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Tentukan himpunan penyelesaian pertidaksamaan berikut. a.

Pertanyaan

Tentukan himpunan penyelesaian pertidaksamaan berikut.

a. log presuperscript 5 space open parentheses fraction numerator x minus 2 over denominator x plus 2 end fraction close parentheses greater than 0 

Pembahasan Soal:

Ingat sifat-sifat bentuk logaritma:

log presuperscript a space a to the power of n equals n 

Pada pertidaksamaan logaritma untuk a greater than 0 jika:

log presuperscript a space f open parentheses x close parentheses greater than log presuperscript a space g open parentheses x close parentheses rightwards arrow f open parentheses x close parentheses greater than g open parentheses x close parentheses comma space f open parentheses x close parentheses greater than 0 comma space dan space g open parentheses x close parentheses greater than 0 

Diketahui log presuperscript 5 space open parentheses fraction numerator x minus 2 over denominator x plus 2 end fraction close parentheses greater than 0 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 5 space open parentheses fraction numerator x minus 2 over denominator x plus 2 end fraction close parentheses end cell greater than 0 row cell log presuperscript 5 space space open parentheses fraction numerator x minus 2 over denominator x plus 2 end fraction close parentheses end cell greater than cell log presuperscript 5 space 5 to the power of 0 end cell row cell log presuperscript 5 space open parentheses fraction numerator x minus 2 over denominator x plus 2 end fraction close parentheses end cell greater than cell log presuperscript 5 space 1 end cell row cell fraction numerator x minus 2 over denominator x plus 2 end fraction end cell greater than 1 row cell fraction numerator x minus 2 over denominator x plus 2 end fraction minus 1 end cell greater than 0 row cell fraction numerator x minus 2 over denominator x plus 2 end fraction minus fraction numerator x plus 2 over denominator x plus 2 end fraction end cell greater than 0 row cell fraction numerator x minus 2 minus x minus 2 over denominator x plus 2 end fraction end cell greater than 0 row cell fraction numerator negative 4 over denominator x plus 2 end fraction end cell greater than 0 end table 

Oleh karena pembilang bernilai negatif maka tanda pertidaksamaan berubah menjadi lebih kecil dari nol sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 2 end cell less than 0 row x less than cell negative 2 end cell end table 

Dengan demikian himpunan penyelesaianya adalah open curly brackets x vertical line x less than negative 2 comma space x element of straight R close curly brackets.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Amamah

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 11 Juli 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai  yang memenuhi pertidaksamaan  adalah...

Pembahasan Soal:

log space open parentheses x minus 2 close parentheses plus log space open parentheses x plus 1 close parentheses less than 2 log space open parentheses x plus 4 close parentheses

Ingat sifat logaritma!

log presuperscript a space b plus log presuperscript a space c equals log presuperscript a space b c 

table attributes columnalign right center left columnspacing 0px end attributes row cell log space open parentheses x minus 2 close parentheses plus log space open parentheses x plus 1 close parentheses end cell less than cell 2 log space open parentheses x plus 4 close parentheses end cell row cell log space open parentheses x minus 2 close parentheses open parentheses x plus 1 close parentheses end cell less than cell log space open parentheses x plus 4 close parentheses squared end cell row cell open parentheses x minus 2 close parentheses open parentheses x plus 1 close parentheses end cell less than cell open parentheses x plus 4 close parentheses squared end cell row cell x squared plus x minus 2 x minus 2 end cell less than cell x squared plus 4 x plus 4 x plus 16 end cell row cell x squared minus x minus 2 end cell less than cell x squared plus 8 x plus 16 end cell row cell negative x minus 8 x end cell less than cell 16 plus 2 end cell row cell negative 9 x end cell less than 18 row x greater than cell fraction numerator 18 over denominator negative 9 end fraction end cell row x greater than cell negative 2 end cell end table 

Jadi, nilai x yang memenuhi pertidaksamaan tersebut adalah x greater than negative 2.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Penyelesaian dari pertidaksamaan adalah ....

