BAB 1 BENTUK PANGKAT, AKAR, DAN LOGARITMA
g)
LKS 1
h)
5 5 5 5 5 5 5
57 5
5
5 kali 14
(0, 5) 5) (0, 5) 5) (0 (0, 5) 5)
1.
D
2.
3 3 3 3 3 3 81 C
7 kali
4
2.
( 1) 5 ( 1) ( 1) ( 1) ( 1) ( 1) 1
D 10
( 1)
) ( 1) 1 ( 1 10kali
4.
A ( 2)7 ( 2) ( 2) 128
a)
03 0 0 0 0
b)
05 0 0 0 0 0 0
c)
04
d)
( 1) 6 ( 1) (1) 1
e)
7kali
6.
0 0 0 0 0 6 kali
5.
9
( 1) (1) ( 1) 1 9 kali
A ( 7) ( 7) 7) ( 7) 343 E
f)
( 1) 5 (1) ( 1) 1 5 kali
3
1 14 14 14 614 7.
B
8.
43 C
9.
343.000 70 70 70 70 3 D 2
3 3 3 3 3 3
3.
a)
52 53 54
b)
x 2 2xy y 2 ( x y )2
B. Evaluasi Pemahama Pemahaman n dan Penguasa Penguasaan an Materi
37 34 35
5 53 5 5 5 5 5 5 5
b)
3 3 34 3 3 3 3 3 3 3 37
c)
2 2 2 2 2 2 2 2 2 2 2 2
d)
15 15 15 15 15 1 15 15
5
8
e)
3 (3 ) 3 3 3 3
5
2
2 3
53 5
3
5 5 5 5 5 5 5 5 5 555
2
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
31
52
2
5 kali
3 3 3 3 1 3
f)
5 kali
(7 ) 3 7 2 7 2 7 2 (3
7 7 7 7 7 7 n ) (3m 2n 3 ) (3m 2 n3 ) (3m 2 n3 ) 27 m n n
2 3 3
27
7
7 7 7
2
3 3 3 3 3
6 kali
4 kali
f)
2
33
3 3 3 3 3 3
5
4
7 777 7
3 3
d)
g)
17 17 17 17 17 1 17 17 75
36
4 kali
3 3 3
4
17
5
c)
15 e)
7 kali
6 kali
a)
2
5 5 5 5
5 kali
b 2 c 2 2ab 2ac 2bc (a b c)2
2
5555 5
3 3 3 3
5
10. A
1.
7
(0,5) ,5) (0,5) ,5) (0,5) ,5)
(0, 5) 5) (0 (0, 5) 5) (0 (0, 5) 5) (0, 5) 5)
A. Evalua Evaluasi si Pengertia Pengertian n atau atau Ingatan Ingatan
3.
5
5 5 5 5
4
h)
( x 3 y 5 ) 2 xy
9 kali
6 9
n
3 5
3 5
( x y ) (x y ) xy
x y y x
6 kali
10kali
xy
9 y y x y x x 5 kali
9 kali
Bab 1 | page 1
i)
4 2
3 3
(2 2 ) (2 2 )
2.
3
2
12kali
2 2
3
( 2 2 ) (2 2 )
4.
3
2
( 3)7 ( 3) ( 3)
b)
b)
( )
c)
(1,75) ,75) 1,75 1,75 1,75
d)
(10, 5) 5)5 10, 5 10, 5
3
y 3z )2 x 2 y 2 9 z2 2 xy
( x
x 3)2 x 4 2 x3 5 x2 6 x 9
c)
( x 2
d)
( a b c) ( a b c) 2 ( a b c) 2 ( a b c) 2 8 ac
7 kali 5
( 13 ) ( 13 )
e)
( x 2a ) 3 x 3 6ax 2 12a 2 x 8a3
f)
( ax by by)3 a3 x3 3 a2 bx2 y
5 kali
3ab xy2 b3 y3
3
3.
5 kali
e)
( 0, 8) 7 0, 8 0, 8 7 kali
f) g) h)
(3 152 )4
( 121 )4 ( 3)
2 )5
(2
5
8 3
41 41 41 12 12
41 41 41 141 2 12
83
83
83
a)
( x
1 2 ) x
x2
1 x2
x2
b)
( x
1 2 ) x
x 2 2x x
1
x2
c)
( x
1 3 ) x
x 3 3 xx 3 x2
d)
b)
( 3b) ( 3b) ( 3b) 729 b
2
2
3
x 1 3
x 3 3 xx 3 x2
3
4.
6
6 kali 2 3
1 x
1 x
3
x3 3 x 3 x x x13
( 2 ) (2a ) ( 2a ) ( 2a) 8a 6
2
x
2
1 3 ) x
( x
1 x2
1
x
3
( 1 71 ) 6 ( 7 ) 6 ( 87 ) ( 87 )
3 4
x
2
x 3 x x3
83
a)
c)
2 x x
2
6 kali
5.
2
6 xz 6 yz
(4 y) 4 y 4 y 4 y 4 y
a)
2
2 c 2bc
8 kali
j)
2
( a b c) a b c 2 ab
2 2
(2 2 4 ) (2 (22 24 )
a)
2 3
(6 y ) (6x y ) (6 x y )
28 3 4
a)
20.736
b)
11.390.6 0.625
c)
161. 161.05 051 1 11
d)
2.744 23 73
a)
32
b)
(0,125)
c)
( 0,12 0,12)) 112 00
d)
(5,13)
5 36
4 kali
1.296 8 y12 d)
4 3
3
4
3
4
3
4
9 12
) (3 a ) (3 a ) (3 a ) 3 a
(3
5.
3
e)
2
2 y2
4
f)
3 x2 y 2
2 y2
3 x2 y2
2 y2
3 x2 y2
8 y6
3 x2 y2
81x8 y8
5
32 243 3
3
3
1102050 81
4
4
4
513 100
4
5112 4
25 39081.625 513) 4 100) 4
C. Evaluasi Kemamp Kemampuan uan Analisis Analisis 1.
a)
( x
y ) x2 2 xy y 2
b)
( x
y ) x 2 2 xy y 2
c)
( x
y ) x3 3x 2 y 3 xy 2 y3
d)
( x
y ) x 3 3x 2 y 3 xy 2 y 3
e)
( x y ) ( x y)2
LKS 2 A. Evalua Evaluasi si Pengertia Pengertian n dan Pemaham Pemahaman an
x 2 2xy y 2 x 2 2xy y 2 4xy f)
1.
3 5
( x y ) ( x y )2 x 2 xy y x 2xy y 2
2
D
2
2
2 x 2 y 2
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
2.
1 3
5
1 243
D 0
7
1 0
7
1 0
tak terdefinisi
Bab 1 | page 2
3.
B ( 1)
4.
5
1
d)
1
1
( 1) 5
2 3
a x
( 1) 1
7
8.
2
1
(0, 00 001) 2
1 (10 3 ) 2
1 10 6
10
p2 p7
10
f)
p p
p
a)
20 1; 21
1
b)
60 1; 6 2
1
D 1
1
0, 0000001
107 1
125 5 C 1 4
3
2.
5
B 1
1
2 3
2
4 9
9
2
13
d)
25 25
10. E 1
3. 1
14
3
1 64
64
a) b)
10
3
p p 1
32
; 64
1
8
p
; 2 7 1
1.296
128
;
1
3
1
;
9
13
5
125
;
2
0
1; 13 0
2
9; 13 4
2
25
1; 25
25
3
27
;
8 2
2
2
256
16 ; 64 64 8 ; 4 2 ;1 2
1
4
1
46.656
53 125
5 3 11. B 1
3a2
p5
; 2 5
36
3
1
6 6
c)
2
a x
x
p8
p
3
5
x2
3a x3
6
3 4
9.
1
2 3
x
x
C 10
7.
p p
e)
D (0, 001)
6.
7
1
5
a x
D
3
2 3
1
8
5.
1
3x 5
;
1
2 4 ;
16
1 64
2 6 ;
0
1 256
2 8
12. C 0,0000 0,000000 0025 256 6 2,56 10 13. C
c)
123. 123.00 000. 0.00 000 0 1, 23 10 14. C 1 1 1
d)
25a
2
2
5 a
2
(5a)
(5 a)
2
7
17 7
1
3
7 3
4.
a)
B. Evaluasi Pemahama Pemahaman n dan Penguasa Penguasaan an Materi c) 1.
a)
a
1
111;
11
1
b)
5
7 ;1 7
1
b) c)
a b
3
4
a
2
b
3
a 1 4
1 a2 1 b
3
1 a5 1 b
4
3
b
b3 b
4
1
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
0
1 121
1
11 2;
1 1.331
113;
1
a 2 ab b 1
a 1 b2 c 3 80
a
1 2
1 3
c
( a b)
( a b)
2
2
2
1 x 2 y 2 z 2 2 xy 2 xz 2 yz
1
2
11
14.641
3
1 3
1
7
15. E 1
3
16.807 7 ; 34 343 7 ; 49 49 7 ;
1 ( y z )
d) 1 2 3 4 2 a c3 d 4
( y z )
1
2b 2 3c 3 4d 4
Bab 1 | page 3
5.
a)
3b
7
7.
6
b) x y c)
1
1 4
nk
8.
x2 2
2
9 3 (3
27x
)
40.095
729 3
55
1
33 y 2 3
27
b c
3m n k 9
x
c
acd
3 2 1
1
8
55
5 7
2
8
11 57 6
a2
2 1
f)
6
(6 3 )
6
a b c
4
4 4
6 3
a bc d
e)
2
x
a b c 1
d)
8
11 57 24
y
2 5 7
2
6
2 5 (6 3 )
3 2
1
E
3
m2 mk 3
m3 k 2
2
3nn k
3n
9.
3
B xa b x
a b
xb c x
x 2 ab xab b 2
xc a x
x b 2 bc xbc c 2
c a
x c 2 ac xac a 2
ab ab ab b2 b 2 bc bc bc c 2 c 2 ac ac ac a 2
x
LKS 3
b c
x0 1
10. D 1
A. Evalua Evaluasi si Pengertia Pengertian n atau atau Ingatan Ingatan 1. 2.
1
3
3 23
3 2
31
9
3 3 3 1
2
3.
3
3
12 a
92 a 7
3
23 3
9 9 3
5a
312
a 5a
12 4 a
3
33 3 2
9 6
( 3)
2
12461
2
3
1
(m
2
am an an
n 1 )1
(32 )2a 7
3
2
54
x
1
x 2
9
2
x 2
3x x 1
x
3 3
x 2
x
3 3 3 x
3
x
3 3
3
9 1
16 25
1 3
8 2 3
12
A
1
81a9 b2 6 5
3a b
27 a9 6 5 2
b
a
a
n
m
n
m
an
n
a
am an
1
1 1
m
1 m n mn
n 1
1 1
1 n
mn n
15. D
33 a3 3
b
3a b
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
1 x
3
2
1 2 x 1 2 2 x x 3 x x 2 9 x
x 2 3 6 b5 9 2 81a b
m
m
1 an
y2
x
54
3 1 x
a
m
1
y 4 ( x 2 y 2 )( x 2 y 2 ) x 2 y 2 y 2
Diketahui : x x
1
12. C
4 a 14 14
D 3
n
11. D
B
3 1 12 14
6.
a m a n am
1
1
a a
312 4 a 4 a14 3 2
5.
