<s>[INST] Sum of all real solutions for $x$ to the equation $\displaystyle (x^2+4x+6)^{{(x^2+4x+6)}^{\left(x^2+4x+6\right)}}=2014.$ Find the sum of all real solutions for $x$ to the equation $\displaystyle (x^2+4x+6)^{{(x^2+4x+6)}^{\left(x^2+4x+6\right)}}=2014.$
$\bf{My\; Try::}$ Let $y=x^2+4x+6 = (x+2)^2+2\geq 2$.
So our exp. equation convert into $\displaystyle y^{y^{y}} = 2014\;,$ where $y\geq 2$
Now at $y=2\;,$ We Get $\displaystyle 2^{2^{2}} = 16<2014$ and $\displaystyle 3^{3^{3}} = 3^{27}>2014$
So $y$ must be lie between $2$ and $3$.
But I did not Understand How can I calculate it..
Help me
Thanks
 [/INST] As you have written if one substitutes $y=x^2 + 4x +6 = (x+2)^2+2\ge 2$, then the resulting equation is $f(y)=2014$ where $f(y)=y^{y^y}$. Due the the fact that $2014>2^{2^2}$, it accepts solutions in ${\mathbb R}$. Now note that $f(y)=y^{y^y}$ is an strictly increasing function on ${\mathbb R}_{\ge 2}$. So for $f(y)=2014$ there exist a unique $y_0\in {\mathbb R}_{\ge 2}$ such that $f(y_0)=2014$. So one have that
$$
x^2 + 4x +6-y_0 = 0,
$$
where the sum of solutions is $-4$.
</s>