How do you solve $|x+1| < |3^x + 5|$?

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:

1. $x+1geq 0$,
2. $x+1<0$.

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1. In the 1st case, the inequality becomes $x+1<3^x + 5$. I claim that this is true for all $xgeq -1$. This follows from well known inequality $$e^x geq x + 1,quad forall xinmathbb R.$$ 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: $$3^x = e^{xln 3} geq xln 3 + 1 implies 3^x + 5 geq xln 3 + 6,$$ and you can easily see that $xln 3 + 6 > x + 1$ for $xgeq - 1$, which proves that our inequality is true on $[-1,infty)$.

2. 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 $$3^x + x + 6 > 0 iff xin (x_0,+infty),$$ and taking into account that we are in our 2nd case, $$|x+1| < |3^x + 5|,quadforall xin (x_0,-1).$$

Taking the union of our two cases, we conclude that the solution to our inequality is $(x_0,+infty)$.

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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 $ageq 0$ 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^{xln 3} /cdot e^{-(x+6)ln 3}\ & iff -(x+6)e^{-(x+6)ln 3} = e^{xln 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 $-6.00137$.

In such questions you have to make some approximations when finding solutions. first we remove the modulus function by splitting the domain in two parts.

Case 1: $xge-1$ The inequality becomes$$x+1<3^x+5$$ $$x-4<3^x$$ At $x=0$ value of $3^x$ is larger than the value of $x-4$. Also the value of slope of $3^x$ at $x=0$ is $ln(3)(>1)$ and it keeps on increasing with further increase in x $big(frac{d}{dx}(3^x)=ln(3)3^xbig)$. hence, $forall x>0$ above inequality is true (as the function $f(x)=3^x$ increases much more rapidly than $f(x)=x-4$). also $x-4$ is -ve for all $-1<x<0$ whereas $3^x$ is +ve. hence this whole set is a part of the solution.

Case 2: $x<-1$ We get $$-x-1<3^x$$ $$3^x>-x-6$$ $$implies 3^x+x+6>0$$ the graph of $y=3^x+x+6$ is monotonically increasing. hence, if the root of $$3^x + x + 6 =0$$ is $alpha$ the soluton is $x>alpha$. Now, if you put $x=-6$ in L.H.S. then we get $3^{-6}$ which is very near to zero. hence our required root is near to $x=-6$ (as it is a continuous function). hence, solution in this case is $-6<x<-1$(approximately)

Final approx solution can be written as: $(-6,-1)cup[-1,infty)$

You say "I'm not sure if this is an approximation or the exact answer". Just check that !

$$|-6.001+1|=5.001$$ and $$|3^{-6.001}+5|=5.0013702359272494899224732425168cdots$$

so there is room in between.

By the way, you can conclude without computation by noting that the LHS is a rational number, while the RHS is a rational times

$$sqrt[1000]3,$$ which is an irrational number.

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The equation is transcendental and you cannot solve it analytically, unless you resort to Lambert's W function. Otherwise a numerical solver is required. You are right to be lost. :)