Pembahasan Soal:

table row 25 row space end table log open parentheses x minus 3 close parentheses plus table row 25 row space end table log open parentheses x plus 1 close parentheses less or equal than 1 half  S y a r a t space open parentheses x minus 3 close parentheses greater than 0 rightwards arrow x greater than 3  open parentheses x plus 1 close parentheses greater than 0 rightwards arrow x greater than negative 1  table row 25 row space end table log open parentheses x minus 3 close parentheses plus table row 25 row space end table log open parentheses x plus 1 close parentheses less or equal than 1 half  table row 25 row space end table log open parentheses x minus 3 close parentheses open parentheses x plus 1 close parentheses less or equal than 1 half table row 25 row space end table log 25  table row 25 row space end table log left parenthesis x squared minus 2 x minus 3 right parenthesis less or equal than table row 25 row space end table log 25 to the power of 1 half end exponent  x squared minus 2 x minus 3 less or equal than 5  x squared minus 2 x minus 8 less or equal than 0  left parenthesis x minus 4 right parenthesis left parenthesis x plus 2 right parenthesis less or equal than 0  x equals 4 space a t a u space x equals negative 2  U j i space t i t i k space u n t u k space x space equals space 0 comma space m a k a space 0 squared minus 2 open parentheses 0 close parentheses minus 3 equals negative 3 less or equal than 5 left parenthesis m e m e n u h i right parenthesis  K a r e n a space minus 2 less or equal than 0 less or equal than 4 comma space m a k a space u n t u k space d a e r a h space minus 2 less or equal than x less or equal than 4 space b e r n i l a i space n e g a t i f  A m b i l space y a n g space n e g a t i f comma space m a k a space d i d a p a t space p e n y e l e s a i a n

0

Roboguru

Jika , nilai x yang memenuhi adalah...

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.

0

Roboguru

Himpunan penyelesaian pertidaksamaan  adalah. . . .

Pembahasan Soal:

Diketahui pertidaksamaan log presuperscript 2 open parentheses x minus 2 close parentheses plus log presuperscript 2 open parentheses x plus 5 close parentheses less or equal than 3. Dengan menggunakan sifat bentuk logaritma berikut.

untuk a comma space b comma space c greater than 0 dan a not equal to 1, berlaku:

  1. log presuperscript a space open parentheses b c close parentheses equals log presuperscript a space b plus log presuperscript a space c
  2.  log presuperscript a space a equals 1
  3. log presuperscript a space b to the power of m equals m cross times log presuperscript a space b

maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 open parentheses x minus 2 close parentheses plus log presuperscript 2 open parentheses x plus 5 close parentheses end cell less or equal than 3 row cell log presuperscript 2 open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell less or equal than cell 3 cross times 1 end cell row cell log presuperscript 2 open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell less or equal than cell 3 cross times log presuperscript 2 space 2 end cell row cell log presuperscript 2 open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell less or equal than cell log presuperscript 2 space 2 cubed end cell row cell log presuperscript 2 open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell less or equal than cell log presuperscript 2 space 8 end cell end table

Kemudian, ingat bahwa, jika log presuperscript a space f open parentheses x close parentheses less or equal than log presuperscript a space pa greater than 1 dan p greater than 0 maka f open parentheses x close parentheses less or equal than p dan f open parentheses x close parentheses greater than 0.

Misal, f open parentheses x close parentheses equals open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses maka dari pertidaksamaan log presuperscript 2 open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses less or equal than log presuperscript 2 space 8 diperoleh:

  • f open parentheses x close parentheses less or equal than 8

 table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell less or equal than 8 row cell open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell less or equal than 8 row cell x squared plus 3 x minus 10 end cell less or equal than 8 row cell x squared plus 3 x minus 10 minus 8 end cell less or equal than 0 row cell x squared plus 3 x minus 18 end cell less or equal than 0 row cell open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses end cell less or equal than 0 end table

Nilai x yang memenuhi saat sama dengan 0 sebagai berikut.

open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses equals 0 table row cell x equals negative 6 end cell atau cell x equals 3 end cell end table

Lalu, nilai x di atas dapat dituliskan pada garis bilangan sebagai berikut.

Berdasarkan garis bilangan di atas, terdapat 3 daerah yaitu x less than negative 6negative 6 less than x less than 3, dan x greater than 3. Uji titik pada setiap daerah tersebut sebagai berikut.

Ketika x less than negative 6, pilih x equals negative 7 maka diperoleh:

open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative 7 plus 6 close parentheses times open parentheses negative 7 minus 3 close parentheses end cell equals cell open parentheses negative 1 close parentheses times open parentheses negative 10 close parentheses equals 10 space greater than 0 end cell end table

Berdasarkan uji titik di atas, ketika x less than negative 6, nilai open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses lebih dari 0.

