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Tentukan himpunan penyelesaian dari pertidaksamaan logaritma berikut ini: 1.

Pertanyaan

Tentukan himpunan penyelesaian dari pertidaksamaan logaritma berikut ini:

1. begin mathsize 14px style log presuperscript 6 space open parentheses x squared plus x minus 6 close parentheses greater or equal than 1 end style 

Pembahasan Video:

Pembahasan Soal:

Gunakan sifat:

begin mathsize 14px style log presuperscript a space b greater or equal than log presuperscript a space c rightwards double arrow b greater or equal than c end style 

Dengan syarat:

begin mathsize 14px style a greater than 0 space comma space space a not equal to 1 b comma space c greater than 0 end style 

Perhatikan perhitungan berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell log presuperscript 6 space open parentheses x squared plus x minus 6 close parentheses end cell greater or equal than 1 row cell log presuperscript 6 space open parentheses x squared plus x minus 6 close parentheses end cell greater or equal than cell log presuperscript 6 space 6 end cell row cell x squared plus x minus 6 end cell greater or equal than 6 row cell x squared plus x minus 12 end cell greater or equal than 0 row cell open parentheses x plus 4 close parentheses open parentheses x minus 3 close parentheses end cell greater or equal than 0 end table x less or equal than negative 4 space atau space x greater or equal than 3 end style 

Berdasarkan syarat, maka diperoleh perhitungan sebagai berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus x minus 6 end cell greater than 0 row cell left parenthesis x plus 3 right parenthesis left parenthesis x minus 2 right parenthesis end cell greater than 0 row cell x plus 3 end cell greater than cell 0 space atau space x minus 2 greater than 0 end cell row x greater than cell negative 3 space atau space x greater than 2 end cell end table end style 

 

Jadi, himpunan penyelesaiannya adalah begin mathsize 14px style open curly brackets x vertical line x less or equal than negative 4 space atau space x greater or equal than 3 close curly brackets end style.

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Terakhir diupdate 03 Mei 2021

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Pertanyaan yang serupa

Tentukan HP dari pertidaksamaan .

Pembahasan Soal:

Gunakan sifat.

begin mathsize 14px style Untuk space a greater than 1 space berlaku colon space 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 end style 

Perhatikan perhitungan berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell log presuperscript 4 space open parentheses 2 x squared plus 24 close parentheses end cell less or equal than cell log presuperscript 4 space open parentheses x squared plus 10 x close parentheses end cell row cell 2 x squared plus 24 end cell less or equal than cell x squared plus 10 x end cell row cell 2 x squared plus 24 minus x squared minus 10 x end cell less or equal than 0 row cell x squared minus 10 x plus 24 end cell less or equal than 0 row cell open parentheses x minus 4 close parentheses open parentheses x minus 6 close parentheses end cell less or equal than 0 row cell 4 less or equal than x less or equal than 6 end cell blank blank end table end style 

 

Jadi, himpunan penyelesaian dari pertidaksamaan begin mathsize 14px style log presuperscript 4 space open parentheses 2 x squared plus 24 close parentheses less or equal than log presuperscript 4 space open parentheses x squared plus 10 x close parentheses end style adalah begin mathsize 14px style open curly brackets x vertical line 4 less or equal than x less or equal than 6 close curly brackets end style.

0

Roboguru

Untuk,  himpunan penyelesaian pertidaksamaan logaritma

Pembahasan Soal:

begin mathsize 14px style log presuperscript 2 left parenthesis x squared minus 3 x plus 2 right parenthesis less than log presuperscript 2 left parenthesis 10 minus x right parenthesis end style

Syarat:

begin mathsize 14px style 0 less than x squared minus 3 x plus 2 less than 10 minus x end style

begin mathsize 14px style x squared minus 3 x plus 2 greater than 0 left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis greater than 0 x less than 1 space atau space x greater than 2 end style  

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 3 plus 2 end cell less than cell 10 minus x end cell row cell x squared minus 2 x minus 8 end cell less than 0 row cell left parenthesis x plus 2 right parenthesis left parenthesis x minus 4 right parenthesis end cell less than 0 end table end style 

begin mathsize 14px style x greater than negative 2 space atau space x less than 4 end style 

Jadi himpunan penyelesaiannya adalah begin mathsize 14px style negative 2 less than x less than 1 space dan space 2 less than x less than 4 end style untuk begin mathsize 14px style x element of R end style.

0

Roboguru

Diketahui fungsi . Supaya  maka haruslah ...

Pembahasan Soal:

Diketahui:

f left parenthesis x right parenthesis equals log presuperscript 3 space x

Ditanya:

Nilia a dan b agar f left parenthesis a right parenthesis greater than f left parenthesis b right parenthesis

Perlu diingat bahwa suatu fungsi logaritma terdefinisi jika x greater than 0.

Jika terdapat bilangan a dan b sehingga memenuhi f left parenthesis a right parenthesis greater than f left parenthesis b right parenthesis, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis a right parenthesis end cell greater than cell f left parenthesis b right parenthesis end cell row cell log presuperscript 3 space a end cell greater than cell log presuperscript 3 space b end cell row a greater than b end table

Karena syarat fungsi logaritma adalah x greater than 0, maka supaya f left parenthesis a right parenthesis greater than f left parenthesis b right parenthesis nilai a dan b haruslah a greater than b greater than 0.

