Let $x$ be the smallest positive solution of the following equation $$x=\frac{1}{\beta}\ln x+a-\frac{1}{\beta}\ln a+c$$ where $\beta>1$ and $a,c\in(0,1)$ are fixed constants.
I need to know the behaviour of $x$ as a function of $a$ when $\beta$ and $c$ are fixed.
I would like to prove that $x(a)$ is an increasing function.
Could someone help me? Thank you
Let
$$f(x)=x-\left(\frac{1}{\beta}\ln x-a+\frac{1}{\beta}\ln a+c\right).$$
We have $$f'(x)=1-\frac{1}{\beta x}\qquad f''(x)=\frac{1}{\beta x^2}$$
so that $f$ is convex with minimum at $x=1/\beta$.
We have
$$f(1/\beta)=\frac{1}{\beta}\left(1+\ln \frac{\beta}{a}\right)+a-c\leq 0$$
if and only if
$$\frac{1}{\beta}\left(1+\ln \frac{\beta}{a}\right)\leq c-a.$$
Since $\ln (\beta/a)>0$, there is no solution if $c\leq a$. If the condition above holds with equality then there is exactly one solution $x=1/\beta$. If it holds with strict inequality then there are exactly two solutions, with the smaller satisfying $x<1/\beta$. Any solution has to be positive (clearly).
Let's suppose there is a solution. Record dependence of $x$ on $a$:
$$x(a)=\frac{1}{\beta}\ln [x(a)]-a+\frac{1}{\beta}\ln a+c.$$
Differentiate both sides with respect to $a$:
$$x'(a)=\frac{1}{\beta}\frac{x'(a)}{x(a)}-1+\frac{1}{\beta}\frac{1}{a}.$$
Then solve for $x'(a)$ to get:
$$x'(a)=\frac{(a\beta-1)x}{a(1-\beta x)}$$
so long as $\beta x\neq 1$ (we need $a>0$ for $\ln a$ to be defined). When there are two solutions, the smaller satisfies $\beta x<1$, so that the denominator is positive. The numerator is also positive so long as
$$a\beta>1.$$
You need to impose this extra condition (it is not implied by the other conditions) to ensure $x'(a)>0$.