Consider the following equation:
$$\sqrt{\frac{\log\left(x + a + n\right)}{a}} - \sqrt{\frac{\log\left(x + a + n\right)}{x}} = \Phi$$
where all the variables belong to $\mathbb{R}$, and:
- $\log$ is the natural logarithm
- $x \ge 1$
- $a \ge 1$
- $n \ge 0$
- $\Phi$ can be positive or negative
And I'm wondering how to solve for $x$. Even for the special case $n = 0$, I'm not sure how to solve it.
COMMENT.- Your equation is equivalent to $$\log(x+a+n)^{\dfrac{\sqrt{ax}}{2\Phi(\sqrt x-\sqrt a)}}=1\iff(x+a+n)^{\dfrac{\sqrt{ax}}{2\Phi(\sqrt x-\sqrt a)}}=e$$ Keeping into account the data, the only thing I can say is the following:
When $x+a+n$ is exactly equal to $2$, you have $\dfrac{\sqrt{ax}}{2\Phi(\sqrt x-\sqrt a)}\approx1.443$.
When $x+a+n=3$, you have $\dfrac{\sqrt{ax}}{2\Phi(\sqrt x-\sqrt a)}\approx 0.91$
When $x+a+n=5$ you have $\dfrac{\sqrt{ax}}{2\Phi(\sqrt x-\sqrt a)}\approx 0.621$
It seems that if $x+a+n$ is big then $\dfrac{\sqrt{ax}}{2\Phi(\sqrt x-\sqrt a)}$ tends to zero.
Maybe someone else can say something more meaningful. I can not.