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Hitunglah eksponen bilangan bulat  jika  dan  !

Pertanyaan

Hitunglah eksponen bilangan bulat x to the power of negative 2 end exponent plus y to the power of negative 2 end exponent jika x equals negative 2 dan y equals 2 !

Pembahasan Soal:

Substitusi nilai x dan y ke persamaan x to the power of negative 2 end exponent plus y to the power of negative 2 end exponent, seperti berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell x to the power of negative 2 end exponent plus y to the power of negative 2 end exponent end cell equals cell 1 over x squared plus 1 over y squared end cell row blank equals cell 1 over open parentheses negative 2 close parentheses squared plus 1 over 2 squared end cell row blank equals cell 1 fourth plus 1 fourth end cell row blank equals cell 2 over 4 end cell row blank equals cell 1 half end cell row blank equals cell 2 to the power of negative 1 end exponent end cell end table 

Dengan demikian, eksponen dari persamaan tersebut adalah negative 1.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

P. Tessalonika

Mahasiswa/Alumni Universitas Negeri Medan

Terakhir diupdate 07 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Hasil pemangkatan  adalah ...

Pembahasan Soal:

Bilangan berpangkat bulat positif dapat didefinisikan sebagai berikut.

a to the power of n equals stack a cross times a cross times a cross times horizontal ellipsis cross times a with underbrace below table row blank cell space space space space space space space space end cell cell n space text faktor end text end cell end table

Ingat sifat bilangan berpangkat berikut.

a to the power of negative m end exponent equals 1 over a to the power of m comma space a not equal to 0

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 comma space b not equal to 0

open parentheses a b close parentheses to the power of m equals a to the power of m times b to the power of m

open parentheses a to the power of m close parentheses to the power of n equals a to the power of m n end exponent

sehingga penyelesaian dari soal di atas, yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator 3 a squared over denominator 4 b end fraction close parentheses to the power of negative 3 end exponent end cell equals cell open parentheses 3 a squared close parentheses to the power of negative 3 end exponent over open parentheses 4 b close parentheses to the power of negative 3 end exponent end cell row blank equals cell open parentheses 4 b close parentheses cubed over open parentheses 3 a squared close parentheses cubed end cell row blank equals cell fraction numerator 4 cubed b cubed over denominator 3 cubed a to the power of 6 end fraction end cell row blank equals cell fraction numerator 64 b cubed over denominator 27 a to the power of 6 end fraction end cell end table

Oleh karena itu, jawaban yang tepat adalah D.

1

Roboguru

Hitunglah hasil dari bilangan berpangkat berikut. d.

Pembahasan Soal:

Bilangan berpangkat bulat positif dapat didefinisikan sebagai berikut.

a to the power of n equals stack a cross times a cross times a cross times horizontal ellipsis cross times a with underbrace below table row blank cell space space space space space space space space end cell cell n space text faktor end text end cell end table

Ingat sifat bilangan berpangkat berikut.

a to the power of negative m end exponent equals 1 over a to the power of m comma space a not equal to 0

sehingga penyelesaian soal di atas, yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell 12 to the power of negative 3 end exponent end cell equals cell 1 over 12 cubed end cell row blank equals cell fraction numerator 1 over denominator 12 cross times 12 cross times 12 end fraction end cell row blank equals cell fraction numerator 1 over denominator 1.728 end fraction end cell end table

Dengan demikian, hasil dari table attributes columnalign right center left columnspacing 0px end attributes row cell 12 to the power of negative 3 end exponent end cell equals cell fraction numerator 1 over denominator 1.728 end fraction end cell end table

0

Roboguru

Hitunglah hasil dari bilangan pecahan berpangkat berikut. a.

Pembahasan Soal:

Ingat!

