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Jika  dan  memenuhi sistem persamaan maka nilai

Pertanyaan

Jika x dan y memenuhi sistem persamaan vertical ellipsis open table attributes columnalign right end attributes row cell 2 to the power of x plus 1 end exponent minus 3 to the power of y equals 15 end cell row cell 2 to the power of x plus 3 to the power of y equals 9 end cell end table close curly brackets maka nilai x plus y equals...

  1. 1

  2. 2

  3. 3

  4. 4

  5. 5

Pembahasan Video:

Pembahasan Soal:

Diketahui sistem persamaan:

open curly brackets table attributes columnalign left end attributes row cell 2 to the power of x plus 1 end exponent minus 3 to the power of y equals 15 end cell row cell 2 to the power of x plus 3 to the power of y equals 9 end cell end table close space atau space open curly brackets table attributes columnalign left end attributes row cell 2 to the power of x times 2 minus 3 to the power of y equals 15 end cell row cell 2 to the power of x plus 3 to the power of y equals 9 end cell end table close

Gunakan pemisalan agar bentuk sistem persamaan eksponen di atas menjadi SPLDV yang dapat lebih mudah untuk dikerjakan.

Misalkan:

  • 2 to the power of x equals a
  • 3 to the power of y equals b

Sehingga sistem menjadi

open curly brackets table attributes columnalign left end attributes row cell 2 a minus b equals 15 space... left parenthesis straight i right parenthesis end cell row cell a plus b equals 9 space space space space space... left parenthesis ii right parenthesis end cell end table close

Lakukan subsitusi dan eliminasi sehingga didapat nilai a space dan space b.

bottom enclose table attributes columnalign right center left columnspacing 2px end attributes row cell 2 a minus b end cell equals 15 row cell a plus b end cell equals cell 9 space space space plus end cell end table end enclose table attributes columnalign right center columnspacing 2px 2px end attributes row cell space space space space space space 3 a end cell equals 24 row a equals 8 end table

Substitusi nilai a ke persamaan (ii) .

table attributes columnalign right center left columnspacing 0px end attributes row cell a plus b end cell equals 9 row cell 8 plus b end cell equals 9 row b equals cell 9 minus 8 end cell row b equals 1 end table

Karena telah dimisalkan 2 to the power of x equals a dan 3 to the power of y equals b, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of x end cell equals a row cell 2 to the power of x end cell equals 8 row cell 2 to the power of x end cell equals cell 2 cubed end cell row x equals 3 end table              table attributes columnalign right center left columnspacing 0px end attributes row cell 3 to the power of y end cell equals b row cell 3 to the power of y end cell equals 1 row cell 3 to the power of y end cell equals cell 3 to the power of 0 end cell row y equals 0 end table

Ditanyakan nilai dari x plus y, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus y end cell equals cell 3 plus 0 end cell row blank equals 3 end table

Dengan demikian, nilai dari x plus y adalah 3.

Oleh karena itu, jawaban yang benar adalah C.

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

Jika  maka .

Pembahasan Soal:

Ingat kembali:

  • Jika a to the power of f open parentheses x close parentheses end exponent equals a to the power of g open parentheses x close parentheses end exponent dengan a>0 dan a=1, maka maka f open parentheses x close parentheses equals g open parentheses x close parentheses
  • 1 over a to the power of m equals a to the power of negative m end exponent dengan a not equal to 0
  • a equals a to the power of 1
  • a to the power of m a to the power of n equals a to the power of m plus 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

Dengan demikian, 

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 over 32 close parentheses to the power of 3 minus 5 x end exponent end cell equals cell 2 times 2 squared times 2 cubed times horizontal ellipsis times 2 to the power of 20 end cell row cell open parentheses 1 over 2 to the power of 5 close parentheses to the power of 3 minus 5 x end exponent end cell equals cell 2 to the power of 1 plus 2 plus 3 plus midline horizontal ellipsis plus 20 end exponent end cell row cell open parentheses 2 to the power of negative 5 end exponent close parentheses to the power of 3 minus 5 x end exponent end cell equals cell 2 to the power of 210 end cell row cell 2 to the power of negative 15 plus 25 x end exponent end cell equals cell 2 to the power of 210 end cell end table

sehingga karena a=2 yang berarti a>0 dan a=1, maka diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 15 plus 25 x end cell equals 210 row cell 25 x end cell equals cell 210 plus 15 end cell row cell 25 x end cell equals 225 row x equals cell 225 over 25 end cell row x equals 9 end table

Oleh karena itu, jawaban yang benar adalah B.space 

0

Roboguru

Penyelesaian persamaan   adalah ....

