Show that $\log( x^\alpha + 2)/\log(x + 2)$, $0 < \alpha < 1$, has a unique critical point on $(0, \infty)$ Fix $0 < \alpha < 1$. Consider the function $f : [0,\infty) \to (0,\infty)$ defined by
$$f(x) = \frac{\log(x^{\alpha} + 2)}{\log(x + 2)}. $$
We clearly have $f(0) = 1$ and, by L'Hospital's rule, $\lim_{x \to \infty} f(x) = \alpha$.
We can also compute, for $0 < x <\infty$,
$$ f'(x) = (\log(x + 2))^{-2} \Big[\frac{\alpha x^{\alpha-1}}{x^\alpha + 2} \log(x + 2) - \frac{1}{x + 2} \log(x^\alpha + 2)\Big]. $$

I would like to show that $f$ has a unique critical point $x_0 \in (0, \infty)$ and that $f' < 0$ on $(x_0, \infty)$.

The first step is to show
$$\alpha x^{\alpha - 1}(x + 2) \log(x+2) = (x^\alpha + 2)\log(x^\alpha + 2),$$
for some unique value $x = x_0$. This seems difficult to solve analytically, so I'm curious to know what asymptotic methods may be available to prove this expression has a unique root.
 A: Clearly $$\lim_{x\rightarrow 0^+}f’(x)=+\infty,f’(1)=\frac 1{(\log 3)^2}\cdot \frac {\log 3}3\cdot(\alpha-1)<0,$$ hence by the Intermediate Value Theorem for continuous functions, there is $0<x_0<1$ such that $f’(x_0)=0.$ Let $$g(x)=\alpha x^\alpha(x+2)\log(x+2)-x(x^\alpha+2)\log(x^\alpha+2),$$ so $$f’(x)=\frac{(\log(x+2))^{-2}}{x(x^{\alpha}+2)(x+2)}\cdot g(x),\quad (1)$$ and $g(x)$ determines the sign of $f’(x).$
Lemma 1. $g(x)<0,\forall x\geq 1.$
Proof. Note that $$g(x)<0,\forall x\geq 1$$
$$\Leftrightarrow \alpha<\frac{x+2x^{1-\alpha}}{x+2}\cdot \frac{\log(x^{\alpha}+2)}{\log(x+2)},\forall x\geq 1.$$ Since the first factor on the RHS of the last inequality is $>1$, it suffices to show that $$\frac{\log(x^\alpha+2)}{\log(x+2)}\geq \alpha,\forall x\geq 1$$
$$\Leftrightarrow \log(x^\alpha+2)-\alpha\log(x+2)\geq 0,\forall x\geq 1$$
$$\Leftrightarrow x^{\alpha}+2\geq (x+2)^\alpha,\forall x\geq 1.$$ The latter is true as shown below. Let $$k(x)=x^\alpha+2-(x+2)^\alpha.\quad ({\rm Note~that~}0<\alpha<1.)$$ Then $k(1)=3-3^\alpha>0$ and $$k’(x)=\alpha x^{\alpha-1}-\alpha(x+2)^{\alpha-1}=\alpha\left(\frac 1{x^{1-\alpha}}-\frac 1{(x+2)^{1-\alpha}}\right)>0,\forall x\geq 1,$$ which shows that $k(x)>0$ for $x\geq 1$ and this concludes the proof of Lemma 1.
Lemma 2. Let $0<x_0<1$ and $g(x_0)=0$. Then $g(x)<0,\forall x\in (x_0,1).$
Proof. Write $g(x)=x^\alpha\cdot h(x),$ where $$h(x)=\alpha(x+2)\log(x+2)-(x+2x^{1-\alpha})\log(x^\alpha+2).$$ Note that $g(x_0)=0$ implies $h(x_0)=0$ and one has $$h’(x)=\alpha\log(x+2)+\alpha-(1+2(1-\alpha)x^{-\alpha})\log(x^{\alpha}+2)-\frac{x+2x^{1-\alpha}}{x^\alpha+2}\cdot\alpha x^{\alpha-1}$$
$$=\alpha\log(x+2)-(1+2(1-\alpha)x^{-\alpha})\log(x^\alpha+2)$$
$$<\alpha\log(x+2)-\log(x^{\alpha}+2)$$
$$< \alpha\log(x+2)-\log(x+2)<0,$$ where one used $0<x<1$ and $0<\alpha<1.$ It follows that $h(x)<0,\forall x\in(x_0,1)$ and so is $g(x)$.
Now as in the beginning sentence, there exists $x_0$ such that $0<x_0<1$ and $f’(x_0)=0,$ and so by (1) $g(x_0)=0$. By Lemma 2, such $x_0$ is unique (by Lemma 1, $g(x)<0,\forall x\geq 1$). Furthermore by Lemma 1 and Lemma 2, one has by (1) that $$f’(x)<0~{\rm for~all~}x\in (x_0,\infty).$$
QED
