f4 mathematics assignment 2010-02-10
TRANSCRIPT
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7/29/2019 f4 Mathematics Assignment 2010-02-10
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F.4 Mathematics Assignment (2010-02-10)
1. Simplify and express
4
3
20
t
sr with positive indices.
2. Simplify and express
11
1
+
y
x
y
xwith positive indices.
3. Solve the equation 32
4 279 =
x
.
4. Find the value of729
1log 3 without using a calculator.
5. Simplify and express1
2
4
3
)(
b
aba
b
awith positive indices.
6. Simplify and express 43243
)()5(25 xyyxx with positive indices.
7. Find the value of2log2
9
1log
36log4
1216log 3
aa
aa
+without using a calculator.
8. Solve the equation 12642 142 = + xx .
9. Given that log 5 =x and log 6 =y, express 288log in terms ofx and y.
10. Solve the simultaneous equations
=
=+
64)2(4
033 12
yx
yx
.
11. Find the value of43log212log27log
31
222 + without using a calculator.
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12. (91-CE-I-2)
13. (99-CE-I-6)
14. (04-CE-I-10)
It is known thaty is the sum of two parts, one part varies as x and the other part varies as the
square ofx. Whenx = 3,y = 3 and whenx = 4,y = 12.
(a) Expressy in terms ofx.
(b) If x is an integer andy < 42, find all possible value(s) ofx.
15. The monthly expenses $E of a boarding school is partly constant and partly varies as the
number of boarders N. The monthly expenses for 120 and 150 boarders are $134000 and
$158000 respectively. If the monthly expenses are $198000, how many boarders are there?
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END
Suggested Solution
1.
4
3
24
3
201
=
t
s
t
sr2.
1
11
1
+=
+
y
xxy
y
x
y
x1
432 )( = ts 1 xyxy +=
)4(3)4(2 = ts xy2= 1 1
128ts=
3. 32
4 279 =x
4.6
33
3
1log
729
1log = 1
3
2
342 )3()3( =x
13
33log=
22
33 22
=
=x
x
3= 1
4=x 1
6. 43243
)()5(25 xyyxx
5.
=
1
12
4
31
2
4
3
)()(b
aba
b
a
b
aba
b
a 1
)()5()25( 34
3
4
22
3
22
1
yxyxx = 1 )1(1)4(123 +
= ba 1
)()5()5( 34
3
4
22
3
22
1
yxyxx = 60ba=
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7/29/2019 f4 Mathematics Assignment 2010-02-10
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3
4)2(
3
4
2
3
2
1
)2(15
+=
+yx 1
6b= 1
3
10
3
1
35 yx=
3
10
3
1
125 yx= 1
7.2log2
9
1log
36log4
1216log 3
aa
aa
+
4log9
1log
36log6log 4
aa
aa
+
=
=
2
2
1
6
1log
)6(6log
a
a
)6(log
)(6log2
2
3
=a
a
2
2
3
= 4
3=
1 1 1 1
8. 12642 142 = + xx 9. 288log2
1288log =
2
3
1)1(2
22
24
126)14(4
12644
12642
)1(2
1
31
12
1)2(2
=
=
=
=
=
=
=
+
+
x
x
x
x
x
xx
xx
5)log310log36log2(2
1
)5log10log6log(2
1
5
106log
2
1
)2log(62
1
332
3
32
32
+=
+=
=
=
)33(22
1xy +=
10.
=
=+
64)2(4033
12
yx
yx
)2()1(
11.
43log212log27log
31
222 +
From (1),
12
12
3312
+==
=
xy
yx
yx
2
223
24
3log12log27log
+=
From (2),
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4
62
22 62
=+
=++
=++
yx
yx
yx
+=16
9log12log3log 222
Consider the simultaneous equations:
=+
+=
4
12
yx
xy
)4(
)3(
=
16
9
123log 2
By substituting (3) into (4), we have
1413
4)12(
==+
=++
xx
xx
62log
64log
6
2
2
==
=
By substitutingx = 1 into (3), we have
3
1)1(2
=+=y
The solution isx = 1,y = 3.