1 a m n
1 10
1
1
1
A 3
4.
1 a n m
A (5 4b 5 )(2a 3b 7 ) 10a 4 3b 5 7 10ab2 A (0, 6) 0 (0,1) 1
1 x
2
2
Jadi, x
9 2 11 1 x2
11.
Bab 1 | page 4
18. C 3
x
5 2 x 2 3 5x 2 x 2 5x 3 0
x
atau x 3
1 2
x 12 4 x 2
x
4
2
1 2
2
(ii) x 3 4x
9
12
4(3)
x
52
(3)2
B. Evaluasi Pemahama Pemahaman n dan Penguasa Penguasaan an Materi a)
6
64 23
b)
32 34
c)
3
3
32 4 3 35
23 2 4 25
d)
2
2 2 2 2 6 (9 32 )3
e)
(32 3 2 )3
(3 27)
8
3 (16 4) (4 8)
3
(2 2 )
(2 2 )
2
2.
2 2
2
212 15
2
1
2
5 ( 16)
52
243
26 (16 160 )
26 15
91
35
8
2
3
3
8 6
2
b b4 b9 4 b5
b)
y 6 y 2 y 6 2 y 4
c)
m 10 m6 m 10 6 m 4
d)
x12 x3 x12 3 x15
e)
(2 ab3 ) 2
(34 )3
f)
(34 ) 2
(3 x
2 4 3
3 3
6 2
(2 )
(2 ab3 ) 2 1
y )
1
(3 x
2 4 3
y )
1 y 1 4
1
2 2 a 2 b6 1
4a2 6
3 6 12
3 x
6
y
27
12
2
2 2 2 2 4 2 2 2 2 x y x y xy
g)
2 x1 y
h)
4m 3 2 2 2 6 2 6 3 (4 n ) 4 m n 16 m n n
i)
6 3
j)
3 3 5 x 5 x 1 1 3 3 12 4 12 3 12 y 5 x y 125 x y
a)
(2 a b) 1
b)
(14 a 8b) 1
c)
m n 1 1 2
d)
7 x
e)
3a 2b2
f)
17 x 3 2 54 y
5 3
(2 )
2
12 15
2 3
3 4 2 5 2 36 5 2 3 32 7 5 5 2 7 5 3 78
a)
35
3128 3 4
4
3
(3 33 ) 2 312
f)
3 4 51 2 6
33
1
34 3 2 1
26 (16 1) 5 ( 1) 1) 2 4
a)
23 4 2
2
3
1
1 2 ( 3) 2 ( 3) 2 ( 3) 1 ( 3) 9 ( 3) 3) (9 9) 12 18 216
d)
1.
6
4.
2
( 16 16) (16 160 )
c)
5
1
35
36 1 35 Jadi, nilai yang memenuhi adalah 35.
1
6
2
9
2
3
60 ) 3 (611 61 )
5
9
1 36 35 2
1
3 3
b) 9
1
1
3 (1 16 ) 3
x (2 x 1)( x 3) 0
(i)
1
3 (6 6 6
3 . a)
4
2
2 8
6 a c
6
8
a
2 6
6 c
8
a
36c
6
3
72 32
d)
35 7 4 7 3
8 5 2 3 4
e)
5 2 4 9
1
2
35 2 74 3 2
37 73
(23 ) 2 52 3 4
5.
(52 )2 24 (32 ) 2
52 34 4 4 4 4 5 2 3 2 2
1
1 6
3 4 4 5 4 2
1 210 5 2
f) 2 4
3 2
(2 ) ( 2 ) 2 2
5 3
(2 ) ( 2 )
2 2
8 4
2
6
2
15 15
1
2
15 8 6
1 25
2
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
0
0
0
0
0
0
8 y 7(1) 8(1) 15 15 0
(3a)0 (2b2 )0
1 1
1
0
(17x 3 )0 1 1 2 0 1 (54 y ) Bab 1 | page 5
6.
a) b)
x
3 3
y z
4 5
6
y
z
x y z
x
0 1 3
(
b )
( a )2 b 5
2
( b )c a
3
0 2
( a) a c
d)
4
3 0
2
3 4
x y z
n
0
m p
p 3
2
n
f)
7.
a)
n
3
a 3b 4 3 c a
z
m
3 x y 3 2 9 x y
2 4
b)
2
d)
a b c 6 3 ab c
a)
a
b) x
1
2
3 x y 2 4 3x y
2
i)
(2 a
x
2
x2
3 x 4 2 y
2 1 3 2
0
(a b )
2
1
c
2
(a b ) 1
(2a
1
b)
1
(a b
2 3
)
1
2
ab 1 2 b
2
2 ab a
1
c2
2
(1 b )
2
1 ( b) c 2
(a b ) 1
2a a
2
2
(2 b)
2
1
a
b
1 b2
3
6
2
3
( ab 1)
C. Evaluasi Kemamp Kemampuan uan Anali Analisis sis 1.
(3 x1 2 y 2 ) 1
a)
1
3
2y
4
2
2
( x 6) ( x 1)
a
x y 2
x2 y3 y
( x 6)
2
2
( x 1)
2
xy
1 3 x
2
y 2 2 x
2( x 6)( x 1)1
2( x 6) x 1
2
( x 6) 2( x 6)( x 1) 2
( x 1)
1 b2 a
x y
3 y 2 2 x xy
b)
2
1 2
b32 20 8 a c
1
3 x
4
y
1 1
3
y24 12 12 x z
b 2 1
y
2
2 3
(a b )
j)
2
1
3
b 1 4 31 a c
b)
m2
2
2 6
1
1
n
1 (1 (1 b) c
c
p2
2
3
y 2 3
2
( a b)
1
m m 2np 2
9 x10 y 4
y3 5 4 4 x z
a
2
(1 b )
h)
2
2
c2
2
a8 4 c b8 c 6
1
1
2
3 2
2 3 2
b 2 c 0
c) x y
6
2
b 5c
8.
b)
3
2
y 4 x z 4 2
x z
(a
6
3 x 4 y3 z0 c) z 4 y 5
7 4
c 3 a a13 a 3 b4 b4 c 3
1
n3 m
2 3
3
m n
2 3
x z y z 1
n 1 p2
b2
p
0 3
4 2
( 1) a a c
2y
3 3
n 3 6
m
1 3 3
2 y
f)
1
x2 y2
b5
c
p
3
g)
m
6
2
4 3 0 4
z
3
e)
x
2
a0 b3 a 2
4
2
x y z
b2
3 3
b
e)
10
a2
b5 3
7 8
y z x
b5
3
6 4
a0 b3 (a ) 2
c)
3 4 3 5
2
( x
6) ( x 6) 2( 2( x 1) ( x 1)
2
( x 6)( x 8) 2
( x 1)
2
1
3
d) m4
n 3 p 0 m 4
1 n3
1
n 1 n 3
4 3
n3
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 6
c)
3(4 x 5)( x 2) 1 x 5 3(4 x 5) ( x 2) 2 x 2 (4 x 5)( x 2)
b
( x 2)
(4 x 5) 1 3( 3( x 2) ( x 2)
2
3( x 4) (2 x 5)
2
4) 2
3( x
(2 x 5)
( x 2)
H
5.
F
Lr ( r 2 L2 )2
2 x 5
( 2 x 5)
( 2 x 5)
( x 1) 2
(3 x 1) 2
a 2 b2 1 a b
4 Lmr 2
2 2
(r L )
1 1
1
f 2 f1 d ( f1 f 2 )
1
1
1
1 2
1 f 1
f 1 f 2
f 1 f 2
f1 f 2 d f 1 f 2
2
f2 d
f1
2
2
( x 1)(3 x 1) 1
x 1 3 x 1
(3 x 1)
LKS 4
2( x 1) 2 ( x 1)(3x 1)
(3x 1) 2
( x 1) 2( x 1) (3x 1)
b13
1 a3 1 3 a
3
b 3 3 b3
( x 1)( x 3)
2
A. Evalua Evaluasi si Pengertia Pengertian n atau atau Ingatan Ingatan
(3x 1) 2
b3a3 a 3b 3
1 3 b
3
b3a 3 a 3b 3
b a
3
1.
A
b a
3
3
2.
b a 3
3
b a3
1
3
b
b3
1
1 a3 b
m
2
m
3
1 x
2
4
m 2 m 1 m 0 m m 1 m4 2
m 2 m 1 m2
1
m
1
1
x y x2
1 m
1
1
b1
b a
a b
4 2
y
1 m2
y ( x2 y )
6.
25 32 5 243 3 6
1
ab
1
1
xb bc xc c a x x
1
m b a ab
b a ab
2
3
y
5 2
y
7
3
5 2
1 2
x2 y2 x y
7 2
52
a )(b a )
31 2
y 2
1
xy
B 1
1
1
x y y z z x
y x
z
y
1
1
1
1
1
1
x 2 y 2 z 2
x z
1
x 2 y 2 z 2
2
8.
a a2
1
ca xb xc xa 1 1 1 1 xaxbxc
1
4
5 24 16 4 81 3
A
1 m14 m3 1 1 m
b a (
x (x y )
4 5
B
x
2
45
12
x 2
2
1
2 x 6 x 3 x 2 x 4 3 x 7
12
7
y
x y
m
b a b ab 1 a
2
5.
1
B
xa b x
2
xy2
m
2
x
2
x y
4.
xy 2
3
(2 2 ) 3 2 2 2
B
3
a b
x y
1
y1
1 x 2
3.
3 1 ab 1
3
y 2 x 2 y 1 x
a3 b3
1 3
243 32
Dengan membagi pembilang dan 3 penyebut di atas dengan diperoleh:
c)
( a b)( a b)
( 4m )
8
a b
2
2( x 1) (3x 1)
b)
a b ab
2
ab
1
2
( x 4)(5x 17)
a)
(x 4)(2x 5)1
(2 x 5)
3.
a b
a b ab
( x 4) 3( x 4) (2 ( 2 x 5)
2.
1
1 b b a
2
x4
2
(4 x 5)(3x 7)
4.
3( x 4) 2 ( x 4)( 2 x 5)
a
1 a a b
2
e)
d)
x 5 3( 3(4 x 5)( x 2)
d)
a1 b1
1 b a
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
A
8 x 2 y 4 23 x 2 1 1 2 53 y 2 4 125 x y
22 x2 2
5 y
4
2
3 23 x3 3 6 5 y
2
3
0,16 x 2 y 4
Bab 1 | page 7
9.
