RoboguruRoboguru
SD

Himpunan penyelesaian dari pertidaksamaan 4log (x−1)≤41​log (2x−1) adalah...

Pertanyaan

Himpunan penyelesaian dari pertidaksamaan log presuperscript 4 space open parentheses x minus 1 close parentheses less or equal than log presuperscript 1 fourth end presuperscript space open parentheses 2 x minus 1 close parentheses 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 

R. Hajrianti

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Pembahasan

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.

 

312

5.0 (1 rating)

Pertanyaan serupa

Jika 2log (x2+x+4)<5log 625, nilai x yang memenuhi adalah...

916

5.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia