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Penyelesaian dari pertidaksamaan: 1 + 2 lo g x 4 ​ − 2 lo g x − 1 3 ​ < 0 adalah ...

Penyelesaian dari pertidaksamaan:  adalah ...

  1. 1 fourth less than x less than 4 

  2. 0 less than x less than 4 

  3. 0 less than x less than 1 fourth 

  4. 1 fourth less than x less than 1 atau 2 less than x less than 4 

  5. 1 fourth less than x less than 2 atau x greater than 4 

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D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah D.

jawaban yang benar adalah D.

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Pembahasan

Diketahui : Ingat kembali bahwa : Misal maka Cari penyelesaiannya dengan uji titik berdasarkan akar-akar sehingga nilai yang memenuhi yaitu Substitusikan nilai ke kedua pertidaksamaan tersebut untuk syarat maka untuk syarat maka Jadi, penyelesaiannya yaitu Oleh karena itu, jawaban yang benar adalah D.

Diketahui : 1 plus fraction numerator 4 over denominator log presuperscript 2 space x end fraction minus fraction numerator 3 over denominator log presuperscript 2 space x minus 1 end fraction less than 0

Ingat kembali bahwa : 

bullet space log presuperscript a space a to the power of m equals m bullet space log presuperscript a space f left parenthesis x right parenthesis less than log presuperscript a space g left parenthesis x right parenthesis comma space a greater than 1 rightwards arrow 0 less than f left parenthesis x right parenthesis less than g left parenthesis x right parenthesis 

Misal y equals log presuperscript 2 space x maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 plus fraction numerator 4 over denominator log presuperscript 2 space x end fraction minus fraction numerator 3 over denominator log presuperscript 2 space x minus 1 end fraction end cell less than 0 row cell 1 plus 4 over y minus fraction numerator 3 over denominator y minus 1 end fraction end cell less than 0 row cell fraction numerator left parenthesis y squared minus y right parenthesis plus 4 left parenthesis y minus 1 right parenthesis minus 3 left parenthesis y right parenthesis over denominator y squared minus y end fraction end cell less than 0 row cell fraction numerator y squared minus y plus 4 y minus 4 minus 3 y over denominator y squared minus y end fraction end cell less than 0 row cell fraction numerator y squared minus 4 over denominator y left parenthesis y minus 1 right parenthesis end fraction end cell less than 0 end table 

Cari penyelesaiannya dengan uji titik berdasarkan akar-akar y 

sehingga nilai y yang memenuhi yaitu 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 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank less than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank y end table table attributes columnalign right center left columnspacing 0px end attributes row blank less than blank 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 space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank atau end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 1 end table table attributes columnalign right center left columnspacing 0px end attributes row blank less than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank y end table table attributes columnalign right center left columnspacing 0px end attributes row blank less than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table 

Substitusikan nilai y equals log presuperscript 2 space x ke kedua pertidaksamaan tersebut

untuk negative 2 less than y less than 0 

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 2 end cell less than cell y less than 0 end cell row cell negative 2 end cell less than cell log presuperscript 2 space x less than 0 end cell row cell log presuperscript 2 space 2 to the power of negative 2 end exponent end cell less than cell log presuperscript 2 space x less than log presuperscript 2 space 2 to the power of 0 end cell row cell log presuperscript 2 space 1 fourth end cell less than cell log presuperscript 2 space x less than log presuperscript 2 space 1 end cell row cell 1 fourth end cell less than cell x less than 1 end cell end table 

syarat x greater than 0 maka 1 fourth less than x less than 1 

untuk 1 less than y less than 2 

table attributes columnalign right center left columnspacing 0px end attributes row 1 less than cell y less than 2 end cell row 1 less than cell log presuperscript 2 space x less than 2 end cell row cell log presuperscript 2 space 2 to the power of 1 end cell less than cell log presuperscript 2 space x less than log presuperscript 2 space 2 squared end cell row cell log presuperscript 2 space 2 end cell less than cell log presuperscript 2 space x less than log presuperscript 2 space 4 end cell row 2 less than cell x less than 4 end cell end table 

syarat x greater than 0 maka 2 less than x less than 4 

Jadi, penyelesaiannya yaitu 1 fourth less than x less than 1 space atau space 2 less than x less than 4 

Oleh karena itu, jawaban yang benar adalah D.

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