Log_2aa=x    then, a=(2a)^x  ......(1)

Log_3a2a=y    then,2a=(3a)^y ......(2)

Log_4a 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.

This picture develops a clear understanding.

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