Solomon Press
E
XPONENTIALS AND
L
OGARITHMS
C2
Worksheet A
1 Express each of the following in the form log
a
b = c.
a 10
3
= 1000 b 3
4
= 81 c 256 = 2
8
d 7
0
= 1
e 3
−3
=
1
27
f
1
5
32
−
=
1
2
g 19
1
= 19 h 216 =
3
2
36
2 Express each of the following using index notation.
a log
5
125 = 3 b log
2
16 = 4 c 5 = log
10
100
000 d log
23
1 = 0
e
1
2
= log
9
3 f lg 0.01 = −2 g log
2
1
8
= −3 h log
6
6 = 1
3 Without using a calculator, find the exact value of
a log
7
49 b log
4
64 c log
2
128 d log
3
27
e log
5
625 f log
8
8 g log
7
1 h log
15
1
15
i log
3
1
9
j lg 0.001 k log
16
2 l log
4
8
m log
9
243 n log
100
0.001 o log
25
125 p log
27
1
9
4 Without using a calculator, find the exact value of x in each case.
a log
5
25 = x b log
2
x = 6 c log
x
64 = 3 d lg x = −3
e log
x
16 =
2
3
f log
5
1 = x g log
x
9 = 1 h lg 10
12
= x
i 2
log
x
7 = 1 j log
4
x = 1.5 k log
x
0.1 =
1
3
− l 3
log
8
x + 1 = 0
5 Express in the form log
a
n
a log
a
4 + log
a
7 b log
a
10 − log
a
5 c 2
log
a
6
d log
a
9 − log
a
1
3
e
1
2
log
a
25 + 2
log
a
3 f log
a
48 − 3
log
a
2 −
1
2
log
a
9
6 Express in the form p
log
q
x
a log
q
x
5
b
1
2
log
q
x
15
c log
q
1
x
d log
q
3
x
e 4
log
q
1
x
f log
q
x
2
+ log
q
x
5
g log
q
2
1
x
+ loq
q
3
1
x
h 3
log
q
x
2
−
1
2
log
q
x
4
7 Express in the form lg n
a lg 5 + lg 4 b lg 12 − lg 6 c 3
lg 2 d 4
lg 3 − lg 9
e
1
2
lg 16 −
1
5
lg 32 f 1 + lg 11 g lg
1
50
+ 2 h 3 − lg 40
8 Without using a calculator, evaluate
a log
3
54 − log
3
2 b log
5
20 + log
5
1.25 c log
2
16 + log
3
27
d log
6
24 + log
6
9 e log
3
12 − log
3
4 f log
4
18 − log
4
9
g log
9
4 + log
9
0.25 h 2
lg 2 + lg 25 i
1
3
log
3
8 − log
3
18
j
1
3
log
4
64 + 2 log
5
25 k
1
2
log
5
(
16
9
1) + 2 log
5
10 l log
3
5 − 2 log
3
6 − log
3
(3
4
3
)
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Solomon Press
E
XPONENTIALS AND
L
OGARITHMS
C2
Worksheet B
1 Express in the form p
log
10
a + q
log
10
b
a log
10
ab b log
10
ab
7
c log
10
3
a
b
d log
10
a b
e log
10
(ab)
2
f log
10
1
ab
g log
10
53
ba
h 3
log
10
2
3
a
b
2 Given that y = log
q
8, express each of the following in terms of y.
