BAB 1 BENTUK PANGKAT, AKAR, DAN LOGARITMA
g)
LKS 1
h)
55 55 5 5 5 3 5 5 5 5 5 5 (0,5)2 (0,5)5 (0,5)7 57 4
(0,5) (0,5) (0,5) (0,5)
A. Evaluasi Pengertian atau Ingatan
5 kali
(0,5) (0,5) (0,5)
14
1. D
3 3 3 3 3 3 81 2. C ( 1) 5 ( 1) ( 1) ( 1) ( 1) ( 1) 1 3. D
7 kali
4
( 1)
(1) ( 1) 1
10
10kali
4. A ( 2)7 (2) ( 2) 128
2. a) b)
05 0 0 0 0 0 0
c)
04 0 0 0 0 0
d)
( 1) 6 (1) ( 1) 1
e)
( 1) (1) ( 1) 1
f)
( 1) 5 (1) ( 1) 1
6 kali
7kali
5 kali
1 1 1 14 14 64 4 4 3
3. a)
7. B
243 3 3 3 3 3 3 8. C 343.000 70 70 70 70 3 9. D 5
b)
B. Evaluasi Pemahaman dan Penguasaan Materi
e)
2 2 2 2 5 2 2 2 2 2 2 2 2 2 8
1 1 1 5 5 5 6 1 5 2 4 17 17 17 17 17 17 17 17 1 5
2
15
4
15 15 15
17
7 kali
4 kali
3 3
36
c) d) e)
33 3 33 3 1 335 32 33 3 33 3 33 55 55 55 5 57 52 53 52 5 5 5 5 5
(32 )3 35 32 32 3 3 5 kali 11 3 3 3 3 3 6 kali
4 kali
f)
37 34 35
52 53 5 5 5 5 5 55
b) 3 32 34 3 3 3 3 3 3 3 37 d)
5 5 5 5 5 52 53 54 5 5 5 5 5 3 3 3 3
5 kali
a 2 b 2 c 2 2ab 2ac 2bc (a b c ) 2 10. A x 2 2 xy y 2 ( x y )2
c)
9
9 kali
5. A ( 7) ( 7) (7) 343 6. E
1. a)
03 0 0 0 0
7
7 7 7 7 7 7 2 3 7 7 7 7 75
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
f)
5 kali
(7 2 )3 7 2 7 2 7 2 7 7 7 7 7 7 76
g)
(3m 2n 3 ) 3 (3m 2n 3 ) (3m 2n 3 ) (3m 2n 3 ) 27 m m n n 6 kali
9 kali
27m 6n 9 ( x y ) ( x y ) ( x 3 y 5 ) 2 xy xy 3 5
h)
3 5
x y y x 6 kali 10kali xy x y y x y x
5 9
5 kali
9 kali
Bab 1 | page 1
i)
(2 2 2 4 ) 2 (2 23 )3
2. a)
2 2 (2 2 2 4 ) (2 2 2 4 ) 4 12kali 2 3 3 2 (2 2 ) (2 2 ) 2
b)
8 kali
j) 4. a)
(4 y) 4 y 4 y 4 y 43 y3 3
( a b c) 2 a 2 b2 c2 2 ab 2ac 2bc ( x y 3z )2 x2 y 2 9 z 2 2 xy 6 xz 6 yz
c)
( x 2 x 3)2 x 4 2 x3 5 x 2 6 x 9
d)
( a b c) 2 ( a b c) 2 ( a b c) 2
( 3)7 (3) ( 3)
( a b c) 2 8 ac
7 kali
b) ( 13 )5 ( 13 ) ( 13 )
e)
( x 2a ) 3 x 3 6ax 2 12a 2 x 8a 3
f)
( ax by)3 a3 x3 3a2 bx2 y
5 kali
c)
3ab2 xy2 b3 y3
(1, 75) 3 1, 75 1, 75 1, 75
d) (10,5)5 10,5 10,5 5 kali
e)
( 0,8) 7 0,8 0,8 7 kali
f)
41 )4 41 41 41 41 (3 125 )4 ( 12 12 12 12 12
3. a)
( x 1x )2 x 2 2xx x12 x 2 x12 2
b)
( x 1x )2 x 2 2xx 12 x 2 12 2
c)
3 ( x 1x )
d)
( x 1x ) x 3 xx 3 x2 13
g) (2 23 )5 ( 83 )5 83 83 83 83 83 h) ( 1 17 ) 6 ( 87 ) 6 ( 87 ) ( 87 )
( 2a )3 ( 2a ) ( 2a ) ( 2a ) 8a 3
b) ( 3b) ( 3b) ( 3b) 729 b 6
c)
6 kali 2 3
6
4. a)
2 3
1.296 x y
d) (3 a ) (3 a ) (3 a ) (3 a ) 3 a
f)
3 4
3 4
9 12
3
e)
3 4
2 y2
3 x2 y2
4
2 y2
3 x2 y2
2 y2
3 x2 y2
2 y2
3 x2 y2
2
x
x
20.736 28 34 161.051 115
d)
2.744 23 73
5. a) b)
86 y
3 x2 y2
3
c)
4 kali 8 12
3 4 3
x 3 x2 13 x x 3 3 1 x 3 x x 3 x
b) 11.390.625 56 36
(6 x y ) (6 x y ) (6 x y ) 2 3 4
3
x
2 3 xx
x3 3 x 3xx x13
6 kali
5. a)
x
3
c)
81x8 y8
d)
32 23 5 243 3 125 3 1 (0,125)3 81 512 1000 12 4 81 ( 0,12)4 100 253 4 390.625 4 4 (513) 513 (5,13) 100 (100) 4 4
C. Evaluasi Kemampuan Analisis 1. a)
( x y )2 x 2 2 xy y 2
b) ( x y )2 x 2 2 xy y 2 c)
LKS 2
( x y )3 x 3 3x 2 y 3 xy 2 y 3
d) ( x y )3 x 3 3x 2 y 3xy 2 y 3 e)
( x y )2 ( x y )2
A. Evaluasi Pengertian dan Pemahaman
x 2 2 xy y 2 x 2 2xy y 2 4 xy f)
( x y )2 ( x y )2 x 2 2 xy y 2 x 2 2xy y 2 2 x 2 2 y 2
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
1. D
1 1 35 5 243 3 2. D 1 1 7 0 7 tak terdefinisi 0 0
Bab 1 | page 2
3. B
( 1) 5
d)
1
1 1 5 1 (1)
2 3
a x
3x
5
x5 x2 2 3 2 3a x 3a
4. D
( 1) 8 1 5. D 1 1 1 (0, 001) 2 10 6 2 3 2 (0, 001) (10 ) 106 6. C 1 7 10 7 0, 0000001 10 7. D 1 1 53 125 53 8. C 1 4 3 3 4 9. B
1
1 9 4 4 9
2 3 2
10. E 1
53 11. B 1
1
p
3
e)
p p p
f)
p p
2. a) b)
c) d)
2
5
7
10
p2 p7 8 p p
p5 p p 10
3
1 8
p
1 1 1 20 1; 21 ; 2 5 ; 2 7 2 32 128 1 1 60 1;62 ;64 ; 36 1.296 1 66 46.656 0 2 3 1 1 2 ; 13 1; 13 9; 13 27 3 9 1 0 2 5 4 2 ; 25 1; 25 ; 5 2 25 3 125 2 5 8
5 125 3
3. a)
14
1 3 1 x5 2 3 5 2 3 a x x a x 3
3
1 64 641
b)
256 16 ;64 8 ; 4 2 ;1 2 1 1 2 1 4 1 6 8 2 ; 2 ; 2 ; 2 4 16 64 256 2
2
2
0
12. C 7
0, 000000256 2,56 10 13. C 123.000.000 1, 23 10 8 14. C 1 1 1 2 2 2 (5 a) 2 2 25a 5 a (5a) 15. E 3 3 1 17 71 7 3 3 17
c)
d)
4. a)
B. Evaluasi Pemahaman dan Penguasaan Materi c)
a
b) a c)
b
5
3
4
1 5 a a
b
2
3
5
3
2
7 7 ;1 7 1 1 1 1 2 3 11 ; 11 ; 11 ; 11 121 1.331 1 4 11 14.641 1
b)
1. a)
16.807 7 ;343 7 ; 49 7 ; 0
1
1 ( a b) 2 a 2 ab b ( a b) 2 2
2
1 1 1 a1 b2 c 3 80 a b2 c3 1 x 2 y 2 z 2 2 xy 2 xz 2 yz 1 ( x y z )2 ( x y z )2
1 1 1 3 2 5 a a a 3
1 1 1 b 1 3 4 3 4 b 4 b b b b b
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
d) 1 2 3 4 a 1 2b 2 3c 3 4d 4 2 3 a b c d4
Bab 1 | page 3
5. a)
3b
7
3 2
x y
b)
7. E
3 7 b
4
6
6
8
2
11 57 24
y2
(6 3 ) 6
8
11 57 6
6 3 40.095 729 36 55 55 6
3
x
a2 5 7 2 2b c b 1 1 4 d) a bc d 4 acd 1 3 2 1 a3 e) ab c 2 4 4b c 3m2 n2 k 1 m2 mk3 m3 k 2 f) 3 1 3 2 9m nk 3nn k 3n 1
c)
4 4
2 5 (6 3 )
2 5 7
a b c
x
8.
x 2
9 3 (3 2 ) 2
3
3y
27
x
2
3
8
1 27
9. B xa xb
ab
xb xc
x a ab xab b 2 2
xa
LKS 3
2
bc
xc x a
x b bc xbc c 2 2
c a
x c ac xac a 2 2
ab ab b 2 b 2 bc bc c 2 c 2 ac ac a 2
x0 1
10. D A. Evaluasi Pengertian atau Ingatan 1. A 5 7 (5a 4b 5 )(2a 3b 7 ) 10a 4 3b 10ab 2 2. A
(0, 6) 0 (0,1) 1
1
3
3 2
1 3
1
9
12a
23 3
33 3 2
9
2a 7
9
35 a
12a 5a
3
3
x 2
3
x
x 1
3 3 x
2
x 2 x 3x 9 1 8 3 3 3 12 x x x 2 3 1 13 3 33 3
6. A
3a6 b5 9 2 81a b
1
a m a n an
a a a m a n a m a n a a 1 a m a n n
11. D 1
1 1
( m n )
1 1 1 1 1 1 m n m n 1 mn m n m n mn
x
16 5 4 3 1 12 14 12461 4 5 25 2
a n
12. C
4. B
5. D
(3 )
1 312 4 a 4 a14 3 2 9
2
1 a m a n am
2 2a 7
312 4 a 34 a14
2
a
m
m
9 3 3 3 3 1 9 3 6 2 2 3 3 3. A
3
1 1 1 n m 1 a m n 1 am 1 a n
1 10
( 3) 3 23
1 1 a n m
1
3 81a9 b2 27 a9 6 33 a3 3a 6 5 5 2 3 3a b b b b
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
4
y
4
x 2 y2
(x
2
2
y )( x
2
2
y )
x2 y2
x2 y2
15. D
1 Diketahui : x 3 x 2
1 2 2x 1 2 x 3 x 2 9 x x x 1 2 x 2 9 2 11 x 1 2 Jadi, x 2 11. x
Bab 1 | page 4
18. C
x 12 atau x 3 (i)
(ii)
3 (61 61 61 60 ) 3 (611 61 ) 5 5 3 (1 16 ) 3 6 2 1 1 1 1 3 5 b) 3 3 3 3 5 3 3 3 34 32 1 91 35 243 2 0 6 (16) (16 16 ) 2 (16 160 ) c) 6 2 5 2 5 ( 16)
3. a)
3 2 x 5 2 x 2 3 5 x 2 x 2 5 x 3 0 x (2 x 1)( x 3) 0
1 2 2
9 x 12 4 x 2 2 4 12 x
9
2
1 36 35 9 x 3 4x2 4(3)2 2 x (3)2 9
26 (16 1) 26 15 3 3 8 6 2 8 4 2 2 5 (1) 24 5
36 1 35 Jadi, nilai yang memenuhi adalah 35. d) B. Evaluasi Pemahaman dan Penguasaan Materi 1. a) b) c) d) e)
1
4. a)
6 4
64
b)
23
2 4 3
4 2
32 4 3 35
33
d)
2.
