Higher Mathematics
3.3 Logs and Exponentials
Paper 1 Section A
Each correct answer in this section is worth two marks.
1. Which of the following diagrams represents the graph with equation log3 y = x ?
y
(1, 3)
A.
x
O
y
(1, 1)
B.
O
x
y
(1, 3)
1
C.
O
x
y
(3, 1)
O
x
1
D.
Key
C
Outcome
3.3
hsn.uk.net
Grade
A/B
Facility
0
Disc.
0
Calculator
CN
Page 1
Content
A7, A34
Source
2011 P1 Q19
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
2. Simplify
logb 9a2
, where a > 0 and b > 0.
logb 3a
A.
2
B.
3a
C.
logb 3a
D.
logb (9a2 − 3a)
Key
A
Outcome
3.3
Grade
A/B
Facility
0
Disc.
0
Calculator
NC
Content
A28
Source
2012 P1 Q20
Calculator
NC
Content
A28
Source
HSN 17
3. Simplify log4 8 + log4 2 − 3 log5 5.
A.
− 12
B.
−1
C.
log4
D.
log4
Key
B
16
5
16
125
Outcome
3.3
hsn.uk.net
Grade
A/B
Facility
0.53
Disc.
0.33
Page 2
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
4. Solve logb x − logb 7 = logb 3 for x > 0.
A.
x = 21
B.
x = 10
C.
x=
7
3
D.
x=
3
7
Key
A
Outcome
3.3
Grade
A/B
Facility
0.64
Disc.
0.59
Calculator
CN
Content
A28, A32
Source
HSN 175
Content
A28, A32
Source
HSN 111
5. Solve loga 5 + loga x = loga 20 for x > 0.
1
4
A.
x=
B.
x=4
C.
x = 15
D.
x = 100
Key
B
Outcome
3.3
hsn.uk.net
Grade
A/B
Facility
0.7
Disc.
0.45
Calculator
CN
Page 3
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
6. The diagram shows the graph of y = 3ekx .
y
y = 3ekx
(4, 18)
3
x
O
What is the value of k?
3
A.
2e
B.
1
4
loge 6
C.
1
4
loge 15
D.
1
18
Key
B
Outcome
3.3
loge
hsn.uk.net
4
3
Grade
A/B
Facility
0.43
Disc.
0.31
Calculator
NC
Page 4
Content
A30
Source
HSN 095
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
7. Solve 3 loga 2 =
A.
a = 64
B.
a = 36
C.
a=
4
9
D.
a=
1
16
Key
A
Outcome
3.3
1
2
for a.
Grade
A/B
Facility
0.56
Disc.
0.77
Calculator
CN
Content
A31
Source
HSN 112
[END OF PAPER 1 SECTION A]
Paper 1 Section B
[SQA]
8. Evaluate log5 2 + log5 50 − log5 4.
Part
Marks
2
1
Level
C
A/B
Calc.
NC
NC
3
Content
A28
A28
•1 pd: use loga x + loga y = loga xy
•2 pd: use loga x − loga y = loga xy
•3 pd: use loga a = 1
Answer
2
U3 OC3
2000 P1 Q9
•1 log5 100 − log5 4
•2 log5 25
•3 2
√
9. Evaluate log4 16 + 2 log3 (3 3) .
Part
Marks
3
Level
A
Calc.
NC
3
Content
A28
•1 ss: use laws of logs
•2 ss: use laws of logs
•3 pd: complete
hsn.uk.net
Answer
5
U3 OC3
OB 11-004
•1 log4 16 = 2
•2 log3 (9 × 3)
√
•3 log4 16 + 2 log3 (3 3) = 5
Page 5
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
10. Evaluate log3 18 + log3 6 − log3 4 + log5
Part
Marks
2
2
Level
B
C
Calc.
CN
CN
√
Content
A28
A28
•1 pd: use loga x + loga y = loga xy
•2 pd: use loga x − loga y = loga xy
•3 pd: use loga xk = k log a x
•4 pd: use loga a = 1
5.
4
Answer
U3 OC3
Ex 3-3-2
7
2
•1
•2
•3
•4
log3 (18 × 6)
log3 ( 184×6 )
1
log3 33 + log5 5 2 = 3 log3 3 + 12 log5 5
3 + 12 = 72
11. The expression 4 loga (2a) − 3 loga 2 can be written in the form 2n + loga n, where
n is a whole number.