a log
q
64 b log
q
2 c log
q
16
q
d log
q
4q
3
3 Given that a = lg 2 and b = lg 3, express each of the following in terms of a and b.
a lg 18 b lg 96 c lg
16
9
d lg 6 − lg 8
e lg 6 f
3
2
lg 16 +
1
2
lg 81 g 4
lg 3 − 3
lg 6 h lg 60 + lg 20 − 2
4 Without using a calculator, evaluate
a
1
3
log
5
1000 −
1
2
log
5
4 b 2
log
12
4 +
1
2
log
12
81 c log
4
12 + log
4
2
3
d
7
7
log 81
log 3
e 3
log
27
12 − 2
log
27
72 f
11
1
11
5
log 25
log
5 Solve each equation, giving your answers correct to 3 significant figures.
a log
3
x = 1.8 b log
5
x = −0.3 c log
8
(x − 3) = 2.1
d log
4
(
1
2
x + 1) = 3.2 e 15 − log
2
3y = 9.7 f log
6
(1 − 5t) + 4.2 = 3.6
6 Express in the form log
2
[f(x)]
a 5
log
2
x b log
2
x + log
2
(x + 4) c 2
log
2
x +
1
5
log
2
x
5
d 3
log
2
(x − 2) − 4
log
2
x e log
2
(x
2
− 1) − log
2
(x + 1) f log
2
x −
1
2
log
2
x
4
+
1
3
log
2
x
2
7 Solve each of the following equations.
a log
3
x + log
3
5 = log
3
(2x + 3) b log
9
x + log
9
10 =
3
2
c log
4
x − log
4
(x − 1) = log
4
3 +
1
2
d log
5
5x − log
5
(x + 2) = log
5
(x + 6) − log
5
x
e 2
log
6
x = log
6
(2x − 5) + log
6
5 f log
7
4x = log
7
1
6x −
+ 1
8 Solve each pair of simultaneous equations.
a log
x
y = 2 b log
5
x − 2
log
5
y = log
5
2
xy = 27 x + y
2
= 12
c log
2
x = 3 − 2
log
2
y d log
y
x =
3
2
log
y
32 =
5
2
−
1
3
x +
1
2
3y = 20
e log
a
x + log
a
3 =
1
2
log
a
y f log
10
y + 2
log
10
x = 3
3x + y = 20 log
2
y − log
2
x = 3
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Solomon Press
E
XPONENTIALS AND
L
OGARITHMS
C2
Worksheet C
1 Find, to 3 significant figures, the value of
a log
10
60 b log
10
6 c log
10
253 d log
10
0.4
2 Solve each equation, giving your answers to 2 decimal places.
a 10
x
= 14 b 2(10
x
) − 8 = 0 c 10
3x
= 49
d 10
x − 4
= 23 e 10
2x + 1
= 130 f 100
x
− 5 = 0
3 Show that log
a
b =
log
log
c
c
b
a
, where a, b and c are positive constants.
4 Find, to 3 significant figures, the value of
a log
2
7 b log
20
172 c log
5
49 d log
9
4
5 Solve each equation, giving your answers to 3 significant figures.
a 3
x
= 12 b 2
x
= 0.7 c 8
−y
= 3 d
1
2
4
x
− 0.3 = 0
e 5
t + 3
= 24 f 16 − 3
4 + x
= 0 g 7
2x + 4
= 12 h 5(2
3x + 1
) = 62
i 4
2 − 3x
= 32.7 j 5
x
= 6
x − 1
k 7
y + 2
= 9
y + 1
l 4
5 − x
= 11
2x − 1
m
1
2
3
4
x +
− 5
1 − 2x
= 0 n 2
3y − 2
= 3
2y + 5
o 7
2x + 5
= 7(11
3x − 4
) p 3
2x
= 3
x − 1
× 2
4 + x
6 Solve the following equations, giving your answers to 2 decimal places where appropriate.
a 2
2x
+ 2
x
− 6 = 0 b 3
2x
− 5(3
x
) + 4 = 0 c 5
2x
+ 12 = 8(5
x
)
d 2(4
x
) + 3(4
−x
) = 7 e 2
2y
+ 1
+ 7(2
y
) − 15 = 0 f 3
2x + 1
− 17(3
x
) + 10 = 0
g 25
t
+ 5
t + 1
− 24 = 0 h 3
2x + 1
+ 15 = 2(3
x + 2
) i 3(16
x
) − 4
x + 2
+ 5 = 0
7 Sketch each pair of curves on the same diagram, showing the coordinates of any points of
intersection with the coordinate axes.
a y = 2
x
b y = 3
x
c y = 4
x
d y = 2
x
y = 5
x
y = (
1
3
)
x
y = 4
x
− 1 y = 2
x + 3
8 A curve has the equation y = 2 + a
x
where a is a constant and a > 1.
a Sketch the curve, showing the coordinates of any points of intersection with the coordinate
axes and the equations of any asymptotes.
Given also that the curve passes through the point (3, 29),
b find the value of a.
9 y
y = 2
x
− 5
O B x
A
The diagram shows the curve with equation y = 2
x
− 5 which intersects the coordinate axes at the
points A and B. Find the length AB correct to 3 significant figures.
