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Verified answer
LogAritMa²log 3 = m
³log 5 = n
6'log 75
= ³log 75 / ³log 6
= ³log (5² × 3) / ³log (2 × 3)
= (³log 5² + ³log 3) / (³log 2 + ³log 3)
= (2n + 1) / (1/m + 1)
= m(2n + 1) / (1 + m)
••
9'log 120
= ³log (2³ × 3 × 5) / (³log 3²)
= (3 . ³log 2 + ³log 3 + ³log 5) / 2
= (3/m + 1 + n) / 2
= (3 + m + mn) / 2m