Find the value of n.
Part
Marks
3
Level
B
3
Calc.
CN
Content
A28
•1 pd: use log laws
•2 pd: process
•3 ic: state n
hsn.uk.net
Answer
n=2
U3 OC3
Ex 3-3-4
•1 4(loga 2 + loga a) − 3 loga 2
•2 4 + 4 loga 2 − 3 log a 2 = 4 + loga 2
•3 n = 2
Page 6
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
12. The graph below shows the curve with equation y = log8 x .
y
(8, 1)
y = log8 x
x
O 1
1
x2
Sketch the graph of y = log8
Part
Marks
3
Level
A
Calc.
CN
.
3
Content
A28, A3
Answer
sketch
•1 ss: use laws of logs
•2 ic: know to reflect and scale
•3 ic: annotate sketch
U3 OC3
OB 11-003
•1 y = −2 log8 x
•2 reflect in x-axis and scale
•3 show (1, 0) and (8, −2)
y
y = log8
O 1
1
x2
x
(8, −2)
13. The diagram shows the curve with equation y = log2 x .
y
y = log2 x
1
O
Sketch the curve y = log2
Part
•1
•2
•3
•4
Marks
4
ss:
ic:
ic:
ic:
Level
A
2
x
Calc.
CN
x
.
4
Content
A29, A28
•1
•2
•3
•4
use law of logs
interpret reflection
interpret translation
annotate sketch
hsn.uk.net
Answer
sketch
Page 7
U3 OC3
AT064
y = log2 2 − log2 x
reflect curve in y-axis
shift 1 unit up
decreasing log curve thro’ (1, 1)
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
14. Sketch and annotate the curve with equation y = loge ( x + 1)2 .
Part
•1
•2
•3
•4
Marks
4
ss:
ic:
ic:
ic:
Level
A
Calc.
CN
Content
A29, A28
use log law
interpret translation
interpret scaling
sketch with points annotated
hsn.uk.net
Page 8
Answer
sketch
•1
•2
•3
•4
4
U3 OC3
WCHS U3 Q10
y = 2 loge ( x + 1)
y = loge x translated 1 unit left
then made twice as tall
sketch, with points (0, 0)
(e − 1, 2)
and
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
15. The diagram below shows the graph of the function f ( x ) = e x .
y
Q
y = f (x) P
x
O
The points P and Q have x -coordinates −1 and 1 respectively.
Straight lines are drawn from the origin to P and Q as shown.
(a) Show that OP and OQ are perpendicular.
(b) Show that the area of triangle OPQ is
3
1 + e2
.
2e
3
(c) The function g is defined by g( x ) = f ( x − 2) + 1.
(i) Sketch the curve with equation y = g( x ) .
(ii) The graphs of y = f ( x ) and y = g( x ) intersect when x = a.
e2
Show that e a = 2
.
e −1
Hence express a in the form A + B loge (e − 1) + C loge (e + 1) , stating the
values of A, B and C .
Part
(a)
(b)
(ci)
(cii)
Marks
3
3
3
6
Level
B
B
C
A
Calc.
CN
CN
CN
CN
Content
G5, G2
G1
A3
A30, A28
•1 ic: obtain coordinates of P and Q
•2 pd: find gradients
•3 ss: use m × m⊥ = −1
•4 pd: compute side length
•5 pd: compute side length
•6 ss: use area formula and complete
•7 ss: translate parallel to x-axis
•8 ss: translate parallel to y-axis
•9 ic: annotations
•10 ss: form equation
•11 ic: complete
•12 ss: know to take logs
•13 pd: use laws of logs
Page 9
•14.uk.net
pd: use laws of logs
•15 ic: state A, B, C
hsn
Answer
proof
proof
sketch
A = 2, B = C = −1
U3 OC3
AT040
•1 P −1, 1e , Q(1, e)
•2 mOP = − 1e , mOQ = e
•3 mOP × mOQ = −1 so OP ⊥ OQ
√
•4 OP = √1 + 1/e2
•5 OQ = e2 + 1
•6 Area = 12 × OP × OQ = (1 + e2 )/2e
•7
•8
•9
•10
•11
•12
•13
•14
•15
shift 2 units to right
shift 1 unit up
P′ (1, 1/e + 1), Q′ (3, e + 1)
f ( x ) = g( x )
e x = e2 / ( e2 − 1)
x = loge e2 /(e2 − 1)
c SQA
Questions
marked
‘[SQA]’ · · · = logeAll
e2 others
− loge
(ce2Higher
− 1) Still Notes
· · · = 2 − loge (e + 1) − loge (e − 1)
A = 2, B = C = −1
9
Higher Mathematics
[SQA]
16. Given x = log5 3 + log5 4, find algebraically the value of x .
Part
[SQA]
Level
C
A/B
Calc.