a)
d) e)
y 6 y 2 y 6 2 y 4
1 y4
1 m 10 m6 m10 6 m4 4 m x12 x3 x123 x15 6
23 2 4 25
2
2 22 26
345 1 26
2
1
(9 32 ) 3
(32 32 )3 (34 )3 2 (3 27) (3 33 ) 2 (34 ) 2
e)
1 1 b (2 ab3 )2 3 2 2 2 6 (2 ab ) 2 a b 4a2
f) 2 4 3
(3 x y )
312 3128 3 4 8 3 f)
b9 b4 b9 4 b5
2
c)
32 34
2 1 2 ( 3) ( 3) ( 3) 2 1 ( 3) 9 ( 3) (9 9) 12 18 216
1 1 x6 2 4 3 3 6 12 12 (3 x y ) 3 x y 27 y 2
(24 22 )2 (26 )2 (4 8)3 (22 23 )3 (25 )3
g)
2 x1 y
212 21215 2 3 215
h)
4m 3 2 2 2 6 2 6 3 (4 mn ) 4 m n 16 m n n
i)
6a4 c 3
8 8 62 a8 a a 6 2 6 6 c 6 c 36c
5x y 4
3
j)
(16 4)2
7 3 34
3 42 5 2 36 52 7 53 78
32 75 52 72 32 3
3 7 7 5
4
1 1 52 432 7 3 7 3 73
82 52 34 252 24 9 2 6
(23 ) 2 52 34 (52 )2 24 (32 ) 2
4
2 5 3 1 1 54 24 34 2 46 3 4 4 5 42 210 52 f) 2
(2 2 ) 4 (2 3 ) 2 2 2
5 3
(2 ) (2 )
2 8 26 1 1 4 15 25 4 15 8 6 2 2 2 2
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
2 2 2 4 2 2 2 2 xy x y x y 2
2
5. a)
2
53 x 3 1 1 12 3 3 12 3 12 y 5 x y 125 x y
(2 a b)0 1
b)
(14 a 8b) 0 1
c)
m n 11 2
d)
7 x0 8 y0 7(1) 8(1) 15
e)
0 1 3a (3a) 2 2 0 1 1 2b (2b )
f)
17x 3 54 y 2
0
0
0
0
(17x 3 )0 1 1 (54 y 2 )0 1 Bab 1 | page 5
4 3 3
6. a) b)
34 35
7 8
x y z y z 64 6 4 5 x y z x
y z 10 x
e)
( ab2 ) c3
x 4 y 3 z 0
f)
g)
3 4
m n p 0
7. a)
m 2n 3
n 3 6 n 3 42 6 m m
m 4n 6 a3b 4 c 3a
6
p p 0 3 3 m n2 m n2
3
2
c 3 a 3 4 a b
a 4 b4 c3 3x 2 y 4 b) 9 x3 y2
2
a13 4 3 b c
2
3
y3 5 4 4 x z y8 x 4 z 4
a 4b 2c d) ab 6 c3
8. a)
a
1
2
y
x
c)
x y 2
3
(2 a
1
2
b)
j)
2
2 3
( a b )
1 1 2 (2a b )
1
2a b
1 2
2
a2 (2 ab)2
1 1 2 3 (a b ) a 12
3
b
2
2
32 1 x 32 4 2 y
6
1
2
ab 1 2 b
3
b 2 ( ab 1)3
C. Evaluasi Kemampuan Analisis
3
1. a)
1 1 (3 x1 2 y 2 ) 1 1 2 3 2 3x 2 y x y2
3
y24 x12 z12
4
b)
4
2
1 3 y 2 x xy2 2
2
xy 3 y 2 2 x
( x 6)2 ( x 1)2 2( x 6)( x 1) 1
( x 6) 2( x 6) 2 x 1 ( x 1) 2
b32 20 8 a c
1 x y 2 2 2 x y x y 1
i)
1 1 (a b ) c c c 2 2 (a b ) (a b )
2
4
2
2
( a b)
2 aab
1 1 ab2 a 2 0 2 b c b 1 a a
b)
3
2
h)
9 x10 4 y
b 26 1 4 31 a c b8 a 5c 2
2 3
z
2
32 x 3 y 2 2 4 3 x y
x 4 y3 z0 4 5 z y
2 3
1 1 1 2 2 n p 2 p n m 2 2 2 n m m np m2 n 1 0 2 2 2 ( a b) c 0 c 2 (a b ) m
2
a8 8 6 b c
3x 5 y 2 c)
x z y z 1
2 2
33
3
1 2 2 x y 3 z
1 1 (1 b) c 2 c 2 2 (1 b ) (1 b )
x 43 z 04 x7 z 4 3 3 2 y 2y
2x z m p
f)
3
( a) a3 c3 ( 1) a1 a3 c3 b2 b2
a3
3 2
e)
2
a0 b3 (a) 2 a0 b3 a 2 5 ( a )2 b 5 b5 b
a 4c3 2 b d)
2
( a0 b1 ) 3
a 0 2 a 2 5 3 2 b b c)
x y z
2
( x 6) 2( x 6)( x 1) 2 ( x 1) 2
( x 6) ( x 6) 2( x 1) ( x 6)(x 8) 2 2 ( x 1) ( x 1)
2
1 x2 y3 1 2 x 3 y3 y
d)
1 m 4n 3 1n 3 m 4 n 3 p 0 m 4 1 n3 n3 Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 6
c)
(4 x 5)( x 2) 2 3(4 x 5)( x 2) 1 4 x 5 3(4 x 5) ( x 2) 2 x 2
1 a b 1b a1 b1 a b a ab 2 2 2 1 1 a b a b ab a b b a a b2 ab
d)
a b 1 ( a b)( a b) a b
4 x 5 3(4 x 5)( x 2) ( x 2)2
(4 x 5) 1 3( x 2) (4 x 5)(3x 7) 2 2 ( x 2) ( x 2)
4.
H
Lr ( r 2 L2 )2
4 Lmr 2 2 2 (r L )
1
(4m )
d) 3( x 4)2 (2 x 5)2 ( x 4)(2 x 5)1
3( x 4)2 x 4 2 2 x 5 (2 x 5)
5.
F
3( x 4)2 ( x 4)(2 x 5) 2 (2 x 5)
1 1 f2
f
1
1
1 1 f1 d ( f1 f2 )
f2
f1 1
f df
1 2
f1 f2 f1 f2 d
1 d 1 f 2 f1 f2
( x 4) 3( x 4) (2 x 5) 2 (2 x 5) ( x 4)(5 x 17) 2 (2 x 5) e)
2( x 1) 2 (3x 1) 2 ( x 1)(3x 1) 1
LKS 4
2( x 1) 2 x 1 2( x 1) 2 ( x 1)(3x 1) (3x 1) 2 3 x 1 (3x 1) 2 ( x 1) 2( x 1) (3 x 1) ( x 1)( x 3) (3x 1) 2 (3x 1) 2
2.
