Konsep Bilangan Berpangkat

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Konsep Bilangan Berpangkat

  1. 1. KONSEP BILANGAN BERPANGKATSIFAT-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 bAPLIKASI KONSEP DALAM SOAL1. 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. 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 a5b3. 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) 2LATIHAN a b1. 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. 3. TUGASTES DAYA SERAP a b1. 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 11. = 5 = 1 = 1= b a (64) (32)5 (64) (25 )5 b a5 2 (1) (8) 8 = = =1 (2) (4) 82. Cari sendiri 

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