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

Question

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?

Answer 1

Your reasoning and calculation are both correct, for real solutions.

Answer 2

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*}

Answer 3

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 $x=-2,-4$