NC
NC
Content
A31
A28
Marks
2
1
2
Level
C
A/B
A/B
Calc.
NC
NC
NC
Content
A32
A3
A28, A3
Answer
U3 OC3
1998 P1 Q19
17.
Part
(a)
(a)
(b)
[SQA]
Marks
1
3
4
Answer
U3 OC3
1994 P1 Q16
18. Find x if 4 logx 6 − 2 logx 4 = 1.
Part
Marks
3
Level
C
Calc.
NC
3
Content
A32, A28, A31
•1 pd: use log-to-index rule
•2 pd: use log-to-division rule
•3 ic: interpret base for logx a = 1 and
simplify
hsn.uk.net
Page 10
Answer
x = 81
U3 OC3
2001 P1 Q8
•1 logx 64 − logx 42
4
•2 logx 642
•3 all processing leading to x = 81
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
19. Solve loga 12 + loga x − 2 loga 2 = 6 for x in terms of a, where a > 1.
Part
Marks
4
Level
B
•1 ss: use log law
•2 ss: use log law
•3 ss:
know to
exponential
•4 pd: complete
Calc.
CN
Content
A32, A28, A31
convert
log
to
Answer
a = 13 a6
•1
•2
•3
•4
4
U3 OC3
WCHS U3 Q6
loga 12 + loga x − loga 22
loga ( 12x
4 )
3x = a6
a = 13 a6
20. Solve log8 x − log6 4 = log6 9 for x > 0.
Part
Marks
3
Level
C
Calc.
CN
Content
A32, A28, A31
•1 ss: use laws of logs
•2 pd: use laws of logs
•3 pd: use laws of logs
[SQA]
3
Answer
x = 64
U3 OC3
Ex 3-3-6
•1 log8 x = log6 9 + log6 4 = log6 36
•2 log8 x = 2
•3 x = 82 = 64
21. The graph illustrates the law y = kx n .
log5 y
If the straight line passes through
A (0·5, 0) and B (0, 1) , find the values of
k and n.
B (0, 1)
4
O
Part
•1
•2
•3
•4
Marks
4
ic:
ss:
ss:
pd:
Level
A/B
Calc.
NC
Content
A33
hsn.uk.net
Answer
y = 5x−2
•1
•2
•3
•4
interpret graph
use log laws
use log laws
solve log equation
Page 11
A (0·5, 0)
log5 x
U3 OC3
2002 P1 Q11
log5 y = −2(log5 x) + 1
log5 y = log5 x−2 + . . .
. . . + log5 5
y = 5x−2
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
22. The graph below shows a straight line in the (log4 x, log4 y) -plane.
log4 y
(7, 10)
3
log4 x
O
Find the equation of the line in the form y = kx n , where k, n ∈ R .
Part
•1
•2
•3
•4
•5
Marks
5
ic:
ic:
ss:
ss:
ic:
Level
A
Calc.
NC
Content
A33, G3, A28
interpret graph (gradient)
interpret graph (complete eqn)
use log laws
use log laws
complete
Answer
y = 64x
•1
•2
•3
•4
•5
5
U3 OC3
OB 11-002
gradient = 1
log4 y = log4 x + 3
log4 y = log4 k + n log4 x
log4 k = 3 ⇒ k = 43
y = 64x
[END OF PAPER 1 SECTION B]
hsn.uk.net
Page 12
c SQA
Questions marked ‘[SQA]’ c Higher Still Notes
All others Higher Mathematics
Paper 2
1. (a) Given that log4 x = P, show that log16 x = 12 P.
3
(b) Solve log3 x + log9 x = 12.
Part
(a)
(b)
Marks
3
3
Level
A
A
Calc.
