Nilai x yang memenuhi persamaan log(1−xx2​)=log x+log(1+x2x​) adalah ...

Pertanyaan

Nilai x yang memenuhi persamaan log left parenthesis fraction numerator x squared over denominator 1 minus x end fraction right parenthesis equals log space x plus log left parenthesis fraction numerator 2 x over denominator 1 plus x end fraction right parenthesis adalah ...

  1. 1 fifth 

  2. 1 fourth 

  3. 1 third 

  4. 1 half 

  5. 1 

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Pembahasan

Diketahui : log left parenthesis fraction numerator x squared over denominator 1 minus x end fraction right parenthesis equals log space x plus log left parenthesis fraction numerator 2 x over denominator 1 plus x end fraction right parenthesis

Ingat kembali bahwa : 

bullet space log presuperscript straight a space straight b plus log presuperscript straight a space straight c equals log presuperscript straight a space left parenthesis bc right parenthesis bullet space log presuperscript straight a space f left parenthesis x right parenthesis equals log presuperscript straight a space g left parenthesis x right parenthesis rightwards arrow f left parenthesis x right parenthesis equals g left parenthesis x right parenthesis space space space space dengan space straight a greater than 0 comma space straight a not equal to comma space f left parenthesis x right parenthesis greater than 0 comma space g left parenthesis x right parenthesis greater than 0     

Maka

table attributes columnalign right center left columnspacing 0px end attributes row cell log left parenthesis fraction numerator x squared over denominator 1 minus x end fraction right parenthesis end cell equals cell log space x plus log left parenthesis fraction numerator 2 x over denominator 1 plus x end fraction right parenthesis end cell row cell log left parenthesis fraction numerator x squared over denominator 1 minus x end fraction right parenthesis end cell equals cell log left parenthesis x times fraction numerator 2 x over denominator 1 plus x end fraction right parenthesis end cell row cell log left parenthesis fraction numerator x squared over denominator 1 minus x end fraction right parenthesis end cell equals cell log left parenthesis fraction numerator 2 x squared over denominator 1 plus x end fraction right parenthesis end cell row cell fraction numerator x squared over denominator 1 minus x end fraction end cell equals cell fraction numerator 2 x squared over denominator 1 plus x end fraction end cell row cell x squared left parenthesis 1 plus x right parenthesis end cell equals cell 2 x squared left parenthesis 1 minus x right parenthesis end cell row cell x squared plus x cubed end cell equals cell 2 x squared minus 2 x cubed end cell row cell x cubed plus 2 x cubed plus x squared minus 2 x squared end cell equals 0 row cell 3 x cubed minus x squared end cell equals 0 row cell x squared left parenthesis 3 x minus 1 right parenthesis end cell equals 0 row x equals cell 0 space atau space x equals 1 third end cell end table  

Perhatikan bahwa untuk x equals 0 maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell fraction numerator x squared over denominator 1 minus x end fraction end cell row cell f left parenthesis x right parenthesis end cell equals cell fraction numerator 0 squared over denominator 1 minus 0 end fraction end cell row blank equals 0 end table  

Karena salah satu syarat adalah f left parenthesis x right parenthesis greater than 0 maka x equals 0 tidak memenuhi, sehingga penyelesaiannya yaitu x equals 1 third.

Oleh karen itu, jawaban yang benar adalah C.

48

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