3
3
1 3 a 3 3 1 a b 3 a
a
b
b 3 a 3 3 3 a3 b 3 b a a 3b 3
13
b 13 b
3
3 3
3. a)
1
2
2
1
x y x
3
y
b)
m
2
m
3
1 x
12
1 x2
y 1y
x y xy2
m
4
m 2 m 1 m 0
2
xa xb 5. A
xy 2
7
2
2
1
y3
5
x2 6. B 1
13 14
1 m2
12
6
6
1
bc xc a x
4 3
4
5 2 4 16 34 81
x
7
7
1
m1 1
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
1
1
3
1
y z x x y z 1
1
1
x2 y 2 z 2 1 1 1 1 x2 y 2 z 2
b a 1 (b a )(b a ) b a
1
x y y z z x
m
m m 1 2 m 1 1 m4 42 2 4 m 2 m 1 m m m m2 1 1 b a a 1 b 1 b a a b ab 2 2 1 1 b a b a a b ab b2 a2 a b ab
1
ca xb x c x a 1 1 1 1 xaxbxc
7 5 3 1 x2 y2 5 1 x 2 2 y 2 2 xy y x2 y2
2
c)
1
2 2 2 4
ab xb c x
x2
x( x y ) 2 x y y( x2 y)
1 3
4. B
x2 y x2 y
m
2
x
x y
1 m2
3 2
x x 3 x 2 x
3 12
x y 2
1 3
4
1 a 3 1 ab b b 3 b3 a3 b3 1 a 3 1 a b b 3
8 (2 )
243 5 32 5 25 32 243 5 3 3. B
Dengan membagi pembilang dan 3 penyebut di atas dengan b diperoleh:
b3 a3
1. A
2. B
b a b 3 a 3 3
A. Evaluasi Pengertian atau Ingatan
8. A 2
2
3 3 8x 2 y 4 23 x 2 1 23 x3 3 24 3 6 125x 1 y 2 5 y 5 y
2
3
22 x2 2 4 2 4 0,16 x y 5 y Bab 1 | page 7
9. E
2 3
3
8 x y
3 26 x 3 y 6 26 x 3 y 6
6
4xy 1 3
(3 )22 3 2 3 2 2 (2 )2 2 1 2
1 3
2
1 2
2
2
2
1 2
2
2
9
e)
814 (3 ) 4 3 27
f)
25 2 (5 ) 2 5 125
g)
32
h)
10 10 10
i)
1 3 4 4 64
2
10. C
1 3 1 2
4
1. a)
3
d)
5 1 8
2. a)
3
2
1
2 (2 ) 2 2 4
5
4 1
3 54
2
3
k)
l)
2 2 2 4 9 2 3
2
2
3 9 2
1
1
1 2
3 3 3 3 27 3
1 12
2
3
2
2
1
2
12
1
27 3 (22 )
12
3
(32 ) 2 (24 )4 3 3 3 3 2 5 2 32 2 2 (23 ) 3
3
1
2
83 h)
34
3
1
25 2 8
23
13
81
1
(52 ) 2 (23 ) 1 (34 ) 3
4 5
1
3
2 3
5 5
4 1 1 5
243 3 3 3 3
3 3 1 9 6 243 5 3 3 6
23
3 23
i)
27
4 5
3
2
13
2
1 3 9 (26 ) 3 (34 ) 4 16 1 3 13 16 3 (3 ) 1
3
1 2 33 22 3 33 32 9 1 2 2 232 32 (26 ) 2
1
(32 ) 2 (8 3 )2 1 2
12
32 (43 ) 9
32
3
1
2
5 1 44 3 3
j)
65
4 2 (2 2 ) 2 2 3
1
814
64
1
3
23
3 19 3 3 3 57 3 4 4
8 3
92 32 32 32 2 20 1
81 13 5
3
1
812 (34 ) 3
1
3
1 1 7 (33 ) 3 3 2 2 3 2 1 1 3 34 32 4 4 3 3 16 8 (2 ) (2 ) 8 4 8 1 1 1 1 7 49 2 36 2 (7 2 ) 2 (62 ) 2 6
4
92 164
12
3 5
2 3
1 3 27
2
g)
3
b) 16 4 (2 4 ) 4 2 3 8 c)
)
2 2 3
3
1
f)
96
3. a)
12 4
9 2 125 3 (32 ) 2 (53 ) 3 3 25 28
e)
0
5
4
1 12
125 3 814 (53 ) 3 (34 ) 4 25 27 2
74
1 3
1 9
2 3
1 2 2 3 9 3
4
1 3
d)
1
4 1 27
h)
10
3 3 (3
c)
3 33
e)
j)
2
3
d)
3
3 2
1 1 3 8 2
b) 16 4 27 3 (24 ) 4 (33 ) 3 8 9 1
3 2
32 2 2 4 2
5
g)
3
2
j)
4. a)
1 32 9
c)
5 35
(2 )
3
1 4
14 8
1 3
i)
53
1
2 2 2 0 1 2
3
3
2
112
b)
f)
1
35
3
1
(2 ) 5 2
g) 16 h)
12
2
3
f)
1
2
3
4 (2 ) 3 2 3
c)
e)
1
3
3
1
(2 ) 2 2
1 2
b)
5
32 (2 ) 2
4
3
B. Evaluasi Pemahaman dan Penguasaan Materi 5 3
3
2 3
1
32 2 2 1 2 422
1 3
1 1 1 (32 ) 2 3
12
d)
5.
2
4
16 2 9 2 4 3 2 2342 3 2 43 1 13 41 B 1 1 1 3 1 2 16 4 9 2 4 6 2 3 23 22 33 108
27 3 (33 ) 3 32 9
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 8
2
6. a) 142 3 27, 22
h)
3
b) 67 5 12, 46 c)
8
1
e) f) g) h)
7. a)
1
2 3
2
25
2 3
1
154
x
23
1 1 1 3 2 3 2 17 4 3 4 3 x x x x 12
3 3 5
y
h)
3
1
3
1
3
4
3
4
3
1
y8y
13
m 21
34
y 8
1
43
y
16
85
i)
1
5
x b
34
1
43
x
35 34
23
1 2
12
x
3 4
j)
1 2
m 3 n2
y
3 4
8. a)
1
1
a b ( a b) 2 2
a
b 2 b a 1 a
1 2
2
b c 2 bc a2 a
a2 1 3 b b3 1 1 3 4 3 b b4 3
x2 x4 a7 a a
1
x 2
1 8
2
b)
3 20
13
43 81
9
x 8
2 1 1 3 2
a 7
12
19
a 42
3
2
1
1
1
1 1 7 2 2 b3b b3 b3 1 1 a 4b 4 1 4 4 a b
c)
1 a 3 a3
1
d) a 2 2 1 3
5 3 5 1 2
a b a b 1 3 4 3 g) a 4 3 a 4 a
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
a 3 a 7 a
1
1 b) ( a b) a b 1 2 1 c) a b 2 a b
f)
2
x 2 y x 3 y 3
x
1
x2
1
e)
3
1
10. a)
b3
j)
1 1 21 2 2 2
1
3
23 2
35
g)
41
m n m n x y x y 1 2
f)
h)
1
34
1 3 1 1 3 3 3 3
d)
14
a 7 a 5 a 7 5 a 35
x i)
c)
y 5 y 3 y 5 3 y 15
13
y8
g)
4 5
e)
5
3 3 4
b)
4 15
3
f)
1
3 32
1 8
d) m 7 m 3 m 7 e)
1
116 1, 49
81
3 b3 b 1 1 3 2 b3 1 3 32 1 1 4 1 a b 4 1 4 4 4 a 4b a b
9. a)
7 7 7 0, 78 18 18 181 0, 46
y c)
12
16
3 4
x4 b)
j)
2 3 0,52 34
113
1 2
i)
1 4
34
1
11
3
14 3 50, 69
24 3 11
1
3 11 59 2,17 10 8 4 5 8 20
3
1,7
a b
3
a2 a a 1 5 b b5
15
1
115
84 d) 6
3 2
x 2 y4 1
d)
a
13
1
3
x 3 y 2
f)
11. a)
1
1 47 a 21
3
1
7
y4
1 2 5 x 4 3 x 12
x 4 y 4
1
m n m 2
1
1
y 2 4
2
1
47 21
a
x 2 y4 x 2 8 x8 3 3 3 1 5 x8 y 8 y8 4 y8
1
( xy ) 8 e)
3 73 31
1 3
2
4
1
m 3 n1
1
2
n3
m 3 n 1
2
2
n3
1
5
n3
1 13 4 m3 m3 1 1
( a b c )( 6 a b c ) a 2 6b 2
2 3
a b c
2 12 2 1
c
5 2
Bab 1 | page 9
2 2
b)
d e
1
2 2
f
2
d3e f 5
b)
d e f2 3 2 5 d e f
c)
x
3
y3 z2 12
1
x 2 d) e)
1
1
x 2
y3 3z3
32
y3 z3
13
3
3
a 2 b( a 2 )
ab 2 ab 3
a2
1 1 1 a 2 b(a 2 b b2 ) a 2 b b2
( x 2 y )( x 3 y 2 ) x5 y 3
1
1
1
2x
0
x 1 2( 1)2 ( 1)1 2( 1)0 2 1 2 5 x 1 2(1)2 (1)1 2(1)0 2 12 3 2 1 0 x 2 2(2) (2) 2(2) 1 1 2 2 2 2
xy 3 z 6 1
a 2b
x
1 1 2 3 2 2
1 9 df 2
512
2
2
x 2 2( 2) 2 ( 2) 1 2(2) 0
2 2
e 3 2 d f
2x
2
x2 y 2 x 2 y 3 5 3 x2y2
14. a) 1
x
2 12 52
y
1 2 3 2 3 2
0 13
x y
2 2 3 3 3 (2 2 ) (27 4 2) 3 353 35
1
1
1
1
y3 f)
a 2 b 2 c2 b23 c2 9 6 3 9 6 2 (a b c ) a b c a ( abc) 2
2 1 3 3
b5 c11 4 a 5
g)
3
x y z x3
16
yz
1 5
1 2
x y z 1 2
1 3
1
1
x y z
1 3
16
y
1 2
13
x
z
b)
(32 42 122 ) 169 13
c)
13 23 33 43 53 63 441 21
15. a)
1 1 3 6
1 1 2 5
1
b)
3 125 x 4 y 3 5 x 3 y 5 x 3 3 5 x2 2 27 x 2 y6 3 x 3 y 2 3 y2 1 3 y
c)
40 a 4 b3 c 16a 6 bc5
y6z 2 3
x 10 h)
(a
3
bc
12
)(ab)
a3 bc
1 3
3
1 2
21
1 3
ab c a b 3 1 a 2b 2 c
a
313 23
c
b
1 3
1 1 1 2 3 2
112
11 6
a b 3 c2
12. a)
1 (2 ) 2 12 2 (2 ) (terbukti) 1 1 3 13 (2 ) 1 (23 ) 3 2
42
b) 8
31
2 12
1
3
x 1 22 2 3 1 2
x 2 23 ( 2) 2 3 8 2 3 3
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
2
1
2 12 52
2
2 c
4c 5 3 2 12 32 ab a b 1
1
1
( x 2 y 2 )
2
1 1 ( x 2 y 2 )2
3
e)
(x 4 z
f)
(m
12
12
1 1
2
1
y ) 3 x 2 z 1
n 2 )( m
1
x y 2 x 2 y 2
12
2x 1 ( 2)x 3
1 1 x 2 2 1 ( 2)2 3 3 3 2 2 1 0 1 x 1 2 ( 2) 3 1 3 2 1 2 2
4
2 a 2 b 2 c 2 22 a 3b 12 c 52
1
(terbukti) 13. a)
4
1 3
d) 1
1 3y 2 (9 x 4 y 2 ) 2 3 x y 2 x
31
1
1
1 x2y 3 y3 1 z3
1 2
1
1 2
n 2 ) ( m 2 ) ( n 2 ) 1
m n
1
1 m n
C. Evaluasi Kemampuan Analisis 1. a)
(2 x 5)( x 1)
14
3
( x 1) 4
3 2 x 5 ( x 1) 4 1 ( x 1) 4 2 x 5 x 1 3 x 4 1 1 ( x 1) 4 ( x 1) 4
Bab 1 | page 10
b) ( x 7)(3x 5)
12
5.
1
(3x 5) 2
2mL H 2 2 3 ( r L ) 2
1 x 7 (3x 5) 2 1 (3 x 5) 2
1
(r 2 L2 ) 2 1 1 ( r2 L2 )( r2 L2 ) 2 ( r2 L2 ) 2 2mL
1
x 7 (9 x2 25) 2 (3x 5)
2
3
6.
23
6( m 4)
4m 1
6( m 4) 2 ( m 4) 3
3
36V 3 36 43 r 2
3 2
16 3r 6 3 643r 6 3 36 9
3 y 7 8( y 2) 5 y 9 3 3 ( y 2)5 ( y 2)5 d) (4 m 1)( m 4)
2
2
(3 y 7)( y 2) 5 8( y 2) 5 2 3 y 7 8( y 2) 5 3 ( y 2) 5
c)
1
2mL (r L ) 2 2 2 2 ( r L )
1 2
3 26 3r 6 4r 2
1 3
Jadi, A 36V 3
2
.
