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3log5 = m <=> 5log3 = 1/m
7log5 = m <=> 5log7 = 1/n
maka caranya adalah
35log15= 5log15 / 5log35 = 5log 5 x 3 / 5log 5 X 7 = 5log5 + 3log5 / 5log5 + 5log7
= 1 + 1/m / 1 + 1/n ( dikali mn)
<=> = mn + n / mn + n
Kurang lebih seperti itu jawabannya