Pembahasan Soal:

Ingat kembali:

  • Jika a to the power of f open parentheses x close parentheses end exponent equals a to the power of g open parentheses x close parentheses end exponent dan a not equal to 0 maka f open parentheses x close parentheses equals g open parentheses x close parentheses
  • a equals a to the power of 1
  • a to the power of m a to the power of n equals a to the power of m plus n end exponent
  • 1 over a to the power of m equals a to the power of negative m end exponent dengan 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

Oleh karena persamaan dapat diubah menjadi bentuk seperti berikut:

3 times 3 to the power of 1 minus 5 x end exponent equals 1 over 9 to the power of 2 x end exponent 3 to the power of 1 times 3 to the power of 1 minus 5 x end exponent equals 1 over open parentheses 3 squared close parentheses to the power of 2 x end exponent 3 to the power of 1 plus 1 minus 5 x end exponent equals 1 over 3 to the power of 4 x end exponent 3 to the power of 2 minus 5 x end exponent equals 3 to the power of negative 4 x end exponent 

sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 minus 5 x end cell equals cell negative 4 x end cell row cell 2 minus 5 x plus 4 x end cell equals 0 row cell negative 5 x plus 4 x end cell equals cell 0 minus 2 end cell row cell negative x end cell equals cell negative 2 end cell row x equals 2 end table

padi, Penyelesaian persamaan  3 times 3 to the power of 1 minus 5 x end exponent equals 1 over 9 to the power of 2 x end exponent adalah 2.

Oleh karena itu, jawaban yang benar adalah D.space 

1

Roboguru

Tentukanlah nilai  supaya:

Pembahasan Soal:

Diketahui:

fraction numerator 2 cross times 2 squared cross times 2 cubed cross times... cross times 2 to the power of x over denominator 4 cross times 4 squared cross times 4 cubed cross times... cross times 4 to the power of 14 end fraction equals 1

Ditanya:

Nilai x

Perlu diingat bahwa:

a to the power of m times a to the power of n equals a to the power of m plus 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 times n end exponent

Perhatikan perhitungan berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 cross times 2 squared cross times 2 cubed cross times... cross times 2 to the power of x over denominator 4 cross times 4 squared cross times 4 cubed cross times... cross times 4 to the power of 14 end fraction end cell equals 1 row cell 2 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end exponent over 4 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus 14 end exponent end cell equals 1 row cell 2 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end exponent over 4 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus 14 end exponent end cell equals 1 row cell 2 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end exponent over open parentheses 2 squared close parentheses to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus 14 end exponent end cell equals 1 row cell 2 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end exponent over 2 to the power of 2 plus 4 plus 6 plus 8 plus 10 plus... plus 28 end exponent end cell equals 1 row cell 2 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end exponent end cell equals cell 2 to the power of 2 plus 4 plus 6 plus 8 plus 10 plus... plus 28 end exponent end cell row cell 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end cell equals cell 2 plus 4 plus 6 plus 8 plus 10 plus... plus 28 end cell row cell 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end cell equals cell 2 open parentheses 1 plus 2 plus 3 plus 4 plus 5 plus... plus 14 close parentheses end cell end table

Dengan menggunakan deret aritmatika yaitu S subscript n equals n over 2 open parentheses a plus U subscript n close parentheses, didapat:

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end cell equals cell 2 open parentheses 1 plus 2 plus 3 plus 4 plus 5 plus... plus 14 close parentheses end cell row cell x over 2 open parentheses 1 plus x close parentheses end cell equals cell 2 times 14 over 2 open parentheses 1 plus 14 close parentheses end cell row cell x open parentheses x plus 1 close parentheses end cell equals cell 2 times 14 times 15 end cell row cell x squared plus x end cell equals 420 row cell x squared plus x minus 420 end cell equals 0 row cell open parentheses x plus 21 close parentheses open parentheses x minus 20 close parentheses end cell equals 0 end table
x plus 21 equals 0 space atau space x minus 20 equals 0 x equals negative 21 space atau space x equals 20                                 

Karena nilai x tidak mungkin bernilai negatif, maka nilai x adalah 20.

