1. KONSEP BILANGAN BERPANGKAT
SIFAT-SIFAT PERPANGKATAN:
5. ao = 1 1. am × an = am+ n
6. a 1 = a am
2. n = am-n
-m 1 a
7. a = m atau
a
1
( )
n
3. am = am×n
m
a = -m 4. (ab)m = (am )(bn )
a
1 1 m
atau  a am
8.
a2 = a a= a2 10.   = m
b b
APLIKASI KONSEP DALAM SOAL
1. Hasil perkalian dari ( 4a) −2 × ( 2a) 3 = ....
Jawab:
(
= ( 4a) −2 × ( 2a) 3 sifat ( ab ) m = am × bn )
= (4-2 )(a-2 )(23 )(a3 ) rubah 4 menjadi 22)*
= ([22 ] -2 )(23 )[a-2 ][a3 ] sifat am ( ) n
= am×n
= (22×[-2] )(23 )[a-2 + 3 ] sifat am × an = am+ n
= (2 -4 )(23 )[a1 ]
= (2 -4 + 3 )[a1 ] )*Contoh lain:
8=23 9 =3 2 125=53
= (2 -1 )[a 1 ]
16=24 27=33 36=62
 1  32=25 81=34 216=63
= 1  × a
2  64=26 243=35 49=72
1 128=27 25=52 dst
= a
2

2. 1
Jika a = 32, b = 27 maka = ....
(2a5 × 3 b)
2.
Jawab:
1 1 1 1 1
= = =
(2a5 × 3 b ) (2a5 × b 3 ) (2[32]5 × [27]3 )
1 1 1 1
5× 3×
=
(2[25 ]5 3 3 =
× [3 ] ) (2[2 5 ] × [3 3 ]) = (2[2] × [3])
=12
a5b
3. Jika a = 16, b = 32 maka = ....
b a
Jawab:
1 1 1 1
5b
=
a
=a b 5 = (16) (32)5 = (16) (25 )5 = (1) (2) = 1
b a (2) (4) 4
b a (32) ( 16 ) (32) (4)
2
LATIHAN
a b
1. Jika a = 32, b = 64 maka = ....
b5a
Jawab: 1
(32)( 64 ) (32)(8)
a b a b 1
= 5 = 1= 1=
b a 5
b a5 (64) (32)5 (64) (2 )5
(1) (8) 8 2
= = =1
(2) (4) 8
1
Jika a = 8, b = 32 maka = ....
(4a3 × 5 b)
2.
Jawab:
1 1 1 1 1
= = =
(4a3 × 5 b ) (4a3 × b 5 ) (4[8]3 × [32]5 )
1 1
= =(4)(2)(2)=16
(4[23 ]3 × [25 ]5 )

3. TUGAS
TES DAYA SERAP
a b
1. Jika a = 32, b = 64 maka = ....
b5a
1
Jika a = 8, b = 32 maka = ....
(4a3 × 5 b)
2.
Jawab:
a b (32)( 64 ) (32)(8)
a b 1
1. = 5 = 1 = 1=
b a (64) (32)5 (64) (25 )5
b a5
2
(1) (8) 8
= = =1
(2) (4) 8
2. Cari sendiri 