E 2 3
3
1
8 x y
10. C
e)
814 (3 ) 4 3 27
f)
25 2 (5 ) 2 5 125
g)
32
(32 )
1
1 1 (32 )2 2 2 2 2 2 3 3 3 3 2 2 4 1 1 ( 22 ) 2 2 2 2 2 2
B. Evaluasi Pemahama Pemahaman n dan Penguasa Penguasaan an Materi 1.
a)
3
d)
5 1 8
3
2.
a)
1
2 2 2 12
1
c)
1 3
d)
5
3
i)
5
j)
3
2
3
3
35
35
(25 )
23
2 3
2 3
4
4 1
3 54
i) j)
k)
1 32
l)
2 2 2 22 9 2 3 3
a)
125 3
b)
16 4
c)
9 2 125 3 (32 ) 2 (53 ) 3 3 25 25 28
d)
4
e)
16
f)
49 2 36
1 3
3 3 3
3
3
4
1 12
3
1
(3 2 ) 4 32 9
1 2
3 9 1
23
2
3
12 43
3
1
2
1
12
32
1
(33 ) 3
1
3
2
1
2
1
(7 2 ) 2 (6 2 )
12
7
3
( 2 4 ) 4 (23 ) 3
12
34
(32 ) 2 (24 ) (23 )
23
13
8
2
1 4
3 8
6 3
3 2
2
2
2 2
2 2
5
3 32
4 13
2 1 5 5
333
2 3
56
3
4 11 5
i)
3
1
8 14 81
64
13
27
4 5
43
3
6 23
(2 )
1
5
(3 ) 23
0
243 3 3 3 3
149 3 3 3
1
6 1 243
3 23
1
2
2
9
3
2 3
(5 ) 2 (2 )
(34 ) 33
15 3
34
1
2 3
25 8
57 3 3 4 1
4
(3 ) 4
3 13
1 3 16 1
16
3
(3 )
9
j) 3
1
(3 ) 2 (8 3 )2
3 2 6
1 2
33 22 3
12
32 (43 ) 9
3
32 (26 )
1 2
12
33
32
3 2
2
2
9 2
3
1 2
(2 ) 2 2
3
b) 16 4 (2 4 ) 4 2 3 8 2
2
h)
9 3 3 3
27 3
2
8
1
1 2
2
3
3
81
81
3
2
27 3 (22 )
9 2 16
12
3
812
1 2
2
27 3 (2 4 ) 4 (33 ) 3 8 9 1
1
74
27 3
1 3 27
3
81 4 (5 3 ) 3 (34 ) 4 25 27 27 2
1
c)
10
64
4
8
3 a)
10 2
3
2
1
1 12
1 2
96
3.
3
2
1
3
1 4
10
1
g) 1
13
h)
1
3
f)
1 9
2
2 (2 ) 2 2
2 2 2 2
4 1 27
3
22
3
e)
g)
1
4.
3 5
3
9
b)
1
0
1 4
14 8
3
3
1
1 2
(2 ) 2
g) 16 h)
3
2
3
f)
4
( 22 ) 2 3
c)
e)
3
5
(2 1) 2
1 2
b)
1
5
3
10
(2 ) 2 3
32
1 2
h)
1 2
32 2 4 2 2 1 2
1
9
3 26 x 3 y 6 26 x 3 y 6 3 4xy 2
6
1
12
d)
5.
B
2
16 2 9 24 3 1 4
1 2
16 9 4
1 6
4
2 23 42 3 2 3 2
1 3
2
2 43 1 13 4 1
3
22 33 108
2
(33 ) 3 32 9
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 8
6.
2
22 a) 142 3 27, 22
b) c)
a b
15
3 2
67 12, 46 46 8
115
3
3
6
84
1
11 5
i)
2,17 10
59
1
3
8 20
1
1,7
1
1 3
3 2 b3
69 14 3 50, 69
4 14
j)
a b
e)
2 3
24
3
f)
11
2 3
11
1
2 3
a)
3
16
g)
9.
12
7
1 6
11 1, 49
b)
4
h)
7.
a)
18
x
81
1
0, 78
1
7
m
c)
3
4
34
b
n
43
35
8.
a) b) c) d) e) f) g)
a
a4 4
b
0, 46
4 15
e) 1
3 2 4 3
x
3 1 5 3
y
m
43
y
85
2
b ( a b)
g)
b a 1
14 14 15
j)
1 2
b c
a2
a 2
c
1
a 3
b
1 5 8
1
y x 3 y 3
2
3
f)
i)
2 1 2
2
17 12
16 16 1
1
b3 1
1 4
b3
m 2 n 2 1
3
3
12
10. a)
3 4
x y
x 2 x 4 x
b)
3
x 20
13
2
x
a7 a a
x
1 8
1
9
x8
2 1 1 7 3 2
12
4 81
19
a 42
3
1
1 3
14
2
b
b
c) d)
x 2 y
b
x x
a
23 4
14
1
y2
1
y4
m 2 n 3 1
1
b
f)
1
3 73 31
4271
a
1
a
( xy )
e)
( a b)
3
1
a3
2 1
a 3 a 7 a
7 3
1 1
a b
1 3 2
1
1
4
1
b b
a 4b
3
1
1 4
3 3 1 3
1
1
1
5
1
d)
41
34
34
b3
j)
34
3
34
3
a 35
3 1 7 3 1
23 2
2
x
35
y8
b
3
1
h)
x y x
1
2 3
y y y
1 2
x
1 3
m m
1
i)
3 5
1 3
e) y 8 y
h)
1 3 4
a5 a7
3 7
1 18
x x
3 5
13
d)
g)
18
3
c)
f)
154
3 4
y y
23
x
b)
25
1 2
a4 b
35
78 2 3
3
b
0, 52
3 4
112 3 4
5
1
34
1 4
a a
1 5
b
1
84
d)
h)
3 5
3 2
x2 y 3
41
3
83
x 3
x8 y 8
y8
2 41
y4
y
41
47 1 1
x8
5
y8
7
14 32
x
5 2
2
1
m 3 n 1
4 1
n
2
m 3 n 1
2
1
1
4
n3
m3
n3 13
m3
b
1 2
a 2
11. a)
1 3
5 35
( a b c
2
1 1
)( 6 a b c ) a 2 6b
1 2
3
2 12 2 1
c
5
b2
b a b 34
1 a
3 4
4a
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 9
b)
d e
2 2
f
3
5
e f
d e f
3 2
e
12
1
x 2 d) e)
3
x y
1
y3
z2
13
2
z3
2
2
2
3
2
1
1
5
3
a2
1
a 2 b( a 2 b b )
y )( )( x 3 y )
1
2
2
2 3
( 1) 1 2( 1)0 2 1 2 5 2 1 0 x 1 2(1) (1) 2(1) 2 1 2 3 2 1 0 x 2 2(2) ( 2) 2( 2)
13
3
1
1
x 1 2( 1)
xy 3 z 6
a 2 b( a 2 )
3
9
df
z 3
1
ab a b 2
f
1
5 2
x y3
a 2b
( x
3 2
2 x 1 2x 0 x 2 2( 2) 2 ( 2) 1 2( 2) 0
2 2
d
5
e f
c)
b)
2 2
1
a 2 b b2
2
1
1
2
2 2
2
x2 y 2 x 2 y 3
5 3
14. a)
x 2 y 2
x y
1
x
2 21 52
y
1 23 2 3 2
0 13
x y
1
32 (2 2 ) 2 3 ( 27 4 2) 3 353 3 35
1 y
f)
( abc)
a 2 b2 c 2
2 1 3 3
6 3 9
(a b c )
g)
c
x y z
16
1 2
x 3 y z
a
11
1 2
3
x y z
6 2
a b c
5
b2 3 c2 9
1 3
x y z
1 3
y
16
13
1 2
x
(32
c)
13
15. a)
11 3 6
z
11 2 5
h)
(
3
bc
3
a
z
)( a b )
1
1 2
3
a b c
21
3 2
bc
1 3
a b
1 3
b
1 1 1 2 3 2
1 12
11 6
a b
4
(2 )
1
2
1
(2 ) 2
8
31
( 23 )
13
1
3
(2 )
1 3
12
1
2
x 1
2
e)
( x z
f)
(m
12
12
x 1 2 ( 2 )
1
1 1
x y 2 x y 2
2 3
1 2
y) x z
31
x 1 2
1 3
y
1 2
y3 1 3
n 2 )( m 2 n 2 ) ( m 2 ) 2 ( n 2 ) 2 1
1
1
1
1
1 2
1
1
1 n
1
3 3 2
1
3 1 2
2
b
m 1n
x 2 2 1 ( 2 ) 2 3
2
5
b
( 2 )x 3
0
4c
( x y 2 )
3 4
3 2 12 32
1 2
1
1
b c
1
2
3 21 2
21 52
2 a
(terbukti) 13. a)
22
1
2
2 3 12
3 2
( x 2 y 2 )
4
a b 2c
1 3
2
x
4 3
(terbukti) b)
3
c)
d) 2 12
y
40 a4 b3 c 6 5 16a bc
a b c 3 23
2
441 21
b)
1 2
c
12. a)
1
(9 x 4 y 2 ) 2 3 x
3
2 1
23 33 43 53 63
2 125 x 4 y 3 5x y 5x3 3 5x 2 6 3 x 23 y 2 3 y2 1 3 y 27 x y
x 10 12
42 122 ) 169 13
1 3
1 6
b)
2
C. Evaluasi Kemamp Kemampuan uan Anali Analisis sis
3
1 2
1.
a)
(2 x 5)( x 1)
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
1
( x 1) 4
2 3 1 2
x 2 23 ( 2 ) 2 3 8 2 3 3
2 x 5
41
3
(x 1) 4 3
( x 1) 4
2 x 5 x 1 x 1)
3x 4 1
( x 1) 4
Bab 1 | page 10
( x 7)(3x 5)
b)
x 7
12
(3x 5)
5.
1 2
H
2mL 2
2
1
(3 x 5) 2
1 2
(3 x 5)
3
(r L ) 2
2mL
2
2
2
2
( r L )( r L )
2
2
2
2
(r L ) 1
(r L )2
1
x (9 x 2 25) 2
(3 x 5)
(3 y 7)( y 2)
c)
3 y 7
4m 1
4 3
1
108
b
1
c 4 3
c)
3 2
1 53
c
5
108a 2 b 3
1 2
27 3 a 2 bc
3 1 2
2
13
a b c
1 2
c
144a 31b 4
2
c
2.
1
3.
1
1
10
51
5
1 2
3 5
65
1 3
2
2 3 (2 ) 15
(2 5) 1
22
5
12 65 1 51
5
5 y 5 z 5
2
z
z
1 4
3 5
53
2
3.
1 x
2 x y xy
625 25
D 10 250 2500 50
2 35
4.
C
6
5. 2 35
2.401 176 49
248
52 144
225 15 15
B
2 3
(2 )
176
248 8 256 16 16
57
4 3
3 5 1
33
x
7
7
5
1 34
51 35 75
1
4.
3 3 5 4
4375
1
4
B 4375
144 b 3 c 3 144a 2 3 b 4c 2 2 3 3 4 4
3025 55 55 552 ; 65 6561 81 8 1 812
32
5
D 12321 111 1 11 11 111 ; 20 2025 45 45 452 ;
2
27 a bc 32
5 3
108a b c
21
2 3
1.
65
3
3
.
A. Evalua Evaluasi si Pengertia Pengertian n atau atau Ingatan Ingatan
108 a b
2 3
36 V
LKS 5
6
1
3
65
c
2
3 64 3r 6
2
7
1
3
6
(m 4) 3
b 3 c 6 24 a 5 3 b c7
a2 b 3c
3 36 43 r 3
2
Jadi, A
1 32 1 2 56
5 2
36 V
10m 23
27 a bc
32 a3 2 b
24 b)
3
1 3
21
2 3
a b c 24 32
2
13
2 2
3 26 3r 6 4 r 2
1
(m 4) 3 2
3
6 ( m 4) 3
m 1 6(m 4)
4 3
2
(r L )
3 3 36 16 9
( y 2) 5
6 ( m 4)
2
( m 4) 3
a)
6.