Ketika negative 6 less than x less than 3, pilih x equals 0, maka diperoleh:

open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 0 plus 6 close parentheses times open parentheses 0 minus 3 close parentheses end cell equals cell 6 times open parentheses negative 3 close parentheses equals negative 18 space less than 0 end cell end table

Berdasarkan uji titik di atas, ketika negative 6 less than x less than 3, nilai open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses kurang dari 0.

Ketika x greater than 3, pilih x equals 4, maka diperoleh:

open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 4 plus 6 close parentheses times open parentheses 4 minus 3 close parentheses end cell equals cell 10 times 1 equals 10 space greater than 0 end cell end table

Berdasarkan uji titik di atas, ketika x greater than 3, nilai open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses lebih dari 0.

Hasil di atas dapat dituliskan pada garis bilangan sebagai berikut.

Berdasarkan garis bilangan di atas, nilai x yang memenuhi pertidaksamaan open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses less or equal than 0 adalah negative 6 less than x less than 3.

  • f open parentheses x close parentheses greater than 0

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell greater than 0 row cell open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell greater than 0 end table

Nilai x yang memenuhi saat sama dengan 0 sebagai berikut.

open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses equals 0 table row cell x equals 2 end cell atau cell x equals negative 5 end cell end table

Lalu, nilai x di atas dapat dituliskan pada garis bilangan sebagai berikut.

Berdasarkan garis bilangan di atas, terdapat 3 daerah yaitu x less than negative 5negative 5 less than x less than 2, dan x greater than 2. Uji titik pada setiap daerah tersebut sebagai berikut.

Ketika x less than negative 5, pilih x equals negative 6 maka diperoleh:

open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative 6 minus 2 close parentheses times open parentheses negative 6 plus 5 close parentheses end cell equals cell open parentheses negative 8 close parentheses times open parentheses negative 1 close parentheses equals 8 space greater than 0 end cell end table

Berdasarkan uji titik di atas, ketika x less than negative 5, nilai open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses lebih dari 0.

Ketika negative 5 less than x less than 2, pilih x equals 0, maka diperoleh:

open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 0 minus 2 close parentheses times open parentheses 0 plus 5 close parentheses end cell equals cell open parentheses negative 2 close parentheses times 5 equals negative 10 space less than 0 end cell end table

Berdasarkan uji titik di atas, ketika negative 5 less than x less than 2, nilai open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses kurang dari 0.

Ketika x greater than 2, pilih x equals 3, maka diperoleh:

open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 3 minus 2 close parentheses times open parentheses 3 plus 5 close parentheses end cell equals cell 1 times 8 equals 8 space greater than 0 end cell end table

Berdasarkan uji titik di atas, ketika x greater than 2, nilai open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses lebih dari 0.

Hasil di atas dapat dituliskan pada garis bilangan sebagai berikut.

Berdasarkan garis bilangan di atas, nilai x yang memenuhi pertidaksamaan open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses greater than 0 adalah x less than negative 5 atau x greater than 2.

Kemudian, numerus dari suatu bentuk logaritma bernilai lebih dari 0, maka dari log presuperscript 2 open parentheses x minus 2 close parentheses plus log presuperscript 2 open parentheses x plus 5 close parentheses less or equal than 3 terdapat syarat x minus 2 greater than 0 dan x plus 5 greater than 0. Jika x minus 2 greater than 0, maka x greater than 2 dan jika x plus 5 greater than 0, maka x greater than negative 5.

Lalu, penyelesaian dari log presuperscript 2 open parentheses x minus 2 close parentheses plus log presuperscript 2 open parentheses x plus 5 close parentheses less or equal than 3 adalah irisan dari penyelesaian f open parentheses x close parentheses less or equal than 8f open parentheses x close parentheses greater than 0x minus 2 greater than 0, dan x plus 5 greater than 0. Irisan dari keempat penyelesaian tersebut dapat ditentukan dengan garis bilangan sebagai berikut.

 

Berdasarkan garis bilangan di atas, himpunan penyelesaian pertidaksamaan log presuperscript 2 open parentheses x minus 2 close parentheses plus log presuperscript 2 open parentheses x plus 5 close parentheses less or equal than 3 adalah open curly brackets x vertical line space 2 less than x less or equal than 3 close curly brackets.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Himpunan penyelesaian dari pertidaksamaan  adalah...

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.

Oleh karena itu, jawaban yang benar adalah C.

 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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