Given that the function f (x) satisfies f (x + 1) = loga (x-1) / (3 + x) (a > 0 and a ≠ 1), the analytic expression and domain of definition of F (x) are obtained

F (x + 1) = loga (x-1) / (3 + x) (a > 0 and a ≠ 1)
Let x + 1 = t
Then x = T-1
∴ f(t)=loga [((t-2)/(t+2)]
∴ f(x)=loga (x-2)/(x+2)
The domain is (X-2) / (x + 2) > 0
That is, x > 2 or x2 or X

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