This was originally a typo in my textbook* and I spent a lot of time trying to solve it.
Lost and confused I turned to desmos which gave me some hope when I saw that this had a real solution, and it was weird.
The answer is apparently that the inequality holds for $x> -6.001$. I'm not sure if this is an approximation or the exact answer and I don't know where to head.
I tried squaring both sides, as one would when normally solving typical inequalities, to remove the mod but that leads nowhere.
$$x^2 + 2x +1 < 3^{2x} + 10\times3^x + 25 $$
And I don't know where to go from here.
I also notice that $|3^x + 5|$ is the same as $3^x + 5$ because it's ever-positive. So then I try to do this by thinking of mod as distances and try to write the inequality as:
\begin{align*}
&|x+1| < 3^x+5
\end{align*}
Now we have to consider two cases
$\bullet~$ Case 1: When $ \lvert x + 1 \rvert > 0 $
\begin{align*}
&x+1 < 3^x + 5 \\
\implies& x-4 < 3^x \\
\implies&\ln(x-4 ) < x\cdot \ln(3)
\end{align*}
$\bullet~$ Case 2: When $\lvert x + 1 \rvert < 0$
\begin{align*}
&-(x+1) < 3^x + 5 \; \quad [\text{No solution}] \\
\end{align*}
And I'm lost again. I don't know what I'm doing wrong or what even is the correct way to do this! Help!
*Originally $|x+1| < |3x+5|$
Best Answer
What I would like to show is that indeed we have only one real solution for $|x+1|=|3^x+5|$, call it $x_0$, and that the solution to the inequality is $(x_0,+\infty)$. Also, $x_0 \approx -6.001$.
Your idea to get rid of absolute values is a good one and your second approach works quite nicely.
You correctly conclude that since $3^x + 5 > 0$, you can just remove absolute value on the RHS of inequality.
For $|x+1|$ it looks like you have correct idea to look at cases, but your notation is incorrect. The correct cases to consider are:
In the 1st case, the inequality becomes $x+1<3^x + 5$. I claim that this is true for all $x\geq -1$. This follows from well known inequality $$e^x \geq x + 1,\quad \forall x\in\mathbb R.$$ If you never saw this, just plot the graphs of $e^x$ and $x+1$. You will see that the line $y = x + 1$ is tangent to the graph of $e^x$ at $x=0$. Let us use it on your problem: $$3^x = e^{x\ln 3} \geq x\ln 3 + 1 \implies 3^x + 5 \geq x\ln 3 + 6,$$ and you can easily see that $x\ln 3 + 6 > x + 1$ for $x\geq - 1$, which proves that our inequality is true on $[-1,\infty)$.
The 2nd case is more complicated. Now the inequality becomes $$-x-1 < 3^x + 5 \iff 3^x + x + 6 > 0.$$ First let us observe that the function $3^x + x + 6$ is strictly increasing. That means that the equation $3^x + x + 6 = 0$ either has no solutions or if it has solutions, it has only one solution. If we plug in $x = -6$, we get $3^{-6}$ which is positive, and if we plug in $x = -7$ we get $3^{-7} - 1$ which is negative. That means that (by continuity) that there exists $x_0 \in (-7,-6)$ such that $3^{x_0} + x_0 + 6 = 0$. Since $x_0< -1$, it is also the unique solution to $|x+1| = |3^x + 5|$ as I claimed at the start. Anyway, we now know that $$3^x + x + 6 > 0 \iff x\in (x_0,+\infty),$$ and taking into account that we are in our 2nd case, $$|x+1| < |3^x + 5|,\quad\forall x\in (x_0,-1).$$
Taking the union of our two cases, we conclude that the solution to our inequality is $(x_0,+\infty)$.
Hopefully, the above is not too hard to follow. What remains is to approximate $x_0$. As we already saw above, $3^x + x + 6$ evaluates to $3^{-6}$ for $x = -6$, which is quite close to $0$, so we expect $x_0$ to be close to $-6$ and more precisely, $x_0$ is a bit less than $-6$.
To get better approximations, we could use numerical methods, but since this is pre-calculus, I don't want to get into that.
I will get into something that is called Lambert W function, which you could argue is even less appropriate, but I think the arithmetic that we will perform with it is not too advanced.
To explain what Lambert W function does we need to observe equation of the form $$xe^x = a.$$ This equation might have $0$, $1$ or $2$ real solutions. If $a\geq 0$ then the equation $xe^x = a$ has unique nonnegative solution, and we will denote it by $x = W(a)$. Compare this to the equation $x^2 = a$ and how we denote one of its solutions with $x = \sqrt a$. We don't really know precise value of $W(a)$ in most cases, but then again, we don't know the precise value of $\sqrt a$ in most cases either. Luckily, we know how to approximate both $\sqrt a$ and $W(a)$.
But, let us return to our equation $3^x + x + 6 = 0.$ I will manipulate it so that we may use Lambert W:
\begin{align} 3^x + x + 6 = 0 & \iff -(x+6) = e^{x\ln 3}\ /\cdot e^{-(x+6)\ln 3}\\ & \iff -(x+6)e^{-(x+6)\ln 3} = e^{x\ln 3-(x+6)\ln 3}\\ & \iff -(x+6)e^{-(x+6)\ln 3} = 3^{-6}\ /\cdot \ln 3\\ & \iff -(x+6)\ln3 \cdot e^{-(x+6)\ln 3} = 3^{-6}\ln 3. \end{align}
Now, if we substitute $y = -(x+6)\ln 3$, the last line becomes $ye^y = 3^{-6}\ln 3$ and we conclude that $y = W(3^{-6}\ln 3)$ (note that $3^{-6}\ln 3>0$). Substituting $x$ back, we can now easily calculate that $$x = -6 - \frac 1{\ln 3} W(3^{-6}\ln 3),$$ which is our $x_0$ from before.
Finally, it happens to be that $W(a) \approx a$ when $a$ is close to $0$. This is analogous to $\sin a \approx a$ when $a$ is close to $0$, if you ever saw that one, without going into reasons why that might be true. Now, since $3^{-6}\ln 3 \approx 0.00150701$, we have $W(3^{-6}\ln 3) \approx 3^{-6}\ln 3$. With this we can now approximate $$x_0 = -6 - \frac 1{\ln 3} W(3^{-6}\ln 3) \approx - 6 - \frac{3^{-6}\ln 3}{\ln 3} = -6 - 3^{-6} \approx -6.0013717.$$
This approximation is quite good. You can check that Wolfram Alpha approximates $x_0$ to be $-6.00137$.