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I mean an equation $$ \log _{2 \sqrt{\sqrt{3}+2}}\left(x^2-2 x-2\right)=\log _{\sqrt{3}+2}\left(x^2-2 x-3\right).$$

The result of

ClearAll["Global`*"]; Reduce[Log[2*Sqrt[2 + Sqrt[3]], x^2 - 2 x - 2] == 
Log[2 + Sqrt[3], x^2 - 2 x - 3], x, Reals]

x == Root[{(Log[2] + Log[2 + Sqrt[3]]/2)*Log[-3 - 2*#1 + #1^2] - Log[2 + Sqrt[3]]*Log[-2 - 2*#1 + #1^2] & , -3.234170902346232549156572876357812204797}] || x == Root[{(Log[2] + Log[2 + Sqrt[3]]/2)*Log[-3 - 2*#1 + #1^2] - Log[2 + Sqrt[3]]*Log[-2 - 2*#1 + #1^2] & , 5.2341709023462325491515728763578122314}]

is numeric whereas the roots can be expressed symbolically.

FullSimplify[Log[2*Sqrt[2 + Sqrt[3]], x^2 - 2 x - 2] == 
Log[2 + Sqrt[3], x^2 - 2 x - 3] /. x -> 1 + Sqrt[11 + 4*Sqrt[3]]]

True

and

FullSimplify[Log[2*Sqrt[2 + Sqrt[3]], x^2 - 2 x - 2] == 
Log[2 + Sqrt[3], x^2 - 2 x - 3] /. x -> 1 - Sqrt[11 + 4*Sqrt[3]]]

True

Is there a way to find symbolic solutions?

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    $\begingroup$ Please do not use the bugs tags for your own questions. See the tag description. Wait until someone confirms the bug and adds the tag. $\endgroup$
    – Szabolcs
    Oct 19 at 16:23
  • $\begingroup$ It's very hard to visually follow the inputs you are showing. Can you assign parts of the expressions to variables, and rewrite everything in terms of those, to make it clear what you are showing? Don't keep writing out the full equation many times without good reason. We can't see if it's really the same equation or not. $\endgroup$
    – Szabolcs
    Oct 19 at 16:24
  • $\begingroup$ @Szabolcs: Sorry, don't understand. Could you present an example of such change? TIA. $\endgroup$
    – user64494
    Oct 19 at 16:29
  • $\begingroup$ @user64494 your first and second equations aren't even the same. The first one starts with 2*Log[2..., the second one with Log[2*... . Put that leading 2 back, and all the True's become False which is consistent with the first reduce. No bug here. $\endgroup$
    – flinty
    Oct 19 at 16:36
  • $\begingroup$ @flinty's comment illustrates perfectly what I meant in both comments. If you had written, "With eq = ..., Reduce[eq, x] gives no solutions but eq /. x -> r returns True", it would have been much easier to follow what you were doing, and you would not have made a mistake. $\endgroup$
    – Szabolcs
    Oct 19 at 16:45
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Not foolproof, but RootApproximant works for this case:

sol = Solve[Log[2*Sqrt[2+Sqrt[3]],x^2-2 x-2]==Log[2+Sqrt[3],x^2-2 x-3],x,Reals] //Values;
ToRadicals@RootApproximant[sol]

{{1 - Sqrt[11 + 4 Sqrt[3]]}, {1 + Sqrt[11 + 4 Sqrt[3]]}}

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  • $\begingroup$ +1. Thank you. As I understand it, RootApproximant produces only a a guess. $\endgroup$
    – user64494
    Oct 19 at 17:23
  • $\begingroup$ Do you mean "whole" instead of "fool"? TIA and deepregard. $\endgroup$
    – user64494
    Oct 19 at 17:27
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    $\begingroup$ Having the answer in hand, it's not hard to verify (sol - ans // FullSimplify). $\endgroup$
    – Michael E2
    Oct 19 at 17:43
  • $\begingroup$ @MichaelE2: A clever move. Thank you. I am able to present a lot of trigonometric equations where Reduce produces numerical solutions instead of symbolic ones. I am not sure whether RootApproximant works in that situations. $\endgroup$
    – user64494
    Oct 19 at 17:51
  • $\begingroup$ I accept your answer together with the valuable MiichaelE2's comment which finishes the work. $\endgroup$
    – user64494
    Oct 19 at 17:55

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