Roboguru

Carilah semua nilai  yang memenuhi pertidaksamaan berikut. a.

Pertanyaan

Carilah semua nilai x yang memenuhi pertidaksamaan berikut.

a. 2 cross times log presuperscript 9 space open parentheses 2 x minus 3 close parentheses less or equal than log presuperscript begin inline style 1 third end style end presuperscript space open parentheses x plus 1 close parentheses plus 1 

Pembahasan Soal:

Ingat sifat-sifat bentuk logaritma:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript a space a end cell equals 1 row cell log presuperscript a to the power of m end presuperscript space b to the power of n end cell equals cell n over m space log presuperscript a space b end cell row cell log presuperscript a space begin inline style 1 over b end style end cell equals cell negative log presuperscript a space b end cell end table  

Pada pertidaksamaan logaritma untuk a greater than 1 jika:

log presuperscript a space f open parentheses x close parentheses less or equal than log presuperscript a space g open parentheses x close parentheses rightwards arrow f open parentheses x close parentheses less or equal 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 2 cross times log presuperscript 9 space open parentheses 2 x minus 3 close parentheses less or equal than log presuperscript begin inline style 1 third end style end presuperscript space open parentheses x plus 1 close parentheses plus 1 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 cross times log presuperscript 9 space open parentheses 2 x minus 3 close parentheses end cell less or equal than cell log presuperscript begin inline style 1 third end style end presuperscript space open parentheses x plus 1 close parentheses plus 1 end cell row cell 2 cross times log presuperscript 3 squared end presuperscript space open parentheses 2 x minus 3 close parentheses end cell less or equal than cell log presuperscript begin inline style 3 to the power of negative 1 end exponent end style end presuperscript space open parentheses x plus 1 close parentheses plus 1 end cell row cell 2 cross times 1 half log presuperscript 3 space open parentheses 2 x minus 3 close parentheses end cell less or equal than cell negative log presuperscript begin inline style 3 end style end presuperscript space open parentheses x plus 1 close parentheses plus log presuperscript 3 space 3 end cell row cell log presuperscript 3 space open parentheses 2 x minus 3 close parentheses end cell less or equal than cell negative log presuperscript begin inline style 3 end style end presuperscript space open parentheses x plus 1 close parentheses plus log presuperscript 3 space 3 end cell row cell log presuperscript 3 space open parentheses 2 x minus 3 close parentheses end cell less or equal than cell log presuperscript begin inline style 3 end style end presuperscript space fraction numerator 1 over denominator open parentheses x plus 1 close parentheses end fraction plus log presuperscript 3 space 3 end cell row cell log presuperscript 3 space open parentheses 2 x minus 3 close parentheses end cell less or equal than cell log presuperscript begin inline style 3 end style end presuperscript space fraction numerator 3 over denominator x plus 1 end fraction end cell row cell 2 x minus 3 end cell less or equal than cell fraction numerator 3 over denominator x plus 1 end fraction end cell row cell open parentheses 2 x minus 3 close parentheses minus fraction numerator 3 over denominator x plus 1 end fraction end cell less or equal than 0 row cell fraction numerator open parentheses 2 x minus 3 close parentheses open parentheses x plus 1 close parentheses over denominator x plus 1 end fraction minus fraction numerator 3 over denominator x plus 1 end fraction end cell less or equal than 0 row cell fraction numerator 2 x squared minus x minus 3 over denominator x plus 1 end fraction minus fraction numerator 3 over denominator x plus 1 end fraction end cell less or equal than 0 row cell fraction numerator 2 x squared minus x minus 6 over denominator x plus 1 end fraction end cell less or equal than 0 row cell fraction numerator open parentheses 2 x plus 3 close parentheses open parentheses x minus 2 close parentheses over denominator x plus 1 end fraction end cell less or equal than 0 end table   

Ingat pada fungsi rasional penyebut tidak boleh sama dengan nol maka x not equal to negative 1, sehingga nilai yang memenuhi negative 3 over 2 less or equal than x less or equal than 2 dan x not equal to negative 1.

 syarat numerus 1):

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 3 end cell greater than 0 row x greater than cell 3 over 2 end cell end table

syarat numerus 2):

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 1 end cell greater than 0 row x greater than cell negative 1 end cell end table  

Berdasarkan syarat numerus dan negative 3 over 2 less or equal than x less or equal than 2 maka irisanya adalah 3 over 2 less than x less or equal than 2

Dengan demikian nilai x yang memenuhi adalah 3 over 2 less than x less or equal than 2.

