Scholar
Higher Mathematics
Homework Session
Thursday 19th March 7:30pm
You will need a pencil, paper and a calculator for some of the activities
Margaret Ferguson
SCHOLAR online tutor for Maths
and
Author of the new SCHOLAR
National 5 & Higher Maths courses
Tonight’s Revision Session
will cover
Logs & Exponentials
and
Recurrence Relations
Logs & Exponentials
things you should know
Logs and exponentials are inverses of each other
If y = ax then loga y = x e.g. 7 = 5x => log5 7 = x
If loga x = y then ay = x
e.g. log3 x = 2 then 32 = x
Exponential functions take the form y = ax
a > 1 for a growth function
0 < a < 1 for a decay function
The exponential function is ex where e is 2.718281828…
Logs & Exponentials
more things you should know
Laws of exponentials
a m x a n = am + n
m
a
m-n
=a
n
a
(am)n = amn
a
-m
1
= m
a
a0 = 1
m
n
a = a
n
m
Laws of logarithms
loga m + loga n = loga (mn)
m
loga m - loga n = log a
n
loga xn = n loga x
loga 1 = 0
loga a = 1
Which of the following sketches shows the
graph of y = log6 2x?
Vote for the correct answer now
y
(a)
y
(6,1)
(c)
(3,1)
x
(½ ,0)
(½ ,0)
x
y
y
(b)
✓ log
The graph crosses the x-axis when y = 0
(6,1)
(1,0)
(d)
x
0 = 2x
6
2x
=
0
=>
6
2x = 1 so
x = ½ and
(½,0) lies on the graph
(3,1)
(1,0)
x
The other point on the graph occurs when y = 1
log6 2x = 1 => 61 = 2x
2x = 6 so x = 3 and the point (3,1) lies on the graph
A = 2 x A0 and t = 1.5
loge = ln
A = A0ekt
2A0 = A0e1.5k
2A0
= e1.5k
A0
2 = e1.5k
loge 2 = 1.5k
log e 2
=k
1.5
k = 0.462
Given that log4 8 + log4 q = 1, what is the value of q?
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A
1
2
B
1
32
C
1
8
D
2
✓
log4 8 + log4 q = 1
log4 8q = 1
41 = 8q
4
=q
8
1
q=
2
Experimental data of the form y = kxn
apply logs to both sides of the equation and choose a base
loga y = loga kxn
loga y = loga k + loga xn
loga y = loga k + n loga x
loga y = n loga x + loga k
This equation looks a bit like y = mx + c but is written as
Y = mX + c where
where Y = loga y, m = n, X = loga x and c = loga k
To find the equation from the graph of loga x against loga y
n = m where m is the gradient
k = ac where c is the value of the y-intercept
The graph illustrates the law y = kxn
log5 y
The straight line passes through
B(0,1)
the points A(0.5,0) and B(0,1).
What are the values of k and n?
Step 1: Find the equation of the line.
0 -1
m=
= -2 and B(0,1)
0.5 - 0
so y – 1 = -2(x – 0)
y = -2x + 1
log5 y = n log5 x + log5 k
log5 y = -2log5 x + 1
Step 2: Identify k and n.
Hence n = -2 and
log5k = 1 => 51 = k
log5 x
A(0.5,0)
Remember y = kxn gives
log5 y = log5 kxn
log5 y = log5 k + log5 xn
log5 y = log5 k + n log5 x
log5 y = n log5 x + log5 k
giving the equation y = 5x -2
Experimental data of the form y = abx
apply logs to both sides of the equation and choose a base
log2 y = log2 abx
log2 y = log2 a + log2 bx
log2 y = log2 a + x log2 b
log2 y = x log2 b + log2 a
This equation looks a bit like y = mx + c but is written as
Y = mx + c where
Y = log2 y, m = log2 b and c = log2 a
To find the equation from the graph of x against log2 y
b = 2m where m is the gradient
a = 2c where c is the value of the y-intercept
The results of an experiment give rise to the graph
shown below.
It is given that P = loge p and Q = q.
Show that p and q satisfy a
relationship of the form p = abq.
1.8 - 0
If p = abq then loge p = loge abq
m=
= 0.6
0 - (-3)
loge p = loge a + loge bq
P = 0.6Q + 1.8
loge p = loge a + qloge b
loge p = 0.6q + 1.8
loge p = qloge b + loge a
loge p = qloge b + loge a
Hence loge b = 0.6 and
e0.6 = b
and
b = 1.82 and
giving p = 6.05 x 1.82q
loge a = 1.8
e1.8 = a
a = 6.05
Given that log10(x) = y log 10(3) + 1, express x in terms
of y.
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✓
A
x = 10 x 3y
B
x = 3010y
log10 x = y log10 3 + log10 10
x = 3y + 10
log10 x = log10 3y + log10 10
x = y3 + 10
log10 x = log10 (3y x 10)
C
D
log10 x = y log10 3 + 1
x = 3y x 10
Recurrence Relations
A sequence is a pattern of numbers that can be defined by a
rule or formula.
A recurrence relation describes a sequence where each term is
a function of the previous term.
A geometric sequence takes the form un + 1 = aun
An arithmetic sequence takes the form un + 1 = un + b
A linear recurrence relation can defined by un + 1 = aun + b (a ≠ 0)
For a linear recurrence relation a limit exists if -1 < a < 1
The limit is given by the formula
b
c
L=
or L =
1- a
1- m
A sequence is defined by the recurrence relation
un+1 = 0.6un + k and u0 = 3.
As n -> ∞, the limit of this sequence is 5.
What is the value of k ?
Vote for the correct answer now
A
B
C
D
0
0.88
✓
2
8
b
c
L=
or L =
1- a
1- m
k
L=
=5
1- 0.6
k
=5
0.4
k = 5 x 0.4
=2
A recurrence relation is defined by
un + 1 = pun + q, where -1 < p < 1 and u0 = 12.
If u1 = 15 and u2 = 16, find the values of p and q.
What is an expression for u1?
What is an expression for u2?
15 = 12 x ⅓ + q
15 = 4 + q
q = 11
15 = 12p + q
16 = 15p + q
1 = 3p
p=⅓
subtract
Hence or otherwise find the limit of this recurrence
relation. un + 1 = ⅓un + 11
a limit exists because -1 < ⅓ < 1
11
L=
1- 1
11
L=
2
3
3
3
L =11´
2
33
L=
2
The floors of a shopping centre are cleaned daily.
Grimex removes 70% of all germs but during the next 24 hours, 300
“new” germs per sq unit are estimated to appear.
Carefree removes 80% of all germs but during the next 24 hours, 350
“new” germs per sq unit are estimated to appear.
For Grimex let un represent the number of germs per unit2 on the floor
immediately before disinfecting for the nth time and vn for Carefree.
(a) Write down a recurrence relation for each product prior to
disinfecting.
un+1 = 0.3un + 300 and vn+1 = 0.2vn + 350
(b) Determine which product is more effective in the long term.
300 300
=
= 428.6
Lu =
1- 0.3 0.7
350 350
Lv =
=
= 437.5
1- 0.2 0.8
Grimex is more effective in the long term as the number of
germs will settle around 428 germs per unit2 which is less
that Carefree which will settle around 437 germs per unit2.
Question Time
• If you have any questions about tonight’s session please ask
• The next session will be on 30th April at 7:30pm
• The session will cover revision for the Exam
• Please give us feedback on tonight’s session
• You will find a link in the chat box