1 3
4m 1 6(m 4) 10m 23 2 2 (m 4) 3 (m 4) 3
a b c 27 a 3 2
4 3
2. a)
2
13
2 3
21
bc
LKS 5
65
2 2 32 a3 2 b 3 1c 26 1
4
2
5
A. Evaluasi Pengertian atau Ingatan
3
5 2
1 3
7 6
2 4 a b c 2 4 a 5 3 b 6 c 7
a b c 27 a 3 2
4 3
b)
13
108a
3 1 2
b
1 2
2 3
12
31 2
135
c
bc
65
3 2
4 3
3
2
13
144a
2 3
5 3
108a b c
5 3
a b c 27 a
1 2
108a b 108 a b 2 3 2 3 c c c)
1. D 12321 111 111 1112 ; 2025 45 45 452 ;
2
21
bc
32
2. B
5
4375
65
b
3.
1
10
15 1 2
1
4. C
176 2.401 176 49 225 15
7
33 5 5 3 3 55 4 5 6 1 3
5. B 2 35
1 4
2 3 (2 ) (2 ) 2 3 4 3 7 15 (2 5) 55 33 5 5 1
22
12 65 1 51
5 1
4.
2
1 1 34
33
15 35 75
2 30 1 10 5 1
25 y 5 z 5 x (5 2 ) y 5 z x 1
4375 625 25 7
10 250 2500 50
c
4
22 33 44
3. D
2 5 31 3 2 2 3 4
1
7
2
144a 2b 3 c 3 144a 2 3 b 4c 1
3025 5555 552 ; 6561 81 81 812
248 52 144 248 8 256 16 6. D
18225 182, 25 1,8225 0,018225 135 13,5 1,35 0,135 149,985
1
2 1 2 1 z z y x y x 2x y 1 xy z xy z 2x y
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
7. B 2
1, 219 12110 9 3 1,1 1,1 121102
Bab 1 | page 11
8. A
0, 004 0, 4 a 0, 4 0, 004 b 4 102 a 16 104 b a
4 102
b
a 0,0016 b
9. D 1, 21 0, 01 1,1 0,1 1 10. A 12 4 3 4 3 2 1, 732 3, 464
B. Evaluasi Pemahaman dan Penguasaan Materi
324 22 34 29 18
1. a) b)
2601 32 17 2 317 51
c)
1681 412 41
d)
1849 432 43
e)
2809 532 53
f)
3844 22 312 2 31 62
0,3136 3136 10 4
2. a)
11. D
0,814 0,9 2 9 0, 0064 6, 25 0, 08 2,5
2
2 7 10 6 2
b)
10,89 1089 10
1
3 11 10 2
12. E
0, 204 42 36 6 0, 07 3, 4
110 14
441 4
3, 24 324 102 22 34 10 1 1,8 0,5184 5184 10 4 26 34 10 2 0, 72
21 10,5 2 e)
1,1025 11025 104
14. E
3 5 7 10 2 2 2
3
0,361 36110 1,9 10 190 5 0, 00169 169 10 0,13 10 13
f)
1, 05
116, 64 11664 10 2 3 10 4 6
3. a)
2
2
15. E
0, 01 0, 0064 0, 09 0,3
3,3
c)
d)
13. C
2
0,56
2
1
10,8
1326 18 45 10521
2 3 5 7 13 24750 2
16. E
4,5 0, 0085 18,9 15 0, 0017 13,5 2,1 17. A
b)
x 1 x 1 13 13
18. C 1
3
1 0,577 1, 732
19. E 20. B
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
2
2
2
169 0, 0009 1, 212025 6
3 5 10 11 13 19,305 6
4. a)
27 x 14 x 1 1 1 169 13 13 13
6
2
2
2
75, 24 x 0, 71 Kuadratkan kedua ruas diperoleh: 75, 24 x 0,5041
x 74, 7359 1872 b) 234 234 x 1872 x Kuadratkan kedua ruas diperoleh: 18722 2 x 8 64 . 2342 5. a) KPK dari 15, 18, dan 25 adalah 450. Jadi, bilangan kuadrat terkecil yang habis dibagi ketiga bilangan di atas adalah 900. b) Bilangan terkecil yang harus dikurangkan dari 549162 agar dapat menjadi bilangan kuadrat adalah 81, karena 549162 – 81 = 549081. 2 549162 741 Bab 1 | page 12
c) Misal bil.yang dicari = a .
352a 2 11a 4 22a Agar menjadi bilangan kuadrat maka nilai x 22 . 5
B. Evaluasi Pemahaman dan Penguasaan Materi 1. a.
36a 2 6a
6. a)
1 1 7 7 7
5 5 5 30 3 c. 20 3 5 20 2 b.
b)
( a 3) a 3
c)
4 4a a 2 (2 a ) 2 2 a
d)
9a2 42a 49 (3a 7)2 3a 7
2
6 1 1 48 3 2 8 8 8 2 e. 14 7 7 6 2 f. 2 6 3 d.
LKS 6 A. Evaluasi Pengertian atau Ingatan
2. a.
1. D
8
3
243 35 32 3 9 3 1 1 c. 320 26 5 4 5 2 2 b.
4,899 1, 633 3 3 3 3
24
2. E
3 2 2 3 3. A
3
3
3 6 1 2 6 12 2, 448 1, 224 6
d. e.
4
4 3
2
32x 5 y 5 3 25 x 5 y 5 2xy 3 4x 2 y 2
5. A 3
6m3 4
1
1
1
1
(6m3 ) 3 2 3 6 3 2 3 m m 3 1 2 12 1 2 (22 )3 23
6. D 3
3 2
212 212 212 4
2 3 32 2 23 6 12 3 3
3
y 6
3
a.
98m4 2.72 m4 7m2 2
b.
104a5 b 23.13a5 b 2a2 26ab
c.
150 x 2 y 5 2.3.5 2 x 2 y 5 5 xy 2 6 y
d.
8. E
4 x
3.
6
7. C 3
4 x y 4 xy 3 3 6
162 m3 n5 y 8 2.3 4 m3 n5 y8 9 mn2 y 4 2 mn 2
e.
9. A
363ab 5c12 3.112 ab5c12 11b 2c 6 3ab
1
375k 3 n8 (3 53 k 3 n8 ) 2 5kn4 15k f.
10. B 5
a
24
4 12 4 2 2 3 8 3
f.
4. B 3
980 22 5 72 14 5
5 2 3 52 25 6 1 1 g. 864 2 5 33 4 6 3 3 2 h. 2 2.000 2 2 5 10 2 40 5
80a b 2 5a b 4a b 5b 4 3
99 32 11 3 11
a3 a a 5
2 4
a
43
a 5
73
a 15
7
96 x5 y8 z2 25.3 x5 y8 z2 4 x2 y4 z 6 x
81a6 b12 34 a6 b12 9a3b4
g. h.
1
3
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
3 4 xz 2 3 3 x 2 y 2 z 2 192 x5 y 2 z 8 2 6.3 x5 y 2 z 8
Bab 1 | page 13
3
128m 2 m 4m 3 3 6 2 3 2 m2 6 27 p 3 p 3p
c.
7a a 7a 2 a 14a . 8 2 2 2 4
72a 7 23 .32 a 7 2a 3 3 9b6 32 b6 b
d.
13n3 n 13n 6m n 78mn . 3 54m 18m2 3m 6m 6m
e.
36 y3 6 y y 3x 2 y 3 xy . 27 x 3x 3 3x 3x
5
i.
3
j.
7
5
3
2a
4. 3
a. 3
b.
y y y 2.3
6
f.
4 a 4 a 16 a 3.2
3
c.
2
108a 5b 3 2 2.3 3 a 5b 3 6a 2b 3ab 2 5 3 5 27c 3c 3c 3c
6
4 x 2.3 4 x 6 4 x
2
4
d.
2a b ab c 2 . c c c
6m 4.2 6m 8 6m
2a2 b abc c3
5. a.
4
6 m3 4 6 m3 4 6 m3 2 4 24 m3 . 4 4 2 2 2 22 64 a 6 2 6 a6 8 a3 3b 8 a3 3b . 27b 3 33 b3 3b 3b 3b 3b 2
b.
4 6 3 3x 2 y 2 3 18 x 2 16 y 6 3 2 y 2 y2 3 2 . 2 3 4 4 2 81x 4 3x 3 3x 3 3 x 3 x 3x 2
c.
3
3
25 4 y2 20 y 25 2 y 5 2 y 5 y2 5 y 4 4 4 2 2
b. 2 3 x 4 3 3 x 4 3 16 9x 24x 16 3 x 2 8 x . 3 3 3 3 3 2
c. 4 m 1 2 4m 1 2 1 16 m2 8 m 1 8 m2 4 m . 2 2 2 2 2
3 4 y x 3 16 y 2 x3 x . 2 4y 3 4 y 3 4y 4 y
2
2
d.
7. a.
d.
4 25a 2 20 a 4 5 a 2 5 5 a 2 5 5a 2 4a . 5 5 5 5 5 2
e. 3 3 n 32 m5 3 25 m5 2 m 3 22 m2 . 4 3 2 4 3 2 2 9n 2 3 n n 3 n 3 3 n 2
3
3
2
4
2
8.
2
2
5 2 cm b. 5 3 cm a. 3
2
2m 2 .3 m n 2m 12m n 2 3n 2 n 3 n 4
f.
2
81a 2 4 34 a 2 3 a 2b 3 2ab . 4 2 2 4b2 2b 2b 2b 2 b
g.
32 y8 4 2 5 y 8 2 y 2 4 2 3 x 2 y 2 4 18 x2 . 4 2 2 9 x2 3x 3x 3x 3 x
4
h. 14m 4
8 n2
4
. 4
8n 14m 2 n 8n 8n 2
2
3
4
9
6
2
3
56mn 4 2n 2 7 m 4 2n 2 8n 2 n
c.
Luas =
alas tinggi 5 2 5 25 2 cm2 2 2 2
9.
6 2 . 3 2 3 1, 414214, 243 2 2 2 3 2 2 b. . 3 1, 73205 1,155 3 3 3 3 6 2 3 c. . 2 6 2 2, 44949 4,899 3 3 a.
6. a.
46a2 b a 46b 3a a 138ab 1 . 138ab 3a 3a 3 3a 3a b.
40 13a
23 .5 13a
.
13a
2 130a 13a 13a
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 14
d.
2.
12 6
1
2 6
3
3 3
x3 5 x4 3 x2 x 7.5.3.2 x19.3 2 x
7
210 x59 x
6 6 3 3 6 2, 44949 1, 73205 3 5, 476
2 1 3 6 3 . 1,394 3 3 3
e.
210 x59.21 210 x119
3. a. 5 x 1 7.5 x 1 x 1
52
2
7
35 x 1 7
f.
