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I've been trying to solve this equation and I'm not getting right results. I've tried rewriting it with the natural logarithm and then tried solving the exponents but I seem to get way complicated results. Could anyone give me any hints on how I should go by. $$x^{1+\sqrt{(1+(2-x)\sqrt{(x^2+4x+3))}}} = x^x$$

The right answer in the book is 1.

Thanks in advance! /Lloyd

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  • $\begingroup$ Hope its correctly edited . please see mathjax and then edit if its wrong $\endgroup$ – Archis Welankar Apr 16 '16 at 15:33
  • $\begingroup$ The second square root is under the first square root. The rest is perfect. Big thanks! $\endgroup$ – Lloyd Kizito Apr 16 '16 at 15:43
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There are two possibilities. Either $x=1$ in which case we do not care what the exponents are - the two sides must be equal. Or the exponents are equal.

Equal exponents requires $1+(2-x)\sqrt{x^2+4x+3}=(x-1)^2$ and hence $-x=\sqrt{x^2+4x+3}$ and so $x=-\frac{3}{4}$. But that must fail as a solution since going back to the original equation the exponent on the lhs is always positive.

So the only solution is $x=1$.

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  • $\begingroup$ Thanks for the explanation man! $\endgroup$ – Lloyd Kizito Apr 16 '16 at 16:45
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$$( x>0 \land x\ne 1) \implies$$ $$\implies 1+\sqrt {1+(2-x)\sqrt {x^2+4 x+3}}=x\implies$$ $$\implies \sqrt {1+(2-x)\sqrt {x^2+4 x+3}}=x-1\implies$$ $$\implies 1+(2-x)\sqrt {x^2+4 x +3}=x^2-2 x+1\implies$$ $$\implies (2-x)\sqrt {x^2+4 x+3}=(x-2)x\implies$$ $$\implies (x=2\lor 0<\sqrt {x^2+4 x+3}=-x<0)\implies x=2. $$ But $x=2$ is not a solution to the original equation. And $x=1$ is a solution, because for any $y, z$ we have $1^y=1^z=1.$

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