Log_{2a}a=x then, a=(2a)^{x}......(1)

Log_{3a}2a=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.

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