Post on 24-Mar-2022
Higher Higher Portfolio Logs and Exponentials
EF1. Logarithms and Exponentials
Section A - Revision Section
This section will help you revise previous learning which is required in this topic.
R1 Revision of Surds and Indices
1. Express each of the following in its simplest form.
(a) 8 (b) 12 (c) 50
(d) 45 (e) 332 (f) 540
2. Express each of the following with a rational denominator.
(a) 1
2 (b)
20
2 (c)
4
5 2
3. Simplify the following writing the answers with positive indices only.
(a) π₯2 Γ π₯5 (b) π¦β3 Γ π¦7 (c) π₯6 Γ· π₯4
(d) π¦β3 Γ· π¦β1 (e) (2π4)3 (f) (πβ4)β2
(g) π₯3Γπ¦5
π₯2Γπ¦2 (h) πβ1Γπ3
πβ2Γπ (i) 5π₯3 Γ 2π₯β3
(j) 3π₯5π¦3
6π₯2π¦5 (k) 3π5 Γ 2π1
2 (l) 4π8 Γ· 2πβ2
(m) πβ1
2 Γ π3
2 (n) πβ1
3 Γ π1
3 (o) π₯2(π₯3 + 1)
4. Write in the form ππ₯π + ππ₯π + β―
(a) π₯β2(π₯ β 3) (b) 1
π₯2(π₯3 + 2π₯) (c)
1
π₯(3π₯2 + 2π₯)
(d) 1
βπ₯(
1
βπ₯β 1) (e) (2π₯5 β 3)(π₯ + 4π₯β2) (f) (
1
π₯+ 1)2
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(g) 1
βπ₯23 (h) 1
2 βπ₯3 (i)
1
5 βπ₯34
(j) 3
5 βπ₯52 (k) 2
7 βπ₯23 (l) 6
βπ₯3
(m) π₯2+3π₯+5
π₯ (n)
2π₯3+π₯2+π₯
βπ₯ (o)
π₯4β6π₯+π₯3
π₯2
(p) π₯+5
βπ₯32 (q) 3 + π₯3
3π₯2 (r)
π₯+2
βπ₯
(s) (π₯+1)(π₯+2)
π₯ (t)
(π₯β1)(π₯+3)
5 βπ₯43 (u) 3π₯2+ 5π₯+1
2π₯2
R2. Revision of Straight Line
1. Find the gradient and equation each of the straight line between the
following points
(a) A(2 , β1) and B(4 , 7) (b) X(β1 , 1) and Y(5 , 13)
(c) R(β2 , β5) and S(2 , β7) (d) Q(β1 , 3) and T(β1 , 7)
2. Write down the gradient and yβintercept of the following lines
(a) π¦ = 3π₯ β 2 (b) π¦ = π₯ + 4 (c) π¦ = 4π₯
(d) π¦ = β2π₯ β 1 (e) π¦ β 2π₯ = 3 (f) π¦ β π₯ + 3 = 0
3. Find the equation of the straight line shown
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4. Find the equation of straight line, in the form π¦ = ππ₯ + π, passing through
each point with the given gradient.
(a) gradient = 5 passing through (4, 7)
(b) gradient = 2
3 passing through (2, β3)
(c) gradient = -2 passing through (β5, 1)
5. Find the equation of the line parallel to π¦ = 3π₯ + 5 which passes through
the point (3, 7).
6. Find the equation of the line parallel to π¦ =1
2π₯ β 7 which passes through
the point (β2, 4).
7. Find the equation of the line parallel to 2π₯ + π¦ = β3 which passes through
the point (β1, β3).
8. Find the equation of the line parallel to 5π₯ β 2π¦ = 7 which passes through
the point (3, 7).
9. Find the equation of the line parallel to 2π₯ β π¦ β 7 = 0 which passes
through the point (β1, 7).
10. Find the equation of the line parallel to 3π₯ + 5π¦ + 1 = 0 which passes
through the point (4, 0).