1 5
1
3
1 2
5 3 2 5 3 2
6 5 10 3 15 2 30 1, 732
5 x 1 b.
a
2
6
2.5 2
15 a2 1 6.5 a2 1 2
2
30 a 2 1 12
5 a 2 1
10.
2
a.
c b a 6 3 27 3 3
b.
a b 2 c 2 13 2 6 2 133
4.
c.
b a 2 c 2 22 32 13
d.
b a2 c2 122 82 208 4 13
1 T 1 450 45000 n 2 L m 40 0, 04 40 4
2
2
2
2
e. c b a 20 10 300 10 3 2
2
2
2
f. a b c 40 25 975 5 39 2
2
2
2
C. Evaluasi Kemampuan Analisis 1. a.
5.
2.32.52.10 2 150 2 15 2 2, 65 80 80 8
T 2
L 40 2 2 2 g 980 7
a b c 4 5 6 15 s 2 2 2
L s s a s bs c
15 7 5 3 . . . 2 2 2 2 2
A. Evaluasi Pengertian atau Ingatan
2
3 .5 .7 15 7 24 4 40 25 35 100 b. s 50 m 2 2 L 50.10.25.15
LKS 7
1. D
2 3 4 3 9
3
2. C
3 2 3 a 8
a
3.54.102 250 3 m 433, 01 m 2
2
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 15
3. D
11 2 5 2 14
2. a.
2
3
4. D
3 10 21 5 8
23.5a 3 2 2.33.5a 3 5a 2.53 a
2 a 3 5 6 a 10 a 25 a 10 a
5
2 a 3 5 31a 10 a
5. A
3 1 3 7 7 7 40 4 5 8
b.
6. E
5a
a 22.10a 3 3 10a5 6a 32.10 2 a2 10 a 3 a2 10 a 18 a 10
2 3.3 2 2.2 2 7 2 6 2 4 2 7 2 3 2
2
10a 18a 10
a 10 5a a 18
7. E
c.
5.3 3 3 3 10 3 32 3 15 3 10 9 3 13 3
3x 2.32 x 3 2 x 5 x 2 25 x 9 x 2 2 x x 2 2 x 4 x 2 2 x
8. A
7 3 4 3 3 3 7 4 3 3 6 3
14 x 2 2 x
9. E
d.
2 x 2 2 x 5 2 x 1 2 5 2 x 2 2 x
42 a 3b 4 62 a 5b 2 4 ab2 a 6 a2 b a
10. A 7a 5 3a.5 5 a 5 7a 15a a 5 23a 5
2 ab a 2 b 3 a e.
24.3 m4 n7 26.3 m4 n7
B. Evaluasi Pemahaman dan Penguasaan Materi
4 m2 n3 3 n 8 m2 n3 3 n 12m2 n3 3n
1. a.
5 2 2 2 3 2 6 2
3. b.
3 3 5 3 4 3 2 3
a.
3
24 3 2.33 2 3 2 3 3 2 3 2
c.
2.4 5 3 5 2.5 5 5
b.
3
34 3 23.34 3 3 3 6 3 3 9 3 3
d.
2 7 3 7 5 7 10 7
c. 7 3 2 2 .3 213 3 4 3 3 17 3 3
e.
3.5 3 4.2 5 4 5 15 3 12 5
d.
3
4
g.
2 6 2.6 6 5.3 6 5 6
3
26 x 3 2 x
2 2 2.3 3 5.3 2 17 2 6 3
e.
6 2 x 4 2.3 x 24 x 2 36 x 2 5
h.
3
5 24 x 4 9 27 x 4 10x 3 2x 36x 3 2x 3
f.
3
9 2.2 6 5 3 9 4 2 3 5 3 9 3 4 2 5
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
2
4
2
60x 2
Bab 1 | page 16
f.
4. A
8m 2 .3m 4m 2.3 m m 3.5 m 3
3
16m 11m
3
3
23
3 12m
23
3 12m
3
3
2 3
2 5m
2 3
2
m 11 3 12 2 2
3
3
3
23
3
3
3
5. B
8 6 2 2 2 2 3 2 4 6 6. B
2 8 2 3 16 2 6 4 2 6
4. a.
7 7 2 7 2 7
2 2 3 a a 3 a a a a
7. D
23 3 2 9 11 2.32 2 2 2 2 a a a a a
8. A
3 a
b.
2
c.
3 2 3 4 1 3 2 6 16 1 5 6 1
9. C
3 3 12 3 3a 12 a 2 2a 2 2 2a a a a a a 3 3 12 a 2 2 a2 a
7 5 7 5 2 7 5
10. B
3 13 2 2
3 1 4 2 3
d.
6x
33 2 2.3 1 x 3x ax x x
18 x 3 2 x 3 3x x x x 18 x 3 x 2 x 3 x 3x x x x 16 x 1 3 x x
11. C
12 2
12 2 2
24 2 14 4 6
12. A
10 8 10 2 2
8.10 8
18 2 24 .5 18 8 5 13. A
7 2 10 a b
LKS 8 A. Evaluasi Pemahaman dan Penguasaaan Materi 1. B
3 3 3 2. D
6 6 6
a b 7 a 5; b 2 ab 10 Penyelesaian : 5 2 14. B
8 2 15 a b a b 8 a 5; b 3 ab 15 Penyelesaian : 5 3
3. B
3 6 3 2 3 3 2 Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 17
15. D
f.
3 5 3
2 7 21 2 2 6
4.5 5 3 2 ab 4 4
2 3 7 2 7 2 3 2 2 3
a b 3 5 1 5 a ; b ab 2 2 4
g. 3 1 2
5 1 Penyelesaian : 2 2
h. 7 5 2 i. 4.3 2 10
B. Evaluasi Pemahaman dan Penguasaan Materi 1. a.
3. a.
6 2 60 10 16 4 15
6 6
b.
b.
25.8 20 80 40 240 20 2 .5 4
21 3.2.31 2 22 6.26 6.33 6
240 80 5
6 2 2.2 6.33 6 108 c.
c.
36.3 84 3 49 157 84 3
21 3.61 2 22 6.63 6 6 2 2.63 6 864
d.
7 4 12 7 8
d.
2
31 4.61 3 33 12.6 4 12 12 33.6 4
84 192
199 8 22 .21 e.
199 16 21
7 1 2.51 3 7 3 6.5 2 6 6 7 3.5 2 e.
f.
45 42 30 294 339 42 30
7 1 5.101 2 7 2 10.10 5 10 10 7 2.10 5
f. 2. a.
16 35 4 42 4 30 6
21 5 3
b.
42 2 60 42 4 15 3
14 4 5
4. a.
x 2 x 2 y 2 y 2
b. c.
2 2 .3 8 30 8 3 8 30 8 3 1 10 4
t 2 t 2 11
c. d.
40 3a 8 6 a 5 18 a 12
21 14 15
120 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 Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 18
d.
7. a.
8 7m16 70m 6 70mn 12 100mn 2
2
1 4 2 8 2 2 2 2
56m 16m 70 6 70mn 120 mn
16 8 2 2 16 8 2 2
8m 7 2 70 mn 6 70 120
4
2 4 4 2
b.
5. a.
Keliling 4 2 4 2 8 2 2 16 cm
20x 2 6 3 2x 7 3 10x 42
3
3 10x 3 2x 6 3 2x 7 3 2x 3 5 42 3 2 x
10 x 6 7 5 42 3
c.
8 2 2 4 Luas
3
32 2 8 2 2 2 4
16 x 2 7 5 8 x 4 5 2 x 28
5 2 x
8 x 7 5
4 4 28
5
16 4 2 4 4 23 cm
c.
8. a.
6 4 40 12 4 96 14 4 30 28 4 72
2 2 3 20 6 48 7 15 14 36 4
4
4
4
4
15 2 5 2 2 5 2 5 5 2 5 2 5
30 15 5 24 55 4 5 30 15 5 2 5
d.
54 3 160 9 3 40.32 6 3 48.4 3 48.32 54 3 2 3.20 9 3 2 6.20 6 3 2 6.3 3 2 9.3 1083 20 36 3 20 123 3 83 3 144 3 20 20 3 3
27 15 5
b.
1002 5 2 52 5 1004 5 2 5 100 4 4 5 5 3
4.25 2 5 2 5
2
6. a.
2
F x 2 x 3 x 2
3 2 5 2 4 4 5 5 6 3 5
F 2 5 2 2 5
2
8 8 5 10 6 3 5
400 5 900 c. 3
2
52 2 5 2 5 3
3
5 2 4 5 5 2
12 5 5
2
9. a.
b.
F x 2 x 3 x
4 2 3 a b
2
6 2 3 2 6 2 3 3 6 2 3 2
F
2
2
b. 5
4
2 6 4 2.32 12 3 6 6 3 12 24 2 24 3 6 6 3
a b 4 a 3; b 1 ab 3 Penyelesaian : 3 1
36 2 24 3 3 6 3
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 19
b.
7 2 12 a b a b 7 a 4;b 3 ab 12 Penyelesaian : 4 3 2 3 c.
10.a. 2 3 5 2 3
5 2 3 5 2 3 5 2 3 5 2 2 6 35 2
2 6
b.
5 24 5 2 6 a b
5 2 6 a b a b 5 a 3; b 2 ab 6
a b 5 a 3; b 2 ab 6
Penyelesaian : 5 2 6 3 2
Penyelesaian : 3 2
7 2 12 a b a b 7 a 4; b 3 ab 12
d.
15 10 2 15 2 25.2 a b
Penyelesaian : 7 2 12 4 3 2 3
a b 15 a 10; b 5 ab 50
6 2 8 a b a b 6 a 4; b 2 ab 8
Penyelesaian : 10 5
Penyelesaian : 6 2 8 4 2 2 2
e.
5 2 6 7 2 12 6 2 8
21 21 5 21 5 4. 5 2 a b 4 4 a b 5 7 3 21 a ; b ab 2 2 4 Penyelesaian :
7 3 2 2
.
3 2 2 3 2 2 2 3 2 2
C. Evaluasi Kemampuan Analisis
1. a.
x y z x y 2 z x y z 2
f. 9 4 4 2 3
2
4 2 3 a b
x y 2 xy 2 xz 2 yz z
a b 4 a 3; b 1 ab 3
x y z 2
Penyelesaian : 4 2 3 3 1 9 4
3 1 9 4
3 4 13 4 3
13 2 4.3 13 2 12 a b a b 13 a 12; b 1 ab 12 Penyelesaian : 9 4 4 2 3 12 1 2 3 1
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
xy xz yz
b.
5 3 2 5 3 2 2
5.3 5.2 3.2
10 60 40 24
Bab 1 | page 20
2.
c.
30 10 5 30 2 25.5 x y
1
1 361 192
1 1 13 169
19 8 3 19 8 3
x y 30 x 25; y 5 xy 125 Penyelesaian : 25 5 5 5 a b a 5; b 5
Jadi,
5.
a.
a 4
3.
