Please Solve this

Question

a= log₂₄12  , b= log₄₈36 , c= log₃₆24 Then how to prove 1+abc = 2bc

Answer 1

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