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Section B - Assessment Standard Section
This section will help you practise for your Assessment Standard Test for
Exponentials and Logarithms (Expressions and Functions 1.1)
1. (a) Simplify πππ45π + πππ47π . (b) Simplify πππ53π₯ + πππ54π¦.
(c) Simplify πππ32π + πππ35π.
2. (a) Express πππππ₯7 β πππππ₯3 in the form ππππππ₯ .
(b) Express πππππ8 β πππππ2 in the form ππππππ .
(c) Express πππππ9 β πππππ4 in the form ππππππ .
3. Solve πππ4(π₯ β 2) = 1 .
4. Solve πππ5(π₯ + 3) = 2 .
5. Solve πππ16(π₯ β 5) =1
2 .
6. (a) Simplify ππππ8 β ππππ2 (b) Simplify πππ52 + πππ550 β πππ54
(c) Simplify 3πππ42 + πππ48
7. Solve πππππ₯ β ππππ7 = ππππ3 for π₯ > 0.
8. Find π₯ if 4ππππ₯6 β 2ππππ₯4 = 1.
9. (a) Simplify ππππ10 + ππππ4 (b) Simplify πππ4320 β πππ45
(c) Simplify 2πππ36 β πππ34
10. Given that πππ48 + πππ4π = 1, what is the value of π?
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Section C β Operational Skills Section
This section provides problems with the operational skills associated with
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O1 I can convert between exponential and logarithmic forms.
1. Given π = ππ‘ which of the following is true:
(a) ππππ‘π = π
(b) πππππ = π‘
2. Given πππππ₯ = π¦ which of the following is true:
(a) ππ¦ = π₯
(b) π₯π¦ = π
O2 I can use the three main laws of logarithms to simplify expressions,
including those involving natural logarithms.
1. Simplify
a. ππππ₯3 + ππππ₯5 β ππππ₯7
b. ππππ32 β 2ππππ4
2. Show that (a) πππ38
πππ32 = 3 . (b)
ππππ9π2
ππππ3π = 2 .
3. If πππ3π₯ = 2πππ3π¦ β 3πππ3π§ find an expression for π₯ in terms of π¦ and π§.
4. Find a if ππππ64 =3
2 .
5. Simplify 3ππππ(2π) β 2ππππ(3π) expressing your answer in the form
π΄ + πππππ΅ β πππππΆ where π΄, π΅ and πΆ are whole numbers.
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O3 I can solve logarithmic and exponential equations using the laws of
logarithms.
1. Given the equation π¦ = 3 Γ 4π₯ find the value of π₯ when π¦ = 10 giving your
answer to 3 significant figures.
2. Given the equation π΄ = π΄0πβππ‘, find, to 3 significant figures:
(a) π΄ when π΄0 = 5, π = 0 β 23 and π‘ = 20.
(b) π when π΄ = 70, π΄0 = 35 and π‘ = 20.
(c) π‘ when π΄ = 1000, π΄0 = 10 and π = 0.01.
3. Solve πππ4π₯ + πππ4(π₯ + 6) = 2, π₯ > 0.
4. Solve the equation πππ5(3 β 2π₯) + πππ5(2 + π₯) = 1, β2 < π₯ <3
2.
5. (a) Given that πππ4π₯ = π, show that πππ16π₯ =1
2π.
(b) Solve πππ3π₯ + πππ9π₯ = 12 .
6. The curve with equation π¦ = πππ3(π₯ β 1) β 2 β 2, where π₯ > 1, cuts the x-
axis at the point (π, 0). Find the value of π.
7. If πππ48 + πππ4π = 1, find the value of π.
8. Solve the equation πππ2(π₯ + 1) β 2πππ23 = 3.
9. Find π₯ if 4ππππ₯6 β 2ππππ₯4 = 1 .
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O4 I can solve for π and π equations of the following forms, given two pairs
of corresponding values of π and π: ππππ¦ = πππππ₯ + ππππ, π¦ = ππ₯π and,
ππππ¦ = π₯ππππ + ππππ, π¦ = πππ₯
1. Each graph below is in the form π¦ = πππ₯.
In each case state the values of π and π.