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:

Ingat sifat bentuk 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 to the power of n end cell equals n row cell log presuperscript a to the power of n end presuperscript space b to the power of m end cell equals cell m over n times log presuperscript a space b end cell end table   

dan ingat pada pertidaksamaan logaritma untuk a greater than 1berlaku:

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

Sehingga:

table row blank cell log presuperscript 4 space open parentheses 4 x minus 3 close parentheses less or equal than 1 plus log presuperscript 16 space open parentheses x minus 3 over 4 close parentheses end cell row left right double arrow cell log presuperscript 4 space open parentheses 4 x minus 3 close parentheses less or equal than log presuperscript 4 space 4 plus log presuperscript 4 squared end presuperscript space open parentheses x minus 3 over 4 close parentheses end cell row left right double arrow cell log presuperscript 4 space open parentheses 4 x minus 3 close parentheses less or equal than log presuperscript 4 space 4 plus 1 half times log presuperscript 4 space open parentheses x minus 3 over 4 close parentheses end cell row left right double arrow cell log presuperscript 4 space open parentheses 4 x minus 3 close parentheses less or equal than log presuperscript 4 space 4 plus log presuperscript 4 space open parentheses x minus 3 over 4 close parentheses to the power of begin inline style 1 half end style end exponent end cell row left right double arrow cell log presuperscript 4 space open parentheses 4 x minus 3 close parentheses less or equal than log presuperscript 4 space 4 open parentheses x minus 3 over 4 close parentheses to the power of begin inline style 1 half end style end exponent end cell row left right double arrow cell 4 x minus 3 less or equal than 4 open parentheses x minus 3 over 4 close parentheses to the power of begin inline style 1 half end style end exponent end cell row left right double arrow cell open parentheses 4 x minus 3 close parentheses squared less or equal than open parentheses 4 open parentheses x minus 3 over 4 close parentheses to the power of begin inline style 1 half end style end exponent close parentheses squared end cell row left right double arrow cell open parentheses 4 x minus 3 close parentheses squared less or equal than 16 open parentheses x minus 3 over 4 close parentheses end cell row left right double arrow cell 16 x squared minus 24 x plus 9 less or equal than 16 x minus 12 end cell row left right double arrow cell 16 x squared minus 24 x minus 16 x plus 9 plus 12 less or equal than 0 end cell row left right double arrow cell 16 x squared minus 40 x plus 21 less or equal than 0 end cell row left right double arrow cell open parentheses 4 x minus 3 close parentheses open parentheses 4 x minus 7 close parentheses less or equal than 0 end cell end table  

sehingga nilai x yang memenuhi 3 over 4 less or equal than x less or equal than 7 over 4.

Syarat numerus:

 table row cell 4 x minus 3 greater than 0 end cell rightwards arrow cell x greater than 3 over 4 end cell row cell x minus 3 over 4 greater than 0 end cell rightwards arrow cell x greater than 3 over 4 end cell end table   

Diperoleh 3 over 4 less or equal than x less or equal than 7 over 4 dan x greater than 3 over 4 maka irisannya adalah 3 over 4 less than x less or equal than 7 over 4. Dengan demikian nilai x yang memenuhi pertidaksamaantersebut adalah 3 over 4 less than x less or equal than 7 over 4.

Oleh karena itu, jawaban yang benar adalah B.

 

0

Roboguru

Himpunan penyelesaian pertidaksamaan  adalah ...

Pembahasan Soal:

Ingat sifat bentuk 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 to the power of n end cell equals n end table 

dan ingat pada pertidaksamaan logaritma berlaku:

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

Sehingga

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

sehingga nilai x yang memenuhi adalah negative 6 less or equal than x less or equal than 3.

Syarat numerus:

table row cell x minus 2 greater than 0 end cell rightwards arrow cell x greater than 2 end cell row cell x plus 5 greater than 0 end cell rightwards arrow cell x greater than negative 5 end cell end table 

Diperoleh x greater than negative 5x greater than 2 dan negative 6 less or equal than x less or equal than 3 maka irisannya adalah 2 less than x less or equal than 3. Dengan demikian himpunan penyelesaian dari pertidaksamaan tersebut adalah open curly brackets x vertical line 2 less than x less or equal than 3 comma space x element of straight R close curly brackets.

Oleh karena itu, jawaban yang benar adalah E.

 

0

Roboguru

Nilai yang memenuhi pertidaksamaan:  adalah ...

Pembahasan Soal:

Ingat pada pertidaksamaan bentuk logaritma jika scriptbase log invisible function application f left parenthesis x right parenthesis end scriptbase presuperscript a less than scriptbase log invisible function application b end scriptbase presuperscript a dan a greater than 0 maka 0 less than f left parenthesis x right parenthesis less than b.

Ingat juga sifat-sifat pada bentuk bentuk logaritma yaitu  

  • scriptbase log invisible function application a end scriptbase presuperscript a equals 1
  • scriptbase log invisible function application b to the power of m end scriptbase presuperscript a equals m times scriptbase log invisible function application b end scriptbase presuperscript a

Sehingga akan diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell scriptbase log invisible function application open parentheses 2 x plus 4 close parentheses end scriptbase presuperscript 2 end cell less than 3 row cell scriptbase log invisible function application left parenthesis 2 x plus 4 right parenthesis end scriptbase presuperscript 2 end cell less than cell 3 times scriptbase log invisible function application 2 end scriptbase presuperscript 2 end cell row cell scriptbase log invisible function application left parenthesis 2 x plus 4 right parenthesis end scriptbase presuperscript 2 end cell less than cell scriptbase log invisible function application 2 cubed end scriptbase presuperscript 2 end cell row cell scriptbase log invisible function application left parenthesis 2 x plus 4 right parenthesis end scriptbase presuperscript 2 end cell less than cell scriptbase log invisible function application 8 end scriptbase presuperscript 2 end cell row cell 2 x plus 4 end cell less than 8 row cell 2 x end cell less than cell 8 minus 4 end cell row cell 2 x end cell less than 4 row x less than 2 end table

