Penyelesaian dari pertidaksamaan 8log (21​log (x2−2xx−3​))>0 adalah ...

Pertanyaan

Penyelesaian dari pertidaksamaan log presuperscript 8 space open parentheses log presuperscript begin inline style bevelled 1 half end style end presuperscript space open parentheses fraction numerator x minus 3 over denominator x squared minus 2 x end fraction close parentheses close parentheses greater than 0 adalah ...

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

penyelesaian dari pertidaksamaan log presuperscript 8 space open parentheses log presuperscript begin inline style bevelled 1 half end style end presuperscript space open parentheses fraction numerator x minus 3 over denominator x squared minus 2 x end fraction close parentheses close parentheses greater than 0 adalah  open curly brackets x space left enclose space space x greater than 3 end enclose close curly brackets.

Pembahasan

Perlu diingat sifat logaritma dan pertidaksamaan logaritma yaitu:

log presuperscript a space 1 equals 0 log presuperscript a space a equals 1 log presuperscript a space f left parenthesis x right parenthesis greater than log presuperscript a space b space rightwards arrow f left parenthesis x right parenthesis greater than b space left parenthesis a greater than 1 right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards arrow f left parenthesis x right parenthesis less than b left parenthesis 0 less than a less than 1 right parenthesis

  • Syarat numerus

                  table attributes columnalign right center left columnspacing 0px end attributes row cell space open parentheses fraction numerator x minus 3 over denominator x squared minus 2 x end fraction close parentheses end cell greater than 0 row cell fraction numerator x minus 3 over denominator x left parenthesis x minus 2 right parenthesis end fraction end cell greater than 0 row blank blank space end table

Pembuat nol pembilang dan penyebutnya adalah

x minus 3 equals 0 space space space space space space space space space space space space space x equals 0 space space space space space space space space space space space space space space space space x minus 2 equals 0 space space space space space x equals 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x equals 2 space space space

Jadi garis bilangan untuk table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator x minus 3 over denominator x squared minus 2 x end fraction close parentheses end cell greater than 0 end table adalah

Jadi penyelesaian untuk table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator x minus 3 over denominator x squared minus 2 x end fraction close parentheses end cell greater than 0 end table adalah 0 less than x less than 2 space atau space x greater than 3 

 

  • Perhatikan perhitungan berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 8 space open parentheses log presuperscript begin inline style bevelled 1 half end style end presuperscript space open parentheses fraction numerator x minus 3 over denominator x squared minus 2 x end fraction close parentheses close parentheses end cell greater than 0 row cell log presuperscript 8 space open parentheses log presuperscript begin inline style bevelled 1 half end style end presuperscript space open parentheses fraction numerator x minus 3 over denominator x squared minus 2 x end fraction close parentheses close parentheses end cell greater than cell log presuperscript 8 space 1 end cell row cell log presuperscript begin inline style bevelled 1 half end style end presuperscript space open parentheses fraction numerator x minus 3 over denominator x squared minus 2 x end fraction close parentheses end cell greater than 1 row cell log presuperscript 1 half end presuperscript space open parentheses fraction numerator x minus 3 over denominator x squared minus 2 x end fraction close parentheses end cell greater than cell log presuperscript 1 half end presuperscript space 1 half end cell row blank blank blank row cell open parentheses fraction numerator x minus 3 over denominator x squared minus 2 x end fraction close parentheses end cell greater than cell 1 half space end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x minus 3 over denominator x open parentheses x minus 2 close parentheses end fraction end cell less than cell 1 half end cell row cell fraction numerator x minus 3 over denominator x open parentheses x minus 2 close parentheses end fraction minus 1 half end cell less than 0 row cell fraction numerator 2 left parenthesis x minus 3 right parenthesis minus x left parenthesis x minus 2 right parenthesis over denominator 2 x left parenthesis x minus 2 right parenthesis end fraction end cell less than 0 row cell fraction numerator 2 x minus 6 minus x squared plus 2 x over denominator 2 x left parenthesis x minus 2 right parenthesis end fraction end cell less than 0 row cell fraction numerator negative x squared plus 4 x minus 6 over denominator 2 x left parenthesis x minus 2 right parenthesis end fraction end cell less than 0 row cell fraction numerator Definit space negatif over denominator italic 2 x left parenthesis x minus 2 right parenthesis end fraction end cell greater than 0 end table

Pembuat nol pada penyebut adalah

2 x equals 0 x equals 0  x minus 2 equals 0 space space space space space x equals 2

Garis bilangan dari table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator negative x squared plus 4 x minus 6 over denominator 2 x left parenthesis x minus 2 right parenthesis end fraction end cell less than 0 end table adalah

Jadi penyelesaian untuk table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator negative x squared plus 4 x minus 6 over denominator 2 x left parenthesis x minus 2 right parenthesis end fraction end cell less than 0 end table adalah

x less than 0 space atau space x italic greater than italic 2

 Penyelesaiannya adalah irisan antara 0 less than x less than 2 space atau space x greater than 3 space dan space x italic less than italic 0 italic space atau italic space x italic greater than italic 2 space

Jadi, penyelesaian dari pertidaksamaan log presuperscript 8 space open parentheses log presuperscript begin inline style bevelled 1 half end style end presuperscript space open parentheses fraction numerator x minus 3 over denominator x squared minus 2 x end fraction close parentheses close parentheses greater than 0 adalah  open curly brackets x space left enclose space space x greater than 3 end enclose close curly brackets.

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