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Himpunan penyelesaian dari pertidaksamaan 4 lo g ( x − 1 ) ≤ 4 1 ​ lo g ( 2 x − 1 ) adalah...

Himpunan penyelesaian dari pertidaksamaan  adalah...

  1. open curly brackets x vertical line 1 half less than x less than 1 close curly brackets 

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

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

  4. open curly brackets x vertical line 0 less or equal than x less than 1 half close curly brackets 

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

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R. Hajrianti

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Pembahasan

Menentukan daerah penyelesaian Syarat: Untuk , jika maka: Untuk , jika maka: Untuk , jika maka: Jadi, himpunan penyelesaian pertidaksamaan tersebut adalah . Oleh karena itu, jawaban yang benar adalah C.

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 4 space open parentheses x minus 1 close parentheses end cell less or equal than cell log presuperscript 1 fourth end presuperscript space open parentheses 2 x minus 1 close parentheses end cell row cell log presuperscript 4 space open parentheses x minus 1 close parentheses end cell less or equal than cell scriptbase log space end scriptbase presuperscript 4 to the power of negative 1 end exponent end presuperscript open parentheses 2 x minus 1 close parentheses end cell row cell log presuperscript 4 space open parentheses x minus 1 close parentheses end cell less or equal than cell scriptbase log space end scriptbase presuperscript 4 open parentheses 2 x minus 1 close parentheses to the power of negative 1 end exponent end cell row cell x minus 1 end cell less or equal than cell open parentheses 2 x minus 1 close parentheses to the power of negative 1 end exponent end cell row cell x minus 1 end cell less or equal than cell fraction numerator 1 over denominator 2 x minus 1 end fraction end cell row cell open parentheses x minus 1 close parentheses open parentheses 2 x minus 1 close parentheses end cell less or equal than 1 row cell 2 x squared minus x minus 2 x plus 1 end cell less or equal than 1 row cell 2 x squared minus 3 x plus 1 minus 1 end cell less or equal than 0 row cell 2 x squared minus 3 x end cell less or equal than 0 row cell x open parentheses 2 x minus 3 close parentheses end cell less or equal than 0 row x less or equal than 0 row cell 2 x minus 3 end cell less or equal than 0 row cell 2 x end cell less or equal than 3 row x less or equal than cell 3 over 2 end cell end table 

Menentukan daerah penyelesaian

Syarat:

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 1 end cell greater than 0 row x greater than cell 1 space open parentheses plus close parentheses end cell row cell 2 x minus 1 end cell greater than 0 row cell 2 x end cell greater than 1 row x greater than cell 1 half space open parentheses plus close parentheses end cell end table 

Untuk x less or equal than 0, jika x equals negative 1 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x squared minus 3 x end cell less or equal than 0 row cell 2 open parentheses negative 1 close parentheses squared minus 3 open parentheses negative 1 close parentheses end cell less or equal than 0 row cell 2 plus 3 end cell less or equal than 0 row 5 greater or equal than cell 0 space open parentheses plus close parentheses end cell end table 

Untuk 0 less or equal than x less or equal than 3 over 2, jika x equals 1 maka:

 table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x squared minus 3 x end cell less or equal than 0 row cell 2 open parentheses 1 close parentheses squared minus 3 open parentheses 1 close parentheses end cell less or equal than 0 row cell 2 minus 3 end cell less or equal than 0 row cell negative 1 end cell less or equal than cell 0 space open parentheses minus close parentheses end cell end table 

Untuk x greater or equal than 3 over 2, jika x equals 2 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x squared minus 3 x end cell less or equal than 0 row cell 2 open parentheses 2 close parentheses squared minus 3 open parentheses 2 close parentheses end cell less or equal than 0 row cell 2 open parentheses 4 close parentheses minus 6 end cell less or equal than 0 row cell 8 minus 6 end cell less or equal than 0 row 2 greater or equal than cell 0 space open parentheses plus close parentheses end cell end table 

Jadi, himpunan penyelesaian pertidaksamaan tersebut adalah open curly brackets x vertical line 1 less than x less or equal than 3 over 2 close curly brackets.

Oleh karena itu, jawaban yang benar adalah C.

 

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