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Diketahui  .  Buktikan bahwa

Pertanyaan

Diketahui 7 open parentheses 8 to the power of p close parentheses equals 9 open parentheses 5 to the power of q close parentheses space dan space 7 open parentheses 16 to the power of p plus 1 end exponent close parentheses equals 12 open parentheses 5 to the power of q close parentheses .  Buktikan bahwa 2 to the power of p equals 1 over 12 

Pembahasan Video:

Pembahasan Soal:

Diketahui 7 open parentheses 8 to the power of p close parentheses equals 9 open parentheses 5 to the power of q close parentheses space dan space 7 open parentheses 16 to the power of p plus 1 end exponent close parentheses equals 12 open parentheses 5 to the power of q close parentheses . 

Ingat!

Sifat bilangan berpangkat:

  • a to the power of m space colon space 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 
  • 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 7 open parentheses 8 to the power of p close parentheses end cell equals cell 9 open parentheses 5 to the power of q close parentheses end cell row cell 7 open parentheses 2 cubed close parentheses to the power of p end cell equals cell 9 open parentheses 5 to the power of q close parentheses end cell row cell 7 open parentheses 2 to the power of 3 p end exponent close parentheses end cell equals cell 9 open parentheses 5 to the power of q close parentheses space.... space left parenthesis 1 right parenthesis end cell end table   

table attributes columnalign right center left columnspacing 0px end attributes row cell 7 open parentheses 16 to the power of p plus 1 end exponent close parentheses end cell equals cell 12 open parentheses 5 to the power of q close parentheses end cell row cell 7 left parenthesis 16 to the power of p.16 to the power of 1 right parenthesis end cell equals cell 12 open parentheses 5 to the power of q close parentheses end cell row cell 112 left parenthesis 16 to the power of p right parenthesis end cell equals cell 12 open parentheses 5 to the power of q close parentheses end cell row cell 112 open parentheses 2 to the power of 4 close parentheses to the power of p end cell equals cell 12 open parentheses 5 to the power of q close parentheses end cell row cell 112 open parentheses 2 to the power of 4 p end exponent close parentheses end cell equals cell 12 open parentheses 5 to the power of q close parentheses space.... space left parenthesis 2 right parenthesis end cell end table 

Bagi persamaan (1) dengan persamaan (2), maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 7 open parentheses 2 to the power of 3 p end exponent close parentheses over denominator 112 open parentheses 2 to the power of 4 p end exponent close parentheses end fraction end cell equals cell fraction numerator 9 up diagonal strike open parentheses 5 to the power of q close parentheses end strike space over denominator 12 up diagonal strike open parentheses 5 to the power of q close parentheses end strike space end fraction end cell row cell fraction numerator up diagonal strike 7 open parentheses 2 to the power of 3 p end exponent close parentheses over denominator 16 cross times up diagonal strike 7 open parentheses 2 to the power of 4 p end exponent close parentheses end fraction end cell equals cell 9 over 12 end cell row cell 2 to the power of 3 p end exponent over 2 to the power of 4 p end exponent end cell equals cell 9 over 12 cross times 16 end cell row cell 2 to the power of 3 p minus 4 p end exponent end cell equals cell 144 over 12 end cell row cell 2 to the power of negative p end exponent end cell equals 12 row cell 1 over 2 to the power of p end cell equals 12 row cell 1 over 12 end cell equals cell 2 to the power of p space left parenthesis terbukti right parenthesis end cell end table 

Jadi, dari uraian yang teah dipaparkan di atas, persamaan 2 to the power of p equals 1 over 12 merupakan persamaan yang benar. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 13 September 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Pembahasan Soal:

Ingat kembali:

a to the power of m over a to the power of n equals a to the power of m minus n end exponent comma space a not equal to 0  

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

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 diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 to the power of negative 3 end exponent a cubed b to the power of negative 4 end exponent over denominator open parentheses begin display style 1 half end style close parentheses to the power of 4 a to the power of 4 b to the power of negative 5 end exponent end fraction end cell equals cell fraction numerator 2 to the power of negative 3 end exponent a cubed b to the power of negative 4 end exponent over denominator open parentheses begin display style 2 to the power of negative 1 end exponent end style close parentheses to the power of 4 a to the power of 4 b to the power of negative 5 end exponent end fraction end cell row blank equals cell fraction numerator 2 to the power of negative 3 end exponent a cubed b to the power of negative 4 end exponent over denominator begin display style 2 to the power of negative 4 end exponent end style a to the power of 4 b to the power of negative 5 end exponent end fraction end cell row blank equals cell 2 to the power of negative 3 minus open parentheses negative 4 close parentheses end exponent a to the power of 3 minus 4 end exponent b to the power of negative 4 minus open parentheses negative 5 close parentheses end exponent end cell row blank equals cell 2 to the power of negative 3 plus 4 end exponent a to the power of negative 1 end exponent b to the power of negative 4 plus 5 end exponent end cell row blank equals cell fraction numerator 2 b over denominator a end fraction end cell end table  

Dengan demikian, nilai dari fraction numerator 2 to the power of negative 3 end exponent a cubed b to the power of negative 4 end exponent over denominator open parentheses begin display style 1 half end style close parentheses to the power of 4 a to the power of 4 b to the power of negative 5 end exponent end fraction  adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 2 b over denominator a end fraction end cell end table.

