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


  • $\begingroup$ The solution is $-4$! $\endgroup$ Jul 21, 2014 at 18:01

1 Answer 1


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 $-4$.


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