How to prove monotonicity of this function?

Question

Let $a>0$. How to prove that the function: $$f(x)=\frac{a x-1}{\log(a x)}\cdot\frac{\log x}{x-1},$$ is monotonic (depending on $a<1$ or $a>1$). I know that we can calculate the derivative and determine its sign, but this needs much of calculation. I'm wondering if we can decide the monotonicity using a simple trick.

Answer 1

Consider $$\begin{align} g&:\mathbb{R}\to \mathbb{R}& g(t)&= \ln \frac{\mathrm{e}^t-1}{t} \end{align}$$ (where the value at $0$ is defined by continuity). Then from $$\ln f(x) = g(\ln x + \ln a) - g(\ln x)$$ we know that $\tfrac{1}{\ln a} f$ is monotone for all $a\neq 1$ iff $g$ is convex. But $g$'s second derivative $$g''(t)=\frac{\cosh t-1-\tfrac{t^2}{2}}{t^2(\cosh t -1)}$$ is nonnegative.

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Admittedly, since $$g(t)=t + \sum_{n=1}^{\infty}\frac{B_n}{n}\frac{t^n}{n!}$$ is itself an important elementary function, using the second derivative to reduce the result to inequalities of other elementary functions seems unsatisfactory. Maybe there's a more intrinsic route?