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Find all real number $x$ satisfying $10^x+11^x+12^x = 13^x+14^x$

My Work
Dividing by $13^x$ we get $$\left( \frac{10}{13} \right)^x + \left( \frac{11}{13} \right)^x + \left( \frac{12}{13} \right)^x = 1 + \left( \frac{14}{13} \right)^x$$
The LHS is a decreasing function of $x$ and the RHS is an increasing function of $x$. So there is only one intersection in their graph. I am looking for a formal way to find the root. I know that $x=2$ works. But how to formally find this root?

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    Doyou wanna solve REAL numbers ? – 2017-02-07
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    @S.C.B. All of the solutions is just showing that $x=2$ works but how to formally find the solution? – 2017-02-07
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    There is no general method to find the solution once one has shown there exists a unique solution. To wit, if one were to solve $$10^x+11^x+12^x=13^x+15^x$$ the same argument would prove the existence and uniqueness of its solution but one would have to resort to numerical estimations to show this solution is approximately $$x=1.68227$$ Yes, this kind of "clever" question is frustrating... – 2017-02-07
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    @Did Does a formal way mean that he wants to solve this analytically? – 2017-02-07
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    @RezwanArefin Kudos for having sticked to your guns, and to your specific question. – 2017-02-07
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    @labbhattacharjee All of the solutions there just showed that $x=2$ works and there is only one solution. But they didn't find the root in hand. Just checked that $2$ is a root – 2017-02-07
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    @labbhattacharjee I already pointed that out above. – 2017-02-07
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    @S.C.B., I've seconded that only. When u've found it, u should post answer in the original one. – 2017-02-07
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    I think my question is different :( – 2017-02-07
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    @RezwanArefin Maybe, but the accepted answer over there does make the point to be made, that I expanded in my comment here. – 2017-02-07
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    @Did the accepted answer just showed that $2$ works. But didn't added any method or technique to find the root other that just plugging that seeing if a value satisfies. – 2017-02-07
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    Yes -- and for good reasons since there is no such general method. Say, did you miss my first comment above by any chance? – 2017-02-07
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    @Rezwan Arefin: Sorry for the last comment. It works for e.g. x_0:=3, there is no problem. But I think that's not what you want. Method: $x_{n+1}=x_n-\frac{10^{x_n}+11^{x_n}+12^{x_n}-13^{x_n}-14^{x_n}}{10^{x_n}\ln 10+11^{x_n} \ln 11+12^{x_n} \ln 12-13^{x_n} \ln 13-14^{x_n} \ln 14}\enspace$ . – 2017-02-07

3 Answers 3

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As you have done, $$\left( \frac{10}{13} \right)^x + \left( \frac{11}{13} \right)^x + \left( \frac{12}{13} \right)^x = 1 + \left( \frac{14}{13} \right)^x$$ has a decreasing function on the left and an increasing function on the right. So there is only one solution.

If you want to find them, just plug in some values. Plug in $x=1,2, \dots$. Now, we see $x=2$ works. $$10^2+11^2+12^2=13^2+14^2=365$$ Problems like this usually have a solution over the integers, or don't have a closed form solution at all. The OP pointed out that one might make a mistake if doing it by hand. However, plugging in certain values is helpful in most cases. This is true since it would help approximate the solution using IVT.

I doubt that there is a truly formal/analytical way to solve this equation. If there was, we could express the roots of $$2^x+3^x+4^x=5^x+6^x$$ using some well known functions. However, this kind of equation is known to have no closed form solution. We just know by plugging the values that $x$ lies somewhere between $0$ and $1$. So this is impossible.

However, you could approximate the roots using Newton's Method or a similar method.

