Let f (x) = (x - 3) log x
                Then, f (1) = - 2 log 1 = 0 and f (3) = (3-3) log 3 = 0. As, (x-3) and log x are continuos and differentiable in [1, 3], therefore (x-3) log x = f (x) is also continuos and differentiable in [1, 3]. Hence, by Rolle's theorem, there exists a value of x in (1, 3) such that
f ' (x) = 0 $\Rightarrow$ log x+(x-3) $\frac{1}{x}$ = 0
$\Rightarrow$ x log x = 3 - x.
Hence (c) is the correct answer.