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Solomon Press
E
XPONENTIALS AND
L
OGARITHMS
C2
Worksheet D
1 Given that a = log
10
2 and b = log
10
3, find expressions in terms of a and b for
a log
10
1.5, (2)
b log
10
24, (2)
c log
10
150. (3)
2 Find, to an appropriate degree of accuracy, the values of x for which
a 4
log
3
x − 5 = 0, (2)
b log
3
x
3
− 5
log
3
x = 4. (3)
3 a Given that p = log
2
q, find expressions in terms of p for
i log
2
q
,
ii log
2
8q. (4)
b Solve the equation
log
2
8q − log
2
q
= log
3
9. (3)
4 An initial investment of £1000 is placed into a savings account that offers 2.2% interest
every 3 months. The amount of money in the account, £P, at the end of t years is given by
P = 1000 × 1.022
4t
Find, to the nearest year, how long it will take for the investment to double in value. (4)
5 y
y = (
1
3
)
x
− 4
O x
y = k
The diagram shows the curve with equation y = (
1
3
)
x
− 4.
a Write down the coordinates of the point where the curve crosses the y-axis. (1)
The curve has an asymptote with equation y = k.
b Write down the value of the constant k. (1)
c Find the x-coordinate of the point where the curve crosses the x-axis. (3)
6 a Solve the equation
log
3
(x + 1) − log
3
(x − 2) = 1. (3)
b Find, in terms of logarithms to the base 10, the exact value of x such that
3
2x + 1
= 2
x − 4
. (3)
7 a Given that t = 2
x
, write down expressions in terms of t for
i 2
x − 1
,
ii 2
2x + 1
. (3)
b Hence solve the equation
2
2x + 1
− 14(2
x − 1
) + 6 = 0. (5)
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Solomon Press
8 Find the values of x for which
a log
2
(3x + 5) + log
5
125 = 7, (3)
b log
2
(x + 1) = 5 − log
2
(3x − 1). (5)
9 Given that log
a
(x + 4) = log
a
4
x
+ log
a
5,
and that log
a
(y + 2) = log
a
12 − log
a
(y + 1),
where y > 0, find
a the value of x, (3)
b the value of y, (4)
c the value of the logarithm of x to the base y. (2)
10 A colony of fast-breeding fish is introduced into a large, newly-built pond. The number of
fish in the pond, n, after t weeks is modelled by
n =
18000
18
t
c
−
+
.
a Find the initial number of fish in the pond. (2)
Given that there are 3600 fish in the pond after 3 weeks, use this model to
b show that c =
3
2
, (3)
c find the time taken for the initial population of fish to double in size, giving your
answer to the nearest day. (4)
11 a Given that y = log
8
x, find expressions in terms of y for
i log
8
x
2
,
ii log
2
x. (4)
b Hence, or otherwise, find the value of x such that
3
log
8
x
2
+ log
2
x = 6. (3)
12 Solve the simultaneous equations
