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Pertanyaan

Jika x 1 ​ , x 2 ​ , x 3 ​ , … , x n ​ adalah barisan geometri dengan rasio r, maka lo g x 1 ​ + lo g x 2 ​ + lo g x 3 ​ + ⋯ + lo g x n ​ = ...

Jika adalah barisan geometri dengan rasio r, maka

  1. 1 half nlog invisible function application straight x subscript 1 superscript straight n straight r to the power of straight n

  2. 1 half nlog invisible function application straight x subscript 1 superscript straight n minus 1 end superscript straight r to the power of straight n

  3. 1 half nlog invisible function application straight x subscript 1 superscript straight n straight r squared

  4. 1 half nlog invisible function application straight x subscript 1 superscript 2 straight r to the power of straight n

  5. 1 half nlog invisible function application straight x subscript 1 superscript 2 straight r to the power of straight n minus 1 end exponent

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N. Ayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Padang

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Pembahasan

Barisan geometri Karena maka hasilnya adalah

Barisan geometri straight U subscript straight n equals ar to the power of straight n minus 1 end exponent

log invisible function application straight x subscript 1 plus log invisible function application straight x subscript 2 plus log invisible function application straight x subscript 3 plus horizontal ellipsis plus log invisible function application straight x subscript straight n equals log invisible function application open parentheses straight x subscript 1. straight x subscript 2. straight x subscript 3. horizontal ellipsis. straight x subscript straight n close parentheses  equals log invisible function application open parentheses straight a. ar. ar squared. horizontal ellipsis. ar to the power of straight n minus 1 end exponent close parentheses  equals log invisible function application open parentheses straight a to the power of open parentheses stack stack 1 plus 1 plus horizontal ellipsis plus 1 with underbrace below with sebanyak straight space straight n below close parentheses end exponent straight r to the power of open parentheses 1 plus 2 plus horizontal ellipsis plus straight n minus 1 close parentheses end exponent close parentheses  equals log invisible function application open parentheses straight a to the power of straight n straight r to the power of fraction numerator straight n open parentheses straight n minus 1 close parentheses over denominator 2 end fraction end exponent close parentheses  equals log invisible function application open parentheses straight a to the power of fraction numerator 2 straight n over denominator 2 end fraction end exponent straight r to the power of fraction numerator straight n open parentheses straight n minus 1 close parentheses over denominator 2 end fraction end exponent close parentheses  equals log invisible function application open parentheses straight a squared straight r to the power of straight n minus 1 end exponent close parentheses to the power of straight n over 2 end exponent  equals straight n over 2 log invisible function application open parentheses straight a squared straight r to the power of straight n minus 1 end exponent close parentheses  equals 1 half nlog invisible function application open parentheses straight a squared straight r to the power of straight n minus 1 end exponent close parentheses

Karena straight a equals straight x subscript 1 maka hasilnya adalah 1 half nlog invisible function application open parentheses straight x subscript 1 superscript 2 straight r to the power of straight n minus 1 end exponent close parentheses

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