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Pertanyaan

Jika log invisible function application left parenthesis 3 x plus 1 right parenthesis greater than 2 maka ...

  1. x greater than 33

  2. negative 1 third less than x less than 33

  3. x greater than negative 1 third

  4. negative 1 third less than x less than 0

  5. negative 1 third less than x less than 3 2 over 3

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.

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

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 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.

 

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.

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.

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.

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.

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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