Find the value of $x$ if $x^{x^4}=4$. Given options are

  1. $2^{1/2}$
  2. $-2^{1/2} $
  3. Both 1. & 2
  4. None of the above

From option verification, we get option 3. as correct one. But is there any real method to do the above problem?

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    $\begingroup$ Can you show any working out you have done? $\endgroup$ – Tazwar Sikder Jun 29 '16 at 12:20
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    $\begingroup$ Can you tell us more about the level where this came from? Given that it looks like multiple choice it is very much possible that it was designed to simply test the students' A) grasp of the definition of powers, B) ability to eliminate wrong alternatives in a multiple choice setting. For example, proving that no other solutions exist may be asking a bit much on some levels. Also I'm a bit worried about apparently allowing $x<0$. At precalculus levels fractional powers of negative numbers are a bit troubling. Here we happen to get an integer exponent, but if we are to find all solutions...? $\endgroup$ – Jyrki Lahtonen Jun 29 '16 at 12:28
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    $\begingroup$ @JyrkiLahtonen If we let $x=-2^{\frac 1 2}=-\sqrt 2$, then $x^4=4$ and thus $x^{x^4}=(-\sqrt 2)^4$, which is not troubling at all in pre-calculus. At least in my pre-calc class, we would probably be expected to realize that there are both positive and negative solutions because of the possible even power that could be $x^4$. $\endgroup$ – Noble Mushtak Jun 29 '16 at 12:29
  • $\begingroup$ I know @NobleMushtak. That was part of my point. But how would a precalculus student prove that there are no other solutions? $\endgroup$ – Jyrki Lahtonen Jun 29 '16 at 12:30
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    $\begingroup$ @JyrkiLahtonen Oh...Good point. EDIT: According to Wolfram Alpha, $\pm \sqrt 2$ are not the only solutions. $\endgroup$ – Noble Mushtak Jun 29 '16 at 12:31

There is no systematic way of solving such transcendental equations. In this particular case, we can get better insight by performing a transformation to get rid of the double exponentiation.



$$t^t=4^4$$ and an obvious solution is $t=4$ corresponding to $x=\pm\sqrt2$.

By the study of the function $t^t$, we can verify that it is increasing where it exceeds $1$, so that the above solution is unique.



$(x^{-4})^{x^4}=4^{-4}$ => $x^{-4}=4^{-1}$ => $x^4=4$ => $x=\pm\sqrt{2}$


Jyrki Lahtonen gave me the advise to carry out the proof accurately.

Therefore: $x^{x^4}=4$ , $(z;a):=(x^{-4};\frac{1}{4})$ => $z^{\frac{1}{z}}=a^{\frac{1}{a}}$

We have $0<a^{\frac{1}{a}}<1$ therefore we have only one positive solution $z=a$.

This means $x^4=4$ and therefore $x=\pm\sqrt{2}$.


$x^\frac{1}{x}$ is strictly increasing (therefore bijective) for $0<x<e$ because of $(x^\frac{1}{x})’=x^{(-2+\frac{1}{x})}(1-\ln x)>0$.

  • $\begingroup$ Why downvote ??? $\endgroup$ – user90369 Jun 29 '16 at 12:40
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    $\begingroup$ I haven't had the time to downvote yet, but how do you justify the first implication? $\endgroup$ – Jyrki Lahtonen Jun 29 '16 at 12:42
  • $\begingroup$ It's the compare of $a^\frac{1}{a}=b^\frac{1}{b}$ and the fact, that we have only one solution for $y=x^\frac{1}{x}$ when $0<y<1$. $\endgroup$ – user90369 Jun 29 '16 at 12:44
  • $\begingroup$ I see. It might be a good idea to include a proof of that (it does work). $\endgroup$ – Jyrki Lahtonen Jun 29 '16 at 12:56
  • $\begingroup$ I meant that you would include a proof of the key result that when $0<x<1$ there is only one positive $y$ such that $x=y^{1/y}$. $\endgroup$ – Jyrki Lahtonen Jun 29 '16 at 13:44

You already know how to verify the options here, but you're asking for a way to solve this equation step-by-step. The issue is that there is no way to find a closed form solution for $x$ using elementary functions. The solution is only expressible using a special function known as the Lambert W. This function is the inverse of the function $f(x) = xe^x$. You can read more about it here.

There are two ways to handle this "from first principles".

The first, which has already been covered by @user90369, is to find solutions by observation (which means inspired guess and check, really), then prove rigorously that no other solutions can exist.

The second is to find a direct solution using the Lambert W.

We start by simplifying the form of the equation with the substitution $x = y^{\frac 14}$

After some elementary simplification, you will end up with $y^y= 256$. This is, of course, very amenable to a solution "by inspection", but let's proceed as originally intended.

$$y^y = 256$$

$$e^{\ln y e^{\ln y}} = 256$$

$$\ln y e^{\ln y} = \ln 256$$

$$\ln y = W(\ln 256)$$

$$y = e^{W(\ln 256)} = \frac{\ln 256}{W(\ln 256)}$$

where a property of the Lambert W is used for the final step.

At this point, you have to use a special calculator or mathematical software to find out the value of $W(\ln 256)$. One such calculator is here.

Using that, we can do the calculation to find that $y$ is very close to $4$. We can now make an "inspired guess" that it is $4$ since no more exact calculation is available to us. We find that it works, so we accept the solution (you should know how to find $x$ after determining $y$).

  • $\begingroup$ Certainly, the Lambert W function should not be used on a multiple choice precalculus question. $\endgroup$ – Mark McClure Jun 29 '16 at 13:55
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    $\begingroup$ @MarkMcClure Of course, I know that. This question obviously asks for a simple solution verification, no more. But the OP wanted to know how to find a "real method", which I interpreted to mean a solution from first principles, and which is what I provided. $\endgroup$ – Deepak Jun 29 '16 at 13:56
  • $\begingroup$ +1 for understanding the request, acknowledging it was outside the scope of the original question, and handling it like a pro anyway. What a great answer. $\endgroup$ – The Count Jun 29 '16 at 13:59
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    $\begingroup$ @Deepak Now that is a fine example of being "nicer to downvoters"! $\endgroup$ – Mark McClure Jul 1 '16 at 2:46
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    $\begingroup$ @Deepak Sorry for the confusion - I meant the comment to be genuine. I noticed your participation in the can we be nicer to downvoters thread so I thought I would point out that you are being quite genuinely nice to a downvoter. $\endgroup$ – Mark McClure Jul 1 '16 at 3:28

I'm not sure if this is rigorous, but you could notice that because $4=x^{x^4}$, we can write $$4=x^{x^4}=x^{x^{x^{x^4}}}$$ and so on, until we get a stack of $x$'s equalling to $4$: $$x^{x^{x^{x^{.^{.^{.}}}}}}=4$$ And so $$x^4=4$$ which yields $x=\pm\sqrt 2$


$x^{x^4}=4$ First of all, we think simple. $x^{x^4}=2^{2^1}$ We did that because we have to make first two steps equal$x$ Later we have to make our third step in our second equation ,4 . $$x^{x^4}=2^(2^(1/4))^4$$ $$x^x=2^{2^{1/4}}$$ We have simplified our problem. Now we should make both steps equal in our second equation. If we take the square of our power number (to make sure they are equal), we must take the square root of our number. Therefore:

$2^{2^{1/4}}= 2^{{1/2}^{2^{1/2}}}$ Our previous power $4$ was even. Solutions are $-2^{1/2}$ and $2^{1/2}$.

  • $\begingroup$ First of all: please use MathJax. $\endgroup$ – The Count Jun 29 '16 at 13:58
  • $\begingroup$ Sorry for that. I am answering via my phone. $\endgroup$ – Ali Berke Korkmaz Jun 29 '16 at 14:02

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