We want to solve \begin{align} 0= c_1 x+c_2 x \ln \left(\frac{1+x}{x} \right)+c_3 \ln\left(\frac{1+x}{x} \right)+c_4 \end{align} for $x \in [0,1]$ and where $c_1,c_2,c_3,c_4$ are non-zero constants.
Here are some alternative way of re-writing this equation: \begin{align} 0&=x+a_1x \ln \left(\frac{1+x}{x} \right)+a_2ln \left(\frac{1+x}{x} \right)+a_3\\ 1&=e^x+\left(\frac{1+x}{x} \right)^{a_1x}+\left(\frac{1+x}{x} \right)^{a_2}+e^{a_3} \end{align}
I know that equations of this type generally don't have nice close form solutions. However, can we give at least an interval where the solution should be, that is if $x^*$ is an solution can we come up with $n$ and $m$ such that \begin{align} 0 \le n \le x^* \le m \le 1 \end{align}
I am also interested in the case when: \begin{align} c_2 \ge 0, \ c_3 \le 0, \ c_4 \ge 0 \end{align} and $c_1$ can be either positive or negative.
Thank you for any help
Choose a point $a \in [0, 1]$ and because of the small range, use Series solution for $x-a$:
$$\ln\left(\frac{1+x}{x}\right) = \ln\left(1 + \frac{1}{a}\right) + \frac{(x-a)}{a(1+a)} + \frac{1+2a}{a^2(1+a)^2}\frac{(x-a)^2}{2} + \cdots $$
Then you may suppose $x-a$ is really small and stop to first order.
Notice that you might write the very first term as $A$, a constant.