Jadi, jawaban yang tepat adalah E.

2

Roboguru

Diketahui , , dan . Nilai  adalah ....

Pembahasan Soal:

Untuk soal persamaan eksponen tersebut berlaku sifat-sifat berikut:

  • a to the power of p over a to the power of q equals a to the power of p minus q end exponent dengan a not equal to 0 
  • a to the power of negative p end exponent equals 1 over a to the power of p dengan a not equal to 0
  • open parentheses a to the power of p close parentheses to the power of q equals a to the power of p q end exponent dengan b not equal to 0

Dengan menggunakan sifat-sifat di atas, maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator a to the power of negative 2 end exponent b c cubed over denominator a b squared c to the power of negative 1 end exponent end fraction end cell equals cell a to the power of negative 2 end exponent over a to the power of 1 times b to the power of 1 over b squared times c cubed over c to the power of negative 1 end exponent end cell row blank equals cell a to the power of open parentheses negative 2 minus 1 close parentheses end exponent times b to the power of open parentheses 1 minus 2 close parentheses end exponent times c to the power of open parentheses 3 minus open parentheses negative 1 close parentheses close parentheses end exponent end cell row blank equals cell a to the power of negative 3 end exponent b to the power of negative 1 end exponent c to the power of 4 end cell end table 

Substitusi nilai a equals 1 halfb equals 2, dan c equals 1 pada persamaan diatas, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell a to the power of negative 3 end exponent b to the power of negative 1 end exponent c to the power of 4 end cell equals cell open parentheses 1 half close parentheses to the power of negative 3 end exponent times open parentheses 2 close parentheses to the power of negative 1 end exponent times 1 to the power of 4 end cell row blank equals cell open parentheses 2 to the power of negative 1 end exponent close parentheses to the power of negative 3 end exponent times open parentheses 2 close parentheses to the power of negative 1 end exponent times 1 end cell row blank equals cell 2 cubed times 1 half end cell row blank equals cell 8 times 1 half end cell row blank equals 4 end table 


Dengan demikian, nilai dari fraction numerator a to the power of negative 2 end exponent b c cubed over denominator a b squared c to the power of negative 1 end exponent end fraction equals 4.

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Pembahasan Soal:

Ingat kembali:

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

a to the power of m over a to the power of n equals a to the power of m minus n end exponent comma space a not equal to 0

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

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 diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator a to the power of negative 4 end exponent times b squared times c over denominator a times b to the power of negative 6 end exponent times c cubed end fraction close parentheses to the power of 4 end cell equals cell open parentheses a to the power of negative 4 minus 1 end exponent times b to the power of 2 minus open parentheses negative 6 close parentheses end exponent times c to the power of 1 minus 3 end exponent close parentheses to the power of 4 end cell row blank equals cell open parentheses a to the power of negative 5 end exponent times b to the power of 2 plus 6 end exponent times c to the power of negative 2 end exponent close parentheses to the power of 4 end cell row blank equals cell open parentheses a to the power of negative 5 end exponent times b to the power of 8 times c to the power of negative 2 end exponent close parentheses to the power of 4 end cell row blank equals cell open parentheses a to the power of negative 5 end exponent close parentheses to the power of 4 times open parentheses b to the power of 8 close parentheses to the power of 4 times open parentheses c to the power of negative 2 end exponent close parentheses to the power of 4 end cell row blank equals cell a to the power of open parentheses negative 5 close parentheses times 4 end exponent times b to the power of 8 times 4 end exponent times c to the power of open parentheses negative 2 close parentheses times 4 end exponent end cell row blank equals cell a to the power of negative 20 end exponent times b to the power of 32 times c to the power of negative 8 end exponent end cell row blank equals cell fraction numerator b to the power of 32 over denominator a to the power of 20 times c to the power of 8 end fraction end cell end table  


Dengan demikian, hasil dari open parentheses fraction numerator a to the power of negative 4 end exponent times b squared times c over denominator a times b to the power of negative 6 end exponent times c cubed end fraction close parentheses to the power of 4  adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator b to the power of 32 over denominator a to the power of 20 times c to the power of 8 end fraction end cell end table.

Jadi, jawaban yang tepat adalah D.

