Analyze the property of $\frac{\log\left(1+p - \frac{1}{1+(1-p)x}\right)}{\log(x)},\;\;x>1,\;\;0Consider the function
$$f(x) = \dfrac{\log\left(1+p - \frac{1}{1+(1-p)x}\right)}{\log(x)},\;\;\;x>1,\;\;0<p<1.$$
I guess the function increases firstly and then decreases. But I don't know how to prove this property. I tried computing the derivative but it's too messy to analyse. I am wondering if there is any simple method to analyse it.

It is easy to see there is an upper bound on the function $f(x)$, for example, $f(x) \leqslant \dfrac{\log(2)}{\log(x)}$ and I want to find a tighter bound on $f(x)$.
Firstly, we can show that
$f(x) \leqslant \dfrac{\log(2x-2\sqrt{x}+1)}{\log(x)}-1,\quad \forall p \in (0, 1)$.
Define $\;g(x) = \dfrac{\log(2x-2\sqrt{x}+1)}{\log(x)},\;\;\forall x >1.\;$ Then we just need to bound $g(x)$.
 A: For the first question:
We have
$$f'(x) = \frac{1}{x\ln^2 x} \left(\frac{(1-p)x\ln x}{(1 + x - px)(p + x - p^2x)} - \ln(p + x - p^2x) + \ln(1 + x - px) \right).$$
Let
$$g(x) := \frac{(1-p)x\ln x}{(1 + x - px)(p + x - p^2x)} - \ln(p + x - p^2x) + \ln(1 + x - px).$$
We have
$$\lim_{x\to 1} g(x) = \ln \frac{2 - p}{1 + p - p^2} > 0 \tag{1}$$
and
$$\lim_{x\to \infty} g(x) = - \ln (1 + p) < 0 \tag{2}$$
and
$$g'(x) = \frac{(1-p)\ln x}{(1 + x - px)^2(p + x - p^2x)^2}[p - (1 + p)(1-p)^2x^2]. \tag{3}$$
Let $p_0\in (0, 1)$ be a real root of $p - (1 + p)(1 - p)^2 = 0$.
If $0 < p \le p_0$, we have
$g'(x) < 0$ on $(1, \infty)$. Thus, $g(x) = 0$ has exactly one real solution on $(1, \infty)$, so does $f'(x) = 0$.
If $p > p_0$, letting $x_1 = \sqrt{\frac{p}{(1 + p)(1 - p)^2}} $, we have $g(x)$ is strictly increasing on $(1, x_1)$,
and strictly decreasing on $(x_1, \infty)$. Thus, $g(x) = 0$ has exactly one real solution on $(1, \infty)$, so does $f'(x) = 0$.
Thus, there exists some $x_0 > 1$ such that
$f'(x) > 0$ on $(1, x_0)$, and $f'(x) < 0$ on $(x_0, \infty)$.
In other words, $f(x)$ is strictly increasing on $(1, x_0)$, and strictly decreasing on $(x_0, \infty)$.
