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I don't know how to find the value of $\alpha$: $$(\sqrt{3}+\sqrt{2})^\alpha + (\sqrt{3}+\sqrt{2})^\alpha = 10$$ I tried to simplify it to $$(\sqrt{3}+\sqrt{2})^\alpha=5$$ then take the ($\ln$) of both sides $$\ln{(\sqrt{3}+\sqrt{2})^\alpha} = \ln{5}$$ then $$\alpha\ln{(\sqrt{3}+\sqrt{2})} = \ln{5}$$ then $$\alpha = \frac{\ln{5}}{\ln{(\sqrt{3}+\sqrt{2})}}$$ is that correct?

zizaaooo
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    Welcome to stackexchange. That said, I’m voting to close this question because you show no work of your own. This is not a "do it for me" site. If you [edit] the question to tell us what you tried and where you are stuck we may be able to help. – Ethan Bolker Sep 22 '23 at 20:45
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    Did you try to divide by $2$ first? – Dietrich Burde Sep 22 '23 at 20:52
  • I voted to reopen because you showed enough work. However, you completely solved your (strange but obvious) exercise, so I think this post can be deleted now. I don't think it can be useful to anyone. – Anne Bauval Sep 22 '23 at 21:11
  • Yes, your solution is correct. For a slightly more compact result, you could take $\log_5$ of both sides in place of the $\ln$ step, giving $\alpha=\frac 1{\log_5(\sqrt 3+\sqrt2)}$, but this is neither expected nor any better or worse than what you have done. – abiessu Sep 22 '23 at 21:16
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    Is it possible that the equation be $(\sqrt{3}+\sqrt{2})^\alpha + (\sqrt{3}-\sqrt{2})^\alpha = 10$? then the answer will be very simple $\alpha = 2$ – Etemon Sep 22 '23 at 21:35
  • But if @Etemon's guess is right, your question is rather a duplicate of this: https://math.stackexchange.com/a/2600381 which was itself a dupe of: https://math.stackexchange.com/a/78607 (and there are two solutions: $2$ and $-2$). – Anne Bauval Sep 22 '23 at 21:47

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