4 b2
2
a b a b a b a b
a 48 a 2 12 b.
b 3 x y b x; y 3 3
x y a....(1) xy 12......(2) Substitusi y 3 ke (2) :
1 3
400 392 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 27 8 4
11
27 2 176
a b
27 2 176 a b
7 3 81 7 3 7 21 27 .3 2 3 2 2 2
Penyelesaian : 27 2 176 4 11
2. A
Jadi, 27 8 11 4 11 4
ii 4 27 8
11
27 2 176
a b
27 2 176 a b
24 2 6 1 1 7 168 2 42 7 7
Penyelesaian : 27 2 176 4 11
4. B
Jadi, 4 27 8 11 4 11
24 54 150 2 6 3 6 5 6 0 4 6 4 6
27 8 11 27 8 11 4
4 11 4 11
5. D
8
b. 5
4 3 5 2 15 2 5 5 5 3. B
a b 27 a 16; b 11 ab 176
4
A. Evaluasi Pemahaman dan Penguasaaan Materi 1. C
a b 27 a 16; b 11 ab 176
1 5
2 3.5 2 3 2 3 2 3 1
5 4 3 1
7 5 15
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
8 2 4 2 7 5 7 5
6. B
14
1
3 20 14 2 .3 20 14 2 20 14 2 20 14 2
35 24
14.35 15.24
14.35 49 7 15.24 36 6
Bab 1 | page 21
7. E 42 35
15 56
14.35 15.24
42.15 35.56
9 28
L K S 10
3 7 3 7 2 7 7 14
A. Evaluasi Pemahaman dan Penguasaaan Materi
8. A
27 2 162 a b a b 27 a 18; b 9 ab 162
1. D
Penyelesaian : 27 2 162 18 9 3 2 3 33 800 33 2 200 a b a b 33 a 25; b 8 ab 200
1 9 8 9 8 . 3 2 2 9 8 9 8 9 8 2. A
7 3 2 21 7 2 21 7 2 . 9 2 7 3 2 3 2
Penyelesaian : 33 800 25 8 5 2 2
3 2 3 1, 4142
27 2 162 33 800
3 2 3 5 2 2
1,5858
3. D
2 8
1 3 2
9. C
3 2
.
3 2 3 2 3 2
3 2 1, 732 1, 414
25 2 126 a b a b 25 a 18; b 7 ab 126
0, 318
Penyelesaian : 25 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 2 4 3 2 2 2 3
.
2 3
8 4 6 4 6 12 2 2 3 4 2 6 2 6 6
Penyelesaian : 19 336 12 7 2 3 7 25 2 126 19 336 2 3 3 2
3
2 7 2 3 7
5. C
4 2 2 1 4 2 4 2 2 . 2 1 2 1 2 1 3 2 2 3 1, 4142 2
2 3 3 2
3 2 2 3 1 2 3 3 2
2, 2426 6. B
2 1 2 1 2 2 2 1 . 3 2 2 2 1 2 1 2 1
10. D
48 2 16.2 4 2 3
3 2 1, 4142 0, 4142
7. E
7 5 2 7 5 2 7 2 35 5 7 2 35 5 12 7 5 2
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 22
13.
8. D 8 7 9 8 6 5 7 6 9 8 87 7 6 65
1 5 5 4
2 3 5 2 3 5 2 3 5 5 3 2 3 5 5 3 2 2 3 10 2 3 5 5 3 2
.
2
9. D
5 13 2 39 25 4.3 5 2 3 5 2 3 13
.
5 2 3
13 5 2 3 13
2 2 6 3 5
2 15 3 6 2 90 3 6 6 10 3 6 2 10 6 . 12 12 4 2 6 6
5 13 2 3 13 13
10 15 6 3 5 15 2 6 10
14. A 8 3 3 5 4 3 4 5 32.3 32 15 12 15 12.5 . 4 3 4 5 4 3 4 5 16.3 16.5
10. D
96 20 15 60 32 36 20 15 32
1 3 1 3 1 3 1 31 3 1 3 1 3 13 1 31 3 2 2 3 2 1 3
9 5 15 8
15.
11. A
2 3 5 2 2 3 10 . 2 3 5 2 3 5 2 3 5 2
2
2 6 10 6 . 2 6 6 2 6 6 2 15 12 3 6 15 6
6
2 3 7
.
5 3 2 5 3 2
5 5 5 3 5 2 50 30 20 15.5 15.3 30 30 18 12 2 5 3 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
1. a.
2
2 6 3 2 42 4 4 3 3 7 2 6 3 2 42 3 . 4 3 3 2 18 3 6 42.3 12
.
B. Evaluasi Pemahaman dan Penguasaan Materi
2 3 7 2 6 18 42 2 3 7 2 37
6 2 3 6 3 14 12 1 2 2 6 14 4
53 2
2 6 10 2 2 6 3 5
12. B
5 10 15 6
1 5 3 53 1 . 5 3 5 3 2 53 53 1 2, 23607 1, 73205 0, 25201 2
b.
3 2 1 6 3 . 6 3 2 1 2 1 2 1 2, 44949 1, 73205 0, 71744
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 23
c.
b.
6 4 2 6 4 2 9 8 3 2 2 3 2 2 6 4 1, 41421 2
.
3 2 2
2 1 2 1 2 2 2 1 . 3 2 2 2 1 2 1 2 1
0,34316
c.
1 3 2 2 3 2 2 . 3 2 2 9 8 3 2 2 3 2 2
d.
5 2 3 2 10 3 10 10 3 10 . 12 4 8 2 3 2 2 3 2 5 3 1 4 5 1, 73205 13, 4150625 4
d.
4 5 1 4 5 4 . 5 1 5 1 5 1 5 1 e.
6 7 3 13 28 13 2 7 13 2 7 13 3
e.
2 2 1 2 2 1 8 2 2 2 2 1 9 4 2 . 8 1 7 2 2 1 2 2 1 9 4 1, 41421 7 2, 0938
.
2 7 13
2 7 13 5 f.
4
3 1. 3
2 4 12 6 12 2 16 3 16 18 16 3 2 4 3 2 4 6 6 6 2 8 3 8
f.
3 1 2 3
2 3
2 3 3 2 3 3 3 1 2 3 2 3 3 3 3 5
2 1, 732053 3 1, 73205 5 6, 4641 10,19615 0, 6339
g.
2 4 7 2 7 1 4 7 56 2 4 7 . 2 28 1 2 7 1 2 7 1 h.
3 1 2 3
2 3
3 2 3 3 2 3
3 1
2. a.
3 2 6 3 12 2 3 1 3 3 2 3 2 6 3 12 2 3 b.
2 3 2 3 3 6 3 2 3 2 . 3 9 3 3 3 3 3 3 3. a.
1 2 3 2 3 . 2 3 4 3 2 3 2 3
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
3 3 1 3 3 3 . 3 1 3 1 2
i.
3 2 6 3 3 12 3 6 . 12 6 6 3 6 3 6 3
6 32 18 12 . 18 12 3 2 3 2 3 2 3 2 6 12 6 18 12 63 3 2
2 12 6 18 4 3 6 3 2
Bab 1 | page 24
j.
b.
3 2 6 3 3 12 3 6 . 12 6 6 3 6 3 6 3
4 3 6 2 4 18 4 6 . 18 6 6 2 6 2 6 2
6 3 2 18 12 . 18 12 3 2 3 2 3 2
3 2 4 3 6 6 3 6 2 3 2
3 2 2 a b a b 3 a 2; b 1 ab 2 Penyelesaian : 3 2 2 2 1
2 1 2 2 1 4 2 2 2 1 . 8 1 2 2 1 2 2 1 5 3 2 7
12 6 18 6 18 12 0
4. a. 47 6 10 47 2 90 a b a b 47 a 45; b 2 ab 90 Penyelesaian : 47 6 10 45 2 10 10 34 5 3 2 10 50 34 5 30 20 102 2 . 5 18 5 3 2 5 3 2 50 2 34 5 60 5 102 2 13 52 2 26 5 4 2 2 5 13 25 5 . 5 5 5 5 10 10 34 1 47 6 10 25 5 5 3 2
45 2 4 2 2 5 5 5 3 5 5 2 3 5 5 2
18 8 2 18 2 32 a b a b 18 a 16; b 2 ab 32 Penyelesaian : 18 8 2 4 2 1 1 1 2 2 1 4 2 2 2 2 . 4 2 4 2 4 2 1 4 2 4 2 2 2 16 2 1 3 2 2 1 2 3 2 2 14 28 1 1 2 2 1 2 2 2 3 2 2 18 8 2
5 3 2 1 3 2 2 7 28 20 12 2 3 2 2 18 15 2 28 28 5. a. 3
2
3 1 3 1 3 1 4 2 2 2 8 2 7 3 2 1 1 3 1 3 1 4 3 1 7 2 2 2 1 1 6 3 10 2 3 4 4 3 4 7 2 2
3 3 5 3 2 4 3 3 10
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 25
b.
5. 2
2
3
3
3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2
7 5 7 5 7 5 7 5
3 2 6 2 3 2 3 2 6 2 3 2 6 2 3 2 3 2 6 2
7 5 7 5 7 2 35 5 . 6 35 7 5 7 5 7 5
5 2 6 5 2 6 1 5 2 6 5 2 6
7 5 7 5 7 2 35 5 . 6 35 7 5 7 5 7 5
5 2 6
3
25 24 5 2 6
2
25 24
3
3 7 5 7 5 6 35 6 35 7 5 7 5
2
3
216 108 35 630 35 3 216 108 35 630 35 3 1692
25 20 6 24 1 25 20 6 24 99
Soal Khusus untuk Penggemar Matematika C. Evaluasi Kemampuan Analisis
1. 3
1. a x a x a x a x . a x a x a x a x
3
a b 3 a a ba b5 3
3
a x a xa x a x a xa x a x a x
a a x x 2
2
L K S 11
2.
3 6 2 2 . 3 6 2 2 3 6 2 2
A. Evaluasi Pengertian atau Ingatan
1
2 2 3 6
4 4
2
2 2 3 6
2 2 3 2 18 6
2
a
log b c b ac
4. D a x b 5. B
2 23 6 3 4 2 6 2
4
2 2 3 6 2 2 3 . 2 2 3 2 2 3 4 2 4 2 6 2 12 6 3 2 3 3 3 6 8 9 2 2 6 3 2 2 6 3 1
4. 7
1 2 9 7 9 7 16 2 63 16 2 63
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
a
log b x
log 64 4 log 43 3
6. A 5 log 625 5 log 5 4 4 7. A log100.000 5 8. A 0,1 log100 m 100 (0,1)
102 (10 1 ) m 102 10 m m 2 9. B 5 log125 b 5 log 53 b b 3 b
7
log8 3 8 23
2. C 3x y 3 log y x 3. B
2 2 3 6 2
1. B
log 27 3 log 27 3 log 33 3
10. B 8 log 8 1
Bab 1 | page 26
B. Evaluasi Pemahaman dan Penguasaan Materi 1.
log 216 3 b) log 81 4 a)
6 3 1
c)
3
log 81 4
1
log 27 3 1 3 16 e) log 8 4 e f) log 7,39 2 2 27 g) log 9 3 1 3 81 h) log 27 4 e i) log 0,05 3 1 1 125 j) log 5 3 d)
2.