log
2
y = log
2
(3 − 2x) + 1
log
4
x + log
4
y =
1
2
(8)
13 a Sketch on the same diagram the curves y = 2
x
+ 1 and y = (
1
2
)
x
, showing the
coordinates of any points where each curve meets the coordinate axes. (4)
Given that the curves y = 2
x
+ 1 and y = (
1
2
)
x
intersect at the point A,
b show that the x-coordinate of A is a solution of the equation
2
2x
+ 2
x
− 1 = 0, (2)
c hence, show that the y-coordinate of A is
1
2
( 5 + 1). (4)
14 a Show that x = 1 is a solution of the equation
2
3x
− 4(2
2x
) + 2
x
+ 6 = 0. (I) (1)
b Show that using the substitution u = 2
x
, equation (I) can be written as
u
3
− 4u
2
+ u + 6 = 0. (2)
c Hence find the other real solution of equation (I) correct to 3 significant figures. (7)
C2 E
XPONENTIALS AND
L
OGARITHMS
Worksheet D continued
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Solomon Press
E
XPONENTIALS AND
L
OGARITHMS
C2
Answers - Worksheet A
1 a log
10
1000 = 3 b log
3
81 = 4 c log
2
256 = 8 d log
7
1 = 0
e log
3
1
27
= −3 f log
32
1
2
=
1
5
− g log
19
19 = 1 h log
36
216 =
3
2
2 a 5
3
= 125 b 2
4
= 16 c 10
5
= 100
000 d 23
0
= 1
e
1
2
9 = 3 f 10
−2
= 0.01 g 2
−3
=
1
8
h 6
1
= 6
3 a = log
7
7
2
b = log
4
4
3
c = log
2
2
7
d = log
3
3
3
= 2 = 3 = 7 = 3
e = log
5
5
4
f = log
8
8
1
g = log
7
7
0
h = log
15
15
−1
= 4 = 1 = 0 = −1
i = log
3
3
−2
j = lg 10
−3
k = log
16
1
4
16 l = log
4
3
2
4
= −2 = −3 =
1
4
=
3
2
m = log
9
5
2
9 n = log
100
3
2
100
−
o = log
25
3
2
25 p = log
27
2
3
27
−
=
5
2
=
3
2
− =
3
2
=
2
3
−
4 a 5
x
= 25 b 2
6
= x c x
3
= 64 d 10
−3
= x
x = 2 x = 64 x = 4 x =
1
1000
e
2
3
x = 16 f 5
x
= 1 g x
1
= 9 h 10
x
= 10
12
x = 64 x = 0 x = 9 x = 12
i log
x
7 =
1
2
j 4
1.5
= x k
1
3
x
−
= 0.1 l log
8
x =
1
3
−
1
2
x = 7 x = 8 x = 1000
1
3
8
−
= x
x = 49 x =
1
2
5 a = log
a
(4 × 7) b = log
a
(10 ÷ 5) c = log
a
6
2
= log
a
28 = log
a
2 = log
a
36
d = log
a
(9 ÷
1
3
) e = log
a
1
2
25 + log
a
3
2
f = log
a
48 − log
a
2
3
− log
a
1
2
9
= log
a
27 = log
a
5 + log
a
9 = log
a
48 − log
a
8 − log
a
3
= log
a
(5 × 9) = log
a
[48 ÷ (8 × 3)]
= log
a
45 = log
a
2
6 a = 5
log
q
x b =
15
2
log
q
x c = log
q
x
−1
d = log
q
1
3
x
= −log
q
x =
1
3
log
q
x
e = 4
log
q
1
2
x
−
f = 2
log
q
x + 5
log
q
x g = log
q
x
−2
+ log
q
x
−3
h = 6
log
q
x − 2
log
q
x
= −2
log
q
x = 7
log
q
x = −2
log
q
x − 3
log
q
x = 4
log
q
x
= −5
log
q
x
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C2 E
XPONENTIALS AND
L
OGARITHMS
Answers - Worksheet A page 2
Solomon Press
7 a = lg (5 × 4) b = lg (12 ÷ 6) c = lg 2
3
d = lg 3
4
− lg 9
= lg 20 = lg 2 = lg 8 = lg 81 − lg 9
