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

Mathematics Asked on December 6, 2021

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.

$$x^2 + 2x +1 < 3^{2x} + 10times3^x + 25 $$
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 ) < xcdot 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!

Desmos sketch

*Originally $|x+1| < |3x+5|$

3 Answers

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. $x+1<0$.

  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 $(x_0,+infty)$.


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 $-6.00137$.

Answered by Ennar on December 6, 2021

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: $(-6,-1)cup[-1,infty)$

Answered by Pranay on December 6, 2021

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.


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. :)

Answered by user65203 on December 6, 2021

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