Karena syarat numerus pada bentuk logaritma scriptbase log invisible function application b end scriptbase presuperscript a yaitu b greater than 0, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 4 end cell greater than 0 row cell 2 x end cell greater than cell negative 4 end cell row x greater than cell fraction numerator negative 4 over denominator 2 end fraction end cell row x greater than cell negative 2 end cell end table

Dari hasil di atas maka penyelesainnya yaitu irisan dari x less than 2 dan x greater than negative 2 yaitu negative 2 less than x less than 2.

Jadi, dapat disimpulkan bahwa nilai x yang memenuhi pertidaksamaan: scriptbase log invisible function application left parenthesis 2 x plus 4 right parenthesis end scriptbase presuperscript 2 less than 3 adalah negative 2 less than x less than 2.

Oleh karena itu, jawaban yang benar adalah D.

1

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

Nilai yang memenuhi pertidaksamaan:  adalah ...

Pembahasan Soal:

Ingat pada pertidaksamaan bentuk logaritma jika scriptbase log invisible function application f left parenthesis x right parenthesis end scriptbase presuperscript a greater than scriptbase log invisible function application b end scriptbase presuperscript a dan 0 less than a less than 1 maka 0 less than f open parentheses x close parentheses less than b.

Ingat juga sifat-sifat pada bentuk bentuk logaritma yaitu  

  • scriptbase log invisible function application a end scriptbase presuperscript a equals 1
  • scriptbase log invisible function application b to the power of m end scriptbase presuperscript a equals m times scriptbase log invisible function application b end scriptbase presuperscript a
  • scriptbase log invisible function application b to the power of m end scriptbase presuperscript a to the power of n end presuperscript equals m over n times scriptbase log invisible function application b end scriptbase presuperscript a

Sehingga akan diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell scriptbase log invisible function application open parentheses x squared minus 1 close parentheses end scriptbase presuperscript 0 , 1 end presuperscript end cell greater than 2 row cell scriptbase log invisible function application left parenthesis x squared minus 1 right parenthesis end scriptbase presuperscript 1 over 10 end presuperscript end cell greater than cell 2 times scriptbase log invisible function application 10 end scriptbase presuperscript 10 end cell row cell scriptbase log invisible function application left parenthesis x squared minus 1 right parenthesis end scriptbase presuperscript 10 to the power of negative 1 end exponent end presuperscript end cell greater than cell scriptbase log invisible function application 10 squared end scriptbase presuperscript 10 end cell row cell negative 1 times scriptbase log invisible function application left parenthesis x squared minus 1 right parenthesis end scriptbase presuperscript 10 end cell greater than cell scriptbase log invisible function application 100 end scriptbase presuperscript 10 end cell row cell scriptbase log invisible function application left parenthesis x squared minus 1 right parenthesis to the power of negative 1 end exponent end scriptbase presuperscript 10 end cell greater than cell scriptbase log invisible function application 100 end scriptbase presuperscript 10 end cell row cell left parenthesis x squared minus 1 right parenthesis to the power of negative 1 end exponent end cell greater than 100 row cell left parenthesis x squared minus 1 right parenthesis to the power of 1 end cell less than cell 100 to the power of negative 1 end exponent end cell row cell x squared minus 1 end cell less than cell 1 over 100 end cell row cell x squared minus 1 end cell less than cell 0 , 01 end cell row cell x squared end cell less than cell 0 , 01 plus 1 end cell row cell x squared end cell less than cell 1 , 01 end cell row x less than cell square root of 1 , 01 end root end cell row x less than cell plus-or-minus square root of 1 , 01 end root end cell end table

Didapatkan x less than square root of 1 , 01 end root atau x greater than negative square root of 1 , 01 end root.

Karena syarat numerus pada bentuk logaritma scriptbase log invisible function application b end scriptbase presuperscript a yaitu b greater than 0, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 1 end cell greater than 0 row cell x squared end cell greater than 1 row x greater than cell square root of 1 end cell row x greater than cell plus-or-minus 1 end cell end table

Didapatkan x greater than 1 dan x less than negative 1

Dari hasil di atas maka penyelesainnya yaitu irisan dari x less than square root of 1 , 01 end rootx greater than negative square root of 1 , 01 end rootx greater than 1, dan x less than negative 1 yaitu negative square root of 1 , 01 end root less than x less than negative 1 atau 1 less than x less than square root of 1 , 01 end root.

Jadi, dapat disimpulkan bahwa nilai x yang memenuhi pertidaksamaan: scriptbase log invisible function application left parenthesis x squared minus 1 right parenthesis end scriptbase presuperscript 0 , 1 end presuperscript greater than 2 adalah negative square root of 1 , 01 end root less than x less than negative 1 atau 1 less than x less than square root of 1 , 01 end root.

Oleh karena itu, jawaban yang benar adalah D.

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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