= lg (81 ÷ 9)
= lg 9
e = lg
1
2
16 − lg
1
5
32 f = lg 10 + lg 11 g = lg
1
50
+ lg 10
2
h = lg 10
3
− lg 40
= lg 4 − lg 2 = lg (10 × 11) = lg
1
50
+ lg 100 = lg 1000 − lg 40
= lg (4 ÷ 2) = lg 110 = lg (
1
50
× 100) = lg (1000 ÷ 40)
= lg 2 = lg 2 = lg 25
8 a = log
3
(54 ÷ 2) b = log
5
(20 × 1.25) c = log
2
2
4
+ log
3
3
3
= log
3
27 = log
5
25 = 4 + 3
= log
3
3
3
= log
5
5
2
= 7
= 3 = 2
d = log
6
(24 × 9) e = log
3
(12 ÷ 4) f = log
4
(18 ÷ 9)
= log
6
216 = log
3
3 = log
4
2
= log
6
6
3
= 1 = log
4
1
2
4
= 3 =
1
2
g = log
9
(4 × 0.25) h = lg 2
2
+ lg 25 i = log
3
1
3
8 − log
3
18
= log
9
1 = lg 4 + lg 25 = log
3
2 − log
3
18
= 0 = lg (4 × 25) = log
3
(2 ÷ 18)
= lg 100 = log
3
1
9
= lg 10
2
= log
3
3
−2
= 2 = −2
j = log
4
1
3
64 + (2 × log
5
5
2
) k =
1
2
log
5
25
16
+ log
5
10
2
l = log
3
5 − log
3
6
2
− log
3
15
4
= log
4
4 + (2 × 2) = log
5
1
2
25
16
() + log
5
100 = log
3
[5 ÷ (36 ×
15
4
)]
= 1 + 4 = log
5
5
4
+ log
5
100 = log
3
1
27
= 5 = log
5
(
5
4
× 100) = log
3
3
−3
= log
5
125 = −3
= log
5
5
3
= 3
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Solomon Press
E
XPONENTIALS AND
L
OGARITHMS
C2
Answers - Worksheet B
1 a = log
10
a + log
10
b b = log
10
a + log
10
b
7
c = log
10
a
3
− log
10
b d = log
10
a + log
10
1
2
b
= log
10
a + 7
log
10
b = 3
log
10
a − log
10
b = log
10
a +
1
2
log
10
b
e = 2
log
10
ab f = −log
10
ab g = log
10
3
2
a + log
10
5
2
b h =
3(log
10
a
2
−
log
10
1
3
b )
= 2log
10
a
+
2log
10
b = −log
10
a − log
10
b =
3
2
log
10
a
+
5
2
log
10
b = 6
log
10
a − log
10
b
2 a = log
q
8
2
b = log
q
1
3
8 c = log
q
16 − log
q
q d = log
q
4 + log
q
q
3
= 2y =
1
3
y = log
q
4
3
8 − 1 = log
q
2
3
8 + 3
=
4
3
y − 1 =
2
3
y + 3
3 a = lg (2 × 3
2
) b = lg (2
5
× 3) c = lg 9 − lg 16 d = lg (2 × 3) − lg 2
3
= lg 2 + 2
lg 3 = 5
lg 2 + lg 3 = lg 3
2
− lg 2
4
= lg 2 + lg 3 − 3
lg 2
= a + 2b = 5a + b = 2
lg 3 − 4
lg 2 = lg 3 − 2
lg 2
= 2b − 4a = b − 2a
e =
1
2
lg 6 f =
3
2
lg 2
4
+
1
2
lg 3
4
g = 4lg3 − 3(lg2+lg3) h =lg(6× 10)+lg(2×10)−2
=
1
2
(lg 2 + lg 3) = 6
lg 2 + 2
lg 3 = lg 3 − 3
lg 2 =
lg
6
+
1
+
lg
2
+
1
−
2
=
1
2
(a + b) = 6a + 2b = b − 3a = lg 2 + lg 3 + lg 2
= 2a + b
4 a = log
5
10 − log
5
2 b = log
12
16 + log
12
9 c = log
4
8
= log
5
5 = log
12
144 = log
4
3
2
4
= 1 = 2 =
3
2
d =
4
7
7
log 3
log 3
e = log
27
3
2
12
72
f =
2
11
11
log 5
log 5−
=
7
7
4log 3
log 3
= log
27
12 12 12
612612
××
×××
=
11
11
2log 5
log 5−
= 4 = log
27
1
3
=
1
3
− = −2
5 a x = 3
1.8
b x = 5
−0.3
c x − 3 = 8
2.1
x = 7.22 x = 0.617 x = 3 + 8
2.1
x = 81.8
d
1
2
x + 1 = 4
3.2
e log