5 y 9
( y 2)
2
2
2
(4 m 1)( m 4)
d)
2
2mL (r L ) 2
8( y 2 ) 5
3 y 7 8( y 2)
2.
3
8( y 2) 5
3 5
( y 2)
1 2
2 30 5
1
1
6.
10
1 z
18225 182, 2 5 1, 82 8225 0, 01 018225 135 13,5 1,35 0,135 ,135 149 149,985 ,985 135
(5 2 ) y 5 z x z
D
1
7.
B
1
1, 21 9
y
x
1,1 1,1
xy
2
121 10
2
9
121 121 102
3
2x y
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 11
8.
B. Evaluasi Evaluasi Pemahaman Pemahaman dan Penguasaa Penguasaan n Materi
A 0, 00 004 0, 4 0, 4 0, 00 004 b 4 102 a 16 10 4 b a
4 1 0 2
b
9.
a b
1.
0, 0016
D 1, 21 21 0, 01 01 1,1 0, 0,1 1
10. A 12
4 3 4 3 2 1, 732 3, 3, 464
2.
11. D 0, 81 4
0, 9 2
0, 0 06 064 6, 2 5
0, 08 08 2, 5
a)
324
22 34 2 9 18
b)
2601
32 17 2 3 17 51
c)
1681
412 41
d)
1849
432 43
e)
2809
532 53
f)
3844
22 312 2 31 62 62
a)
0, 31 3136
3136 10
26 72 10 2 0, 56
b)
2
10, 89 89 1089 10
32 112 101 3, 3
12. E 0,204 42
c)
36 6
0,07 3,4
3, 24 24
d)
13. C 110 14
441 4
324 10
0, 51 5184
e)
1,1025 11025 10
14. E 0, 361 0, 00169
361 10
169 10 5
5184 10
26 34 10 2 0, 72
221 10, 5
3
22 34 10 1 1, 8
1, 9 10 0,13 10
32 52 72 10 2 1, 05
190
f)
13
116, 64 64 11664 10
2
24 36 10 1 10, 8
15. E 0, 01 01 0, 00 0064 0, 09 09 0, 3
3.
a)
13 26 18 45 105 21
22 36 52 7 2 132 24750
16. E 4,5 0,0085 ,008518,9 18,9 0,0017 0,0017 13,5 13,5 2,1 2,1
b)
15
17. A
36 52 106 112 132 19, 305 4.
1
13
18. C 1 3
a)
27 14 x 1 1 169 13 13 13
x
1 13
1,732
x 74,7359
b)
0,577
19. E 20. B
5, 24 24 x 0, 71
Kuadratkan kedua ruas diperoleh: 75, 24 24 x 0, 50 5041
x 1
1
169 169 0,0009 0,0009 1,21 2025 2025
1872
234 234 x 1872 x Kuadratkan kedua ruas diperoleh: x
5.
18722 2
82 64 .
234 a) KPK dari 15, 18, dan 25 adalah 450. Jadi, bilangan kuadrat terkecil yang habis dibagi ketiga bilangan di atas adalah 900. b) Bilangan Bilangan terkecil terkecil yang harus dikurangkan dikurangkan dari 549162 agar dapat menjadi bilangan bilangan kuadrat adalah 81, karena 549162 – 81 = 549081. 2
5491 549162 62 741 741
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 12
c) Misal bil.yang dicari = a . 5
352 2 11 a 4 22a Agar menjadi bilangan bilangan kuadrat maka
6.
a)
36a 2 6a
b)
( a 3) a 3
c)
4 4 a2
b.
2
(2 a ) 2 2 a
c.
1
7
5
5
5 30
20 6
d.
8 6 2
f.
0 3 5
2
8
8 2
7
3
1
8
e.
LKS 6
7
7
42a 49 (3a 7)2 3a 7
9a2
d)
1
1. a.
2.
nilai x
B. Evaluasi Pemahaman dan Penguasaan Materi
1 2
3
14
7
6
3
A. Evalua Evaluasi si Pengertia Pengertian n atau atau Ingatan Ingatan 1.
D 8
3
3
2.
6
12 2,448 1,224
1 2
4 3
4
b
4 3
2
32 y 5 6m
25 x 5 y 5 2xy 3 4x 2 y 2
g.
3
3
(6m ) (2 )
1 3
1 3
2 2
1 3 1 3
1 3
1 3
6 2 m 2
h.
m 3 12 2
3 2 12
2
6 212 4
C 2 3
32
2
2 3 12 6
3
4 x y
6
3
64
25 6
1
2 5 33 4 6
3
2.000 2 2 2 5 10 2 40 5
3. a.
8m4
b.
104a5 b
c.
150 2 y 5 2.3.5 2 x 2 y 5 5 xy 2 6 y
2.7 2 m 4 7 m 2 2 23.13a5b 2a 2 26ab
162 m3 n5 y 8 2.3 4 m3 n5 y8 9 mn2 y 4 2 mn
3 4 x 3 y 6 4xy 2 e.
A 3 8
3 3 8
1 2
4
375k n (3 5 k n ) 5 kn
63ab5c12 3.112 ab5 c12 11b2 c6 3ab 15k
10. B 4
1
2
d.
E 3 3
5
12 4 2 2 3 8 3
f. 5 2 3 5
2
212
22 5 7 2 14 5
2 5a b 4a b 5b
D
3
980
d.
A
3
9.
1,633
e.
4
8.
243 35 3 2 3 9 3 1 1 c. 320 26 5 4 5 2 b.
B
3
7.
3
3 6
3
3
6.
3
3
2 3 A
3
5.
4, 899
E
80
4.
24
3
3 2
3.
9 3 2 11 3 11
2. a.
a3a a 5
2 4
a
4 3
5 73 15 7 a a
f.
6 5 y8 z2 25.3 x5 y8 z2 4 x2 y4 z 6 x
g. h.
81a6 b12 34 a6 b12 9a3 b4 3
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
192 x y z 2 .3 x y z 2 8
6
5
2 8
1 3
4 xz2 3 3 x2 y2 z 2
Bab 1 | page 13
i.
128m5
3
3
27 p 6 72a
j.
2 7 m5
7
2
2 .3 a
6
9b
33 p 6 3
2
4m 2a
6
c.
2
3 p2
7
3b
3
3 3
4.
2.3 y 6 y
3
3
c.
4
54m
3
36 y
3
6
a 14a
y
3 3 x
4
6m
.
3m 6m
2
n 13n
27 108a 5b 3 27c
4 x 4 x
2.3
3
2
.
n 78mn
18m2
6m
3x
.
2 y 3 xy
3 x
3x
f.
a 3.2 4 2 a 6 16 a
3
b.
a 7a
13
e. a.
3
2 2
d.
2
3
b
7a
6
2 2.3 3a 5b 3
5
3 5
3c
6a 2b 3ab 3c 2 3c 2
4
d.
6m 6m
4.2
8
5. a.
m
4
4 a6 27b 3
16 y
6
3
x
3
4
2 y
4
3
3
.
4 y
3
2y
23
4 y
2
4 y
2
3b
3
3b
2y
23
18 x
5
y 5 y 2
3 x 2 8 x
3
3
25 m5
3
32 n 4
9n 4
n 3 32 n 2
1a
4
f.
4b
2
4
2
4
3
3
16 y
d.
. 3
4
2
2 .3 m n n 3 n
3 a
2
1
2
2
2 b
2
2
2b
2
2
32
4
9 x
2
5
8
4
2 y
4
3 2 x 2
2y
2 4
2
3 x
.
2b
2y
2 4
18 x
2
4
a.
8 n2
. 4
8n 8n 2
2
3
8n 2
56mn 4 2n 2 8n 2
7m 4 2n 2 n
2 3
3
1 4 m 4 29 n 6
6 2
3x
3x
b. 4
2
4
2y 5 2
2
3 x 4
16 m2 8 m 1
4 m 1
3
3 x 4 3 3 3
.
2
2
2
2
25a 20 a 4 2
4
5
.
5a 2
5
5
2
4m 1
2 2
.
2
2
5 5
a 2 5 5
alas tinggi 2
5 2 5 2
25 2
2 cm
9.
h. 1 4m
9 x 24x 16 3
3 2ab
2b
3x
4
2 y 5
b. 5 3 cm c.
g. 8
3
a. 5 2 cm
2m 12m n 3n 2
2b
.
2a2 b abc
8.
3
20 y 25
2
2
5a 2 4a
uas =
3 a
2
2
2
4
2
3 n 3 n
c
c. 8 m2 4 m
4 y
2 m 3 2 2 m2
4 y
16 3
e. 32 m5
c
c
.
b. 2
3 x 2
2
2
7. a.
8 a 3 3b
3b
3x
3
24 m3 2
3 x
.
3x 3 3x
2
3
2
4
2
.
3b 3b
6
2
.
8a3
34 x 4
x
6 m3 2
33 b 3
3
3
2
4
2
2 6 a6
3
81 x 4
d.
6 m3 4
b.
c.
4
2a b ab ab
c.
2
.
2 3
.
6 2 3
41421 4, 24 243 3 2 3 1, 41
3
3 .
2
3 3
3
2 3
1, 73205 1,155
44949 4, 899 2 6 2 2, 44
6. a.
46a2 b 3a b.
40 13
a 46b 3a
23 .5 13a
3a
.
a 138ab 3a
3a
.
13 13a
1 3
138 ab
2 130a 13a
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 14
d.
2.
12
1
6
2 6
3
3
7
3 5
x
3
4 3
x
x
x 7.5.3.2 x
2
x
210 x59 x
6 6 3
210 x 5 9.9. 2 1 210 x119
3 ,44949 9 1,732 ,73205 2,4494
19.3 2
3
5, 476 21
e.
3
.
3
6 3
3
3
1, 394
3. a. 5 x 1 5 x 1 7.5 x 1
7
35 x 1
f.
1
5
1
3
1
5
5
2
3 3
2 2
5 10 3 15 2
b.
a
2
6
1
5
a
2
2
1 6.5 a 1
1,732
2
a. c
b2 a 2
b. a
2 2 2 2 b c 13 6 133
c. b
a2 c2
d. b e. c f. a
32 27 3 3 32 13
2
2
n
2
10 3 b 2 a 2 202 102 300 10
1
T
2 L
m 2
2
1
450
40
0, 04
2.3 .5 .10
2
a
2
1
12
2
5. T 2
45000
150 2
80
b 2 c 2 40 2 252 975 5 39
C. Evaluas Evaluasii Kemampu Kemampuan an Analis Analisis is
5
2.5 2
4.
a c 12 8 208 4 13 2
2
30 a 1
10.
2
7
5 x 1
30
2
2
80
40 4
15 2 8
2,65
L 0 2 2 2 g 980 7
1. a. s
a b c
4 5 6 15 2 2
s s a s b s c
L
15 7 5 3 . . . 2 2 2 2 2
2
3 .5 .7 4
15
LKS 7 A. Evaluasi Pengertian atau Ingatan
7
2 4 40 25 35 100 b. s 50 m 2 2 L 50.10.25.15
1. D
2 3 4
39 3
2. C
2 3
a 8 a
3.54.10 2 2 ,01 m 250 3 m 2 433,01
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 15
3. D
11 2 5
2. a.
2 14 2
3
4. D
21 3 10 21
2 a 3 5 6 a 10 a 25 a 10 a
5 8 5
3 1a 10 a 2 a 3 5 31
5. A
b.
