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.

X=log_2a a= loga/log2a

y=log_3a 2a= log2a/log3a

z= log_4a 3a= log3a/log4a

So,

xyz = (loga.log2a.log3a) / (log2a.log3a.log4a)

Or, xyz + 1 = loga/log4a +1

Or, xyz +1 = (loga+log4a)/log4a

Or, xyz +1= log(2a)² /log4a

Or, xyz +1= 2(log2a/log4a)

Or, xyz +1=2. (log2a/log3a)×(log3a/log4a)

Or, xyz +1 = 2yz