Нужно! log5(15)=a; 5-основ. log12(24)=b; 12-осн. - вопрос №47315915155122412125125 от Poniy11 17.02.2023 10:43

Question

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

Poniy11 · Answer

Log5(15) = Log5(3) + 1       =   Log5(3) = a-1Log12(24) = Log12(2) + 1   =   Log12(2)=b-1Log125(48) = (Log5(3*16)) / 3 = (Log5(3) + 4Log5(2)) / 3Log5(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...