Roboguru

Jika a=27, b=32, maka nilai dari ​​3(a)−31​×4(b)32​​ adalah ...

Pertanyaan

Jika a=27, b=32, maka nilai dari 3(a)31×4(b)32 adalah ...

  1. 25 

  2. 16 

  3. 0 

  4. 16 

  5. 25 

Pembahasan Soal:

Ingat!

Sifat bilangan berangkat:

  • (am)n=am×n   
  • am×an=am+n   
  • a0=1 
  • an1=na 

Sehingga:

3(a)31×4(b)32============3×(27)31×4×(32)323×(33)31×22×(25)323×33×(31)×22×25×323×31×22×25×3231+(1)×22×231030×22+3101×236+102316253125+3125×2313232 

Oleh karena itu, jawaban tidak ada pada opsi, jawaban yang benar adalah 3232.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 30 Agustus 2021

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Pertanyaan yang serupa

Hasil dari  adalah ...

Pembahasan Soal:

Ingat sifat bilangan berpangkat berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight a to the power of straight m cross times straight a to the power of straight n end cell equals cell straight a to the power of straight m plus straight n end exponent space dan space open parentheses straight a to the power of straight m close parentheses to the power of straight n equals straight a to the power of straight m cross times straight n end exponent end cell end table

Berdasarkan sifat bilangan tersebut maka penyelesaiannya adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell 243 to the power of 1 half end exponent cross times 3 to the power of 1 half end exponent end cell equals cell open parentheses 3 to the power of 5 close parentheses to the power of 1 half end exponent cross times 3 to the power of 1 half end exponent end cell row blank equals cell 3 to the power of 5 over 2 end exponent cross times 3 to the power of 1 half end exponent end cell row blank equals cell 3 to the power of 6 over 2 end exponent end cell row blank equals cell 3 cubed end cell row blank equals 27 end table 


Dengan demikian, hasil dari 243 to the power of 1 half end exponent cross times 3 to the power of 1 half end exponent adalah 27.

0

Roboguru

Sebutkan sifat-sifat eksponen.

Pembahasan Soal:

Sifat-sifat eksponen atau bilangan berpangkat diantaranya sebagai berikut:

  • a to the power of m cross times a to the power of n equals a to the power of m plus n end exponent 
  • a to the power of m divided by a to the power of n equals a to the power of m minus n end exponent 
  • open parentheses a to the power of m close parentheses to the power of n equals a to the power of m cross times n end exponent 
  • open parentheses a cross times b close parentheses to the power of m equals a to the power of m cross times b to the power of m 
  • open parentheses a over b close parentheses to the power of m equals a to the power of m over b to the power of m 
  • 1 over a to the power of m equals a to the power of negative m end exponent 
  • n-th root of a to the power of m end root equals a to the power of m over n end exponent 
  • a to the power of 0 equals 1 
  • 1 to the power of m equals 1 

Dengan demikian, diperoleh sifat-sifat eksponen seperti di atas.

1

Roboguru

Hasil pemangkatan dari

Pembahasan Soal:

Berdasarkan sifat bilangan berpangkat yaitu a to the power of m divided by a to the power of n equals a to the power of m minus n end exponent maka open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent diperoleh sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell equals cell open parentheses a to the power of 9 over 4 minus 3 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell row blank equals cell open parentheses a to the power of 6 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell end table

Berdasarkan sifat bilangan berpangkat open parentheses a to the power of m close parentheses to the power of n equals a to the power of m cross times n end exponent maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell equals cell open parentheses a to the power of 6 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell row blank equals cell a to the power of 6 over 4 cross times negative 5 end exponent end cell row blank equals cell a to the power of negative 30 over 4 end exponent end cell end table

Berdasarkan sifat bilangan berpangkat a to the power of negative n end exponent equals 1 over a to the power of n comma space a not equal to 0 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell equals cell a to the power of negative 30 over 4 end exponent end cell row blank equals cell 1 over a to the power of begin display style 30 over 4 end style end exponent end cell row blank equals cell 1 over a to the power of begin display style fraction numerator 28 plus 2 over denominator 4 end fraction end style end exponent end cell row blank equals cell 1 over a to the power of begin display style 28 over 4 end style plus begin display style 2 over 4 end style end exponent end cell row blank equals cell 1 over a to the power of 7 plus begin display style 2 over 4 end style end exponent end cell end table

Berdasarkan sifat bilangan berpangkat a to the power of m cross times a to the power of n equals a to the power of m plus n end exponent maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell equals cell 1 over a to the power of 7 plus begin display style 2 over 4 end style end exponent end cell row blank equals cell fraction numerator 1 over denominator a to the power of 7 cross times a to the power of begin display style 2 over 4 end style end exponent end fraction end cell end table

Berdasarkan sifat bilangan berpangkat pecahan yaitu a to the power of m over n end exponent equals n-th root of a to the power of m end root comma space a not equal to 0 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell equals cell fraction numerator 1 over denominator a to the power of 7 cross times a to the power of begin display style 2 over 4 end style end exponent end fraction end cell row blank equals cell fraction numerator 1 over denominator a to the power of 7 cross times fourth root of a squared end root end fraction end cell end table

Jadi hasil pemangkatan dari open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent adalah fraction numerator 1 over denominator a to the power of 7 space fourth root of a squared end root end fraction.

Oleh karena itu, jawaban yang benar adalah A.

9

Roboguru

Diketahui a,b,danc adalah bilangan real positif. Jika a4b5​bc​​=ab, maka nilai c adalah ...

Pembahasan Soal:

Ingat : 

  • a=a21
  • am×an=am+n 
  • am÷an=amn 
  • (am)n=am×n 

perhatikan perhitungna berikut 

a4b5bc(a4b5)21(bc)21a2b25b21c21c21c21c21c21c21(c21)2c==========abababb21aba2b25a1+2b1+2521a3b22+51a3b36a3b2(a3b2)2a6b4 

Dengan demikian, c=a6b4

0

Roboguru

Hitung , Beri jawaban dalam bentuk indeks!

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 5 cubed cross times 8 a squared close parentheses to the power of 1 third end exponent cross times 2 to the power of negative 5 end exponent end cell equals cell open parentheses 5 cubed cross times 2 cubed a squared close parentheses to the power of 1 third end exponent cross times 2 to the power of negative 5 end exponent end cell row blank equals cell 5 cross times 2 a to the power of 2 over 3 end exponent cross times 2 to the power of negative 5 end exponent end cell row blank equals cell 5 cross times 2 to the power of negative 4 end exponent cross times a to the power of 2 over 3 end exponent end cell row blank equals cell fraction numerator 5 a to the power of begin display style 2 over 3 end style end exponent over denominator 2 to the power of 4 end fraction end cell row blank equals cell fraction numerator 5 a to the power of begin display style 2 over 3 end style end exponent over denominator 16 end fraction end cell end table

Jadi, hasil dari open parentheses 5 cubed cross times 8 a squared close parentheses to the power of 1 third end exponent cross times 2 to the power of negative 5 end exponent adalah fraction numerator 5 a to the power of begin display style 2 over 3 end style end exponent over denominator 16 end fraction 

0

Roboguru

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