1

Roboguru

Pembahasan Soal:

Ingat kembali:

a to the power of n equals stack stack a cross times a cross times a cross times... cross times a with underbrace below with a space s e b a n y a k space n below comma space a not equal to 0

a to the power of m over a to the power of n equals a to the power of m minus n end exponent comma space a not equal to 0

open parentheses a over b close parentheses to the power of n equals a to the power of n over b to the power of n comma space a not equal to 0 comma space b not equal to 0 

a to the power of m times a to the power of n equals a to the power of m plus n end exponent. space a not equal to 0 

a to the power of 0 equals 1 comma space a not equal to 0

Sehingga, kita peroleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 over 3 close parentheses squared times open parentheses negative 3 over 4 close parentheses squared minus 3 over 2 cubed colon 5 over 8 end cell equals cell 2 squared over 3 squared times open parentheses 3 over 4 close parentheses squared minus 3 over 2 cubed colon 5 over 2 cubed end cell row blank equals cell 2 squared over 3 squared times 3 squared over 4 squared minus fraction numerator 3 over denominator up diagonal strike 2 cubed end strike end fraction times fraction numerator up diagonal strike 2 cubed end strike over denominator 5 end fraction end cell row blank equals cell 2 squared over 3 squared times 3 squared over open parentheses 2 squared close parentheses squared minus 3 over 5 end cell row blank equals cell 2 squared over 3 squared times 3 squared over 2 to the power of 2 times 2 end exponent minus 3 over 5 end cell row blank equals cell 2 squared over 3 squared times 3 squared over 2 to the power of 4 minus 3 over 5 end cell row blank equals cell 2 squared over 2 to the power of 4 minus 3 over 5 end cell row blank equals cell 2 to the power of negative 2 end exponent minus 3 over 5 end cell row blank equals cell 1 over 2 squared minus 3 over 5 end cell row blank equals cell 1 fourth minus 3 over 5 end cell row blank equals cell fraction numerator 5 times 1 minus 4 times 3 over denominator 4 times 5 end fraction end cell row blank equals cell fraction numerator 5 minus 12 over denominator 20 end fraction end cell row blank equals cell negative 7 over 20 end cell row blank equals cell negative 0 comma 35 end cell end table    

Dengan demikian, hasil dari open parentheses 2 over 3 close parentheses squared times open parentheses negative 3 over 4 close parentheses squared minus 3 over 2 cubed colon 5 over 8  adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 0 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank comma end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 35 end table

Jadi, jawaban yang tepat adalah D

0

Roboguru

Pembahasan Soal:

Ingat kembali:

a to the power of n equals stack stack a cross times a cross times a cross times... cross times a with underbrace below with a space s e b a n y a k space n below comma space a not equal to 0

a to the power of m over a to the power of n equals a to the power of m minus n end exponent comma space a not equal to 0

open parentheses a over b close parentheses to the power of n equals a to the power of n over b to the power of n comma space a not equal to 0 comma space b not equal to 0 

a to the power of m times a to the power of n equals a to the power of m plus n end exponent. space a not equal to 0 

a to the power of 0 equals 1 comma space a not equal to 0 

Sehingga, kita peroleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 4 cubed times open parentheses 3 comma 5 close parentheses squared over denominator 4 squared times 7 squared end fraction end cell equals cell fraction numerator 4 cubed times open parentheses begin display style 35 over 10 end style close parentheses squared over denominator 4 squared times 7 squared end fraction end cell row blank equals cell fraction numerator open parentheses 2 squared close parentheses cubed times open parentheses begin display style 7 over 2 end style close parentheses squared over denominator open parentheses 2 squared close parentheses squared times 7 squared end fraction end cell row blank equals cell fraction numerator 2 to the power of 2 times 3 end exponent times begin display style 7 squared over 2 squared end style over denominator 2 to the power of 2 times 2 end exponent times 7 squared end fraction end cell row blank equals cell fraction numerator 2 to the power of 6 times begin display style 7 squared over 2 squared end style over denominator 2 to the power of 4 times 7 squared end fraction end cell row blank equals cell fraction numerator 2 to the power of 6 times 7 squared times 2 to the power of negative 1 end exponent over denominator 2 to the power of 4 times 7 squared end fraction end cell row blank equals cell 2 to the power of 6 plus open parentheses negative 1 close parentheses minus 4 end exponent times 7 to the power of 2 minus 2 end exponent end cell row blank equals cell 2 to the power of 1 times 7 to the power of 0 end cell row blank equals cell 2 times 1 end cell row blank equals 2 end table

Dengan demikian, hasil dari table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 4 cubed times open parentheses 3 comma 5 close parentheses squared over denominator 4 squared times 7 squared end fraction end cell end table  adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table

Jadi, jawaban yang tepat adalah D

1

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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