7 3 1 3 7 7 4 5 8 4 0
a 22.10a 3 3 10a 5 6a 32.10 10 a2 10 a 3 a2 10 a 18a 10
6. E
5a
2 .3 2.2 2 7 2 6 2 4 2 7 2 3 2 3
2
c.
5.3 3 3 3 10 3 32 3 15 3 10 1 0 9 3 13 3
8. A
3
2.32 x 3 2 x 5 x 2 25 x
9
2
2 x x
2x 4x
2
2
2x
14 x 2 2 x
3 6 3
9. E 2 x 2 2 x 5 2 x 1 2 5
10 a 18a 10 10
2
a 10 5a a 18
7. E
3 4 3 3 3 7 4 3
2 2.33.5a 3 5a 2.5 3 a
3
.5
d.
2 x 2 2 x
10. A 5 3a .5 5 a 5 7a 15a a
4 a 3b 4 62 a 5b 2
4 ab 5 23a 5
a 6a b a
2
2
2 ab a 2 b 3a e.
2 .3 m n 2 .3 m n 4
B. Evaluasi Pemahaman Pemahaman dan Penguasaan Penguasaan Materi
7
6
4
7
m 2 n3 3 n 8 m 2 n 3 3n 2 3 12m n 3n
1. a. 5 2
4
2 2 3 2 6 2
3. b. 3 3 5 3 4 3
2 3
c.
.4 5 3 5 2.5 5 5
d.
7 3 7 5 7 10 7
3
4
b.
3
4
3
g.
2.33 2 3 2 3 3 2 3 2
3 23.34 3 3 3 6 3 3 9 3 3 4
2 3 23.3 213 3 4 3 3 17 3 3
4
3 9 27 x 4 10x 3 2x 36x 3 2x
d. 3
f.
3
c. 7 3
.5 3 4.2 5 4 5 15 3 12 5
e.
a.
5 24
6 2.6 6 5.3 6 5 6
26 x 3 2 x
2 2.3 3 5.3 2 17 2 6 3
e. 5
2
h.
2. 2 6 5 3 9 4 2 3 5 3 9 3 4 2 5
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
2
4 2 4 2.3 x 24 x 2 36 x 2
60 x 2
Bab 1 | page 16
f.
8
4. A
2 .3m 4m 2. 2 .3 m m 3.5 m
3
3
3
3
3
3
3
3
7
3
16m 2 3 3 12m2 3 2 5m2 3 3 11m
3 12m
2 3
2 3
7 2 7 2 7
5. B
2
6 2 2 2 2 3 2 4 6
m 2 113 3 12 3 2
6. B
8 2 3 16 2 6 4 2 6
4.
a
23
b.
a
2
a. 3 a
3
2
3 a
2.32
a
2
a 3
a
2 a
2
9 a
2
a
a
2
11 a
7. D
2
c.
2 3 2 3 4 1
8. A
6 1
6 1 6 1 5
9. C
3 3 a a
12
2 2a
a
3 3a 12 a 2 2a 2 a a
7 5
3 12 a 2 2 2 a a
10. B
1
11. C
7 5 7 5 2
2
1 3 2 3 1 4 2 3
d.
33 ax
18
2 2.3
x 3
x 2 x 3
x
18
x 3 x
2x 3x x
16 x 1
3x
x
12 2
2
14 4 6 12 2 24 2 14
3x x
12. A
3x x
10
8
2
10 2 8.10 8 18 2 24 .5
3x
x
18 8 5 13. A
7 2 10
LKS 8 A. Evaluasi Pemahaman dan Penguasaaan Materi
a b
a b 7
a 5; b 2
ab 10
enyelesaian : 5 2 14. B
1. B
3 3
8 2 15 15
a b
a b 8
a 5;b 3
2. D
6 6 3. B
ab 15
enyelesaian : 5 3
6 3 2 3 3 2 Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 17
15. D
f.
3 5 a b 3
5
4
2
5 a
ab
4.5
3
4 ;b 5
Penyelesai Penyelesaian an :
2
5
3 2
4
2 7 21 2 2 6
a b
7 2 3 2 2 3 2 3 7 2
1 2
1 2
g.
1
h. 7 5
2
2 10
i. 4.3 B. Evaluasi Pemahaman dan Penguasaan Materi
2
3. a.
6 0 10 16 4 15 2 60
1. a. 6 6 b.
b. 13
12
2 .2.3
5.8 20 80 8 0 40 240 20 2 .5 4
66 36 .2 .3 .3 22 6.2
240 80 5
6 2 2.2 6.3 3 6 108 c.
c. 13
12
2 .6
2 .6 26
36
6.3 84 3 49 157 84 3
2 .6 864 6
2
3
6
d. d. 14
13
.6
33 12.6 4 12 12 3 3.6 4
7 4 12
2
7 8 84 192 199 8 22.21
e. 12
.5
13
199 16 21
7 3 6.5 2 6 6 7 3.5 2 e.
f. 15
.10
12
7
2 10
.1 .10
5 10
7 .10 10
2
5 42 30 30 294 339 42 30 30
5
f. 2. a.
16 35 35 4 42 42 4 30 30 6
21 5 3
4. a.
b.
42 2 60
42 4 15 3
14 4 5
2
x2 y 2 y 2
b. c.
2 2 .3 8 30 30 8 3 8 30 30 8 3 1 10 4
t2
t
2
1 1
c. d.
0 3a 8 6 a 5 18 a 12
1 14 15 15
20 a 8 6 a 15 2 a 2 3
e. 6 15 12 30 6 15 2 3 30
3 2 5 2 10
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 18
d.
7. a.
7m 16 70m2 6 70mn 12 100mn
2
4 2
56m 16m 70 6 70mn 120 mn
8m 7 2 70 mn 6 70 120
4
2
2 2 16 8 2 2
2 4 2 4
b.
5. a.
eliling 4 2 4 2 8 2 2 16 cm
20 x 2 6 3 2x 7 3 10x 42
3
16 8
1 8 2 2 2
42 3 10 x 3 2x 6 3 2x 7 3 2x 3 5 42 c.
3 2 x 3 10 x 6 7 3 5 42
8 2 2 4 Luas
2
2
b. 5
4
16 x 2 7 5 8 x 4 5 2 x 28
2 x 8 x 7 4 4 28 5
5
5
8. a.
6 4 40 12 4 96 9 6 14 4 30 3 0 28 4 72 72 4
4
15 2 5 2 2 5 2 5 5 2 5 2 5
2 0 6 48 7 15 15 14 36 36 2 2 3 20 4
2
16 4 2 4 4 23 cm
c.
4
2 4 2 8 2 2
4
15 5 2 4 5 5 4 5 30 15
30 15 5 2 5
d.
54 3 16 160 9 3 40 4 0.32 6 3 48 4 8.4 3 48.32
54 3 2 3.20 9 3 2 6.20 6 3 2 6.3 3 2 9.3 2 0 36 20 12 3 8 3 108 20 3
3
3
3
144 3 20 20 3 3 6. a.
27 15 5
b. 3
100 2 5 2 5 2 5 100 4 5 2 5 100 4 4 5 5 4.25 2 5 2 5
2
2
F x 2 x 3 x 2
F
2
5 2 2 5 3 2 5 2 4 4 5 5 6 3 5 8 8 5 10 6 3 5
400 5 900 c. 3
52 2 5 2 5
3
2
2
3
5 4 5 5 2 2
12 5 5
9. a.
b.
42 3
F x 2 x 3x 2
F
6 2 3 2
6 2 3
2
3 6 2 3
2 2.3 12 3 6 6 3 2 6 4 2.
12 24 2 24 3 6 6 3
a b
a b 4 ab 3
a 3; b 1
enyelesaian : 3 1
36 2 24 3 3 6 3
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 19
b.
10.a.
2 3 5 2 3 5
7 2 12 12 a b a b 7
2
2 3 5 2 3 5 2 3 5
a 4;b 3
ab 12
2 2 6 3 5
Penyelesaian : 4 3 2 3
2 6
c.
b.
5 24 5 2 6 a b
5 2 6
a b
a b 5
a b 5
a 3; b 2 ab 6 Penyelesaian : 3
ab 6
a 3;b 2
Penyelesaian : 5 2 6 3 2
2
7 2 12 12 a b a b 7
d.
a 4; b 3
ab 12
15 10 10 2 15 2 25 25.2 a b
Penyelesaian : 7 2 12
a b 15
a 10; b 5
ab 50
4 3 2 3
6 2 8 a b a b 6
Penyelesaian : 10 10 5
ab 8
a 4; b 2
Penyelesaian : 6 2 8 4 2 2 2
e.
5 2 6 7 2 12 6 2 8
21
5 21 2 1 5 4. 4. a b 5 ab
7
2
21 a 4
Penyeles Penyelesaian aian :
;b
7
4
52
21
4
b
.
3 2 2 3 2 2 2 3 2 2
3 2
2
C. Evaluas Evaluasii Kemampu Kemampuan an Analis Analisis is
3 2
1. a.
f. 9 4 4 2 3
42 3 a b
2
2
y 2 z x y z y 2 xy 2 xz 2 yz z
a b 4
a 3; b 1 ab 3
y z 2 xy xz yz
Penyelesaian : 4 2 3 3 1
b.
9 4 3 1 9 4 3 4 13 4 3 13 2 4. 4.3 13 2 12
x y z
5 3 2
b
a b 13
a 12; b 1 ab 12 Penyelesaian : 9 4 4 2 3 12 1 2 3 1
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
5 3 2 2
5.3 5.2 3.2
10 60 40 24
Bab 1 | page 20
2.
c.
30 10 10 5 30 2 25 25.5
1
y
19 8 3 19 8 3
x y 30 x 25; y 5 xy 125
1
361 192 192 361
5; b 5
5.
4
4 b2
2
a b
a
b
a b
x yb
b.
x; y 3
20 14 2 . 20 20 14 1 4 2 20 14 2 20 14 14 2
3
x y a....(1)
3
1
xy 12......(2)
400 392 3 2
Subs Substi titu tusi si y 3 ke (2) (2) : x 3 12 x 4 b 4 2 Substitusi x 4 dan y 3 ke (1) :
LKS 9
4 3 7 a 7 4.
a.
i 4 27 8
11
27 2 176
a b
A. Evaluasi Evaluasi Pemahaman Pemahaman dan Penguasaaa Penguasaaan n Materi 1. C
27 2 17 176 a b a b 27
7 3 81
ab 176
2
a 16; b 11
Penyelesaian : 27 2 176 4 11 Jadi,
ii
4
4
4 3
27 2 176
a b
27 2 17 176 a b
2
168
7 2
21
.3
2
2 5
15
2 6
1
2 42
7
1 7
7
4. B
27 8 11 4 11
24 54 150
27 8 11 27 8 11 4
2 6 3 6 5 6
4 6
4 11 4 11
4 6
0
5. D
8
b. 5
5
24
Penyelesaian : 27 2 176 4 11
4
27
3
3. B
a 16; b 11 ab 176
4
7
5
2 5
a b 27
Jadi,
3
2. A
27 8 11 4 11
27 8 11
b
a 2 12 b 3
13
a.
