Найти log35(7), если log35(25) =a

log(a) b = 1/ log(b) a

log(35) 25 = a

1/ log(25) 35 = 1/ log(5^2) 35 = 2/ log(5) 5*7 = 2/(log(5) 5 + log(5) 7) = 2/( 1 + log(5) 7)  = 1/a

log(35) 7 = 1/log(7) 35 = 1/log(7) 5*7 = 1/(log(7) 7 + log(7) 5) = 1/(1 + 1/log(5) 7)

надо найти log(5) 7

2/( 1 + log(5) 7) = 1/a

2a = 1 + log(5) 7

log(5) 7 = 2a - 1

log(35) 7 = 1/(1 + 1/log(5) 7 ) = 1/(1 + 1/(2a - 1)) = 1 : 2a/(2a - 1) = (2a - 1)/2a = 1 - 1/2a