Jadi, nilai yang tepat adalah C.

0

Roboguru

Tentukan himpunan penyelesaian dari pertidaksamaan logaritma berikut:

Pembahasan Soal:

Ingat bahwa

begin mathsize 14px style scriptbase log space f open parentheses x close parentheses plus scriptbase log space g open parentheses x close parentheses equals scriptbase log space f open parentheses x close parentheses times g open parentheses x close parentheses end scriptbase presuperscript a end scriptbase presuperscript a end scriptbase presuperscript a end style 

dan 

begin mathsize 14px style scriptbase log space b to the power of n equals n times scriptbase log space b end scriptbase presuperscript a end scriptbase presuperscript a end style.

Sehingga diperoleh pertidaksamaan logaritma:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell table attributes columnalign right center left end attributes row cell scriptbase log open parentheses x minus 3 close parentheses plus scriptbase log open parentheses x plus 3 close parentheses end scriptbase presuperscript 2 end scriptbase presuperscript 2 end cell greater or equal than 4 row cell scriptbase log open parentheses x minus 3 close parentheses open parentheses x plus 3 close parentheses end scriptbase presuperscript 2 end cell greater or equal than cell 4 times scriptbase log space 2 end scriptbase presuperscript 2 end cell row cell scriptbase log open parentheses x squared minus 9 close parentheses end scriptbase presuperscript 2 end cell greater or equal than cell scriptbase log space 2 to the power of 4 end scriptbase presuperscript 2 end cell row cell x squared minus 9 end cell greater or equal than 16 row cell x squared minus 9 minus 16 end cell greater or equal than 0 row cell x squared minus 25 end cell greater or equal than 0 row cell open parentheses x minus 5 close parentheses open parentheses x plus 5 close parentheses end cell greater or equal than 0 row cell x greater or equal than 5 end cell logical or cell x greater or equal than negative 5 end cell end table end cell row blank blank blank end table

Garis bilangannya adalah sebagai berikut:

Syarat numerus

begin mathsize 14px style 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 x minus 3 end cell greater than 0 row x greater than 3 end table end style   dan   begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell g open parentheses x close parentheses end cell greater than 0 row cell x plus 3 end cell greater than 0 row x greater than cell negative 3 end cell end table end style

Garis bilangannya menjadi:

Dengan demikian, diperoleh begin mathsize 14px style HP equals open curly brackets right enclose x space x greater or equal than 5 space close curly brackets end style.

0

Roboguru

Jika  maka ...

Pembahasan Soal:

Diketahui : open vertical bar log space g left parenthesis x right parenthesis close vertical bar less than 1

Ingat kembali bahwa :

bullet space open vertical bar straight x close vertical bar less than straight a comma space straight a greater than 0 rightwards arrow negative straight a space less than straight x less than straight a bullet space log presuperscript straight a space straight f left parenthesis straight x right parenthesis greater than log presuperscript straight a space straight b comma space straight a greater than 1 rightwards arrow straight f left parenthesis straight x right parenthesis greater than straight b greater than 0 bullet space log presuperscript straight a space straight f left parenthesis straight x right parenthesis less than log presuperscript straight a space straight b comma space straight a greater than 1 rightwards arrow 0 less than straight f left parenthesis straight x right parenthesis less than straight b  

Misal : log space g left parenthesis x right parenthesis equals x maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar log space g left parenthesis x right parenthesis close vertical bar end cell less than 1 row cell negative 1 end cell less than cell log space g left parenthesis x right parenthesis less than 1 end cell end table 

Syarat (i) log space g left parenthesis x right parenthesis greater than negative 1  

table attributes columnalign right center left columnspacing 0px end attributes row cell log space g left parenthesis x right parenthesis end cell greater than cell negative 1 end cell row cell log space g left parenthesis x right parenthesis end cell greater than cell log space 10 to the power of negative 1 end exponent end cell row cell log space g left parenthesis x right parenthesis end cell greater than cell log space 1 over 10 end cell row cell g left parenthesis x right parenthesis end cell greater than cell 1 over 10 end cell row cell g left parenthesis x right parenthesis end cell greater than cell 0 comma 1 end cell end table   

Syarat (ii) log space g left parenthesis x right parenthesis less than 1 

table attributes columnalign right center left columnspacing 0px end attributes row cell log space g left parenthesis x right parenthesis end cell less than 1 row cell log space g left parenthesis x right parenthesis end cell less than cell log space 10 end cell row cell g left parenthesis x right parenthesis end cell less than 10 end table 

Jadi, penyelesaiannya adalah irisan dari g left parenthesis x right parenthesis greater than 0 comma 1 dengan g left parenthesis x right parenthesis less than 10 yaitu 0 comma 1 less than g left parenthesis x right parenthesis less than 10.

Oleh karena itu, jawaban yang benar adalah E.

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