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Pertanyaan

Himpunan penyelesaian pertidaksamaan 2 to the power of 2 x plus 3 end exponent minus 17 cross times 2 to the power of x plus 2 less or equal than 0 adalah. . . .

  1. open curly brackets x vertical line space 1 less than x less or equal than 3 close curly brackets  

  2. open curly brackets x vertical line space 1 third less or equal than x less than 3 close curly brackets  

  3. open curly brackets x vertical line space 0 less or equal than x less or equal than 1 close curly brackets  

  4. open curly brackets x vertical line space minus 1 less than x less than 3 close curly brackets 

  5. open curly brackets x vertical line space minus 3 less or equal than x less or equal than 1 close curly brackets 

R. Hajrianti

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Pembahasan

Dengan menggunakan sifat bilangan berpangkat berikut.

  1.  a to the power of p cross times a to the power of q equals a to the power of p plus q end exponent
  2.  open parentheses a to the power of p close parentheses to the power of q equals a to the power of p cross times q end exponent

pertidaksamaan 2 to the power of 2 x plus 3 end exponent minus 17 cross times 2 to the power of x plus 2 less or equal than 0 dapat diuraikan menjadi sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of 2 x plus 3 end exponent minus 17 cross times 2 to the power of x plus 2 end cell less or equal than 0 row cell 2 to the power of 2 x end exponent cross times 2 cubed minus 17 cross times 2 to the power of x plus 2 end cell less or equal than 0 row cell open parentheses 2 to the power of x close parentheses squared cross times 8 minus 17 cross times 2 to the power of x plus 2 end cell less or equal than 0 end table

Misal, 2 to the power of x equals p maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 to the power of x close parentheses squared cross times 8 minus 17 cross times 2 to the power of x plus 2 end cell less or equal than 0 row cell p squared cross times 8 minus 17 cross times p plus 2 end cell less or equal than 0 row cell 8 p squared minus 17 p plus 2 end cell less or equal than 0 row cell open parentheses 8 p minus 1 close parentheses open parentheses p minus 2 close parentheses end cell less or equal than 0 end table

Kemudian, nilai p yang memenuhi pertidaksamaan di atas sebagai berikut.

  • Nilai p saat sama dengan 0

open parentheses 8 p minus 1 close parentheses open parentheses p minus 2 close parentheses equals 0

table attributes columnalign right center left columnspacing 0px end attributes row cell 8 p end cell equals 1 row blank equals cell 1 over 8 end cell end table atau p equals 2

  • Garis bilangan

Berdasarkan garis bilangan di atas, terdapat 3 interval yaitu p less or equal than 1 over 81 over 8 less or equal than p less or equal than 2, dan p greater or equal than 2. Uji titik pada setiap interval tersebut sebagai berikut.

  • Ketika p less or equal than 1 over 8, pilih p equals 0 maka diperoleh:

open parentheses 8 p minus 1 close parentheses open parentheses p minus 2 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 8 times 0 minus 1 close parentheses open parentheses 0 minus 2 close parentheses end cell equals cell open parentheses negative 1 close parentheses times open parentheses negative 2 close parentheses end cell row blank equals cell 2 space greater than 0 end cell end table

Berdasarkan uji titik di atas, ketika p less or equal than 1 over 8, nilai open parentheses 8 p minus 1 close parentheses open parentheses p minus 2 close parentheses positif atau lebih dari 0.

  • Ketika 1 over 8 less or equal than p less or equal than 2, pilih p equals 1, maka diperoleh:

open parentheses 8 p minus 1 close parentheses open parentheses p minus 2 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 8 times 1 minus 1 close parentheses open parentheses 1 minus 2 close parentheses end cell equals cell open parentheses 8 minus 1 close parentheses times open parentheses negative 1 close parentheses end cell row blank equals cell 7 times open parentheses negative 1 close parentheses end cell row blank equals cell negative 7 less than 0 end cell end table

Berdasarkan uji titik di atas, ketika 1 over 8 less or equal than p less or equal than 2, nilai open parentheses 8 p minus 1 close parentheses open parentheses p minus 2 close parentheses negatif atau kurang dari 0.

  • Ketika p greater or equal than 2, pilih p equals 3, maka diperoleh:

open parentheses 8 p minus 1 close parentheses open parentheses p minus 2 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 8 times 3 minus 1 close parentheses open parentheses 3 minus 2 close parentheses end cell equals cell open parentheses 24 minus 1 close parentheses times 1 end cell row blank equals cell 23 times 1 end cell row blank equals cell 23 space greater than 0 end cell end table

Berdasarkan uji titik di atas, ketika p greater or equal than 2, nilai open parentheses 8 p minus 1 close parentheses open parentheses p minus 2 close parentheses positif atau lebih dari 0.

Hasil di atas dapat dituliskan pada garis bilangan sebagai berikut.

Berdasarkan garis bilangan di atas, interval nilai p yang memenuhi pertidaksamaan open parentheses 8 p minus 1 close parentheses open parentheses p minus 2 close parentheses less or equal than 0 adalah 1 over 8 less or equal than p less or equal than 2, maka nilai x yang memenuhi sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over 8 end cell less or equal than cell space p less or equal than 2 end cell row cell 1 over 8 end cell less or equal than cell 2 to the power of x less or equal than 2 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of x end cell greater or equal than cell 1 over 8 end cell end table  dan  table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of x end cell less or equal than 2 row cell 2 to the power of x end cell less or equal than cell 2 to the power of 1 end cell end table

Dengan menggunakan sifat bilangan berpangkat a to the power of negative p end exponent equals 1 over a to the power of p comma space a not equal to 0, maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of x end cell greater or equal than cell 1 over 8 end cell row cell 2 to the power of x end cell greater or equal than cell 1 over 2 cubed end cell row cell 2 to the power of x end cell greater or equal than cell 2 to the power of negative 3 end exponent end cell end table dan table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of x end cell less or equal than 2 row cell 2 to the power of x end cell less or equal than cell 2 to the power of 1 end cell end table

Kemudian, ingat bahwa untuk a greater than 1, jika a to the power of f open parentheses x close parentheses end exponent greater or equal than a to the power of g open parentheses x close parentheses end exponent, maka f open parentheses x close parentheses greater or equal than g open parentheses x close parentheses dan jika a to the power of f open parentheses x close parentheses end exponent less or equal than a to the power of g open parentheses x close parentheses end exponent, maka f open parentheses x close parentheses less or equal than g open parentheses x close parentheses, sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of x end cell greater or equal than cell 2 to the power of negative 3 end exponent end cell row x greater or equal than cell negative 3 end cell end table  dan  table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of x end cell less or equal than cell 2 to the power of 1 end cell row x less or equal than 1 end table

x greater or equal than negative 3 space dan space x less or equal than 1 space rightwards double arrow space minus 3 less or equal than x less or equal than 1

Nilai x yang memenuhi yaitu negative 3 less or equal than x less or equal than 1, maka himpunan penyelesaian dari pertidaksamaan 2 to the power of 2 x plus 3 end exponent minus 17 cross times 2 to the power of x plus 2 less or equal than 0 adalah open curly brackets x vertical line space minus 3 less or equal than x less or equal than 1 close curly brackets.

Oleh karena itu, jawaban yang benar adalah E.

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