TITLE: Solving : $5^{x^2+6x+8}$ = 1
QUESTION [5 upvotes]: Solve for $x$:
$$5^{x^2+6x+8} = 1$$

So, I took the natural logarithm on both sides,
$$(x^2+6x+8)\ln(5) = \ln(1)$$
then I divide both sides by $\ln(5)$ to set the polynomial to zero because we know $\ln(1) = 0$.
I will be left with:
$$x^2+6x+8 = 0$$
Factoring this will give:
$$(x+2)(x+4) = 0 \implies x = -2, -4 $$
Then I checked my $x$ values I got $1$. So my question is did I do it correctly?

REPLY [3 votes]: You reasoning is correct, and the other answers here work wonderfully for real solutions, but If you're interested in complex analysis then a proof for all solutions goes as follows.
Consider
$$\forall k\in \mathbb{Z}$$
$$e^{2\pi ki}=1$$
$$\implies e^a=1\iff a=2\pi ki$$
Then we can reformulate
$$5^{x^2+6x+8}=(e^{\ln(5)})^{x^2+6x+8}=e^{\ln(5)(x^2+6x+8)}=1$$
then by $e^a=1\iff a=2\pi ki$ we find that all solutions must be in the form
$$\ln(5)(x^2+6x+8)=2\pi ki$$
And by the quadratic formula,
$$x=-3\pm\sqrt{1+\frac{2\pi ki}{\ln5}}$$
Note that when $k=0$, we get the real solutions $x=-2,-4$