2
3y = 5.3 f log
6
(1 − 5t) = −0.6
x = 2(4
3.2
− 1) 3y = 2
5.3
1 − 5t = 6
−0.6
x = 167 y =
1
3
× 2
5.3
t =
1
5
(1 − 6
−0.6
)
y = 13.1 t = 0.132
6 a = log
2
x
5
b = log
2
(x
2
+ 4x) c = log
2
x
2
+ log
2
x
= log
2
x
3
d = log
2
(x − 2)
3
− log
2
x
4
e = log
2
2
1
1
x
x
−
+
f = log
2
x − 2
log
2
x +
2
3
log
2
x
= log
2
3
4
(2)x
x
−
= log
2
(1)(1)
1
xx
x
+−
+
=
1
3
− log
2
x
= log
2
(x − 1) = log
2
1
3
x
−
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C2 E
XPONENTIALS AND
L
OGARITHMS
Answers - Worksheet B page 2
Solomon Press
7 a log
3
5x = log
3
(2x + 3) b log
9
10x =
3
2
5x = 2x + 3 10x =
3
2
9 = 27
x = 1 x = 2.7
c log
4
1
x
x −
= log
4
3 + log
4
2 = log
4
6 d log
5
5
2
x
x +
= log
5
6x
x
+
1
x
x −
= 6
5
2
x
x +
=
6x
x
+
x = 6x − 6 5x
2
= (x + 2)(x + 6) = x
2
+ 8x + 12
x =
6
5
x
2
− 2x − 3 = 0
(x + 1)(x − 3) = 0
x = −1, 3
log
5
x not real for x = −1 ∴ x = 3
e log
6
x
2
= log
6
5(2x − 5) f log
7
4x − log
7
1
6x −
= 1
x
2
= 5(2x − 5) log
7
4x(x − 6) = 1
x
2
− 10x + 25 = 0 4x(x − 6) = 7
(x − 5)
2
= 0 4x
2
− 24x − 7 = 0
x = 5 x =
24 576 112
8
±+
= 3 ±
1
2
43
log
7
4x not real for x = 3 −
1
2
43
∴ x = 3 +
1
2
43 [ = 6.28 (3sf)]
8 a log
x
y = 2 ⇒ y = x
2
b log
5
x − 2
log
5
y = log
5
2 ⇒
2
x
y
= 2
sub. x
3
= 27 ⇒ x = 2y
2
x = 3 sub. 3y
2
= 12
∴ x = 3, y = 9 y
2
= 4
for real log
5
y, y > 0 ∴ y = 2
∴ x = 8, y = 2
c log
y
32 =
5
2
− ⇒
5
2
y
−
= 32 d log
y
x =
3
2
⇒
3
2
y = x
⇒ y =
2
5
32
−
=
1
4
⇒
1
2
y =
1
3
x
sub. log
2
x = 3 − 2
log
2
1
4
sub. 4
1
3
x = 20
log
2
x = 3 − (−4) = 7
1
3
x = 5
x = 2
7
= 128 x = 5
3
= 125
∴ x = 128, y =
1
4
∴ x = 125, y = 25
e log
a
x + log
a
3 =
1
2
log
a
y ⇒ 3x =
1
2
y f log
10
y + 2
log
10
x = 3 ⇒ x
2
y = 10
3
⇒ y = 9x
2
log
2
y − log
2
x = 3 ⇒
y
x
= 2
3
sub. 3x + 9x
2
= 20 ⇒ y = 8x
9x
2
+ 3x − 20 = 0 sub. 8x
3
= 1000
(3x + 5)(3x − 4) = 0 x
3
= 125
for real log
a
x, x > 0 ∴ x =
4
3
x = 5
∴ x =
4
3
, y = 16 ∴ x = 5, y = 40
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Solomon Press
E
XPONENTIALS AND
L
OGARITHMS
C2
Answers - Worksheet C
1 a 1.78 b 0.778 c 2.40 d −0.398
2 a x = lg 14 = 1.15 b 10
x
= 4 c 3x = lg 49
x = lg 4 = 0.60 x =
1
3
lg 49 = 0.56
d x − 4 = lg 23 e 2x + 1 = lg 130 f (10
2
)
x
= 10
2x
= 5
x = 4 + lg 23 = 5.36 x =
1
2
(lg 130 − 1) = 0.56 2x = lg 5
x =
1
2
lg 5 = 0.35
3 let y = log
a
b ⇒ a
y
= b
y log
c
a = log
c
b
y =
log
log
c
c
b
a
∴ log
a
b =
log
log
c
c
b
a
4 a =
lg 7
lg 2
= 2.81 b =
lg172
lg 20
= 1.72 c =
lg 49
lg5
= 2.42 d =
lg 4
lg9
= 0.631
5 a x lg 3 = lg 12 b x lg 2 = lg 0.7 c −y lg 8 = lg 3 d
1
2
x lg 4 = lg 0.3
x =
lg12
lg3