3
33 27 7 b) 2 128 3 c) 6 216 a)
2 3
343 49 1 2 e) 6 36 d)
2
3 x 4 3 x 4 x 8 d) 10 10 x 8 6 x e) e e x 6 2
x
1 f) 4 2 x
2
1 1 2 2 x 2 3 3 g) x 12 x 3 x 4 h) e e x 4 1 x i) e 2 e x e e 2 x 2 10 x 3 j) e e x 3 10 x 7
2
8 3 9 f) 27 4 3 g) ln 20,1 h) ln 2,72 5 i) ln 148,41 3
16 4 27 j) 81 8 3.
2 x 64 2 x 2 6 x 6 3 b) x 8 512
L K S 12 A. Evaluasi Pengertian atau Ingatan 1. E log(a 2 b 2 ) log(a b )(a b )
a)
= log(a b ) log(a b ) 2. B
log x3 log
3
1 x log log x2 x x
3. A c)
x
2
16 9
2
log 4 2 log12 2 log 6 2 log
4 12 6
2 log8 2 log 23 3
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 27
4. D
15.
6 36 log log log 4 5 log 54 100 100 6 36 1 log 100 100 5 log 4 log10 5 4 54 10 log
5
10
log150 log 2 3 52 log 2 log 3 log 52 a b 2c
log10
10 4
B. Evaluasi Pemahaman dan Penguasaan Materi
5. D
Diket: 2 log 3 1,585 log 6 log 2 3 log 2 log 3 1 1,585 6. D 2
a
2
log pq log a
2
1.
2
log pq log pq p q
a
a
q p
p q
2
a
7. D
log bc a log
c 2
a log c 2 a log c 8. A log10 log100 log1.000 log10.000
log100.000 1 2 3 4 5 15 9. D 11 14 3 22 log 5 log 115 log 143 log 15 log 7 22 15
10. A 3 5 5 36 1 12
log1 0
2.
log125 2 log 4 log 32 1
log125 3 log 4 log 32
12. D log x log y z log xy z z
10 xy 10 x y z
13. A
a 2 log bc log
2
b 2 c 2 log log ac ab 2
2
a b c log1 0 bc ac ab
14. A
Diket:log x 1,9675 x 101,9675
x x log log 2 2 x 6 x 9 x 3
b)
x3 log 3 y 1
2
5 32 log 5 log16 log 32 log log10 1 16
a)
11. C 1 3
bc b a log b c
5 log 35 log 36 log 121 log
2 log16 2 log 8 2 log(2.2.2.2)2 log(2.2.2) 2 log 22 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 2 2 b) log 4 log 64 log16 2( 2 log 2) 6 2 log 2 4 2 log 2 2 6 4 4 4 4 c) log 246 log 16 4( 4 log 4) 2( 4 log 4) 4 2 2 3 3 3 d) log 81log 27 log 3 8( 3 log 3) 8.1 8 a)
log q 2 log q a
a
Diket:log 2 a, log 3 b, log 5 c
c)
x 1 x 3 2 log log x 1 x 3 1 x 1
3
d)
log 3 x 1 x 2 log(3x 3 x 2 )
x xz 3 z3 y e) log log 3 yx 3 log yx 2 x z 2 3 x b f) log a 2 y 3
log(1.000 x) log1.000 log x 3 1,9675 4,9675
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 28
3.a)
log p log q log r
log a log b log( a.a ) log b log a log a log b 2 log a log b 2 c) log a log b log a log(b.b) log a log b log b log a 2 log b b)
j)
2
5. a)
d)
log a 2 log(ab) 2 log a log a log b log a log b 2 e) log a 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)
log(2.2.3) log 2 log 2 log 3 2 log 2 log 3 2(0,3010) 0,4771 1,0791 b) log(10.2) log 10 log 2 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
log(3. 2 ) log 3 log 2 2 1 log 3 log 2 0,6276 2 f) log(100.3) log 100 log 3 log(10.10) log 3 2 log 10 log 3 2,4771 g) log(1000.2) log(10.10.10) log 2 3 log10 log 2 3,3010 6 2.3 h) log log 10 10 log 2 log 3 log 10 0,2219 2 2 i) log log 1000 10.10.10 log 2 3 log 10 2,699
2.18.5 2.18.5 log log 6 .3 18 log( 2.5) log 10 1
8.9.10 7 720 7 b) log log 10 720 5.14,4. 7 7 7 log 7 1
4. a)
9 3.3 log log 2 2 2 log 3 log 2 0,6532
6.
xy 3 xy 3 a) log log 1 x 5 x 5 2 x x 2 x x 1 b) log 2 log y ( y 3) y 3 3 y 2 c)
1 y 3 log 2 log 2 y 3 y y 3y
d)
x 1 3 log x 1log x 9 log 9 x 3
1
e)
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
7. a)
log 5 log 2 0,6990 0,3010 0,398 b) log 5 log 5 log 3 2(0,6990) 0,4771 = 1,8751 c) log 75 – log 4 = 1,8751 – 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. B x
log 12 p
2
x
log(4.3) p
log log log log 2 log log log16
x
log 4x log 3 p
x
log 4 q p
x
log 4 p q
x
2
2
2
log log log log 65.536 2
2
2
2
16
2
2
2
log log 4 log 2 1 2 5. E 2
2
2
0
log log log x 1 log log x 101 10
log x 1010 x 1010 6. A
x x x log log x log 4 4 1 ( p q)
log 3 log 2 log x 1 3 log 2 log x 21
2
2 log x 32 x 29 512 7. B 10 log 2 x 1 2 x 101 x 5 2
1 p q
25
9. a)
2
log x 25 log 5 5 log 5 12 5 log 5 0,5
8. C 8
10. a)
3
1 2 3 98 99 1 log . . ........ . log 2 99 100 2 3 4 100 b) 6
1
log 2 6 log 216 3 log81 5 log(625) 3
214 215 6 1 1 2 3 log . . ........ . log 3 215 216 2 3 4 216
4
2 log 2 6 log 63 3 log 34 5 log 5 3 13 3 4 34 0 9. C log y y log x log x y 2
log xy 3 2
1
log y log x 2 log x 2 y log xy 3
1
5
log y 2 x2 x2 y log( xy ) 2 5 log xy log xy 2 10. B Diket:a 0,909090... dan b 1,331
100a 90,909090... a 0,909090...
L K S 13
______________ _ A. Evaluasi Pengertian atau Ingatan 1. A 2
log 4 log 2 4 2. 1 3
2
2
m 34 (31 ) m
log 81 m 81 13
m 4 3. C 2 3
b 1,331 1.331 1.000 113 103 3 1 1 11 3 10 3 10 a
log( 5 log 625) 2 log( 5 log 54 ) 2
99a 90 a 90 10 99 11
log1.728 m 1.728 (2 3)m m
26 33 2m 32 26 2 m m 6 m 6 m 3 2 3 3 m 6
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
a
11
log b log a
1 a3
a log a 3 3
12. B
Diket: n log1,5 a ; n log 3 b log1,5 log1,5 a log n log n a diperoleh: log 3 log 3 b log n log n b log1,5 log 3 3 log 32 ab a a b x 3 b 3x 3 log 32 x 2 Bab 1 | page 30
13.
c) 4 log 3 – 2 log 5 = 0,5104
2
a log 4 b
a a 1 4 4 4 10 4 4 b b 10 2
3 . log 2 0,4515 2 1 e) (2 log 10 + log 2) = 1,1505 2 1 f) ( 2 log 2 + 2 log 3) = 0,7781 2
2
d)
a
1 2 b2 100 a b 10 2
log
4
b2 log a a
b2
1
1
4 100a 4 1 log a 2
3. a)
14. E
Diket: 5 log 3 a
b)
log 75 log 3 52 5 log 3 5 log 52 a 2 15. C 5
log 45log 5 log 3 2
3
3
2
3
3
2
3
3
log 15
log 5 3
1 3
log 5 2
3
2
1
log 3 log 53 3
2 log 5log 5 4(1 log 5) 12 2
3
3
2
3
1 1 3 log 5 3 3
1 (1 3 log 5) 3
B. Evaluasi Pemahaman dan Penguasaan Materi
a = 0,636363 100a = 63, 6363
a – 100a = -63 -99a = -63
63 7 99 11 23 2 7 b 2 log a 49 log 11 121 7 49 log 11 a=
2 7 11
1.
log(34 ) 2 3 log 38 8.3 log 3 8 2 4 3 2 12 2 b) log( 2 ) log 2 12. log 2 12 2 2 5 3 2 21 2 c) log(2 .2 ) log 2 21. log 2 21 a)
7 1 11 2
3
d)
3
e)
3
1 2
log(38 ) 3 log 3 4 4.3 log 3 4 7
7 7 log(37 ) 2 3 log 3 2 .3 log 3 2 2 2
4 3 4 f) log 2 log 4 1 4 4
1 3
2 2 2 2 2 2 3 g) log 4 2 log 2 3 . log 2 3 6
2
3
h.
log
2
5
5 5 log(2 4 .2 2 ) 2 log 2 2 .2 log 2 2 2
2. a)
log 15 2 =
b)
log 5 log log 3 2 log 2
1
1 log 5 3 2 1 = log 5 log 3 2
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 log 25.7 3 log 4 4 log 25.7 3 4 log 4 log 25 4 log 7
1 1 4 2 4 log 4 log 5 log 7 1 3 log 42.4 log 5 4 log 7 b 2 a c
3
= 0,58805 2
= 2 log 5 + log 3 – 2 log2 = 0,2731 Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 31
9. A 1 1 1 log bc 1 b log ac 1 c log ab 1 1 1 1 a log bc a log a b log ac b log b c log ab c log c 1 1 1 a b c log abc log abc log abc log a log b log c log abc 1 log abc log abc log abc log abc
L K S 14
a
A. Evaluasi Pengertian atau Ingatan 1. C
1 2
log 36
3
1
log 2 log 3 log 36 log 36 log 36
log 2 log 3 log 6 log 6 1 log 36 log 36 log 6 2 2
10. C
2. C
1
1 1 log 2 log 5 5 2 log100 log100 log100 log100 log10 1 log100 2
3. D Diket: 4 log 3 a dan 3 log 5 b
log( xyz)
3
4
xz
Diket: log 2 a dan log 3 b 7
2
log 6
98
4
2 2
log 6
log 98
2 2
1b 1 2 7 1
log 2 2 log 3
log 2 2 log 7 2
1 b a(b 1) 1 2a a 2
a ab a(1 b)
log 2
12. B
4
log 39 ada di antara 5 dan 6. 5. B 1 1 log 25 20 64 log10 log100 log 20 log 64 log 25 log10 log100
2
m 2
n
log 25 log 20 12 log 64 log
1 log( xyz)
11. B
4. E
25 8 log10 1 20
3 3
log a
3
3
log b
log a 2 log b 2
log a3 log 2 log b3 log 2
loga 2 log 3 2
log b log 3
3
log a 3log a log a 3 3log b log b log b log a 2 2 log a log a log b 2 2 log b log b
Jadi, m n.