(a) (b)
2. Each graph below is in the form π¦ = ππ₯π.
In each case state the values of π and π.
(a) (b)
y
x
π¦ = ππ₯π
(1, 3)
(2, 96)
(2, 56)
(3, 189)
y
x
π¦ = ππ₯π
y
x
π¦ = πππ₯
(1, 12)
(3, 108)
y
x
π¦ = πππ₯
(2, 20)
(4, 80)
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3. Given that π¦ = ππ₯π, and that π₯ = 3 when π¦ = 162 and that π₯ = 5 when π¦ =
1250, find π and π.
4. An investment (Β£π΄) grows according to the relationship π΄ = πππ‘ where π‘ is
the time after the investment is made in years. If after 3 years the
investment is worth Β£1157β63 and after 10 years it is worth Β£1628β89, find
the values of π and π.
O5 I can plot and extract information from straight line graphs with
logarithmic axes (axis).
1. Given that π¦ = ππ₯π, where π and π are constants, what would you plot in
order to get a straight line graph?
2. Given that π¦ = π΄πππ₯ where π and π΄ are constants, what would you plot in
order to get a straight line graph?
3. Variables π₯ and π¦ are related by
the equation π¦ = ππ₯π
The graph of πππ2π¦ against πππ2π₯ is
a straight line through the points
(0, 5) and (4, 7), as shown in the
diagram.
Find the values of π and π.
4. Two variables π₯ and π¦ satisfy the equation π¦ = 3 Γ 4π₯.
(a) Find the values of π if (π, 6) lies on the graph with equation π¦ = 3 Γ 4π₯.
(b) If (β1
2, π) also lies on the graph, find π.
(c) A graph is drawn of πππ10π¦ against π₯. Show that its equations will be of
the form πππ10π¦ = ππ₯ + π and state the gradient of this line.
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5. Variables π₯ and π¦ are related by
the equation π¦ = π΄πππ₯
The graph of πππππ¦ against π₯ is
a straight line through the
points (0, 3) and (6, 5), as shown
in the diagram.
Find the values of π and π΄.
O6 I can solve logarithmic and exponential equations in real life contexts.
1. Radium decays exponentially and its half-life is 1600 years.
If π΄0 represents the amount of radium in a sample to start with and π΄(π‘)
represents the amount remaining after π‘ years, then (π‘) = π΄0πβππ‘ .
(a) Determine the value of π, correct to 4 significant figures.
(b) Hence find what percentage, to the nearest whole number, of the
original amount of radium will be remaining after 4000 years.
2. The concentration of the pesticide, Xpesto, in soil can be modelled by the
equation
π = π0πβππ‘
Where
π0 is the initial concentration;
ππ‘ is the concentration at time π‘;
π‘ is the time, in days, after the application of the pesticide.
(a) Once in the soil, the half-life of pesticide is the time taken for its
concentration to be reduced to one half of its initial value.
If the half-life of Xpesto is 25 day, find the value of π to 2 significant
figures.
(b) Eighty days after the initial application, what is the percentage
decrease in the concentration of Xpesto?
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3. Before a forest fire was brought under control, the spread of the fire was
described by a law of the form π΄(π‘) = π΄0πβππ‘ where π΄0 is the area
covered by the fire when it was first detected and π΄ is the area covered
by the fire π‘ 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 π.