x =
lg 0.7
lg 2
y = −
lg3
lg8
x =
2lg0.3
lg 4
x = 2.26 x = −0.515 y = −0.528 x = −1.74
e (t + 3)
lg 5 = lg 24 f (4 + x)
lg 3 = lg 16 g (2x + 4)lg 7 = lg 12 h 2
3x + 1
= 12.4
t =
lg 24
lg5
− 3 x =
lg16
lg3
− 4 x =
1
2
(
lg12
lg 7
− 4) (3x + 1)
lg 2 = lg 12.4
t = −1.03 x = −1.48 x = −1.36 x =
1
3
(
lg12.4
lg 2
− 1)
x = 0.877
i (2 − 3x)
lg 4 = lg 32.7 j x lg 5 = (x − 1)
lg 6
x =
1
3
(2 −
lg32.7
lg 4
) x
(lg 6 − lg 5) = lg 6
x = −0.172 x =
lg 6
lg 6 lg5−
= 9.83
k (y + 2)
lg 7 = (y + 1)
lg 9 l (5 − x)
lg 4 = (2x − 1)
lg 11
y
(lg 9 − lg 7) = 2
lg 7 − lg 9 x
(2
lg 11 + lg 4) = 5
lg 4 + lg 11
y =
2lg7 lg9
lg9 lg 7
−
−
= 6.74 x =
5lg4 lg11
2lg11 lg4
+
+
= 1.51
m (
1
2
x + 3)
lg 4 = (1 − 2x)
lg 5 n (3y − 2)
lg 2 = (2y + 5)
lg 3
x
(
1
2
lg 4 + 2
lg 5) = lg 5 − 3
lg 4 y
(3
lg 2 − 2
lg 3) = 5
lg 3 + 2
lg 2
x =
1
2
lg5 3lg 4
lg 4 2 lg 5
−
+
= −0.652 y =
5lg3 2lg2
3lg2 2lg3
+
−
= −58.4
o 7
2x + 4
= 11
3x − 4
p 3
x + 1
= 2
4 + x
(2x + 4)
lg 7 = (3x − 4)
lg 11 (x + 1)
lg 3 = (4 + x)
lg 2
x
(3
lg 11 − 2
lg 7) = 4
lg 7 + 4
lg 11 x
(lg 3 − lg 2) = 4
lg 2 − lg 3
x =
4lg7 4lg11
3lg11 2lg7
+
−
= 5.26 x =
4lg2 lg3
lg3 lg 2
−
−
= 4.13
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C2 E
XPONENTIALS AND
L
OGARITHMS
Answers - Worksheet C page 2
Solomon Press
6 a (2
x
+ 3)(2
x
− 2) = 0 b (3
x
− 1)(3
x
− 4) = 0 c 5
2x
− 8(5
x
) + 12 = 0
2
x
= −3 [no sols], 2 3
x
= 1, 4 (5
x
− 2)(5
x
− 6) = 0
x = 1 x = 0,
lg 4
lg3
= 0, 1.26 5
x
= 2, 6
x =
lg 2
lg5
,
lg 6
lg5
= 0.43, 1.11
d 2(4
2x
) − 7(4
x
) + 3 = 0 e 2(2
2y
) + 7(2
y
) − 15 = 0 f 3(3
2x
) − 17(3
x
) + 10 = 0
(2(4
x
) − 1)(4
x
− 3) = 0 (2(2
y
) − 3)(2
y
+ 5) = 0 (3(3
x
) − 2)(3
x
− 5) = 0
4
x
=
1
2
, 3 2
y
= −5 [no sols],
3
2
3
x
=
2
3
, 5
x =
1
2
− ,
lg3
lg 4
=
1
2
− , 0.79 y =
3
2
lg
lg 2
= 0.58 x =
2
3
lg
lg3
,
lg5
lg3
= −0.37, 1.46
g 5
2t
+ 5(5
t
) − 24 = 0 h 3(3
2x
) − 18(3
x
) + 15 = 0 i 3(4
2x
) − 16(4
x
) + 5 = 0
(5
t
+ 8)(5
t
− 3) = 0 3(3
x
− 1)(3
x
− 5) = 0 (3(4
x
) − 1)(4
x
− 5) = 0
5
t
= −8 [no sols], 3 3
x
= 1, 5 4
x
=
1
3
, 5
t =
lg3
lg5
= 0.68 x = 0,
lg5
lg3
= 0, 1.46 x =
1
3
lg
lg 4
,
lg5
lg 4
= −0.79, 1.16
7 a y b y
y = 5
x
y = 2
x
y = (
1
3
)
x
y = 3
x
(0, 1) (0, 1)
O x O x
c y d y
y = 4
x
(0, 8)
y = 4
x
−
−−
− 1 y = 2
x + 3
y = 2
x
(0, 1)
x O x
8 a y 9 x = 0 ⇒ y = −4
y = 0 ⇒ 2
x
= 5
x =
lg5
lg 2
AB
2
= 4
2
+ (
lg5
lg 2
)
2
= 21.391
AB = 4.63
O x
b (3, 29) ⇒ 29 = 2 + a
3
a
3
= 27
a = 3
(0, 1)
(0, 0)
(0, 3)
y = 2
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Solomon Press
E
XPONENTIALS AND
L
OGARITHMS
C2
Answers - Worksheet D
1 a = log
10
