder of height
10m and diameter
20m. It was only ¾
full when the leak
began.
a)
b)
c)
Find an expression for the amount of oil
that has leaked after time t hours.
How much oil is spilt by 2.10 am?
A spillage of more than 5000L must be
reported to the environmental
Protection agency. By what time must
the leak be fixed to avoid notification?
10m
a)
Find an
expression for
the amount of
oil that has
leaked after time
t hours.
b) How much oil is
spilt by 2.10
am?
c) A spillage of more
than 5000L must
be reported to
the
environmental
Protection
agency. By
what time must
the leak be fixed
10m
To calculate the volume of oil spilt, we need the
height of the oil above the leak.
Since it starts at 3m above the leak and each hour
the level is 99% of the level of the previous hour
Height of oil above leak H1 = 3  0.99t
Height oil has dropped H2 = 3 - 3  0.99t
a)
Find an
expression for
the amount of
oil that has
leaked after time
t hours.
b) How much oil is
spilt by 2.10
am?
c) A spillage of more
than 5000L must
be reported to
the
environmental
Protection
agency. By
what time must
the leak be fixed
To calculate the volume of oil spilt, we need the height
of the oil above the leak.
Since it starts at 3m above the leak and each hour the
level is 99% of the level of the previous hour
Height of oil above leak H1 = 3  0.99t
Height oil has dropped H2 = 3 - 3  0.99t
a) V = πr2
= π  102  (3 - 3  0.99t) m3
= 100π  (3 - 3  0.99t)  1000L
after t hours the amount of oil leaked
= 300 000π  (1 - 0.99t) L
a) V = πr2
= π  102  (3 - 3  0.99t) m3
c) Time at which 5000L has spilt
= 100π  (3 - 3  0.99t)  1000L
5000 = 300 000π  (1 - 0.99t)L
after t hours the amount of oil leaked
= 300 000π  (1 - 0.99t) L
5000
300 000
0.99t
b) Oil spilt by 2.10 am.
V = 300 000π  (1 - 0.99t) L
= 300 000π  (1 - 0.991/6) L
≈ 1577L
= 1 - 0.99t
=1–
5000
300 000
5000
log 0.99t = log (1 – 300 000 )
5000
t  log 0.99 = log (1 – 300 000 )
t≈
log( 0.99469)
 0.52926 hours  32min
log 0.99
The leak should be fixed
before about 2.32 am
Exercise 10.5 Page 353