The equation in question is $$\log_5x*(\log_5(2*\log_{10}a-x)*\log_x5+1)=2$$
Tried working this down with the rules of logarithms, got it down to a quadratic equation of $x$ with $a$ as one of its parameters, but I'm sure that's not the right way to do it because ultimately I would get a single value and not an interval.
Tried getting it down to a single logarithm but they get nested because of different bases. Trivial statements such as $a>0$ don't help (me at least).
I'm not really sure I have an idea of what I should do with this, any hints about what should I aim to get?
Let's start by simplify your equation : \begin{align} \log_5 x\left [ \log_5\left(2\log_{10}a - x\right)\log_x 5 + 1 \right] = 2 &\Leftrightarrow \log_5\left(2\log_{10}a - x\right) + \log_5 x = 2\\ & \Leftrightarrow x \left(2\log_{10}a - x\right) = 25\\ & \Leftrightarrow -x^2 + 2\log_{10} (a) x -25 = 0 \end{align}
The discriminant of the last equation is : $$\Delta = 4 \log_{10}^2 a - 100.$$ The equation admits real solutions iff $\Delta \geq 0$ hence : $$4\log_{10}^2 a -100 \geq 0 \Leftrightarrow a\geq 10^5.$$ Edit : As @aschepler said the case $\log_{10} a \leq -5$ is rejected because one must have $2\log_{10}a -x >0$.