Jadi, nilai x yang tepat adalah 20.

1

Roboguru

Nilai  yang memenuhi Penyelesaian Persamaan Eksponensial  adalah ...

Pembahasan Soal:

Ingat kembali jika a to the power of f open parentheses x close parentheses end exponent equals a to the power of p dan a not equal to 0 maka f open parentheses x close parentheses equals p dan ingat sifat bilangan berpangkat:

a to the power of m plus n end exponent equals a to the power of m times a to the power of n 

Misalkan 2 to the power of x equals y maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of x minus 2 to the power of x plus 2 end exponent end cell equals cell negative 12 end cell row cell 2 to the power of x minus 2 to the power of x times 2 squared end cell equals cell negative 12 end cell row cell 2 to the power of x minus 2 to the power of x times 4 end cell equals cell negative 12 end cell row cell y minus 4 y end cell equals cell negative 12 end cell row cell negative 3 y end cell equals cell negative 12 end cell row y equals cell fraction numerator negative 12 over denominator negative 3 end fraction end cell row y equals 4 end table

maka nilai x yang memenuhi adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of x end cell equals 4 row cell 2 to the power of x end cell equals cell 2 squared end cell row x equals 2 end table 

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

Nilai  yang memenuhi persamaan adalah ....

Pembahasan Soal:

Ingat kembali:

  • Jika a to the power of f open parentheses x close parentheses end exponent equals a to the power of g open parentheses x close parentheses end exponent dan a not equal to 0 maka f open parentheses x close parentheses equals g open parentheses x close parentheses
  • a equals a to the power of 1
  • 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
  • n-th root of a to the power of m end root equals a to the power of m over n end exponent dengan a greater than 0
  • a to the power of m a to the power of n equals a to the power of m plus n end exponent

Oleh karena persamaan dapat diubah menjadi bentuk seperti berikut:

2 square root of 2 to the power of 3 x minus 1 end exponent end root equals 4 square root of 8 to the power of 1 minus 3 x end exponent end root 2 to the power of 1 square root of 2 to the power of 3 x minus 1 end exponent end root equals 2 squared square root of open parentheses 2 cubed close parentheses to the power of 1 minus 3 x end exponent end root 2 to the power of 1 square root of 2 to the power of 3 x minus 1 end exponent end root equals 2 squared square root of 2 to the power of 3 minus 9 x end exponent end root 2 to the power of 1 times 2 to the power of fraction numerator 3 x minus 1 over denominator 2 end fraction end exponent equals 2 squared times 2 to the power of fraction numerator 3 minus 9 x over denominator 2 end fraction end exponent 2 to the power of 1 plus fraction numerator 3 x minus 1 over denominator 2 end fraction end exponent equals 2 to the power of 2 plus fraction numerator 3 minus 9 x over denominator 2 end fraction end exponent 2 to the power of 2 over 2 plus fraction numerator 3 x minus 1 over denominator 2 end fraction end exponent equals 2 to the power of 4 over 2 plus fraction numerator 3 minus 9 x over denominator 2 end fraction end exponent 2 to the power of fraction numerator 3 x minus 1 plus 2 over denominator 2 end fraction end exponent equals 2 to the power of fraction numerator 4 plus 3 minus 9 x over denominator 2 end fraction end exponent 2 to the power of fraction numerator 3 x plus 1 over denominator 2 end fraction end exponent equals 2 to the power of fraction numerator 7 minus 9 x over denominator 2 end fraction end exponent 

maka diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 3 x plus 1 over denominator 2 end fraction end cell equals cell fraction numerator 7 minus 9 x over denominator 2 end fraction end cell row cell 3 x plus 1 end cell equals cell 7 minus 9 x end cell row cell 3 x end cell equals cell 7 minus 9 x minus 1 end cell row cell 3 x plus 9 x end cell equals cell 7 minus 1 end cell row cell 12 x end cell equals 6 row x equals cell 6 over 12 end cell row x equals cell fraction numerator 6 divided by 6 over denominator 12 divided by 6 end fraction end cell row x equals cell 1 half end cell end table 

Oleh karena itu, jawaban yang benar adalah B.space 

1

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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