Нужно! log5(15)=a; 5-основ. log12(24)=b; 12-осн. log125(; 125-осн

Log5(15) = Log5(3) + 1       =>   Log5(3) = a-1
Log12(24) = Log12(2) + 1   =>   Log12(2)=b-1

Log125(48) = (Log5(3*16)) / 3 = (Log5(3) + 4Log5(2)) / 3

Log5(2) = Log12(2) / Log12(5)

Log12(5) = Log3(5) / 4 = 1 / 4*Log5(3)   ==> Log5(2) = (b-1)+4(a-1)

Log125(48) = [(a-1) + 4(b-1)+16(a-1)] / 3 = (17a+4b-21) / 3