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Himpunan penyelesaian pertidaksamaan 5log (x+2)+5log (x−2)≤1 adalah...

Pertanyaan

Himpunan penyelesaian pertidaksamaan log presuperscript 5 space open parentheses x plus 2 close parentheses plus log presuperscript 5 space open parentheses x minus 2 close parentheses less or equal than 1 adalah...

  1. open curly brackets x vertical line x greater or equal than 2 close curly brackets 

  2. open curly brackets x vertical line x greater or equal than 3 close curly brackets 

  3. open curly brackets x vertical line x greater or equal than negative 2 close curly brackets 

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

  5. open curly brackets x vertical line minus 2 less than x less than 2 close curly brackets 

R. Hajrianti

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah D.

Pembahasan

log presuperscript 5 space open parentheses x plus 2 close parentheses plus log presuperscript 5 space open parentheses x minus 2 close parentheses less or equal than 1

Ingat sifat logaritma!

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript a space b plus log presuperscript a space c end cell equals cell log presuperscript a space b c end cell row cell log presuperscript a space a end cell equals 1 end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 5 space open parentheses x plus 2 close parentheses plus log presuperscript 5 space open parentheses x minus 2 close parentheses end cell less or equal than 1 row cell log presuperscript 5 space open parentheses x plus 2 close parentheses open parentheses x minus 2 close parentheses end cell less or equal than cell log presuperscript 5 space 5 end cell row cell open parentheses x plus 2 close parentheses open parentheses x minus 2 close parentheses end cell less or equal than 5 row cell x squared minus 4 end cell less or equal than 5 row cell x squared minus 4 minus 5 end cell less or equal than 0 row cell x squared minus 9 end cell less or equal than 0 row cell open parentheses x minus 3 close parentheses open parentheses x plus 3 close parentheses end cell less or equal than 0 row cell x minus 3 end cell less or equal than 0 row x less or equal than 3 row cell x plus 3 end cell less or equal than 0 row x less or equal than cell negative 3 end cell end table 

Syarat:

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

Untuk  negative 3 less or equal than x less or equal than 3, jika x equals 0 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 9 end cell less or equal than 0 row cell 0 squared minus 9 end cell less or equal than 0 row cell negative 9 end cell less or equal than cell 0 space open parentheses minus close parentheses end cell end table 

Untuk x less or equal than negative 3, jika x equals negative 4 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 9 end cell greater or equal than 0 row cell open parentheses negative 4 close parentheses squared minus 9 end cell greater or equal than 0 row cell 16 minus 9 end cell greater or equal than cell 0 space end cell row 7 greater or equal than cell 0 space open parentheses plus close parentheses end cell end table 

Untuk  x greater or equal than 3, jika x equals 4 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 9 end cell greater or equal than 0 row cell 4 squared minus 9 end cell greater or equal than 0 row cell 16 minus 9 end cell greater or equal than cell 0 space end cell row 7 greater or equal than cell 0 space open parentheses plus close parentheses end cell end table 

Jadi, himpunan penyelesaiannya adalah open curly brackets x vertical line 2 less than x less or equal than 3 close curly brackets.

Oleh karena itu, jawaban yang benar adalah D.

1rb+

1.0 (1 rating)

Tertius Ivan

Pembahasan tidak menjawab soal

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