1. (a)
bxayba
abyaby
x
x
x
logloglogloglog)log(log
+=
+=
=
=
(b) When x = 1, log y = 0.8. 0.8 = log a + log b ……(1) When x = 3, log y = 0.5. 0.5 = log a + 3 log b ……(2)
(2) − (1):
d.p.) 1 to(cor. 7.010
15.0 log log 23.0
15.0
=
=
−=
=−
−bb
b
By substituting log b = −0.15 into (1), we have
d.p.) 1 to(cor. 9.810
95.0 log15.0 log8.0
95.0
=
=
=
−=
aa
a
(c) From (b), xy 0.150.95 log −=
When y = 100,
71.050.15
0.150.9520.150.95100 log
−=
−=
−=
−=
xx
xx
2. (a) 36 = 62 236log6 =
(b) 23
3 3327 ==
2327log3 =
3. (a)
33log
27log7189log7log189log
33
3
333
=
=
=
⎟⎠
⎞⎜⎝
⎛=−
(b)
52log2log 5
2log2log
2log32log
2log)84(log
2log8log4log
6
6
6
56
6
6
6
6
6
66
=
=
=
=
×=
+
(c)
41221
7log7log
49log7log
7log
27
21
7
7
749
=
=
=
=
4. (a)
5124820
48244
863
4
163
4log
1)63(log)4(log)63(log1)4(log
8
88
88
=
=
−=
=−
=⎟⎠
⎞⎜⎝
⎛−
=−−
−=−
x
xxx
xx
xxxx
xx
(b)
42)2(
8)(
8
8
5.18log
2
32
3
32
32
23
23
5.1
=
=
=
=
=
=
=
x
x
x
xx
5. (a) (i) From the graph, when x = 3, y = −1.6. 6.13log0.5 −=
(ii) 5.3log5.3log 0.52 −= From the graph, when x = 3.5, y = −1.8.
8.15.3log 5.0 −= 8.1)8.1(5.3log2 =−−= (b) (i) From the graph, when y = 0.5, x = 0.7. The solution of 5.0log0.5 =x is x = 0.7.
(ii)
3.1log3.1log3.1log
0.5
0.5
2
=
−=−
−=
xxx
From the graph, when y = 1.3, x = 0.4. The solution of 3.1log2 −=x is x = 0.4. 6. (a)
210log
5021
log
50log3216log
50log32log16log
2
−=
=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
=
−⎟⎠
⎞⎜⎝
⎛=
−−
−
(b)
110log
)52log(5log2log
5log)2log(
5log2log2
221
=
=
×=
+=
+=
+
(c)
29436
2log34
2log62log
2log)2log(
2log16log64log
34
6
31
4
6
3
=
×=
=
=
=
7.
3)log()log(3)log()log()log()log()log(
loglog
2
2
2
32
2
63
2
63
=
=
=
=
+
abababababbaab
ba
8.
(rejected)204234233log)423(log
3loglog)423(log3loglog2)423(log
22
22
22
2
−=
=−−
=−−
=−−
+=−−
+=−−
xx
xxxxxxxxxxxx
The equation has no solutions. 9.
fig.) sig. 3 to(cor. 723.03log22log5
60log60log)3log22log5(60log3log22log560log3log2log60log)32log(
6032
25
25
25
=
+=
=+
=+
=+
=
=
•
•
x
xxx
xx
xx
xx
Alternative Solution
fig.) sig. 3 to(cor. 723.0288log60log
60log288log60log288log
6028860)932(609326032 25
=
=
=
=
=
=
=
=
•
•
•
x
x
x
x
x
xx
xx
10.
1 log log
log log
log log
logloglog
=
= ••
••
ca
bc
ab
acb cba
11.
33log271log
15620log
15log6log20log15log6log220log
33
3
23
32
33333
−=
=
⎟⎠
⎞⎜⎝
⎛=
⎟⎠
⎞⎜⎝
⎛×
=
−−=−−
−
12.
62log
64log169123log
169log12log3log
43log12log27log
43log 212log27log
31
62
2
2
222
2
223
2222
=
=
=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛×
=
⎟⎠
⎞⎜⎝
⎛−+=
⎟⎠
⎞⎜⎝
⎛−+=−+
13.
17log
7
7log
2372727log
272log21log14log
272log
2121log314log
17
23
21
7
337
7
3
77777
−=
=
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
=
⎟⎟⎠
⎞⎜⎜⎝
⎛
××
××=
−−=⎟⎠
⎞⎜⎝
⎛−−
−
14.
432
23
)6(log)(6log
61log
)6(6log
4log91log
36log6log
2log 291log
36log41216log
2
23
2
21
43
−=
−=
=
⎟⎠
⎞⎜⎝
⎛
×=
−⎟⎠
⎞⎜⎝
⎛
+=
−⎟⎠
⎞⎜⎝
⎛
+
−a
a
a
a
aa
aa
aa
aa
15.
121
log2
log21
31
logloglog
1log
loglog2
21
31
2
3
=
−
⎟⎠
⎞⎜⎝
⎛−
=
−=
⎟⎠
⎞⎜⎝
⎛
−−
x
x
xxx
x
xx
a
a
a
aa
a
aa
16.
y
y
yx
x
yx
x
yx
x
yx
x
231
)32(21
33log
21
33log
21
279log
21
279log
36
26
3
)2(3
)13(2
3
2
13
32
13
3
+=
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎥⎦
⎤⎢⎣
⎡=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
+
−
+
−
+
−
+
17.
3loglog
log23)log2(
loglog3
loglog)log3)((log
22
=
⎟⎠
⎞⎜⎝
⎛
=
×=
ba
ab
ba
abab ba
18.
2
log
log31
log21log3
loglog
loglog3
loglog3
33
=
×=
÷=
b
a
a
b
ab
ab
bb
a
a
19.
(a) a1
4log1
9log4log1
4log9log9log
94 ====
(b) a1
4log9log
4log21
9log21
2log3log3log2 ====
20.
(a) 24log
5log5log
2
24
b==
(b) 1215log
15log4log)54(log20log4
5555 +=+=+=×=b
21.
(a) yx
==7log4log
4log3
37
(b) xy32
4log23
7log
2log
7log8log7log
7log3
3
232
3
3
3
38 ====
×
22.
3
34
3
3
33
3
33
93
81
3
34log
2log23
2log21log
29log
loglog
2loglog
=
=
=
=
=+
=+
=+
x
x
x
xx
xx
xx
23.
82
3log41log
121
41log
41log
31
41
16loglog
8loglog
41loglog
32
2
22
2
2
2
2
168
=
=
=
=
=−
=−
=−
xx
x
xx
xx
xx
24.
⎩⎨⎧
=+−−
=−
(2) 01log)2(log(1) )6(223
33
1
…………
yx
yxx
From (1), we have
yx
yxx
66)6(22)23(
1
11
=
=−
−−
yx =−1 …… (3) From (2), we have
yxyxyx
yx
=−
=−
⎟⎠
⎞⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛ −
−=−−
63312
31log2log
1log)2(log
33
33
……(4)
(4) − (3):
0520)1()63(
=−
=−−−
xxx
25
=x
By substituting 25
=x into (3), we have
23
125
=
=−
y
y
The solution is 25
=x , 23
=y .
25.
⎪⎩
⎪⎨⎧
+=
= +−
......(2) log42)(log(1) ...... 3)9(3
33
3
326
yxy
yx
From (1), we have
xyyx
yx
yx
−=
+=−
=
=−+−
+−
522212
333]3[3
132212
32)6(2
……(3)
From (2), we have
yx
yyxyyxy
9)0( 9
)3(log)(log43
423
33
=
≠=
•=
……(4)
By substituting (4) into (3), we have
2195
=
−=
y
yy
By substituting 21
=y into(4), we have
29
=x
The solution is 29
=x , 21
=y .
26. From the question, we have
⎩⎨⎧
=
+=
......(2) log)1......( )30(log5.6
5
2
ENE
From (1), we have
302
2305.6
5.6
−=
=+
EE
……(3)
By substituting (3) into (2), we have
fig.) sig. 3 to(cor. 55.25log
)302log(
)302(log5.6
5.65
=
−=
−=N
The magnitude on Scale B is 2.55. 27. (a)
8.0log
04log5
3
3
−=
=+
xx
From the graph, when y = –0.8, x = 0.4. The solution of 04log5 3 =+x is x = 0.4
(cor. to the nearest 0.1).
(b)
5.0log41
9loglog
41log
3
3
3
9
=
=
=
x
x
x
From the graph, when y = 0.5, x = 1.7.
The solution of 41log9 =x is x = 1.7
(cor. to the nearest 0.1).
28. (a) xx 2
21 loglog −=
∴ The graphs of xy21log= and xy 2log= are
symmetrical about the x-axis.
(b) 1log2loglog2
log 2222 −=−=⎟⎠
⎞⎜⎝
⎛ xxx
MC D B C C C A C D A A B B C A B A C D B C B A A D D D C C D A D D C B D D A D
