_{24}12 , b= log

_{48}36 , c= log

_{36}24 Then how to prove 1+abc = 2bc

a= log_{24}12 , b= log_{48}36 , c= log_{36}24 Then how to prove 1+abc = 2bc

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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

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