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Pertanyaan

Jika f ( x ) = 5 lo g ( x + 1 ) + 5 lo g ( x − 2 1 ​ ) , maka f − 1 ( 5 lo g 2 ) = . . .

Jika  maka 

  1. 3

  2. 4

  3. 5

  4. 6

  5. 7

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N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Pembahasan

Cara - 1 : Maka dapat diperoleh : Cara - 2 : Maka Substitusikan x = 5

Cara - 1 :

table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell log presuperscript 5 open parentheses fraction numerator x plus 1 over denominator x minus 2 end fraction close parentheses end cell row cell open parentheses fraction numerator x plus 1 over denominator x minus 2 end fraction close parentheses end cell equals cell y to the power of 5 end cell row cell x plus 1 end cell equals cell x.5 to the power of y plus x end cell row cell 2.5 to the power of y plus 1 end cell equals cell x.5 to the power of y plus x end cell row cell 2.5 to the power of y plus 1 end cell equals cell x open parentheses 5 to the power of y minus 1 close parentheses end cell row x equals cell fraction numerator 2.5 to the power of y plus 1 over denominator 5 to the power of y minus 1 end fraction end cell row cell f to the power of negative 1 end exponent end cell equals cell fraction numerator 2.5 to the power of x plus 1 over denominator 5 to the power of x minus 1 end fraction end cell end table

 

Maka dapat diperoleh :

table attributes columnalign right center left columnspacing 0px end attributes row cell f to the power of negative 1 end exponent open parentheses log presuperscript 5 space 2 close parentheses end cell equals cell fraction numerator 2.5 space log presuperscript 5 space 2 plus 1 over denominator 5 space log presuperscript 5 space 2 minus 1 end fraction end cell row blank equals cell fraction numerator 2.2 plus 1 over denominator 2 minus 1 end fraction equals 5 end cell end table

Cara - 2 :

f left parenthesis x right parenthesis equals log presuperscript 5 space open parentheses fraction numerator x plus 1 over denominator x minus 2 end fraction close parentheses

Maka

f to the power of negative 1 end exponent open parentheses log presuperscript 5 space open parentheses fraction numerator x plus 1 over denominator x minus 2 end fraction close parentheses close parentheses equals x

 

Substitusikan x = 5

f to the power of negative 1 end exponent open parentheses log presuperscript 5 space open parentheses fraction numerator 5 plus 1 over denominator 5 minus 2 end fraction close parentheses close parentheses equals 5 f to the power of negative 1 end exponent open parentheses log presuperscript 5 space 2 close parentheses equals 5

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