CN
CN
3
Content
A28
A32
Answer
proof
x = 38 (= 6561)
•1 ss: convert from log to exponential
form
•2 ss: know to and convert back to log
form
•3 pd: process and complete
•4 ss: use appropriate strategy
•5 pd: start solving process
•6 pd: complete process via log to expo
form
U3 OC3
2010 P2 Q7
•1 x = 4P
•2 log16 x = log16 4P
•3 log16 x = P × log16 4 and complete
•4 log3 x + 12 log3 x = 12
•5 log3 x = 8
•6 x = 38 (= 6561)
2. Solve log2 x2 − 3 log2 x = 4 − log2 7 for x > 0.
Part
Marks
3
Level
A
Calc.
CN
Content
A28, A31
•1 ss: use log law
•2 ss: use log law
•3 ss:
know to convert
exponential and complete
hsn.uk.net
log
3
Answer
7
x = 16
to
Page 13
U3 OC3
OB 11-001
•1 log2 x2 = 2 log2 x
•2 log2 7 − log2 x = log2 ( 7x )
7
•3 7x = 24 , so x = 16
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
3. The graph of the function f ( x ) = loge x is shown in the diagram below.
y
y = f (x)
B (e2 , b )
O
x
A (1, 0)
The point B has coordinates (e2 , b) .
(a) Write down the value of b.
1
(b) The function g is defined by g( x ) = − f ( x − 2) . Sketch the graph of y = g( x ) .
3
(c) The graphs of y = f (√
x ) and y = g( x ) intersect at C. The x -coordinate of C is
of the form x = m + n .
Determine the values of m and n.
Part
(a)
(b)
(c)
Marks
1
3
6
Level
C
C
A
Calc.
CN
CN
CN
Content
A2
A29
A32, A34
6
Answer
2
sketch
m = 1, n = 2
U3 OC3
AT010
•1 ic:
interpret graph
•1 b = 2
•2 ic:
•3 ic:
•4 ic:
reflection
horizontal translation
annotate sketch
•2 reflect in x-axis
•3 shift 2 units to right
•4 show A′ (3, 0) and B′ (e2 + 2, −2)
•5
•6
•7
•8
•9
•10
pd:
ss:
ss:
ss:
pd:
ic:
•5
•6
•7
•8
•9
•10
expression for g( x)
equate
use log law
convert from log
solve quadratic equation
interpret solution
hsn.uk.net
Page 14
g( x) = − log2 ( x − 2)
loge x = − loge ( x − 2)
loge x = loge ( x − 2)−1
x = 1/( x√− 2)
x = 1± 2
m = 1, n = 2
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
[SQA]
[SQA]
4.
Part
(a)
(b)
(b)
Marks
1
1
3
Level
C
C
A/B
Calc.
CR
CR
CR
Content
A6
A30
A30
Answer
Part
(a)
(a)
(b)
Marks
1
3
1
Level
C
A/B
A/B
Calc.
CR
CR
CR
Content
A2
A30
A7
Answer
U3 OC3
1989 P1 Q20
5.
hsn.uk.net
Page 15
U3 OC3
1993 P1 Q15
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
[SQA]
6.
Part
[SQA]
Marks
4
Level
A/B
Calc.
CR
Content
A30
Answer
U3 OC3
1994 P1 Q20
7. Before a forest fire was brought under control, the spread of the fire was described
by a law of the form A = A0 ekt 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.
Part
Marks
3
Level
A/B
Calc.
CR
Content
A30
•1 ic: form exponential equation
•2 ss: express exp.
equ.
as log
equation
•3 pd: solve log equation
hsn.uk.net
Page 16
Answer
k = 0·46
U3 OC3
2001 P2 Q9
•1 2A0 = A0 ek×1·5
•2 e.g. 1·5k = ln 2
•3 k = 0·46
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
3
Higher Mathematics
8. A population of bacteria is growing in such a way that the number of bacteria N
present after t minutes is given by the formula N (t) = 32e0·01225t .
(a) State N0 , the number of bacteria present when t = 0.
1
(b) The “e -folding” time, l minutes, is the length of time until N (l ) = eN0 .
Find the e -folding time for this population correct to 3 decimal places.
Part
(a)
(b)
Marks
1
3
•1 ic:
Level
C
B
Calc.
