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Pertanyaan

Nilai x yang memenuhi pertidaksamaan: ∣ ∣ ​ 1 − 3 lo g ( 2 x + 1 ) ∣ ∣ ​ > 2 adalah ...

Nilai  yang memenuhi pertidaksamaan:  adalah ...

  1. x greater than 13 

  2. x less than negative 1 third atau x greater than 13 

  3. negative 1 half less than x less than negative 1 third 

  4. negative 1 half less than x less than negative 1 third atau x greater than 13 

  5. x greater than negative 1 half 

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D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah D.

jawaban yang benar adalah D. 

Pembahasan

Diketahui : Ingat kembali bahwa : Maka Syarat (i) atau atau Syarat (ii) Jadi, penyelesaiannya adalah irisan dari dengan yaitu atau . Oleh karena itu, jawaban yang benar adalah D.

Diketahui : open vertical bar 1 minus log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis close vertical bar greater than 2

Ingat kembali bahwa :

bullet space open vertical bar f left parenthesis x right parenthesis close vertical bar greater than a comma space a greater than 0 rightwards arrow f left parenthesis x right parenthesis less than negative a space atau space f left parenthesis x right parenthesis greater than a bullet space log presuperscript a space a to the power of m equals m bullet space log presuperscript a space f left parenthesis x right parenthesis less than log presuperscript a space g left parenthesis x right parenthesis rightwards arrow 0 less than f left parenthesis x right parenthesis less than g left parenthesis x right parenthesis      

Maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 1 minus log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis close vertical bar end cell greater than 2 row cell 1 minus log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis end cell less than cell negative 2 space atau space 1 minus log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis greater than 2 end cell end table 

Syarat (i) 1 minus log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis less than negative 2 atau 1 minus log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis greater than 2

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 minus log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis end cell less than cell negative 2 end cell row cell 1 plus 2 end cell less than cell log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis end cell row 3 less than cell log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis end cell row cell log presuperscript 3 space 3 cubed end cell less than cell log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis end cell row cell log presuperscript 3 space 27 end cell less than cell log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis end cell row 27 less than cell 2 x plus 1 end cell row cell 27 minus 1 end cell less than cell 2 x end cell row 26 less than cell 2 x end cell row cell 26 over 2 end cell less than x row 13 less than x row x greater than 13 end table    

atau

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 minus log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis end cell greater than 2 row cell 1 minus 2 end cell greater than cell log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis end cell row cell negative 1 end cell greater than cell log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis end cell row cell scriptbase log space 3 to the power of negative 1 end exponent end scriptbase presuperscript 3 end cell greater than cell log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis end cell row cell scriptbase log space 1 third end scriptbase presuperscript 3 end cell greater than cell log presuperscript 3 left parenthesis 2 x plus 1 right parenthesis end cell row cell 1 third end cell greater than cell 2 x plus 1 end cell row cell 1 third minus 1 end cell greater than cell 2 x end cell row cell negative 2 over 3 end cell greater than cell 2 x end cell row cell fraction numerator negative begin display style 2 over 3 end style over denominator 2 end fraction end cell greater than x row cell negative 1 third end cell greater than x row x less than cell negative 1 third end cell end table   

Syarat (ii) 2 x plus 1 greater than 0 

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 1 end cell greater than 0 row cell 2 x end cell greater than cell negative 1 end cell row x greater than cell negative 1 half end cell end table

Jadi, penyelesaiannya adalah irisan dari x less than negative 1 third space atau space x greater than 13 dengan x greater than negative 1 half yaitu negative 1 half less than x less than negative 1 third atau x greater than 13.

Oleh karena itu, jawaban yang benar adalah D. 

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