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    Plugging in some values cannot yield **all** the solutions. – 2017-02-07
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    Is there any formal way to find the solution? – 2017-02-07
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    @Did the OP has shown that there is only one solution. So I feel further elaboration wasn't necessary. – 2017-02-07
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    I asked for a formal way to find the solution :( – 2017-02-07
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    @RezwanArefin I'm not sure on the formal way. Wait a moment then. You had not said that you were looking for the formal way in your question, you had just asked how to find the roots. So could you edit it in? – 2017-02-07
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    What does "a formal way" mean? S.C.B.'s work is valid. Unless you are looking for a numerical method (which gives only approximations of the solutions), I don't think this type of questions has any solution without trials. – 2017-02-07
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    @Batominovski I think he is refering to a way to solve it analytically. – 2017-02-07
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    @Batominovski Though it beats me how to do it. I'm going to research some papers. – 2017-02-07
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    I think just plugging some values and check is not a good idea .. One can make mistake when doing in hand not calculator – 2017-02-07
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    @RezwanArefin One can make a mistake when doing it by calculator too...that doesn't strike me as that much a valid reason. As far as I know, there is no closed form solution in general. – 2017-02-07
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    Plugging in some values is a VERY GOOD idea. If you get something close to zero, you know the root is (probably) nearby. You can try numerical methods to get a convergent sequence to a root, of course. – 2017-02-07
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    @RezwanArefin Also, plugging in values helps us find a bound on $x$ by IVT. – 2017-02-07
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    "Problems like this usually have a solution over the integers." One could argue that this is the reason why they tend to be not very interesting... The existence and uniqueness proof has some mathematical value, not the identification of the root. – 2017-02-07
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    @Did What defines a formal solution Anyway? Or a closed form? – 2017-02-07
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    @Did Should I just say use Newton's Algorithm? – 2017-02-07
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    @S.C.B. "What defines a formal solution Anyway? Or a closed form?" ?? There might be some variations in how one defines a closed form solution but in the present case we know there is no such solution, whatever definition one refers to. – 2017-02-07
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    @S.C.B. I just found the general solution. See my answer and see if I missed something :) – 2017-02-07
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If you know that the solution must be an integer, this type of equation is known as exponential Diophantine, and there is no known formal procedure to solve it in the general case and probably none exists (this was proven for ordinary Diophantine equations).

If the solution is allowed to be real, there is no systematic procedure either as this is a transcendental equation. You need to resort to numerical methods to estimate the roots, and this requires a step of root separation (finding intervals that are guaranteed to contain exactly one root). Unfortunately, root separation can require the resolution of an even more difficult equation to get the extrema. In the given case, you are lucky as the function is easily shown to be monotonic.

After root isolation, you can evaluate the root to arbitrary precision (at least in theory) using some unidimensional root solver (dichotomy, secant, Newton...)

If it turns out that the root seems to have a simple form (integer, rational or some other closed expression), it may possible to prove or disprove it formally, but here again, no systematic method.

In the given case, this is straightforward:

$$11^2+12^2+13^2=365=14^2+15^2.$$

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    **there is no known formal procedure to solve it in the general case and probably none exists (this was proven for ordinary Diophantine equations)** I don't mean to force you to do anything, but could you, by any chance, offer sources on this? It seems interesting. – 2017-02-07
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    @S.C.B. https://en.wikipedia.org/wiki/Hilbert's_tenth_problem – 2017-02-07
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I just solved my problem. A well known fact that

$$a^2 +(a+1)^2 + \cdots + (a+k)^2 = (a+k+1)^2 +(a+k+2)^2+ \cdots (a+2k)^2$$
Satisfies when $a = k(2k+1)$.
So in my question $a = 2(2\cdot2+1)$, So without any thinking $x=2$ is an answer.
This helps suppose the equation was - $$55^x+56^x+57^x+58^x+59^x+60^x=61^x+62^x+63^x+64^x+65^x$$
Then it is hard to just substitute $x = 1,2,3,..$ and check? But using this fact without any thinking we can say $x=2$, as $55 = 5(5\cdot2+1)$.

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    This solution is not "fair" if you don't state that you are concerned with equations having this very particular pattern. – 2017-02-07
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    Well, how did you know to use this theorem? I would not say this actually a formal method. Also, one could argue how you knew to do it for squares in the first place. – 2017-02-07
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    But this helps .... – 2017-02-07
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    This will not solve all problem like this ... But this helps here – 2017-02-07
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    With this equation you are already assuming that the power is 2. – 2017-02-07
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    There is a misunderstanding on the terminology. Solving *formally* means in a way that can be proven. In this case, substituting the known root $2$ is quite sufficient as a formal solution. I guess that your question is more how to solve *systematically*. Then the answer is essentially "there is no way", except in a few opportunistic cases where a method is known for some equation patterns. With a pinch of bad faith, you could invoke that "it is a well-known fact that $11^2+12^2+13^2=14^2+15^2$". – 2017-02-07
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    How did you find this formula. It is necessary to give the link where it came from. – 2017-02-07
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    I found this in "101 Number Theory Problems" book... By Titu Andrescu – 2017-02-07
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    Give the full link to it. We decided there. https://twitter.com/int_cl/status/800333340230328320 – 2017-02-07