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Jika 2log(x2+x+4)<5log625, nilai x yang memenuhi adalah...

Pertanyaan

Jika log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses less than log presuperscript 5 space 625, nilai x yang memenuhi adalah...

  1. x less than negative 4 

  2. x greater than 3 

  3. negative 4 less than x less than 3 

  4. x less than negative 4 atau x greater than 3 

  5. x less than negative 3 atau x greater than 4 

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses end cell less than cell log presuperscript 5 space 625 end cell row cell log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses end cell less than cell log presuperscript 5 space 5 to the power of 4 end cell row cell log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses end cell less than cell 4 cross times log presuperscript 5 5 end cell row cell log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses end cell less than cell 4 cross times 1 end cell row cell log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses end cell less than 4 row cell log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses end cell less than cell log presuperscript 2 space 16 end cell row cell x squared plus x plus 4 end cell less than 16 row cell x squared plus x plus 4 minus 16 end cell less than 0 row cell x squared plus x minus 12 end cell less than 0 row cell open parentheses x minus 3 close parentheses open parentheses x plus 4 close parentheses end cell less than 0 row cell x minus 3 end cell less than 0 row x less than 3 row cell x plus 4 end cell less than 0 row x less than cell negative 4 end cell end table 

Menentukan daerah penyelesaian

Untuk x greater than negative 4 dan x less than 3, jika x equals 0 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus x minus 12 end cell less than 0 row cell 0 squared plus 0 minus 12 end cell less than 0 row cell negative 12 end cell less than cell 0 space open parentheses negatif close parentheses end cell end table 

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

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus x minus 12 end cell greater than 0 row cell 4 squared plus 4 minus 12 end cell greater than 0 row cell 16 plus 4 minus 12 end cell greater than cell 0 space end cell row cell 20 minus 12 end cell greater than 0 row 8 greater than cell 0 space open parentheses positif close parentheses end cell end table 

Untuk x less than negative 4, jika x equals negative 5 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus x minus 12 end cell greater than 0 row cell open parentheses negative 5 close parentheses squared plus open parentheses negative 5 close parentheses minus 12 end cell greater than 0 row cell 25 minus 5 minus 12 end cell greater than 0 row cell 25 minus 17 end cell greater than 0 row 8 greater than cell space open parentheses positif close parentheses end cell end table 

Karena tanda pertidaksamaannya adalah kurang dari " less than " maka daerah penyelesaiannya adalah yang bernilai negatif.

Jadi, nilai x yang memenuhi adalah negative 4 less than x less than 3.

Oleh karena itu, jawaban yang benar adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Hajrianti

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 07 Oktober 2021

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

Pembahasan Soal:

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.

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Pembahasan Soal:

Ingat sifat logaritma!

log presuperscript a space a equals 1   

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 6 space open parentheses x squared minus x minus 6 close parentheses end cell less than 1 row cell log presuperscript 6 space open parentheses x squared minus x minus 6 close parentheses end cell less than cell log presuperscript 6 space 6 end cell row cell x squared minus x minus 6 end cell less than 6 row cell x squared minus x minus 6 minus 6 end cell less than 0 row cell x squared minus x minus 12 end cell less than 0 row cell open parentheses x plus 3 close parentheses open parentheses x minus 4 close parentheses end cell less than 0 row cell x plus 3 end cell less than 0 row x less than cell negative 3 end cell row cell x minus 4 end cell less than 0 row x less than 4 end table  

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Pembahasan Soal:

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.

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Penyelesaian pertidaksamaan logaritma 3log(2x−1)<2 adalah...

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 3 space open parentheses 2 x minus 1 close parentheses end cell less than 2 row cell log presuperscript 3 space open parentheses 2 x minus 1 close parentheses end cell less than cell log presuperscript 3 space 3 squared end cell row cell 2 x minus 1 end cell less than 9 row cell 2 x end cell less than 10 row x less than 5 end table 

Jadi, nilai x less than 5.

Oleh karena itu, jawaban yang benar adalah E.

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Nilai x yang memenuhi 2logx−xlog2>0 adalah...

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 space x minus log presuperscript x space 2 end cell greater than 0 row cell log presuperscript 2 space x minus fraction numerator 1 over denominator log presuperscript 2 space x end fraction end cell greater than 0 row cell log presuperscript 2 space x open square brackets log presuperscript 2 space x minus fraction numerator 1 over denominator log presuperscript 2 space x end fraction close square brackets end cell greater than 0 row cell log presuperscript 2 superscript 2 space x minus 1 end cell greater than 0 row cell open parentheses log presuperscript 2 space x minus 1 close parentheses open parentheses log presuperscript 2 space x plus 1 close parentheses end cell greater than 0 row cell log presuperscript 2 space x minus 1 end cell greater than 0 row cell log presuperscript 2 space x end cell greater than 1 row x greater than cell 2 to the power of 1 end cell row x greater than 2 row cell log presuperscript 2 space x plus 1 end cell greater than 0 row cell log presuperscript 2 space x end cell greater than cell negative 1 end cell row x greater than cell 2 to the power of negative 1 end exponent end cell row x greater than cell 1 half end cell end table 

Syarat: 

x greater than 0 comma space x not equal to 1 

Untuk x greater than 2, jika x equals 3 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 superscript 2 space x minus 1 end cell greater than 0 row cell log presuperscript 2 superscript 2 space 4 minus 1 end cell greater than 0 row cell open parentheses log presuperscript 2 space 4 close parentheses squared minus 1 end cell greater than 0 row cell 2 squared minus 1 end cell greater than 0 row cell 4 minus 1 end cell greater than 0 row 3 greater than cell 0 space open parentheses plus close parentheses end cell end table 

Untuk x less than 1 half, jika x equals 1 fourth maka:

 table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 superscript 2 space x minus 1 end cell greater than 0 row cell log presuperscript 2 superscript 2 space 1 fourth minus 1 end cell greater than 0 row cell open parentheses log presuperscript 2 space 4 to the power of negative 1 end exponent close parentheses squared minus 1 end cell greater than 0 row cell 2 to the power of negative 2 end exponent minus 1 end cell greater than 0 row cell 1 fourth minus 1 end cell greater than 0 row cell negative 3 over 4 end cell less than cell 0 space open parentheses minus close parentheses end cell end table 

Untuk 1 half less than x less than 2, jika x equals 0 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 superscript 2 space x minus 1 end cell greater than 0 row cell log presuperscript 2 superscript 2 space 0 minus 1 end cell greater than 0 row cell open parentheses log presuperscript 2 space 0 close parentheses squared minus 1 end cell greater than 0 row cell 0 minus 1 end cell greater than 0 row cell negative 1 end cell less than cell 0 space open parentheses minus close parentheses end cell row blank blank blank end table 

Jadi, nilai x yang memenuhi pertidaksamaan tersebut adalah 1 half less than x less than 1 atau x greater than 2.

Oleh karena itu, jawaban yang benar adalah E.

 

0

Roboguru

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