a= log_{24}12 , b= log_{48}36 , c= log_{36}24 Then how to prove 1+abc = 2bc
a = (log 12)/(log 24)
b = (log 36)/(log 48)
c = (log 24)/(log 36)
1 + abc = 2bc
1 +(log 12)(log 24)(log 36)/(log 24)(log 36)(log 48) = 2(log 24)(log 36)/(log 36)(log 48)
Cancel out between the numerator & denominator on both sides
1 + (log 12)/(log 48) = 2(log 24)/(log 48)
(log 48)/(log 48) + (log 12)/(log 48) = 2(log 24)/(log 48)
(log 48 + log 12)/(log 48) = (log 24^2)/(log 48)
(log (48*12))/(log 48) = (log 576)/(log 48
(log 576)/(log 48) = (log 576)/(log 48)
They are equal