Solving : $5^{x^2+6x+8}$ = 1 
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?

Another way to reach the same equation consists in noticing that the exponential function $a^{x}$ is injective:
\begin{align*}
5^{x^{2} + 6x + 8} = 1 = 5^{0} \Longleftrightarrow x^{2} + 6x + 8 = 0 \Longleftrightarrow \ldots
\end{align*}
