# algebraic equation involving natural logarithm

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

• Hope its correctly edited . please see mathjax and then edit if its wrong – Archis Welankar Apr 16 '16 at 15:33
• The second square root is under the first square root. The rest is perfect. Big thanks! – Lloyd Kizito Apr 16 '16 at 15:43

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 $x=1$.
$$( x>0 \land x\ne 1) \implies$$ $$\implies 1+\sqrt {1+(2-x)\sqrt {x^2+4 x+3}}=x\implies$$ $$\implies \sqrt {1+(2-x)\sqrt {x^2+4 x+3}}=x-1\implies$$ $$\implies 1+(2-x)\sqrt {x^2+4 x +3}=x^2-2 x+1\implies$$ $$\implies (2-x)\sqrt {x^2+4 x+3}=(x-2)x\implies$$ $$\implies (x=2\lor 0<\sqrt {x^2+4 x+3}=-x<0)\implies x=2.$$ But $x=2$ is not a solution to the original equation. And $x=1$ is a solution, because for any $y, z$ we have $1^y=1^z=1.$