Solution
= Solution
Differentiating on the given domain yields
$$
j'(x)=\frac1{x\ln x},\qquad j''(x)=-\frac{\ln x+1}{x^2(\ln x)^2}<0\quad(x>1).
$$
The denominator and $\ln x+1$ are positive there. Thus \b[$j$ is strictly <concave>], and is not <convex>.