Iklan

Iklan

Pertanyaan

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 

Iklan

R. Hajrianti

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Iklan

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.

 

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

2

Iklan

Iklan

Pertanyaan serupa

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

8

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

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia