I stumbled upon this problem as I was going through the Algebra 2 course of https://brilliant.org. It goes as follows:

Starting from

$x = \sqrt{\sqrt{3 \sqrt{\sqrt{3 \sqrt{\sqrt{3 \dots}}}}}}$,

I manage on my own to reach the following equation:

$x^4 = 3x$.

Personally I took it from there as follows:

$\frac{x^4}{x} = 3 \leftrightarrow x^3 = 3 \leftrightarrow x=\sqrt[3]{3}$.

The course accepts my final answer of $x = \sqrt[3]{3}$ as correct, however they reason as follows:

$x^4 = 3x \leftrightarrow x^4 - 3x = 0 \leftrightarrow x(x^3 - 3) = 0$, since in the original equation $x > 0$, we only worry about the root $x^3 - 3$:

$x^3 - 3 = 0 \leftrightarrow x^3 = 3 \leftrightarrow x = \sqrt[3]{3}$.

They do reach to the same answer but that is not important when learning. What bothers me is that I do not seem to understand on why the $x > 0$ part is important here, and how they use that to form their conclusion. It also makes me wonder about how it impacts my approach to the solution. And in general it makes me wonder if there are guidelines that can help me decide on what properties of an original equation I need to take into account when transforming equations while solving a problem.

I suppose it is similar as how when $x$ is in the denominator of the original equation than $x \neq 0$, no matter how you can transform the equation (e.g. cancel out the denominator). I do however not see at the moment why it would apply here.

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    $\begingroup$ You just divided both sides of your equation by $x$ and so you got the right answer but you need to justify that you can do this and the justification is that clearly $x$ is non zero. $\endgroup$
    – Ben
    May 3 '19 at 13:28
  • $\begingroup$ Oh, right. Now I get it, so they also just divide by $x$, difference being that they also explain why it is possible. Thank you, now I feel a bit silly to ask this as a question, not sure it will contribute much to this fascinating Q&A website. Feel free to put it as an answer @Anon, such that I can accept it as an answer. $\endgroup$
    – glendc
    May 3 '19 at 13:29

In your solution, when you perform the step $$ x^4 = 3x \iff \frac{x^4}{x} = x^3 = 3, $$ you are using the law of multiplicative cancelation, which states

Theorem: If $a$, $b$, and $c$ are any real numbers with $c \ne 0$, then $$ a = b \iff ac = bc. $$

In other words, we can cancel common factors from both sides of an equation (that is, we can "divide" both sides of an equation by something), assuming that this factor is nonzero. In your computation, you are applying this theorem. In order for this to be a valid step, the hypotheses of the theorem much hold, hence you need it to be true that $x \ne 0$. Therefore you are implicitly assuming that $x \ne 0$.

Since you ask about "guidelines" for working problems like this, it might be worthwhile to get extremely pedantic, and very carefully reason about every step. For example, your solution could be expanded to something akin to the following:

Let $$ x = \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}}, $$ assuming that $x$ exists.[1] Then $$ x^4 = \left( \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}} \right)^4 = 3 \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}} = 3x. $$ Since $x > 0$,[2] the law of multiplicative cancelation gives $$ x^4 = 3x \iff x^3 = 3. $$ The function $f(x) = x^3$ is invertible on $\mathbb{R}$, and has inverse $f^{-1}(x) = \sqrt[3]{x}$. Thus $$ x^3 = 3 \iff x = \sqrt[3]{3}. $$ Therefore $x = \sqrt[3]{3}$.

This kind of solution is overly pedantic, but each and every step of the computation is justified by appealing to some mathematical principle. Often we skip writing these justifications out explicitly, as we can eventually start assuming that we all speak a common language and are familiar with the same rules. However, it is sometimes helpful to give all the details, if for no other reason than to confirms one's one understanding.


There are a couple steps in the computation which should probably be justified, but the justifications require more advanced mathematics (calculus, at least). You can skip the footnotes and be okay, but they are here for completeness.

[1] One can actually write down funny expressions like this which don't "converge" to real numbers. However, rigorously checking that such an expression gives an actual number requires a firm grasp of the concept of a limit, which is not generally introduced until calculus. So we just have to assume that $x$ really exists.

[2] It seems obvious that $x \ne 0$, but this, also, has the potential to be a little delicate. Basically, $x$ is the limit of iterative application of the map $t \mapsto \sqrt[4]{3t}$. This map has two fixed points (one at zero, and one somewhere else—the second fixed point is the thing that we are trying to find). However, the fixed point at zero is "unstable", in the sense that repeated application of the map won't ever give a sequence converging to zero, unless the very first thing we plug in is zero.


For $ab=0$, either $a=0$, or $b=0$ or both $a=b=0$ (if possible). Here, you see, $x>0$, as $x$ is an even root (so $x$ is definitely not zero). This implies the other term in the product zero. So $x^3-3=0$. Whenever you have equation of the form $a_1a_2a_3...a_n=0$, always make cases.

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    $\begingroup$ Thank you very much for your feedback as well. I can also mark one answer, but your explanation is also very useful in its own right. $\endgroup$
    – glendc
    May 3 '19 at 14:32

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