Roboguru

Jika diketahui  adalah ....

Pertanyaan

Jika diketahui begin mathsize 14px style space squared log space 3 space equals straight p space dan space cubed log 5 space equals straight q comma space maka space nilai space dari space to the power of 6 log space 45 end style adalah ....

  1. fraction numerator 2 straight p plus pq over denominator 1 minus straight p end fraction

  2. fraction numerator 2 straight p minus pq over denominator 1 plus straight p end fraction

  3. fraction numerator 2 straight p plus pq over denominator 1 plus straight p end fraction

  4. fraction numerator straight p minus 2 pq over denominator 1 plus straight p end fraction

  5. fraction numerator straight p plus 2 pq over denominator 1 plus straight p end fraction

Pembahasan Soal:

begin mathsize 14px style blank squared log 3 equals straight p left right double arrow fraction numerator log 3 over denominator log 2 end fraction equals straight p left right double arrow log 2 equals 1 over straight p log 3  to the power of space 3 end exponent log 5 equals straight p left right double arrow fraction numerator log 5 over denominator log 3 end fraction equals straight q left right double arrow log 5 equals qlog 3  to the power of space 6 end exponent log 45 equals fraction numerator log 45 over denominator log 6 end fraction  space space space space space space space space space space equals fraction numerator log left parenthesis 3 squared space straight x space 5 right parenthesis over denominator log left parenthesis 3 space straight x space 2 right parenthesis end fraction  space space space space space space space space space space equals fraction numerator log 3 squared plus log 5 over denominator log 3 space plus log 2 end fraction  space space space space space space space space space space equals fraction numerator 2 log 3 plus log 5 over denominator log 3 space plus log 2 end fraction  space space space space space space space space space space equals fraction numerator 2 log 3 plus qlog 3 over denominator log 3 space plus begin display style 1 over straight p end style log 3 end fraction  space space space space space space space space space space equals fraction numerator left parenthesis 2 plus straight q right parenthesis up diagonal strike log 3 end strike over denominator open parentheses 1 plus begin display style 1 over straight p end style close parentheses up diagonal strike log 3 end strike end fraction equals fraction numerator 2 plus straight q over denominator 1 end fraction straight x fraction numerator straight p over denominator 1 plus straight p end fraction equals fraction numerator 2 straight p plus pq over denominator 1 plus straight p end fraction end style

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 22 Desember 2020

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