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Pertanyaan

Supaya log invisible function application left parenthesis x plus 5 right parenthesis less than log invisible function application left parenthesis 2 x minus 7 right parenthesis maka haruslah ...

  1. x greater than 12

  2. 3 , 5 less than x less than 12

  3. negative 5 less than x less than 12

  4. x greater than negative 5

  5. negative 5 less than x less than 3 , 5

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 g left parenthesis x right parenthesis end scriptbase presuperscript a dan a greater than 0 maka 0 less than f left parenthesis x right parenthesis less than b. Sehingga akan diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell log invisible function application left parenthesis x plus 5 right parenthesis end cell less than cell log invisible function application left parenthesis 2 x minus 7 right parenthesis end cell row cell x plus 5 end cell less than cell 2 x minus 7 end cell row cell 5 plus 7 end cell less than cell 2 x minus x end cell row 12 less than x 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 x plus 5 end cell greater than 0 row x greater than cell negative 5 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 7 end cell greater than 0 row cell 2 x end cell greater than 7 row x greater than cell 7 over 2 end cell row x greater than cell 3 , 5 end cell end table

Dari hasil di atas maka penyelesainnya yaitu irisan dari x greater than 12x greater than negative 5, dan x greater than 3 , 5 yaitu x greater than 12.

Jadi, dapat disimpulkan bahwa supaya log invisible function application left parenthesis x plus 5 right parenthesis less than log invisible function application left parenthesis 2 x minus 7 right parenthesis maka haruslah x greater than 12.

Oleh karena itu, jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Iqbal

Mahasiswa/Alumni Universitas Negeri Semarang

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 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

Supaya  maka haruslah ...

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 g left parenthesis x right parenthesis end scriptbase presuperscript a dan a greater than 0 maka 0 less than f left parenthesis x right parenthesis less than b. Sehingga akan diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell log invisible function application left parenthesis x plus 5 right parenthesis end cell less than cell log invisible function application left parenthesis 2 x minus 7 right parenthesis end cell row cell x plus 5 end cell less than cell 2 x minus 7 end cell row cell 5 plus 7 end cell less than cell 2 x minus x end cell row 12 less than x 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 x plus 5 end cell greater than 0 row x greater than cell negative 5 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 7 end cell greater than 0 row cell 2 x end cell greater than 7 row x greater than cell 7 over 2 end cell row x greater than cell 3 , 5 end cell end table

Dari hasil di atas maka penyelesainnya yaitu irisan dari x greater than 12x greater than negative 5, dan x greater than 3 , 5 yaitu x greater than 12.

Jadi, dapat disimpulkan bahwa supaya log invisible function application left parenthesis x plus 5 right parenthesis less than log invisible function application left parenthesis 2 x minus 7 right parenthesis maka haruslah x greater than 12.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Jika  maka ...

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 a greater than 0 maka f open parentheses x close parentheses greater than b greater than 0.

Sehingga akan diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell log invisible function application left parenthesis 3 x plus 1 right parenthesis end cell greater than 2 row cell log invisible function application left parenthesis 3 x plus 1 right parenthesis end cell greater than cell log invisible function application 100 end cell row cell 3 x plus 1 end cell greater than 100 row cell 3 x end cell greater than cell 100 minus 1 end cell row cell 3 x end cell greater than 99 row x greater than cell 99 over 3 end cell row x greater than 33 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 3 x plus 1 end cell greater than 0 row cell 3 x end cell greater than cell negative 1 end cell row x greater than cell fraction numerator negative 1 over denominator 3 end fraction end cell row x greater than cell negative 1 third end cell end table

Dari hasil di atas maka penyelesainnya yaitu irisan dari x greater than 33 dan x greater than negative 1 third yaitu x greater than 33.

Jadi, dapat disimpulkan bahwa jika log invisible function application left parenthesis 3 x plus 1 right parenthesis greater than 2 maka penyelesaiannya yaitu x greater than 33.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Penyelesaian dari  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  

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.

 

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

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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