3.
a 48
1
169
Penyelesaian : 25 25 5 5 5 a b Jadi, a
1
2 3 .5 2 3 2 3
2 3
1
1
4 3 5 1
7 5
7 5
8 75
2
6. B
14 15
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
35 24
14.35 15.24
14.35 15.24
49 36
7 6
Bab 1 | page 21
1
7. E 42
35
15
14.35
56
42.15
15.24
35.56
9
28
3
2 7
7 7
L K S 10
3 14
A. Evaluasi Pemahaman dan Penguasaaan Materi
8. A
27 2 16 162 a b a b 27
1. D
a 18; b 9 ab 162
1
Penyelesaian : 27 2 162 18 9 3 2 3
33 800 33 2 200 a b a b 33
9 8
9 8
.
3 2
3 2 3 2
Penyelesaian : 33 33 800
9 8
9 8
3 2 2
2. A
7
a 25; b 8
ab 200
9 8
.
25 8 5 2 2
21 7 2 21 7 2 92 7
3 2 3 1, 4142
27 2 162 33 800
1,5858
3 2 3 5 2 2
3. D
28
1 3 2
9. C
3 2
.
3 2
25 2 1 26 26 a b
3 2
3 2
3 2
1,732 ,732 1,414 ,414
a b 25
0,318
a 18; b 7
ab 126
Penyelesaian : 25 2 5 2 126 18 7 3 2 7
19 336 19 2 84 a b a b 19
a 12; b 7 ab 84
4. C
4 24 3 2 22 3
.
2 3
8 4 6 4 6 12
2 3
4 2 6 2 6 6
2
Penyelesaian : 19 336 12 7 2 3 7 25 2 126 19 336 2 33 2
3
2 7 2 3 7
5. C
4 2 2 1
2 1
.
2 1
2 33 2
2,2426
1 6. B
2 1
10. D
48 2 3
4 2 4 2 2 2 1
3 2 2 3 1, 41 4142 2
2 33 2 3 2 2 3
16.2 4 2
2 1
.
2 1 2 1
2 2 2 1 2 1
3 2 2
1,414 42 3 2 1,41
0,4142 7. E 7 5 2 7 5 2 7 2 35 5 7 2 35 35 5 12 7 5 2
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 22
8. D
13.
8 7 9 8 6 5 7 6 1 9 8 8 7 7 6 6 5 5 4
5 3 2
2 3 5
.
2 3 5 3 5 5 3 2 3 5 2 3 5
3 2
.
5 2 3
5 2 3 5 2 3
5 13 2 39
10 15
2 3 10
6 3 5 15 2 6 10 2 2 6 3 5
25 4.3
2 15 3
5 13 13 2 3 13
2
9. D
13
2
2 6
6
.
2 90 3 6 12
6
6 10 3 6 12
2 10 6 4
13 13 5 2 3 13
14. A 8 3 3 5 4 3 4 5 32.3 32 15 15 12 15 15 12.5 . 4 3 4 5 4 3 4 5 16.3 16.3 16.5 16.5
10. D
1 3 1 3 1 3 1 1 3 1 3
36 20 15 15 32
9 5 15 15 8
15.
11. A
. 3 5
2 2
32
2 2 3 2 1 3
96 20 15 15 60
3 1 3 1 3 1 3
3
3
2 3 5 2
5
3 5
2 3 10
2
2 2 6 3 5 6
2
2 6 10 2 6 10
5 10 15 6
6
.
6
2 6 6 2 15
5 3 2
.
5 3 2
5 3 2
5 5 3 5 2 50 30 20 15.5 15.3 30 30 18 12
5 3
2
2
5 5 5 3 30 2 5 5 3 3 5 3 2 2 3 5 2 15 3 2 3 2 8 3 6 5 30 6 2 15 15
12
3 6 15 6
B. Evaluasi Pemahaman dan Penguasaan Materi
12. B 6
2 3
7
.
2 3 2 3
7
2 6 18 42
7
2 3
2
1. a.
7
1
2 6 3 2 42 4 4 3 3 7 6 3 2 42
3
3
.
3
.
5 3
5 3 1 5 3 2
5 3 5 3 1 23607 7 1,73205 ,73205 0, 2520 25201 1 2, 2360
5 3
18 3 6 42.3
12
6 2 3 6 3 14
b.
12
1
2 4
2 6 14
3 2 1
.
2 1 2 1
6 3 6 3 2 1
2,44949 2,44949 1, 73205 73205 0,71744
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 23
c.
b.
2
.
3 2 2
3 2 2 3 2 2
6 4 2
2 1
64 2
98
21
2 1
.
2 2 2 1 3 2 2 2 1
2 1
6 4 1, 41421 1421 0,34316
c.
3 2 2
1
d.
2 3 2
5
. 2 32 2 32
10 3 10 10 12 4 5 4 5 4
3 1
3 2 2 3 2 2
10 3 10 10 8
9 8
32 2
d.
4 5 1
,73205 5 1 3,4150 ,4150625 625 1,7320
.
5 1
4 5 4
5 1
5 1
5 1
e. 3
e.
.
2 7 13
2 7 13 2 7 13
2 2 1 2 2 1 8 2 2 2 2 1 9 4 2 . 8 1 7 2 2 1 2 2 1
32 2
9 4 1,41421 ,41421
6 7 3 13 28 13 2 7 13 5
f.
7 2,0938
3 1 3 2 4 12 6 12 2 16 3 16 16 . 18 16 3 2 4 3 2 4
4
6 6 6 2 8 3 8
f.
3 1 2 3
2 3
3 1
2 33 2 3 2 3 3
2 3 3 3 3 5 g.
2 1,73 ,73205 3
2 4 7 2 7 1 . 2 7 1 2 7 1
3 1,732 ,73205 5
4 7 56 2 4 7 28 1
6,4641 h.
10,19615
3 1 2 3
0,6339
2 3
3 1
3 2 3 3 2 3 3
2.
3 1
a.
3 3 3
2 2 3 2
2
6 3 12 2 3 6 3 12 12 2 3
1
3 2 6 3 6 3 2 3 2 6 3
3.
3 1 3 1
3 3 3 2
i.
b.
2 3 2 3 3 6 3 2 3 2 . 3 9 3 3 3 3 3 3
.
. .
6 3 6 3 3 2 3 2 6 3 2
3 12 3 6
18 12 12
6 3 3 2
12 6
18 12
12 6 18 12 2 12 6 18
a.
1
2 3
2 3 2 3
2 3 4 3
2 3
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
4 3 6 3 2
Bab 1 | page 24
j.
b.
3 2
6 3 4 3
6 3 2 3 2 6 3
6 2 3 2
3 2 4 3 6 2
3 12 3 6
6 2
.
6 3
.
6 2
6 3
.
63 4 18 4 6 62 18 12 12 32
12 6
a b 3
18 6
18 12
a 2;b 1
ab 2
Penyelesaian : 3 2 2
2 2 1 4 2 2 2 1 . 8 1 2 2 1 2 2 1
3 2
12 0 12 6 18 6 18 12
a. 47 6 10
47 2 90
a b
a 45; b 2
ab 90
10 10 34 5 3 2
.
5 3 2 53 2
10 50 50 34 5 30 20 20 102 2
50 2 34 5 60 5 102 2 13
52 2 26 5 4 2 2 5 13
25 5
.
5 5
1 1 1 2 2 1 4 2 2 2 2 . 4 2 4 2 4 2
45 2
7
Penyelesaian : 18 18 8 2 4 2
a b 47
Penyelesaian : 47 6 10
5 18
10 10 34 53 2
25
1 4 2 42 2 2 16 2 1 3 2 2 1 2 3 22 14 28 1 1 2 2 1 2 2
5 5
47 6 10
5 3 2
18 8 2 18 2 32 a b a b 18 a 16; b 2 ab 32
4.
2 1
2 1
6
3 2 2 a b
1 5
2 3 2 2
45 2 4 2 2 5 5 5 3 5 5 2 3 5 5 2
5 3 2
7
1 28
18 8 2
3
2 2
20 12 2 3 2 2
18 15 15 2
28
28
5. a. 3
2
3 1 3 1 3 1 2 8 7 2 2 2
1 2 1 2
3
1
1 2
1
2 2
3 10
3 1
2
4 3 1 7
3 4
2
4 3 4 7
10 3 3 5 3 2 4 3 3 10
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 25
b.
5. 2
3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 6 2
3 2 6 2 5 2 6
5 2 6
7 5
7 5
3 2 6 2
52 6
3
7 5 7 5
. .
7 5 7 5 7 5 7 5
7 2 35 5 6 35 7 5
7 2 35 5 6 35 7 5
3
3
3 3 7 5 7 5 6 35 6 35 7 5 7 5
2
5 2 6
3 2
3 2 6 2
52 6
1 2
3 2
3
7 5 7 5 7 5 7 5
25 24 5 2 6
25 24
16 108 35 3 5 630 35 3 216 108 35 3 5 630 35 3 1692
25 20 6 24 1 25 20 6 24 99
Soal Khusus untuk Penggemar Matematika C. Evaluas Evaluasii Kemampu Kemampuan an Analis Analisis is
1. 3
1. a x a x a x a x x
.
2
a x a x
2
L K S 11
1
2 2
3
. 2 2 6 2 2
A. Evalua Evaluasi si Pengertia Pengertian n atau atau Ingatan Ingatan
6
3 6 3
1.
2
3
2.
3 6
B og 8 3 8 2
2 2 3 6
a b a a b a b 5
a x a x
2.
2 2
C 3 x
2
3.
log b
2 2 3 2 18 6
4. 5.
3 4 2 6 2
6.
89 1
7. 8.
log b x
4
log 4 3
3
5
og 62 625
log 54 4
A og100 og100.0 .000 00 5 A log 100 m 100 (0,1) 102 (101 ) m 102 10 m m 2
9. 4.
16 2 63 16 2 63
og 64 64
0,1
2 2 6 3
a
A 5
2 4 2 6 2 12 6 3 2 3 3 3 6
b
B 4
2 2 3 6 2 2 3 . 2 2 3 2 23
c
c ba
D x
2 2 3 6
2 2 6 3
3 log y x
B
2 2 3 6
4 4
3
3
a x a x a x a x a x a x a x
a a x x
3 3
7 9 7
9 7
1 2
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
B 5
og 125 b log 5 b b 3
5
b
log 27 27
3
log 27 27
3
3
3
log 3 3
10. B 8
og 8 1
Bab 1 | page 26
B. Evaluasi Pemahaman dan Penguasaan Penguasaan Materi 1.
a)
6
log 216 3
b)
3
log 81 4
3 4
x 2
3
x
4
1 3
c)
log 81 4
x
8 6 x e) e e x 6
log 27 3
3
e)
16
f)
e
log
1
3
x
f)
log 7,39 2
g)
27
log 9
h)
81
log
1
2
x
3
27 4 i) log 0,05 3
2.
125
log
1
5
x
1
h) e
3 i) e
27 7 2 128 6 3 216 3
c)
2
e) 6
49 1
8
6
3
4 1
e2 e x e 2 x 2 10 x 3 j) e e x 3 10 x 7
2
d) 343 3
x
2
e4
x
x
a) 3 b)
1 4 2
1 1 2 2 x 2 3 3 g) x 12
e
j)
108
x
d) 10
1
d)
2
2
9
3
4 27 3 g) ln 20,1
f)
h) ln 5
i) ln j)
3.