CN
CR
Content
A6
A30
U3 OC3
Ex 3-3-3
•1 N0 = N (0) = 32
interpret formula
•2 ic: interpret N (l )
•3 ss: form equation
•4 pd: solve
[SQA]
Answer
N0 = 32
l = 81·633 (3 d.p.)
3
•2 N (l ) = eN0 = 32e
•3 32e0·01225l = 32e
•4 0·01225l = 1 ⇒ l = 81·633 (3 d.p.)
9.
Part
(a)
(b)
Marks
1
2
hsn.uk.net
Level
C
C
Calc.
CR
CR
Content
A30, A34
A6
Page 17
Answer
U3 OC3
1991 P1 Q4
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
[SQA]
[SQA]
10.
Part
(a)
(a)
(b)
Marks
1
2
2
Level
C
A/B
A/B
Calc.
CR
CR
CR
Content
A30
A30, A34
A30
Answer
Part
Marks
1
4
Level
C
A/B
Calc.
CR
CR
Content
A30
A30, A34
Answer
U3 OC3
1995 P1 Q18
11.
hsn.uk.net
Page 18
U3 OC3
1996 P1 Q19
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
[SQA]
12.
Part
(a)
(a)
(b)
(b)
Marks
2
3
1
3
hsn.uk.net
Level
C
A/B
C
A/B
Calc.
CR
CR
CR
CR
Content
A30, A34
A30, A34
A30
A34
Page 19
Answer
U3 OC3
1991 P2 Q7
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
[SQA]
13.
Part
(a)
(b)
(b)
(c)
(c)
Marks
2
1
2
1
2
hsn.uk.net
Level
C
C
A/B
C
A/B
Calc.
CR
CR
CR
CR
CR
Content
A30, A34
A30
A30, A34
A2
A2
Page 20
Answer
U3 OC3
1992 P2 Q4
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
[SQA]
14. The size of the human population, N , can be modelled using the equation
N = N0 ert where N0 is the population in 2006, t is the time in years since 2006,
and r is the annual rate of increase in the population.
(a) In 2006 the population of the United Kingdom was approximately 61 million,
with an annual rate of increase of 1·6%. Assuming this growth rate remains
constant, what would be the population in 2020?
2
(b) In 2006 the population of Scotland was approximately 5·1 million, with an
annual rate of increase of 0·43%.
Assuming this growth rate remains constant, how long would it take for
Scotland’s population to double in size?
Part
(a)
(b)
Marks
2
3
Level
B
A
Calc.
CR
CR
Content
A30, A34
A30, A34
Answer
76 million
t = 161·2 years
•1 ic: substitute into equation
•2 pd: evaluate exponential expression
•1 61e0·016×14
•2 76 million
•3 ic: interpret info and substitute
•4 ss: convert expo. equ. to log. equ.
•5 pd: process
•3 10·2 = 5·1e0·0043t
•4 0·0043t = ln 2
•5 t = 161·2 years
hsn.uk.net
Page 21
U3 OC3
2009 P2 Q6
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
3
Higher Mathematics
[SQA]
15.
Part
(a)
(a)
(b)
Marks
1
2
3
hsn.uk.net
Level
C
A/B
A/B
Calc.
CR
CR
CR
Content
A30, A34
A30, A34
A30
Page 22
Answer
U3 OC3
1997 P2 Q8
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
[SQA]
[SQA]
16.
Part
(a)
(a)
(b)
(b)
Marks
2
2
1
2
Level
C
A/B
C
A/B
Calc.
CR
CR
CR
CR
Content
A30, A34
A30, A34
A30
A30
Answer
Part
Marks
4
Level
A/B
Calc.
CR
Content
A31
Answer
U3 OC3
1999 P2 Q7
17.
hsn.uk.net
Page 23
U3 OC3
1997 P1 Q17
c SQA
Questions marked ‘[SQA]’ c
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Higher Mathematics
[SQA]
18. Find the x -coordinate of the point where the graph of the curve with equation
y = log3 ( x − 2) + 1 intersects the x -axis.
Part
Marks
2
1
Level
C
A/B
Calc.
CN
CN
Content
A31
A32
•1 ss: know to isolate log term
•2 pd: express log equation as exp. equ.