6. A
13. B
Diket: 5 log 3 a dan 2 log 5 b . 15
1 log( xyz)
2
log15 log 3 5 log 3 log 5 4
yz
log xy log yz log xz log(xyz ) 2 log( xyz) log( xyz) log( xyz) log( xyz)
4 log 3 a 3 4 log 3 log 5 log 5 ab log 5 b
4
4
xy
5
log 40 5
log 40 log15
5
log 2 log 5
5
3
5
log 3 log 5
a
52
log 33 a
5
log 3 23a log 5 23a
log
b c 1 1 1 b log c log a
14. D
3 5 log 3 2
a
3
b c a log b1 log c 1 log a 1
7. B
1 1 1 3 log 60 4 log 60 5 log 60 log 3 log 4 log 5 log 60 log 60 log 60 log 3 4 5 log 60 1 log 60 log 60 8. A 3 4 5 6 8 log 4 log 5 log 6 log 7 7 log 8 log 9
log 27 a
5
1 3 log 2 1 3 b 1 3 b 5 a 1 b( a 1) log 3 1 5
25
( 1) a log b (1) b log c (1) c log a a log a 1 15.
1
1
1
log x log x log x p q
q r
r p
log q log q log rp log q r rp log1 r 0 log x log x log x log x log x p
p
q
3 log 32 2 Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 32
B. Evaluasi Pemahaman dan Penguasaan Materi
c)
1. a)
3
log 88 3 3
3
log 88 log 22
log 11 3.3 log 2 3 log 11 3 log 2 b 3a b a 3
3.
10 2 20 2 8 12 2 20 log 4 16 2 2 log log 4 4 2
2 2 log 2 2 4 1
4. a)
2
1 log 2 log 2 log 2 log 2 log 9 log 3 2
2
3
d)
2
log 3. 3 log 6.5 log 2 3 5 log 6
3.5 log 2.2 log 3.3 log 6 3.5 log 6 5 3 5 log 6 log 6 4 5 1 5 log log 100 25 b) 1 2 1 2 log log 8 8 5 2 log 5 2 log 2 3 5 log 5 2 1 22 log 2 3 2 1 6 3 Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
32
3
1
32
log 27 log 3 log 2 22
2
2
3 log 2 4 2 2 = log 27 log 3 3.3 log 2 2 3 log 2 2 log 27.3 3 log 2 2
log 3 log 2 4 3
2
3
3
4. log 3. log 2 4 2
3
6. a) log 6 = log (2.3) = log 2 + log3 =a+b b) log 600 = log (100.6) = log 100 + log 6 =2+a+b c) log 5 = log (10:2) = log 10 – log 2 =1–a d) log
5. a)
22
1 3 2 log 3 2 log 2 1 2 1 3 . log 3. log 2 2 2
log 11 3 log 8 log 11 3 log 2
3 log 11 3 log 23 3 log 11 3 log 2
2
3 3 2 9 1 log 2 log 2 log 2 3
2. 22
log 9 log 3
1 = log 1 – log 15 15 = log 1 – log (5.3) = log 1 – log 5 – log 3 = 0 – (1-a) – b =a–b– 1
75 100 3 = log 4
e) log 0,75 = log
= log 3 – = log 3 – = log 3 – = log 3 – = b – 2a
log 4 ( log (2.2)) log 2 – log 2 2 log 2
Bab 1 | page 33
7. a)
2 30 = log 30 1 = log (10.3) 2 1 = (log 10 + log 3) 2 1 = (1 + b) 2 2 2 g) log 12 log 6.2 2 log 6 2 log 2 log 6 = 1 log 2 a b 1 a 2a b a 1
f) log
a
2 a log a 6 a log x 4 a log y 2 a log z = 2.1 + 6.p – 4.q – 2.r = 2 + 6p – 4q – 2r = 2 ( 1 + 3p – 2q – r)
1
1 2
32
1 5 .3 log 4 100 5 log 1 100 . 4 log 3 1 log 5 log 100 . 4 log 3 1 1 a 2 . 4 b 1 a 2 . 4 b
i)
6
5 log 1, 25 6 log 4 6 log 5 6 log 4
log 5 2. log 2 log 5 log 2 2. log 6 log 6 log 5 2 log 2 log 6 1 a 2a 1 3a a b a b 6
1 1 2 4 b) log xy z 1 a 1 a a log x . log y . log z 2 4 1 1 p q r 2 4 a log y a x a c) log z . log y log z a log x rq p a
2 5 h) log 0,05 log 100 1 9
a 2 x 6 a a 6 a 4 a 2 log y 4 z 2 log a log x log y log z
6
8.
1 3 a log 2 2 3 2 3 3 9 b) . log 3 . 2 2 2a 4a 2 3 c) 3. log 2 3. a 2a 3 a)
3
10. a)
d)
2
4
3 log 2 log 3 2 log 2 2 2 = log 3 1 100 3 log 25.1,5 4 log . 4 2
4
4
log 100 3 log 4 1 2 = ( log 100 – log 4 – 1) . = ( 2 – log 4 – 1 ) = ( 1 – log 4)
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
32
log 100 log 4 .
3 2
3 2
3 2 Bab 1 | page 34
C. Evaluasi Kemampuan Analisis
xyz 100 10 xyz log xyz 2 ..... (1) log x 3 3 log x log y ... (2) log y 4 4
L K S 15
2
1.
A. Evaluasi Pengertian atau Ingatan 1. B (Berdasarkan sifat logaritma)
log y 4 5 log z log y ... (3) log z 5 4
2. C
5
log (xyz) = log x + log y + log z = 2 Substitusi:
3
5
3 log 32 5
5
=
5
log 325
5 32 3
3 5 log y + log y log y = 2 4 4
2 3 5
2 5
3 log y = 2 2 3
2 log y y 10 3 1 2 Substitusi log y ke (2), didapat x 10 2 3 5 2 Substitusi log y ke (3), didapat z 10 6 3
3
3. E 3
3
3
2 log 5
3
5
log 5
2
2
4. B 32
3
log 2 2
3
3
log 2
21 5. B
2.
5
3.
a 64 2 log a 6 x
6
2
y
6
5
5
log 81
=9
x
6. A
log a 6 y b 64 2 log b 6 2 y . log b 6 z 6 2 z c 64 2 log c 6 2 z . log c 6 2 log a 2 log b 2 log c 6 6 6 6 1 1 1 6 1 x y z x y z x.
15 log 81 2
2
7. B
2
2 2
log 5
2
2
log 5
5 8. A 3
3
2
2
log 3 4
3
1 log 2
2 3
1 2 11 81 80 8 3 24 9. E 10. D
9
9
2 4 16 4 5
log 2 5
B. Evaluasi Pemahaman dan Penguasaan Materi
2. 3
a)
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
3
log 9
1 2
1 9
Bab 1 | page 35
4 log 5
2
b)
2 2
11. C 12. B
4 5
1
32 2 3 10 4 100 0,03
2 25 d) 5,8 58 log 25
c)
5, 8
2
e)
8
log
log 58
121 . 2 3
13. B 2
3
=
3
2
2
=
log
2
log
8 2 15 5 3 2 5.3
121 . 2 3
3
5 3
121. 2 3
3
14. C
3
3
2 9.3 25.3 4.3 6 3 5 3 2 3 3 3
3
3
= 121. 2 = ( 121 . 2 ) = 242 f)
10
log mlog n
10
15. 16. 17. 18.
log( mn )
= mn.
C C D D 1 2 3 3 4
x2
UKAB 1
23
x 24 24 x 23
19. E
10 2 21 7 3 2 7.3 A. Pilihan Ganda
7 3
1. B
20. B 3
18 a 2 4 b b a 2 6 a2 2 b 2 2 2 3ab 2ab 5ab 2. C
3
13 2 22 11 2 2 11.2
4
1 2 3 1 1 2 3 4 6
3
3 4
11 2 21. A 22. B
1 2 3 2 2 3 2 . 1 3 3 3 3 2 3 2 3 3 3 3 3 3 4 2 3 3 6 3
3. B 4. B 6. D
32 n 5 9 1 16 33 2n + 5 + 9 – 1 – 16 = 3 2n = 6 n=3 7. E 5
3
4
3 2 3 2 5 3 4 2 2 5 8. C 9. D 10. A
24. 25. 28. 29. 30. 31. 33. 36.
B D D B E D B B 2
4 5
3 3 25 = 4 2 5
4
2 2 log log a a
2 log 2 2 log a 2 log a 1 0 1
2 4 16 4 3 81 Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
Bab 1 | page 36
37. D
4. b)
log 4 2 log 4 log 2 2 log 2 2 2
1 4 13 5 2 log log 20449 log log 7 35 11 4 5 20449 7 11 log 13 35 143 4 5 7 11 log 13 35 260 log 7 13 35 log100 2
38. B
B. Bentuk Uraian
5 3 2
a
1.
2
5 a 2 3 2 2 25 2 5 2 6 35 4 6 5 2 6 5 2 6 5 2 6 127 50 6 2
5. a)
2. a)
3
42
74 4 210 74 16 210
log 56 log 42 3 log 7 8 3 log 7 6
log 56 3
74 2 840 60 14 2 60 14
3 log 7 3 log 23 3 log 7 3 log 6
60 14 3. a) 3ab – 2bc + abc
5 2 5 2 5 15 2 5 2( 1) 5(1)
log 7 33 log 2 log 7 3 log 2 3 log 3 3 b a ab 3 1 ab a 1 b 1 a 3
3 5 2 5 2 2 5 2 5
30 15 5 2 5
b)
27 15 5
3 9 5 5 b)
3
66
a .bc 5 2 5 2 5 1 3 2
2
1 3 2
3
2
3
5 2 (4 5) 2 3
5 2 1 125 1
Kunci Penyelesaian Matematika SMA Jl.1A – Sukino
log 44 log 66 3 log 11.4 3 log(11.6)
log 44 3
log 11 2 2 log 2 log 11 3 log 2 3 log 3 2 c a ac 2 1 c 1 ac a 1 a 3
3
Bab 1 | page 37
