Transcript Standard Grade - Prestwick Academy

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Logarithms & Exponentials
Higher Mathematics
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Logarithms
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Reminder
All the questions on this topic will depend
upon you knowing and being able to use,
some very basic rules and facts.
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Logarithms
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Three Rules of logs
loga x  loga y  loga xy
x
log a x  log a y  log a
y
loga x  p loga x
p
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Logarithms
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Two special logarithms
loga a  1
loga 1  0
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Logarithms
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Relationship between log and exponential
loga x  y  a  x
y
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Logarithms
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Graph of the exponential function
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Logarithms
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Graph of the logarithmic function
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Logarithms
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Related functions of y  f ( x)
y  f ( x  a)
y  f ( x  a)
y   f ( x)
y  f ( x)
y  f ( x)  a
y  f ( x)  a
Move graph left
a
Move graph right
units
a
units
Reflect in x axis
Reflect in y axis
Move graph up a units
Move graph down a units
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Logarithms
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Calculator keys
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ln
=
log e
log
=
log 10
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Logarithms
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Calculator keys
loge 2.5
=
log10 7.6
=
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ln
2
.
5
=
= 0.916…
log 7
.
6
=
= 0.8808…
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Logarithms
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Solving exponential equations
2.4  3.1e
x
Use log ab = log a + log b
loge 2.4  loge 3.1ex
loge 2.4  loge 3.1  loge ex
Use log ax = x log a
loge 2.4  loge 3.1  x loge e
Use loga a = 1
loge 2.4  loge 3.1  x
Take loge both sides
x  loge 2.4  loge 3.1  0.25593...  0.26 (2dp)
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Logarithms
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Solving exponential equations
60  80e
Take loge both sides
Use log ab = log a + log b
Use log ax = x log a
Use loga a = 1
k  loge 80  loge 60
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k
loge 60  loge 80ek
k
loge 60  loge 80  loge e
loge 60  loge 80  k loge e
loge 60  loge 80  k
 0.2876...
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 0.29 (2dp)
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Logarithms
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Solving logarithmic equations
log3 y  0.5
0.5
y 3
Change to exponential form

y 3
Change to exponential form
1
2

1
3
1
2
1

3
y  0.577....  0.58 (2dp)
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Logarithms
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3loge (2e)  2loge (3e)
A  loge B  loge C
expressing your answer in the form
Simplify
where A, B and C are whole numbers.
loge (2e)  loge (3e)
3
 loge 8e3  loge 9e2
2
8e3
log e 2
9e
8e
 log e
9
 loge 8  loge e  loge 9
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 1  loge 8  loge 9
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Logarithms
Simplify
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log5 2  log5 50  log5 4
2  50
 log 5
4
 log5 5
2
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 log5 25
 2
 2 log5 5
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Logarithms
Find x if
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4 log x 6  2 log x 4  1
64
 log x 2  1
4
 log x 64  log x 42  1
9
 log x
36  36
1
4 4
1
9
 log x 81  1
1
 x  81
 x1  81
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Logarithms
Given
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x  log5 3  log5 4
find algebraically the value of x.
 x  log5 3 4
 x  log5 12
 5x  12
 x log10 5  log10 12
 log10 5x  log10 12
log10 12
 x
log10 5
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 x  1.5439..
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 x  1.54 (2dp)
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Logarithms
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Find the x co-ordinate of the point where the graph of the curve
with equation
y  log3 ( x  2)  1
intersects the x-axis.
When y = 0
0  log3 ( x  2)  1
Re-arrange
1  log3 ( x  2)
Exponential form
Re-arrange
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31  x  2
1
x  23
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x2
1
3
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Logarithms
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The graph illustrates the law
y  kxn
If the straight line passes through A(0.5, 0)
and B(0, 1).
Find the values of k and n.
log5 y  log5 kx
n
Gradient
log5 y  log5 k  n log5 x
1
0.5
y-intercept
log5 k  1
Y  log5 k  nX
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
 k 5
n  2 (gradient)
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Logarithms
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Before a forest fire was brought under control, the spread of fire was described
by a law of the form
A  A0ekt
where
A0
is the area covered by the fire when it was first detected
and A is the area covered by the fire t hours later.
If it takes one and a half hours for the area of the forest fire to double,
find the value of the constant k.
2 A0  A0ek1.5
 loge 2  1.5k
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 2  ek1.5
 loge 2  1.5k loge e
log e 2
 k
 k  0.462..  0.46
1.5
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(2 dp)
Logarithms
The results of an experiment
Revision
give rise to the graph shown.
a) Write down the equation of the line
in terms of P and Q.
It is given that
P  loge p
and
Q  loge q
b) Show that p and q satisfy a relationship of the form
p  aqb
stating the values of a and b.
0.6
Gradient
y-intercept 1.8
P  0.6Q  1.8
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loge p  loge aqb
loge p  loge a  b loge q
P  loge a  bQ
loge a  1.8  a  e1.8
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b  0.6
 a  6.05
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Logarithms
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The diagram shows part of the graph of
y  logb ( x  a)
Determine the values of a and b.
.
Use (7, 1)
1  logb (7  a)
...(1)
Use (3, 0)
0  logb (3  a)
...(2)
Hence, from (2)
and from (1)
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3 a 1
a  2
b5
1  logb 5
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Logarithms
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The diagram shows a sketch of
part of the graph of
y  log2 ( x)
a) State the values of a and b.
b) Sketch the graph of
a 1
y  log2 ( x  1)  3
b3
Graph moves
1 unit to the left and 3 units down
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Logarithms
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a) i) Sketch the graph of
y  a x  1, a  2
ii) On the same diagram, sketch the graph of
y  a x1 , a  2
b) Prove that the graphs intersect at a point
where the x-coordinate is
a x  1  a x 1
 1 
log a 

 a 1 
 1  a x1  a x
 loga 1  loga ax  loga  a 1
 x   loga  a 1
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 1  ax  a 1
 0  x  loga  a 1
 x  log a  a  1
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1
 1 
 x  log a 

  a  1 
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Logarithms
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Part of the graph of
y  5log10 (2 x  10)
is shown in the diagram.
This graph crosses the
x-axis at the point A and
y 8
the straight line
at the point B. Find algebraically the x co-ordinates of A and B.
8  5log10 (2 x  10) 
 101.6  2x  10
0  5log10 (2 x  10)
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8
 1.6  log10 (2x  10)
 log10 (2 x  10)
5
 101.6 10  2x  x  14.9 B (14.9, 8)
 2 x  10  1
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 x  4.5
A (4.5, 0)
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Logarithms
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The diagram is a sketch of part of
y  ax , a  1
the graph of
a) If (1, t) and (u, 1) lie on this curve,
write down the values of t and u.
b) Make a copy of this diagram and on it
sketch the graph of
y  a 2x
c) Find the co-ordinates of the point of intersection of
y  a 2x
with the line
a)
t a
c)
y  a21
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x 1
u0
y  a2
b)
2
1,
a
 
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Logarithms
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The diagram shows part of the graph with equation
y  3x
and the straight line with equation
These graphs intersect at P.
Solve algebraically the equation
y  42
3x  42
and hence write down, correct to 3 decimal places, the co-ordinates of P.
log10 3x  log10 42
log10 42
 x
log10 3
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 x log10 3  log10 42
 x  3.40217...
 P (3.402, 42)
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