2,72
148,41
16 81
a) 2
x
L K S 12
3 4
27 8
A. Evalua Evaluasi si Pengertia Pengertian n atau atau Ingatan Ingatan 1.
64
x
6
2 x 6 3 b) x 8 512
= log( b ) log(a b )
2.
2
16 9
B og x log
3. c) x
E log(a 2 b 2 ) log(a b )(a b )
x
log
x
3
x
log x2
A 2
log 4
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
1
2
2
log 12 2 log 6 2 log
og 8
2
4 12 6
3
log 2 3
Bab 1 | page 27
4.
D
15.
og
6 100
og log 5.
log
36 100
6 36 100 10 0
5
10
og 150 log 2 3 5 log 2 log 3 log 52
5 log
4 54
Diket: lo log 2 a, log 3 b, log 5 c
log 4 5 log 54 1 10
4
a b 2c
log 10 5
log 10
10 4
B. Evaluasi Pemahaman dan Penguasaan Materi
D Dike Diket: t: 2 log3 log3 1,585 2
6. a
log 6 D
2
a
log pq log
log 2 3
p
a
2
2
pq
log
p q a
og bc a log
2
og
bc a log
8.
a log pq pq
2 a log q
D a
a)
log 2 log 3 1 1, 58 585
7.
1.
a
og
b)
c 2
2 a log c
c)
A og10 log10 log100 0 log1. log1.00 000 0 log10. log10.00 000 0 log 100.000 1 2 3 4 5 15
9.
D og 151 log 134 log
22 15
log
d)
11 14 5 3 22 15
log 7
10. A og
3 5
log
5 36
log 112
log
3 5 36 5 1 12
log 1 0
2.
a)
11. C 1 3
log12 g125 2log 2log 4 log32 og32
b)
1
2 125 3 log 4 log 32 32 log 12
og 5 log 16 log 32 32 lo log
5 32 16
2 log16 2 log 8 2 log(2.2.2.2) 2 log(2.2.2) 2 log 2 2 log 2 2 log 2 2 log 2 2 log 2 2 log 2 2 log 2 7 2 log 2 7 .1 7 2 log 4 2 log 64 2 log16 2( 2 log 2) 6 2 log 2 42 log 2 26 4 4 4 log 246 4 log 16 4( 4 log 4) 2( 4 log 4) 2 4 3 3 log 81 log 27 3 log 3 8( 3 log 3) 8.1 8 x x log log 2 2 x 6 x 9 x 3 x 3 log y 13
c)
log 10 1
x 13 x 3 log x 12 x 3 log 1 x 1
12. D log x
log y z log xy z
xy 10 z x
10
13. A
2 b2 c2 log log log c ac ab og
a
2
bc
b
2
ac
c
2
ab
log 1 0
14. A
d) log 3 x 1 x
2
log(3 x
3
x 2 )
x xz 3 z 3 y e) log log 3 log 2 3 x yx yx z x 2 b3 f) log a 2 y 3
Diket: log x 1, 96 9675 x 101,9675 log( log(1. 1.00 000 0 x ) log1. log1.00 000 0 log log x ,9675 4,967 ,9675 3 1,967
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 28
3.a) 3.a) log p b) log a
log q 2
log r
j)
log b log( a.a ) log b log a log a log b
9 3.3 log log 2 2 2 log 3 log 2 0,6532
2 log a log b c) log a log b
2
log a log(b.b) log a log b log b log a 2 log b
5.
2.18.5 2.18.5 log 6.3 18 log( 2.5) log 10 1
a) log
d)
log a log(ab) 2 log a log a log b log a log b 2
log b 2 log a log b 3 2 f) log a log b 3 log a 2 log b g) log 1 log a 0 log a log a 2 2 h) log 1 log a ab log(a ab) e) log a
2
4. a) log(2.2.3)
b)
6.
log 2 log 2 log 3 2 log 2 log 3 2(0,3010 ) 0,4771 1,0791 log(10.2) log 10 log 2
a)
b)
1 log 2 1,3010 c) log(10.3) log 10 log 3 = 1 log 3 1.4771 d) = log(10.2.3) log10 log 2 log 3 2,7781 1
e)
1
f) log(100.3)
h)
i)
xy 3 xy 3 log log 1 x 5 x 5 2 x x 1 x 2 x log 2 log 3 2 y ( y 3 ) y 3 y y 3 1 log 2 y 32 y y 3 y x 13 3 9 log x 1 log x log x 9
c) log
d)
log(3. 2 ) log 3 log 2 2
log 3 log 2 0,6276
g)
b)
8.9.10 720 7 7 log log 10 720 5.14,4. 7 7 7 log 7 1
log 100 log 3 log(10.10) log 3 2 log 10 log 3 2,4771 log(1000 .2) log(10.10.10) log 2 3 log10 log 2 3,3010
6 2.3 log log 10 10 log 2 log 3 log 10 0,2219 2 2 log log 1000 10.10.10 log 2 3 log 10 2,699
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
7. a) log 5
log 2 b) log 5 log 5
0,6990 0,3010 0,398 log 3 2(0,6990) 0,4771
= 1,8751 c) log 75 – log 4 = 1,8751 – log log 2 – log 2 = 1,8751 – 2(log 2) = 1,8751 – 2(0,3010) = 1,2731 d) log 3 + log 3 + log 3 + log 3 – log 5 – log 5 = 4(log 3) – 2(log5) = 4(0,4771) – 2(0,6990) = 0,5104 e) log (100.2) = log 100 + log 2 = 2 + 0,3010 = 2,3010 f) log 2 + log 2 + log 2 + log 2 – log 3 = 4 (log 2) – log 3 = 4(0,3010) – 0,4771 = 0,7269
Bab 1 | page 29
8.
4.
log 12 p
2
x
log(4.3) p
2 2 2 2 16 2 2 2 l og lo l og lo l og 2 log lo l og lo log 16 16 log lo
x
log 4 log 3 p
x
log 4 q p
x
log 4 p q
x
2
2
2
log log log log65.5 og65.536 36
2 2 2 0 log 4 log 2 1 2 log lo
x
5.
E 1
log lo log lo log x 1 log lo log x 10
log x 1010 x 1010 6.
x x x log log x log 4 4 1 ( p q) 1 p q
0
A og 3 log 2 log x 1
3
log 2 log x 21
2 log x 32 x 29 512 7.
B log 2 x 1 2 x 101 x 25
9. a) 8. 10. a)
98 99 1 2 3 1 log . . ........ . log 2 2 3 4 99 100 100 b) 6
B
x
214 215 6 1 2 3 1 log . . ........ . log 3 215 216 2 3 4 216
og x
25
log 5
52
log 5
10 2
5
1 5 log 5 2
0, 5
C 1
8
og 2 6 log 2 16 16 3 log 81 81 5 log(625) 3
23
4
log 2 6 log 63 3 log 34 5 log 5 3
13 3 4 34 0 9.
C 2
log y y log x log x y log xy
3 2
1
log y log x 2 log x 2 y log xy 3
1
og y 2 x 2 x 2 y log xy
5
log( xy ) 2
log xy
5 2
10. B Diket iket::a 0,9090 ,909090 90.. .... dan b 1,331 ,331 100 100a 90,90 90,9090 9090 90.. ....
L K S 13
a 0,909090...
______________ ______________ _ A. Evalua Evaluasi si Pengertia Pengertian n atau atau Ingatan Ingatan
99a
90 a
90 99
10 11
b 1,331 ,331 1.3 1.331 1.000 11 10
1.
A 5
log( log62 log625 5)
2
5
2
2
2
og 4 log 2 4
2. 1 3
log 81 m 81
13
m
34 (31 ) m 4
3.
a
og b
log 1.728 m 1.728 (2 (2 3 ) 26 33 2m 3 2
26 2 m m 6 m
m 6
33 3 2 m 6
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
a
3
1011
log
1 a3
1
a
3
a log a 3 3
12. B n
n
Diket: log 1, 1, 5 a; log 3 b log 1, 5
C 2 3
1
3
11 10
4
log( log5 log5 )
3
a log n
log 1, 5
diperoleh: log 3 log 3 b log n log n b log1 log1,5 ,5 log3 log3 3 log 32 ab a b x 3 x 3 3 3 log 2 x 2 log n
a
Bab 1 | page 30
13.
c) 4 log 3 – 2 log 5 = 0,5104 0,5104
a b4 2
log a b
1
2
a a 1 4 4 4 10 4 4 b b 10 2
log
d)
b2 100 a
2
10
b2 log a a
b2
4
3 . log 2 0,4515 2 1 e) (2 log 10 + log 2) = 1,1505 2 1
2
1
1
4 100a 4 1 l o g a 2
f)
3. a)
14. E 5
Dike Diket: t: log log 3 a 5
( 2 log 2 + 2 log 3) = 0,7781
log 75 log 3 52
5
b)
log 3 5 log 52 a 2
15. C
3
og 45
2
2
3
log 5
3
log log 15
3
log 5
2
3
2
1 3
3
log 5
3
3 2
1 3 log 5 3
2
log 3 log 5 2
3
1 3
3
log 5
2
a – 100a = -63 -99a = -63 -63
1 3
log 3 log 5 3
4(1 log5) 3
1 (1 3 log 5) 3
a = 0 ,6 ,636363 100a = 63, 6363
a=
12
b 2
63 99
23
log a 49
2
7 11
log
121
49 log
B. Evaluasi Pemahaman dan Penguasa Penguasaan an Materi
1.
log 3 8. log 3 8 2 log( 2 4 ) 3 2 log 212 12. 2 log 2 12 2 log(2 2 .25 )3 2 log 2 21 21.2 log 2 21 3
4
a) log(3 ) b) c)
2
3
8
3
8
3
7
3 log 3 4 4.3 log 3 4 7
e) log(3 )
7
log 3 2 .3 log 3
2
3
7 2
2
f)
4
4 3 4 log 2 log 4 1 4 1
2 2 2 2 3 2 log 4 log 2 3 . 2 log 2 3 3 2 6
g)
2
h.
2
3
5
log(2 .2 2 ) log 2 2 4
2
5
.2 log 2
5
2. 1
a) log152 = =
1 2 1 2
11 7
11
log
7 11
1 2
3
1
d) log(3 ) 2
2 7 11
7
log5 3
4. 3
log 4 4 log 1575 2
3 log 4. 4 log 1575 3 log 9 3 log 1575 3 log 9 1575 3 3 log log 175 9 3 log25.7 3 log 44 log25.7 3 log 44 log 25 4 log 7 1 3 4 2 4 log 7 log 4 log 5 1 4 3 log 4 2. log 5 4 log 7 b 2 a c 1
log 5 log 3
= 0,58805 b) log 5 log log 3 2 log 2 = 2 log 5 + log 3 – 2 log2 = 0,2731 2
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 31
9.
A
L K S 14
1
1
C 1 log 36 36
1
3
log 36
og36
log 6 log 36
log 3
log6 2
1
1
5
log1 og100
log 2
log10 log100 0
log1 og100
log 10
log 5
log10 g100
xy
1
log xy log( xyz)
2
4
5.