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Answers
Section A
R1
1. (a) 2β2 (b) 2β3 (c) 5β2
(d) 3β5 (e) 12β2 (f) 10β10
2. (a) β2
2 (b) 10β2 (c)
2β2
5
3. (a) π₯7 (b) π¦4 (c) π₯2 (d) 1
π¦2 (e) 8π12
(f) π8 (g) π₯π¦3 (h) ππ2 (i) 10 (j) π₯3
2π¦2
(k) 6π11
2 (l) 2π10 (m) π (n) 1 (o) π₯5 + π₯2
4. (a) π₯β1 + 3π₯β2 (b) π₯ + 2π₯β1 (c) 3π₯ + 2 (d) π₯β1 β π₯β1
2
(e) 2π6 + 8π₯3 β 3π₯ β 12π₯β2 (f) π₯β2 + 2π₯β1 + 1 (g) π₯β2
3 (h) 1
2π₯β
1
3
(i) 1
5π₯β
3
4 (j) 3
5π₯β
5
2 (k) 2
7π₯β
2
3 (l) 6π₯β1
3
(m) π₯ + 3 + 5π₯β1 (n) 2π₯5
2 + π₯3
2 + π₯1
2
(o) π₯2 β 6π₯β1 + π₯ (p) π₯β1
2 + 5π₯β3
2
(q) π₯β2 +π₯
3 (r) π₯
1
2 + 2π₯β1
2 (s) π₯ + 3 + 2π₯β1
(t) 1
5π₯
2
3 +2
5π₯β
1
3 β3
5π₯β
4
3 (u) 3
2+
5
2π₯β1 +
1
2π₯β2
R2
1. (a) π = 4, π¦ = 4π₯ β 9 (b) π = 2, π¦ = 2π₯ + 3
(c) π = β1
2, π¦ = β
1
2π₯ β 6 (d) undefined, π₯ = β1
2. (a) π = 3, y-intercept (0, -2) (b) π = 1, y-intercept (0, 4)
(c) π = 4, y-intercept (0, 0) (d) π = β2, y-intercept (0, -1)
(e) π = 2, y-intercept (0, 3) (f) π = 1, y-intercept (0, -3)
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3. π¦ = 18 β 2π₯
4. (a) π¦ = 5π₯ β 13 (b) π¦ =2
3π₯ β
13
3 (c) π¦ = β2π₯ β 9
5. π¦ = 3π₯ β 2 6. π¦ =1
2π₯ + 5 7. π¦ = β2π₯ β 5
8. π¦ =5
2π₯ β
1
2 9. π¦ = 2π₯ + 9 10. π¦ = β
3
5π₯ +
12
5
Section B Answers
1. (a) πππ435ππ (b) πππ512π₯π¦ (c) πππ310ππ
2. (a) 4πππππ₯ (b) 6πππππ (c) 5πππππ
3. π₯ = 6 4. π₯ = 22 5. π₯ = 9
6. (a) ππππ4 (b) 2 (c) 3
7. π₯ = 21 8. π₯ = 81
9. (a) ππππ40 (b) 3 (c) 2
10. π =1
2
Section C
O1
1. (b) 2. (a)
O2
1. (a) ππππ₯15
7 (b) ππππ2 2. (π), (π)πππππ 3. π₯ =
π¦2
π§3
4. π = 16 5. 1 + ππππ8 β ππππ9
O3
1. π₯ = 0 β 868 2. (a) π΄ = 0 β 0503 (b) π = β0.0347 (c) π‘ = 461
3. π₯ = 2 (discard π₯ = β8) 4. π₯ = β1 and π₯ =1
2
5. (a) πππππ (b) π₯ = 38 6. π = 32β2 + 1
7. π =1
2 8. π₯ = 71 9. π₯ = 81
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O4
1. (a) π = 4, π = 3 (b) π = 5, π = 2
2. (a) π = 3, π = 5 (b) π = 7, π = 3 3. π = 2, π = 4
4. π = 1000, π = 1 β 05
O5
1. πππππ¦ against πππππ₯ 2. πππππ¦ against π₯
3. π = 32, π =1
2
4. (a) π =1
2 (b) π =
3
2 (c) π = πππ104 and (0, πππ103)
5. π΄ = π3, π =1
3
O6
1. (a) π = 0 β 0004332 (b) 18% 2. (a) π = 0 β 028 (b) 89 β 4%
3. π = β0 β 462