Definisi bilangan berpangkat:

  •  a to the power of n equals stack stack a cross times a cross times... cross times a with underbrace below with n below 

Sifat bilangan berpangkat:

  • a to the power of negative n end exponent equals 1 over a to the power of n 

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative 1 fifth close parentheses to the power of negative 3 end exponent end cell equals cell 1 over open parentheses negative begin display style 1 fifth end style close parentheses cubed end cell row blank equals cell fraction numerator 1 over denominator open parentheses negative begin display style 1 fifth end style close parentheses cross times open parentheses negative begin display style 1 fifth end style close parentheses cross times open parentheses negative begin display style 1 fifth end style close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator negative begin display style 1 over 125 end style end fraction end cell row blank equals cell 1 cross times open parentheses negative 125 over 1 close parentheses end cell row blank equals cell negative 125 end cell end table 

Dengan demikian, hasil dari table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses negative 1 fifth close parentheses to the power of negative 3 end exponent end cell end table adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell negative 125 end cell end table.

0

Roboguru

Tentukan hasil dari setiap operasi aljabar bilangan berpangkat berikut: f.

Pembahasan Soal:

Bilangan berpangkat bulat positif dapat didefinisikan sebagai berikut.

a to the power of n equals stack a cross times a cross times a cross times horizontal ellipsis cross times a with underbrace below table row blank cell space space space space space space space space end cell cell n space text faktor end text end cell end table

Ingat sifat bilangan berpangkat berikut.

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

a to the power of negative m end exponent equals 1 over a to the power of m comma space a not equal to 0

sehingga penyelesaian soal di atas, yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell 6 squared divided by 6 to the power of 4 end cell equals cell 6 to the power of 2 minus 4 end exponent end cell row blank equals cell 6 to the power of negative 2 end exponent end cell row blank equals cell 1 over 6 squared end cell row blank equals cell fraction numerator 1 over denominator 6 cross times 6 end fraction end cell row blank equals cell 1 over 36 end cell end table

Dengan demikian, hasil dari table attributes columnalign right center left columnspacing 0px end attributes row cell 6 squared divided by 6 to the power of 4 end cell equals cell 1 over 36 end cell end table 

0

Roboguru

Tentukan bentuk sederhana dari  !

Pembahasan Soal:

Ingat!

Definisi bilangan berpangkat:

  •  a to the power of n equals stack stack a cross times a cross times... cross times a with underbrace below with n below 

Sifat bilangan berpangkat:

  • a to the power of negative n end exponent equals 1 over a to the power of n 

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator open parentheses begin display style 2 over 5 end style close parentheses to the power of negative 3 end exponent cross times open parentheses begin display style 1 fourth end style close parentheses to the power of negative 2 end exponent over denominator open parentheses begin display style 5 over 8 end style close parentheses cubed end fraction end cell equals cell fraction numerator begin display style 1 over open parentheses begin display style 2 over 5 end style close parentheses cubed end style cross times begin display style 1 over open parentheses begin display style 1 fourth end style close parentheses squared end style over denominator open parentheses begin display style 5 over 8 end style close parentheses cubed end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 1 over denominator begin display style 2 over 5 end style cross times begin display style 2 over 5 end style cross times begin display style 2 over 5 end style end fraction end style cross times begin display style fraction numerator 1 over denominator begin display style 1 fourth end style cross times begin display style 1 fourth end style end fraction end style over denominator begin display style 5 over 8 end style cross times begin display style 5 over 8 end style cross times begin display style 5 over 8 end style end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 1 over denominator begin display style 8 over 125 end style end fraction end style cross times begin display style fraction numerator 1 over denominator begin display style 1 over 16 end style end fraction end style over denominator begin display style 125 over 512 end style end fraction end cell row blank equals cell fraction numerator 1 cross times begin display style fraction numerator up diagonal strike 125 over denominator 8 end fraction end style cross times 1 cross times begin display style 16 over 1 end style over denominator begin display style fraction numerator up diagonal strike 125 over denominator 512 end fraction end style end fraction end cell row blank equals cell 2 cross times 512 over 1 end cell row blank equals cell 1.024 end cell end table 

Dengan demikian, bentuk sederhana dari table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator open parentheses begin display style 2 over 5 end style close parentheses to the power of negative 3 end exponent cross times open parentheses begin display style 1 fourth end style close parentheses to the power of negative 2 end exponent over denominator open parentheses begin display style 5 over 8 end style close parentheses cubed end fraction end cell end table adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1.024 end cell end table.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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