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Tentukan bentuk sederhana dari: 102:10103×102​

Pertanyaan

Tentukan bentuk sederhana dari: fraction numerator 10 cubed cross times 10 squared over denominator 10 squared space colon space 10 end fraction 

Pembahasan Soal:

Ingat bahwa:

bullet space a to the power of b cross times a to the power of c equals a to the power of b plus c end exponent bullet space fraction numerator a over denominator b over c end fraction equals fraction numerator a cross times thin space c over denominator b end fraction bullet space x to the power of a over x to the power of b equals x to the power of a minus b end exponent 

Mencari fraction numerator 10 cubed cross times 10 squared over denominator 10 squared space colon space 10 end fraction:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 10 cubed cross times 10 squared over denominator 10 squared space colon space 10 end fraction end cell equals cell fraction numerator 10 cubed 10 squared over denominator 10 squared over 10 end fraction end cell row blank blank cell fraction numerator 10 cubed 10 squared over denominator 10 squared over 10 end fraction end cell row blank equals cell fraction numerator 10 to the power of 5 10 over denominator 10 squared end fraction end cell row blank equals cell 10 times 10 to the power of 5 minus 2 end exponent end cell row blank equals cell 10 cubed 10 end cell row blank equals cell 10 to the power of 3 plus 1 end exponent end cell row blank equals cell 10 to the power of 4 end cell row blank equals cell 10.000 end cell end table 

Jadi, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 10 cubed cross times 10 squared over denominator 10 squared space colon space 10 end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 10 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank. end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 000 end table.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 07 Oktober 2021

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Pertanyaan yang serupa

Hasil pemangkatan dari (a49​÷a43​)−5=....

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=4, b=2, dan c=21​. Nilai (a−1)2×c−3b4​=...

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 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 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 open parentheses a to the power of negative 1 end exponent close parentheses squared cross times b to the power of 4 over c to the power of negative 3 end exponent end cell equals cell open parentheses 4 to the power of negative 1 end exponent close parentheses squared cross times 2 to the power of 4 over open parentheses begin display style 1 half end style close parentheses to the power of negative 3 end exponent end cell row blank equals cell 4 to the power of negative 2 end exponent cross times 2 to the power of 4 over open parentheses 2 to the power of negative 1 end exponent close parentheses to the power of negative 3 end exponent end cell row blank equals cell open parentheses 2 squared close parentheses to the power of negative 2 end exponent cross times 2 to the power of 4 over 2 cubed end cell row blank equals cell 2 to the power of negative 4 end exponent cross times 2 to the power of 4 minus 3 end exponent end cell row blank equals cell 2 to the power of negative 4 end exponent cross times 2 to the power of 1 end cell row blank equals cell 2 to the power of negative 4 plus 1 end exponent end cell row blank equals cell 2 to the power of negative 3 end exponent end cell row blank equals cell 1 over 2 cubed end cell row blank equals cell 1 over 8 end cell end table

Dengan demikian, nilai dari table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of negative 1 end exponent close parentheses squared cross times b to the power of 4 over c to the power of negative 3 end exponent end cell equals cell 1 over 8 end cell end table 

0

Roboguru

Nyatakanlah bilangan berikut dalam pangkat positif. a. 43×4−5

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 cubed cross times 4 to the power of negative 5 end exponent end cell equals cell 4 to the power of 3 minus 5 end exponent end cell row blank equals cell 4 to the power of negative 2 end exponent end cell row blank equals cell 1 over 4 squared end cell row blank equals cell 1 over 2 to the power of 4 end cell row blank equals cell 1 over 16 end cell end table 

Jadi, bilangan tersebut dalam pangkat positif adalah 1 over 2 to the power of 4 space atau space 1 over 16.

0

Roboguru

Nilai 365×256−18−2×244​ adalah ...

Pembahasan Soal:

Ingat kembali beberapa sifat bilangan berpangkat berikut:

  1. 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
  2. a to the power of negative m end exponent equals 1 over a to the power of m
  3. 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
  4. 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
  5. a to the power of m over a to the power of n equals a to the power of m minus n end exponent


Sehingga soal tersebut dapat kita hitung sebagai berikut:


table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 8 to the power of negative 2 end exponent cross times 24 to the power of 4 over denominator 36 to the power of 5 cross times 256 to the power of negative 1 end exponent end fraction end cell equals cell fraction numerator open parentheses 2 cubed close parentheses to the power of negative 2 end exponent cross times open parentheses 2 cubed cross times 3 close parentheses to the power of 4 over denominator open parentheses 2 squared cross times 3 squared close parentheses to the power of 5 cross times open parentheses 2 to the power of 8 close parentheses to the power of negative 1 end exponent end fraction end cell row blank equals cell fraction numerator 2 to the power of negative 6 end exponent cross times 2 to the power of 12 cross times 3 to the power of 4 over denominator 2 to the power of 10 cross times 3 to the power of 10 cross times 2 to the power of negative 8 end exponent end fraction end cell row blank equals cell fraction numerator 2 to the power of 12 minus 6 end exponent cross times 3 to the power of 4 over denominator 2 to the power of 10 minus 8 end exponent cross times 3 to the power of 10 end fraction end cell row blank equals cell fraction numerator 2 to the power of 6 cross times 3 to the power of 4 over denominator 2 squared cross times 3 to the power of 10 end fraction end cell row blank equals cell 2 to the power of 6 minus 2 end exponent cross times 3 to the power of 4 minus 10 end exponent end cell row blank equals cell 2 to the power of 4 cross times 3 to the power of negative 6 end exponent end cell row blank equals cell 2 to the power of 4 over 3 to the power of 6 end cell row blank equals cell 16 over 729 end cell end table


Jadi, fraction numerator 8 to the power of negative 2 end exponent cross times 24 to the power of 4 over denominator 36 to the power of 5 cross times 256 to the power of negative 1 end exponent end fraction equals 16 over 729.

Oleh karena itu, jawaban yang benar adalah E.

0

Roboguru

Nyatakanlah bilangan berikut dalam pangkat positif. b. (41​)−3×44

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 fourth close parentheses to the power of negative 3 end exponent times 4 to the power of 4 end cell equals cell 1 over open parentheses 1 fourth close parentheses cubed cross times 4 to the power of 4 end cell row blank equals cell 4 cubed cross times 4 to the power of 4 end cell row blank equals cell 4 to the power of 3 plus 4 end exponent end cell row blank equals cell 4 to the power of 7 end cell row blank equals cell open parentheses 2 squared close parentheses to the power of 7 end cell row blank equals cell 2 to the power of 14 end cell row blank equals cell 16.384 end cell end table  

Jadi, bilangan tersebut dalam pangkat positif 2 to the power of 14 space atau space 16.384.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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