I tried differentiating the equation to get minima and maxima, but failed to find the roots even there. Trial and error provides the answer=2, however, I'm searching for a proper method. Thanks in advance. :)
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Rewrite as $$a^{x/2}+b^{x/2}=1,$$ with $$a=\frac{2}{5+2\sqrt2}\text{ and }b=\frac{\left(\sqrt2+1\right)^2}{5+2\sqrt2}=\frac{3+2\sqrt2}{5+2\sqrt2}.$$
As $0<a,b<1$, the sum of exponentials is strictly decreasing and there is at most one root.
The solution $x=2$ is quasi-obvious, as $a+b=1$.
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The part with the derivative is not needed. To establish the variation of $a^x+b^x$ as a function of $x$, simply note that $0\lt a,b\lt1$. – Did Oct 29 '14 at 11:14
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Thanks Yves for the answer, but, can you please elaborate the derivative part? :) – Harshal Gajjar Oct 29 '14 at 11:17
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@Did: indeed. I am simplifying my answer. – Oct 29 '14 at 11:39
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"Exactly one root" should not be too difficult to deduce now... :-) – Did Oct 29 '14 at 11:58
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Thanks Yves for the feedback. – Harshal Gajjar Oct 29 '14 at 12:02
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@Did: interestingly, the equation can be rewritten as $X^\alpha+X=1$, with $\alpha=\ln a/\ln b$, an irrational number, and it has an algebraic solution. – Oct 29 '14 at 12:03