98
1 20
64
og 10 10 log 2 0 log10 log10
log
25 8 20
1
log( xyz )
log xz log( xyz)
2
log(xyz )
log( xyz )
2
2 2
og 6
2
2
log 98 98 1 b
7
1 log2
log 2 2 log 3
log 2 2 log 7 2
1 b 1
2 a
a (b 1) a 2
3
og a
2
3
log10 log100 0
n
3 3
log a3 log2 3
log b
log b log2
2
loga 2 log3
og
log b 2
2
3
log a
3
3 log a
2 log a
log b
logb log3
log a 2 log b 2
3 log
2 lo log
log a
log a
log b
log b
Jadi, m n.
log 10 1
13. B
15
2
log40
5
log 40 40
5
5
5
3 log 2 1 5
log 2 log 5
5
log 15
log 3 1
25
3
5
log 3 log 5 3
1 1 b
a 1
1
log 60 60 log 3 og 6 0
1 4
log 4 log 6 0
log 3 4 5 log 60 60
5
a
3b b( a 1)
log 6 0
log 60 60 log 60 60
1
A og 4 4 log 5 5 log 6 6 log 7 7 log 8 8 log 9
2a 3
3 log 5
3 5 2
log 3 a
3 2a
log
1b b log 1c c log 1a
a log b1 b log c 1 c log a 1 a b c 1) log b ( 1) 1) log c (1) 1) log a ( 1)
a log a 1
log 60 60
log 5
5 log 33 a
14. D
1
log 60 60
2
log 27
5 log 3
5
B
3
2
m
Diket: log 3 a dan lo log 5 b.
log abc
12. B
log 100
5
log abc
2
1 2
A
3
log( xyz)
log 2 0 12 log 64 64 og 2 5 lo
8.
log yz
log 6 4
1 xz
log( xyz)
4
1
log abc
B
log 25 log
7.
log6
log 3 5 log 3 log 5
log39 ada di antara 5 dan 6.
log log 25
6.
yz
E 4
log abc
log c
7
a ab a(1 b) 4.
log abc
log ab c log c
Diket: lo log 2 a dan log 3 b
3 4 log 3 log 5 log 5 b 3 log5 b 4
c
log abc
c
1
log b
1
11. B
log3 a 4
log 15
b
1
log( xyz)
log10 log100 0
D Diket: 4 lo log 3 a dan 3 lo log 5 b
4
log ac log b
b
1 b
log ab 1
10. C
2
1 1
4
log abc
log abc
log36
log 6
log a
C 2
3.
log 2
log 36 36
og 2 log 3
2.
a
c
1
a
1
log ac 1
og c log a
a
A. Evalua Evaluasi si Pengertia Pengertian n atau atau Ingatan Ingatan
2
b
1
1.
1
log bc 1
a
15. 1
log x p q
log
p q
log x
1
log x q r
log
q r
log x
log
1
log x r p
r p
log x
log
p q
qr r p
log x
log1 log x
0
3 log 32 2 Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 32
B. Evaluasi Pemahaman dan Penguasaan Materi
c)
2
22
log 3
log 2
2
3
32
1 log 2
log 88
3
log 88
3
log 22
3
log 11 3 log 8
3
log 11 log 2
3
log 11 3 log 23
d)
log 11 3 log 2 log 11 3 log 2
a
10 2 20 2 8 12 2 20 4 16 2 log log 4 4
log
2
2 2
1
log 2 4
5
log 3. log 6. log 2 5
5
3
log 6 2
3
3. log 2. log 3. log 6 5
5
2
3
3
2
b) log 600 = log (100.6) = log 100 + log 6 =2+a+b
1 15
= log 1 – log 15 = log 1 – log (5.3) = log 1 – log 5 – log 3 = 0 – (1-a) – b =a–b–1
5. a)
log 2
3 log 2 4 2 2 3 3 log 27 log 33. log 2 2 log 2 log 3
6. a) log 6 = log (2.3) = log 2 + log3 =a+b
d) log
3
22
c) log 5 = log (10:2) = log 10 – log 2 =1–a
2
4. a)
2
3
4
3. 2
2
log 27 2
1
. log 3. log 2
2 log3 3 log 2 4.2 log 3. 3 log 2 4
b 3a
2 log27.3 3 log 2
3
3
1
=
log 11 3. log 2
3
3
3
2
3 1 2 9 3 log log 2 log 2 2 3 1 3 2 log 3 log 2 2
2.
b)
log 9
2 log 9 2 log 3 3 log 2 3 log 2 1
1. a)
22
5
3. log 6
log 6
5
log 6
3
4 5 1 log 100 25 1 1 2 2 log log 8 8
e) log 0,75 = log
log
5
log 5
2
2
log 2 3
5
log 5 2
= log
75 100 3 4
= log 3 – log 4 = log 3 – ( log (2.2)) = log 3 – log 2 – log 2 = log 3 – 2 log 2 = b – 2a
1
log 2 3 2 1
2 2
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 33
7. a)
1
f) log
30 = log 30 2 1 =
2 1
=
2 1
= 2
g) log 12
2
= 2.1 + 6.p – 4.q – 2.r = 2 + 6p – 4q – 2r = 2 ( 1 + 3p – 2q – r)
(1 + b)
log 6
log 2 a b
a 2 x 6 a a 6 a 4 a 2 log 4 2 log a log x log y log z y z
2 a log a 6 a log x 4 a log y 2 a log z
log6.2 2 log 6 2 log 2
1
1
1
p
log0,05 2
3
1
2 1
q
4 log 3 1 log 5 log 100 . 4 log 3 1 1 a 2 . 4 b 1 a 2 . 4 b
c)
log z . log y log z a
a
x
a
5 log 1, 25 6 log 4 6 log 5 6 log 4 6 log 5 2. 6 log 2
log 5 log 2 2. log 6 log 6
log 5
2 log 2 log 6
1 a 2a 1 3a a a
log x
rq p
a)
1 3
log 2
3
a
2
3 3 9 . 2 a a 2 3 c) 3. log 2 3. a 2a 3 b)
3
. log 3 2
10. a)
2
3 log 2 log 3 2 log 2 2
d)
4
2
log 3 1
100 3 . 4 2
log 25.1,5 log 4
4
log 100 log 4 . 4
3 2
log 100 3 1 log 4 2 = ( log 100 – log 4 – 1) . = ( 2 – log 4 – 1 ) = ( 1 – log 4)
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
log y
8.
= 6
r
5 2 log 100
1 5 .3 log 4 100 5 log 1 100 .
i)
4
a
1
2
1
a log x . a log y .a log z
a 1
1 1 log xy 2 z 4 a
b)
a 2a b
1 9
(log 10 + log 3)
2
=
h)
a
log (10.3)
3 2
3
3 2 Bab 1 | page 34
C. Evaluas Evaluasii Kemampu Kemampuan an Analis Analisis is 1. xyz 100
log x log y
3
10 xyz log xyz 2 ..... (1) 3 log x log y ... (2) 4
4
L K S 15
2
log y 4 log z 5
A. Evalua Evaluasi si Pengertia Pengertian n atau atau Ingatan Ingatan
5 log z log y ... (3) 4
1.
B (Berdasarkan sifat logaritma)
2.
C 3
Substitusi:
3
log y + log y
5
5
log 32
5
5
log (xyz (xyz)) = log x + log y + log z = 2
3
5
=
3
325
log y = 2
log y
3
2
3.
y 10 3
Substitusi log y Substitusi log y
2
3 2
3
3
ke (2), didapat x 10
4.
32
3
6
5
3. 2
x.
b 64 2 y
6
2
y .
c 64 2 z
6
2
z .
log a 6 log a 6 y
6.
A
7.
B 2
2
log b 6
2
1
5
log 81
log5
2
log 5
5
z
8.
log c 6
A 3
log a log b log c 6 6 6 6 1 1 1 6 1 x y z x y z 2
log 81
log c 6 2
log 2
=9
log b 6 2
3
3
x
2
2
B 2
log 2 2
15
6
log 5
B
5
5.
x
23
52
2
2.
a 64 2
3
3
3
1
E 2 log 5
ke (3), didapat z 10
3 5 5
2
3 log y = 2
2
5
log 325
2
2
2
log 3 4
81
1
3
1 log 2
2
3
80
2
11 24
9. E 10. D 4
9
9
5
log 2 5
2 4 16
B. Evaluasi Pemahaman dan Penguasaan Materi
2. 3
a)
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
3
log 9
1 2
1 9
Bab 1 | page 35
2
b)
2 2
4 log 5
11. C 12. B
4
1
32 2 3 4 0,03 10 100
log 25
25 2 d) 5,8 58
c)
5, 8
2
e)
8
log
log 58
3 121.3
13. B 2
2
3
=
log
2
3 121.3
8 2 15
2
5 3 2
5.3
5 3 14. C
2
=
log
3
3
3
121. 2
3
2 9.3 25.3
6 3 5 3 2 3 3 3
3
3
= 121. 2 = ( 121 . 2 ) = 242 f)
log m log n
10
15. 16. 17. 18.
log( mn )
10 = mn.
C C D D 1 2 3
x 2
UKAB 1
3 4
23
x 24 24 x 23
19. E
10 2 21 A. Pil Piliha ihan n Gand Ganda a 1.
2 3 1
1
3
2 3 4 6
3
3
2 3 2 2 32 1 3 3 3 3 1
4
.
2
3 5
4
3
3 2
4
3
5
8. C 9. D 10. A
24. 25. 28. 29. 30. 31. 33. 36.
3 5 2
4 5
=
3
4
34
2 2 log log a a
2 log 2 2 log a 2 log a 1 0 1
4
24
3 3
B D D B E D B B 2
5
3 3 3 3 4 2 3 3 6
3
2n + 5 + 9 – 1 – 16 = 3 2n = 6 n=3 E 5
11.2
21. A 22. B
2 32
2 n 5 9 1 16
11 2 2
11 2
B B D
3
7.
13 2 22
2
C 1
3. 4. 6.
7 .3
20. B
4 b4 b a 2 6 2 2 b 2 3ab 2ab 2 5ab 2
3
7 3 2
7 3
B
18 a 3
2.
4.3
16 1
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
Bab 1 | page 36
37. D
4.
log 4 2 log 4 log 2
38. B
B. Ben Bentuk tuk Urai raian an
5
a
3 2 2
5 a 2 2 3 2 2
25 52 6
2
35 4 6 52 6
5 2 6
5 2 6
127 50 6 2.
a)
74 4 210
60 14 2
5.
42
74 16 210
log 56
60 14
3 52 5 22 5 2 5
.bc 5 2 5 2 5 2
1
3
3
5 (4 5) 2
log 42
3
log 7 8
3
log7 6
3
log 7 3 log 23
3
5 11
log 7 log 6 3
3
log 7 33 log 2
log 7 3 log 2 3 log 3
3
b 1
3 a
ab 3
ab a 1
1
log 44
3
log 44
3
log 66
3
log 11.4
3
log(11.6)
2
2
3
b
66
39 5 5 b) a
35
log
b)
27 15 5 1 3 2
log 56
152 5 2( 1) 5( 1) 30 15 5 2 5
3
a) 3ab – 2bc + abc
5 2 5 2 5
13
a)
60 14 3.
7
log
4 5 20449 7 11 log 13 35 143 4 5 7 11 log 13 35 260 7 log 13 35 log100 2
74 2 840
4
1
log20449 2 log
2 log 2 2
1.
b)
2
125 1
Kunci Pe Penyelesaian Matematika SM SMA Jl Jl.1A – Su Sukino
3
log 11 2 log 2 2
log 11 log 2 log 3 3
c
3
52 1
3
c
1
2 a
1
3
ac 2 ac a 1
Bab 1 | page 37
