Sum of all the possible real solutions of $(x^2+4x+5)^{(x^2+4x+5)^{(x^2+4x+5)}}=2018$
My try
I knew that the answer is $-4$ (according to wolfram alpha) but i tried myself solving it by hand.
I tried making $x^2+4x+5=u$, solving for $u$, and then replacing $(x^2+4x+5)$. Then, the equation is $u^{u^u}=2018$, but after that i didn't find a way of solving this.
Any hints?
Best Answer
Part 1: show there is exactly one value of $u$ that solves $u^{u^u} = 2018$.
Part 2: what is the sum of the roots of the quadratic equation $x^2 + 4x +(5-u) = 0$? Hint: the value of $u$ doesn't matter.