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Penyelesaian dari 2log2(1−x)−8>2log(1−x)2 adalah ...

Pertanyaan

Penyelesaian dari log presuperscript 2 superscript 2 space open parentheses 1 minus x close parentheses minus 8 greater than log presuperscript 2 space open parentheses 1 minus x close parentheses squared adalah ...

  1. x less than negative 2 

  2. negative 2 less than x less than 1 

  3. negative 15 less than x less than 3 over 4 

  4. x less than negative 15 atau 3 over 4 less than x less than 1   

  5. x less than negative 15 atau x greater than 3 over 4 comma space x not equal to 1   

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 space b to the power of m end cell equals cell m log presuperscript a space b end cell end table  

Ingat sifat bilangan berpangkat:

x to the power of negative m end exponent equals 1 over x to the power of m 

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

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 g open parentheses x close parentheses greater than 0  

Misalkan log presuperscript 2 space open parentheses 1 minus x close parentheses equals p maka 2 to the power of p equals 1 minus x, sehingga:

table row blank cell log presuperscript 2 superscript 2 space open parentheses 1 minus x close parentheses minus 8 greater than log presuperscript 2 space open parentheses 1 minus x close parentheses squared end cell row left right double arrow cell log presuperscript 2 superscript 2 space open parentheses 1 minus x close parentheses minus log presuperscript 2 space open parentheses 1 minus x close parentheses squared minus 8 greater than 0 end cell row left right double arrow cell log presuperscript 2 superscript 2 space open parentheses 1 minus x close parentheses minus 2 times log presuperscript 2 space open parentheses 1 minus x close parentheses minus 8 greater than 0 end cell row left right double arrow cell p squared minus 2 p minus 8 greater than 0 end cell row left right double arrow cell open parentheses p minus 4 close parentheses open parentheses p plus 2 close parentheses greater than 0 end cell end table 

sehingga nilai p yang memenuhi p less than negative 2 atau p greater than 4.

Untuk nilai p less than negative 2 maka:

table row blank cell log presuperscript 2 space open parentheses 1 minus x close parentheses less than negative 2 end cell row left right double arrow cell log presuperscript 2 space open parentheses 1 minus x close parentheses less than log presuperscript 2 space 2 to the power of negative 2 end exponent end cell row left right double arrow cell 1 minus x less than 2 to the power of negative 2 end exponent end cell row left right double arrow cell 1 minus x less than 1 fourth end cell row left right double arrow cell negative x less than 1 fourth minus 1 end cell row left right double arrow cell negative x less than negative 3 over 4 end cell row left right double arrow cell x greater than 3 over 4 end cell end table 

dan untuk 

table row blank cell log presuperscript 2 space open parentheses 1 minus x close parentheses greater than 4 end cell row left right double arrow cell log presuperscript 2 space open parentheses 1 minus x close parentheses greater than log presuperscript 2 space 2 to the power of 4 end cell row left right double arrow cell 1 minus x greater than 2 to the power of 4 end cell row left right double arrow cell 1 minus x greater than 16 end cell row left right double arrow cell negative x greater than 16 minus 1 end cell row left right double arrow cell negative x greater than 15 end cell row left right double arrow cell x less than negative 15 end cell end table 

diperoleh nilai x yang memenuhi x less than negative 15 atau x greater than 3 over 4.

Syarat numerus:

 1 minus x greater than 0 rightwards arrow x less than 1   

maka irisannya adalah x less than negative 15 atau 3 over 4 less than x less than 1. Dengan demikian penyelesaian dari pertidaksamaan tersebut adalah x less than negative 15 atau 3 over 4 less than x less than 1  .

Oleh karena itu, jawaban yang benar adalah D.

 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Amamah

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Penyelesaian pertidaksamaan 2×log(x+1)≤log(x+4)+log4 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 space b to the power of m end cell equals cell m 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 

Jika bilangan pokok logaritma tidak dituliskan, hal itu berarti logaritma tersebut berbasis 10 ,maka:

table row blank cell 2 cross times log space open parentheses x plus 1 close parentheses less or equal than log space open parentheses x plus 4 close parentheses plus log space 4 end cell row left right double arrow cell log space open parentheses x plus 1 close parentheses squared less or equal than log space 4 open parentheses x plus 4 close parentheses end cell row left right double arrow cell open parentheses x plus 1 close parentheses squared less or equal than 4 x plus 16 end cell row left right double arrow cell x squared plus 2 x plus 1 less or equal than 4 x plus 16 end cell row left right double arrow cell x squared plus 2 x minus 4 x plus 1 minus 16 less or equal than 0 end cell row left right double arrow cell x squared minus 2 x minus 15 less or equal than 0 end cell row left right double arrow cell open parentheses x minus 5 close parentheses open parentheses x plus 3 close parentheses less or equal than 0 end cell end table 

sehingga nilai x yang memenuhi adalah negative 3 less or equal than x less or equal than 5.  

Syarat numerus:

 table row cell x plus 1 greater than 0 end cell rightwards arrow cell x greater than negative 1 end cell row cell x plus 4 greater than 0 end cell rightwards arrow cell x greater than negative 4 end cell end table  

Diperoleh x greater than negative 4x greater than negative 1, dan negative 3 less or equal than x less or equal than 5 maka irisannya adalah negative 1 less than x less or equal than 5. Dengan demikian penyelesaian dari pertidaksamaan tersebut adalah negative 1 less than x less or equal than 5.

Oleh karena itu, jawaban yang benar adalah D.

 

0

Roboguru

Himpunan penyelesaian pertidaksamaan 2log(x−2)+2log(x+5)≤3 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 x yang memenuhi pertidaksamaan 4log(4x−3)≤1+16log(x−43​) 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 2log(x−2)+2log(x+5)≤3 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 x yang memenuhi pertidaksamaan: 2log(2x+4)<3 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

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Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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