3
2
2 a log
3
x =
5
4
= log
10
3 − log
10
2 x =
5
4
3 = 3.95 (3sf)
= b − a b 3 log
3
x − 5 log
3
x = 4
b = log
10
(2
3
× 3) log
3
x = −2
= 3
log
10
2 + log
10
3 x = 3
−2
=
1
9
= 3a + b
c = log
10
(1.5 × 100)
= log
10
1.5 + log
10
100
= b − a + 2
3 a i = log
2
1
2
q =
1
2
log
2
q =
1
2
p 4 2000 = 1000 × 1.022
4t
ii = log
2
8 + log
2
q = 3 + p 2 = 1.022
4t
b 3 + p −
1
2
p = 2 4t lg 1.022 = lg 2
p = log
2
q = −2 t =
lg 2
4lg1.022
= 7.96
∴ q = 2
−2
=
1
4
∴ 8 years
5 a (0, −3) 6 a log
3
1
2
x
x
+
−
= 1
b k = −4
1
2
x
x
+
−
= 3
c (
1
3
)
x
− 4 = 0 x + 1 = 3x − 6
(
1
3
)
x
= 4 x =
7
2
x =
1
3
lg 4
lg
= −1.26 (3sf) b (2x + 1) lg 3 = (x − 4) lg 2
x (lg 2 − 2 lg 3) = lg 3 + 4 lg 2
x =
lg3 4lg 2
lg 2 2lg 3
+
−
7 a i = 2
−1
(2
x
) =
1
2
t 8 a log
2
(3x + 5) + 3 = 7
ii = 2(2
2x
) = 2(2
x
)
2
= 2t
2
3x + 5 = 2
4
= 16
b 2t
2
− 7t + 6 = 0 x =
11
3
(2t − 3)(t − 2) = 0 b log
2
(x + 1) + log
2
(3x − 1) = 5
t = 2
x
=
3
2
, 2 (x + 1)(3x − 1) = 2
5
= 32
x =
3
2
lg
lg 2
, 1 = 0.585 (3sf), 1 3x
2
+ 2x − 33 = 0
(3x + 11)(x − 3) = 0
x =
11
3
− , 3
for real log
2
(3x − 1), x >
1
3
∴ x = 3
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C2 E
XPONENTIALS AND
L
OGARITHMS
Answers - Worksheet D page 2
Solomon Press
9 a x + 4 =
5
4
x 10 a t = 0 ⇒ n = 2000
x = 16 b 3600 =
3
18000
18c
−
+
b y + 2 =
12
1y +
1 + 8c
−3
= 5
(y + 2)(y + 1) = 12 c
−3
=
1
2
y
2
+ 3y − 10 = 0 c
3
= 2
(y + 5)(y − 2) = 0 c =
3
2
y > 0 ∴ y = 2 c 4000 =
18000
18
t
c
−
+
c log
y
x = log
2
16 = 4 1 + 8c
−t
=
9
2
c
−t
=
7
16
−t =
7
16
3
lg
lg 2
t = 3.578 weeks = 25 days
11 a i log
8
x
2
= 2 log
8
x = 2y 12 log
2
y − log
2
(3 − 2x) = 1 ⇒
32
y
x−
= 2
ii y = log
8
x ⇒ x = 8
y
= 2
3y
⇒ y = 6 − 4x
∴ log
2
x = 3y log
4
xy =
1
2
⇒ xy =
1
2
4 = 2
b 3(2y) + 3y = 6 sub. x(6 − 4x) = 2
y = log
8
x =
2
3
2x
2
− 3x + 1 = 0
∴ x =
2
3
8 = 4 (2x − 1)(x − 1) = 0
x =
1
2
, 1
∴ x =
1
2
, y = 4 or x = 1, y = 2
13 a y 14 a when x = 1,
LHS = 8 − 4(4) + 2 + 6 = 0
∴ x = 1 is a solution
b 2
3x
= (2
x
)
3
= u
3
y = (
1
2
)
x
y = 2
x
+ 1 2
2x
= (2
x
)
2
= u
2
∴ (I) ⇒ u
3
− 4u
2
+ u + 6 = 0
c x = 1 ⇒ u = 2 ∴ (u − 2) is a factor
O x
b at A, 2
x
+ 1 = (
1
2
)
x
(2
x
)
2
+ 2
x
= 1
2
2x
+ 2
x
− 1 = 0
c 2
x
=
114
2
−± +
2
x
=
15
2
−−
[no sols] or
15
2
−+
(u − 2)(u
2
− 2u − 3) = 0
∴ 2
x
=
1
2
5
−
1
2
(u − 2)(u − 3)(u + 1) = 0
∴ y = (
1
2
5
−
1
2
) + 1 =
1
2
( 5 + 1) u = 2
x
= −1 [no sols], 2 or 3
x = 1 (given) or
lg3
lg 2
= 1.58
u
2
−
2u
−
3
u − 2
u
3
−
4u
2
+ u + 6
u
3
−
2u
2
−
2u
2
+ u
−
2u
2
+ 4u
−
3u + 6
−
3u + 6
(0, 2)
(0, 1)
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