Question

Выразить log 120 по основанию 300 через a и b, если a = log 3 по основанию 2, b = log 5 по основанию 3

Answer 1

log300 (120)=log2 120/log2 300=log2 (2^3*3*5)/log2 (2^2*3*5^2)=( log2 2^3+log2 3 +log2 5 )/(log2 2^2 +log2 3 + log2 5^2)=(3+a+log2 5)/(2+a+2*log2 5)=(3+a+(log3 5)/(log3 2))/(2+a+2*(log3 5)/(log3 2))=(3+a+(log3 5)*(log2 3))/(2+a+2*(log3 5)*(log2 3))=(3+a+a*b)/(2+a+2*a*b)

Answer 2

Ответ: log₃₀₀120 = log₂120 / log₂300 = (log₂3 + log₂40) / (log₂3 + log₂100) = (a + log₂8 + log₂5) / (a + log₂25 + log₂4) = (a + 3 + log₂5) / (a + 2 + 2 * log₂5) = (a + 3 + b / log₃2) / (a + 2 + 2 * b / log₃2)