•3 pd: process
[SQA]
Answer
U3 OC3
2002 P2 Q7
x = 2 13
•1 log3 ( x − 2) = −1
•2 x − 2 = 3−1
•3 x = 2 13
19.
Part
Marks
3
hsn.uk.net
Level
A/B
Calc.
CN
Content
A31, A2
Page 24
Answer
U3 OC3
1995 P1 Q19
c SQA
Questions marked ‘[SQA]’ c
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3
Higher Mathematics
[SQA]
20.
Part
Marks
3
Level
A/B
Calc.
CN
Content
A31, A2
Answer
U3 OC3
1999 P1 Q15
21. A sequence is defined by the recurrence relation un+1 = 12 un + 3 with u0 = log3 4.
(a) Show that u1 = log3 54.
4
(b) Find an expression for u2 in the form log3 a.
3
(c) Find the value of u2 correct to two decimal places.
4
Part
(a)
(b)
(c)
•1
•2
•3
•4
Marks
4
3
4
ic:
ss:
ss:
ss:
Level
A
A
A
Calc.
CN
CN
CR
Content
A11, A28
A11, A28
A31, A28
•1
•2
•3
•4
find u1
use law of logs
convert constant to log
use law of logs
•5 ic: find u2
•6 ss: use law of logs
•7 pd: complete
•8
•9
•10
•11
ss:
ss:
pd:
ic:
U3 OC3
AT051
u1 = 12 log3 4 + 3
· · · = log3 (41/2 ) + 3
· · · = log3 2 + log3 33
· · · = log3 54
•5 u2 = 12 log3 54 + 3
•6 · · · = log3 (541/2 ) + log3 33
√
√
•7 · · · = log3 (27 54) (= 81 6)
know to change base
use law of logs
process
state value
hsn.uk.net
Answer
proof √
log3 (81 6)
4·82 to 2 d.p.
•8
•9
•10
•11
Page 25
√
3u2 = 81 6.
√
u2 loge 3 = loge 81 6
√
u2 = (loge 81 6)/(loge 3)
· · · = 4·82 to 2 d.p.
c SQA
Questions marked ‘[SQA]’ c
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Higher Mathematics
[SQA]
22.
Part
[SQA]
Marks
1
2
Level
C
A/B
Calc.
CR
CR
Content
A31
A31, A34
Answer
U3 OC3
1989 P1 Q21
23.
(a) (i) Show that x = 1 is a root of x3 + 8x2 + 11x − 20 = 0.
(ii) Hence factorise x3 + 8x2 + 11x − 20 fully.
4
(b) Solve log2 ( x + 3) + log2 ( x2 + 5x − 4) = 3.
Part
(a)
(b)
Marks
4
5
Level
C
B
Calc.
CN
CN
Content
A21
A32
5
Answer
( x − 1)( x + 4)( x + 5)
x=1
U3 OC3
2009 P2 Q3
•1 ss:
know
and
use
f ( a) = 0 ⇔ a is a root
•2 ic: start to find quadratic factor
•3 ic: complete quadratic factor
•4 pd: factorise fully
•1 f (1) = 1 + 8 + 11 − 20 = 0 so x = 1
is a root
2
• ( x − 1)( x2 · · ·
•3 ( x − 1)( x2 + 9x + 20)
•4 ( x − 1)( x + 4)( x + 5)
•5 ss: use log laws
•6 ss:
know to & convert to
exponential form
•7 ic: write cubic in standard form
•8 pd: solve cubic
•9 ic: interpret valid solution
•5
•6
•7
•8
•9
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Page 26
log2 ( x + 3)( x2 + 5x − 4)
( x + 3)( x2 + 5x − 4) = 23
x3 + 8x2 + 11x − 20 = 0
x = 1 or x = −4 or x = −5
x = 1 only
c SQA
Questions marked ‘[SQA]’ c
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Higher Mathematics
[SQA]
24.
Part
(a)
(b)
Marks
3
4
hsn.uk.net
Level
C
A/B
Calc.
CR
CR
Content
G2, G3
A33
Page 27
Answer
U3 OC3
1993 P2 Q10
c SQA
Questions marked ‘[SQA]’ c
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Higher Mathematics
[SQA]
25. The results of an experiment give rise to the graph shown.
P
(a) Write down the equation of the line in
terms of P and Q.
