# 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?

• It looks good to me. Note that there are more solutions when taking into account all the branches of the $\log$ function, but these are only two real solutions. – Graviton Jul 31 '20 at 1:46
• Yes, provided you are looking for real solutions only. You can also verify by plotting a graph: desmos.com/calculator/5w51j5xmlk – an4s Jul 31 '20 at 1:47
• @PrestonLui consider $x=-3+\sqrt{1+\frac{2\pi i}{\log5}}$, then $x^2+6x+8=\frac{2\pi i}{\log5}$ and $5^{x^2+6x+8}=1$. The power is non-zero yet equates to $1$. – Graviton Jul 31 '20 at 1:52
• @PrestonLui $e^{2\pi k i} =1$. So $x^2 + 6x + 8 =2\pi k i$ will be solutions. $x= \frac {-6 \pm\sqrt {36-4(8-2k\pi i)}}{2}$ will be an infinite number of solutions. But they'll only be real solutions if $k =0$. – fleablood Jul 31 '20 at 1:55
• "I think by the property of exponential function it was only 1 if and only if the power is 0" Not in complex analysis. $b^m = 1 \iff m= 2\pi k i$ for some integer $k$. But $k=0$ will be the only time $m= 2\pi k i$ is real. – fleablood Jul 31 '20 at 1:57

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*}

• ahahahha this is more elegant. – EM4 Jul 31 '20 at 1:52
• I'm sure it's unneeded for the purpose of this question, but this ignores complex solutions. the first $\iff$ is only true if $x\in\mathbb{R}$ – Graviton Jul 31 '20 at 1:55
• Indeed, you are right. But I assume the OP is only interested in the real solutions. – APCorreia Jul 31 '20 at 2:02
• this is only for the real solutions. – EM4 Jul 31 '20 at 2:05

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

• Note that a 'real solution' is just one way of saying a solution that lies on the number line that you are familiar with. Actually, there are also 'complex solutions' that lie on a different plane (you may have heard of $i=\sqrt{-1}$, but if you haven't heard of complex numbers before then don't worry about it). – Joe Jul 31 '20 at 1:49
• @Joe I have indeed learnt about complex numbers just last year. I just was thinking along the same lines as user Graviton, that other values may emerge from complex valued logarithms, which I'm not too comfortable with, and hence to be on the safer side, I wrote real solutions😅 – AryanSonwatikar Jul 31 '20 at 1:54
• Yes, complex-valued logarithms are indeed rather tricky! In fact, in order to have a working definition of $\log(x)$, you must first define what $e^x$ means when $x$ is not a real number. Often this is done using a Taylor series. – Joe Jul 31 '20 at 1:58

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

$$\ln(5)(x^2+6x+8)=2\pi ki$$

$$x=-3\pm\sqrt{1+\frac{2\pi ki}{\ln5}}$$
Note that when $$k=0$$, we get the real solutions $$x=-2,-4$$