Mattrab Community - SXC asked a question

Log2aa=x    then, a=(2a) ......(1)

Log3a2a=y    then,2a=(3a)y ......(2)

Log4a 3a=z  then, 3a=(4a)z ......(3)

So, 

a=(2a)x  [from (1)]

Or, a=(3a)xy    [from(2)]

Or, a=(4a)xyz     [from(3)]

Multiplying both sides by 4a,

4a.a=4a.(4a)xyz  

Or,(2a)² =(4a)xyz + 1 

Or,(3a)2y =(4a)xyz+1 

Or,(4a)2yz =(4a)xyz+1 

Or, 2yz = xyz+1 .proved.



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