1·8
−3 O
2
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.
Part
(a)
(b)
Marks
2
4
•1 ic:
•2 ic:
Level
A/B
A/B
Calc.
CR
CR
Content
G3
A33
Answer
P = 0·6Q + 1·8
a = 6·05, b = 0·6
U3 OC3
2000 P2 Q11
•1 m = 13·8 = 0·6
•2 P = 0·6Q + 1·8
interpret gradient
state equ. of line
•3 ic: interpret straight line
•4 ss: know how to deal with x of
x log y
•5 ss: know how to express number as
log
•6 ic: interpret sum of two logs
Method 1
•3
•4
•5
•6
loge p = 0·6 loge q + 1·8
loge q0·6
loge 6·05
p = 6·05q0·6
Method 2
ln p = ln aqb
•3 ln p = ln a + b ln q
•4 ln p = 0·6 ln q + 1·8 stated or implied
by •5 or •6
•5 ln a = 1·8
•6 a = 6·05, b = 0·6
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c SQA
Questions marked ‘[SQA]’ c
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4
Higher Mathematics
26. Variables x and y are related by the equation
y = kx n .
The graph of log2 y against log2 x is a log2 y
straight line through the points (0, 5) and
(4, 7) , as shown in the diagram.
(4, 7)
(0, 5)
Find the values of k and n.
log2 x
O
Part
•1
•2
•3
•4
•5
Marks
5
ss:
ic:
ic:
ic:
ic:
Level
A
Calc.
CN
Content
A33
introduce logarithms to y = kxn
use laws of logarithms
interpret intercept
solve for k
interpret gradient
hsn.uk.net
Page 29
Answer
k = 32, n =
•1
•2
•3
•4
•5
5
U3 OC3
1
2
2011 P2 Q5
log2 y = log2 kxn
log2 y = n log2 x + log2 k
log2 k = 5 or log2 y = 5
k = 32 or 25
n = 12
c SQA
Questions marked ‘[SQA]’ c
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Higher Mathematics
[SQA]
27.
Part
(a)
(b)
Marks
3
6
hsn.uk.net
Level
A/B
A/B
Calc.
CR
CR
Content
A33
A30
Page 30
Answer
U3 OC3
1998 P2 Q11
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
28. The results of an experiment were noted as follows.
x
log10 y
1·70
2·14
2·10 2·50 2·90
1·96 1·75 1·53
The relationship between these data can be written in the form y = abx where a
and b are constants.
Find the values of a and b and hence state a formula relating the data.
Part
•1
•2
•3
•4
•5
•6
[SQA]
Marks
6
ss:
pd:
ic:
pd:
pd:
ic:
Level
A
Calc.
CR
Content
A33, A28
know to take logs
use laws of logs
interpret equation
find gradient
start to find other constant
complete, and state equation
Answer
y = 1023·29 × (0·31) x
6
U3 OC3
WCHS U3 Q13
•1 log10 y = log10 ( ax )
•2 log10 y = (log10 b) x + log10 a
•3 gradient of line is log10 b
1·53 − 2·14
•4 m =
= −0·51 (2 d.p.),
2·90 − 1·70
so b = 10−0·51 = 0·31 (2 d.p.)
5
• (1·70, 2·14) : 2·14 = −0·51 × 1·70 + log10 a
•6 a = 1023·29 (2 d.p.)
so y = 1023·29 × (0·31) x
29.
Part
Marks
2
4
hsn.uk.net
Level
C
A/B
Calc.
CN
CN
Content
G3
A33, A34
Page 31
Answer
U3 OC3
1990 P1 Q14
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
[SQA]
30.
Part
(a)
(b)
Marks
3
4
hsn.uk.net
Level
C
A/B
Calc.
CR
CR
Content
G3
A33, A34
Page 32
Answer
U3 OC3
1996 P2 Q9
c SQA
Questions marked ‘[SQA]’ c
All others Higher Still Notes
Higher Mathematics
eplacements
[SQA]
31.
Part
Marks
1
3
Level
C
A/B
Calc.
CN
CN
Content
A32
A34
Answer
U3 OC3
1993 P1 Q20
[END OF PAPER 2]
hsn.uk.net
Page 33
c SQA
Questions marked ‘[SQA]’ c Higher Still Notes
All others