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What conditions must $a$ and $b$ satisfy for the equation to have at least one real solution? Find all the solutions of this equation: $1+\log_b(2\log(a)-x)\log_x(b)=2\log_b(x)$

I have tried restricting the domain of $2\log(a)-x>0$.

$1 + \log_b(2 \cdot \log(a) - x) \cdot \log_x b = \frac{2}{\log_b x}$

$1 + \log_b(2 \cdot \log(a) - x) \cdot \frac{1}{\log_b x} = \frac{2}{\log_b x}$

$log_b x + \log_b (\log(a^2) - \log(10^x)) = 2$

$log_b \left(x \cdot \log\left(\frac{a^2}{10^x}\right)\right) = 2$

$b^2 = x \cdot \log\left(\frac{a^2}{10^x}\right)$

$\frac{b^2}{x} = \log\left(\frac{a^2}{10^x}\right)$

$10^{\frac{b^2}{x}} = \frac{a^2}{10^x}$

Then what should I do? I've tried a lot, that's the reason i ask for help ,isn't because i want some of you to do my "homework" it's because i'm done.

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    $\begingroup$ Can you please improve the formatting for your question? It would help potential answerers. Also, write your efforts in solving as well. Thanks. $\endgroup$
    – GSmith
    Commented Jul 8 at 2:17
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    $\begingroup$ I think you may want to edit your post to include your thinking: What have you tried? Your thinking needn't be correct: It's natural to make mistakes. Just show you have made an effort before asking. $\endgroup$
    – Simon
    Commented Jul 8 at 2:17
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    $\begingroup$ As an FYI: MSE is not a "do my (home)work for me" site. Such posts (e.g. "PSQs") are often in poor taste, downvoted, & closed; we expect users to actively put in effort! If you want meaningful help, edit your post to add relevant context (more), e.g. your work/attempts, the problem's source, where specifically you're stuck, what you do/don't understand & have learned recently, & so on. Also: this is helpful for formatting LaTeX, & Approach0 can help find past questions. $\endgroup$
    – PrincessEev
    Commented Jul 8 at 2:17
  • $\begingroup$ What is the base for $\log(a)$? $\endgroup$
    – Lucenaposition
    Commented Jul 8 at 2:22
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    $\begingroup$ What happened when you tried restricting the domain? Please write that in the question. $\endgroup$
    – GSmith
    Commented Jul 8 at 3:15

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