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Is it possible to find the value of n (n belongs to Real number) such that it satisfies the equation: $2^n = n^8$ without any help of computer or graph generator (i.e. only manually). If possible, please explain and if it requires explanation of some large number of pre-requisite concepts, atleast please mention the concepts that I need to study.

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    Short answer: no. – Ethan Bolker Jul 09 '22 at 15:56
  • Plus in some values to see what you get. $n=-2,n=-1,n=0,n=1,n=2$. You find there are solutions in the following ranges: $(-1,0)$ and $(1,2)$. You can then use numerical analysis to get close to solutions. Is that what you mean? – SlipEternal Jul 09 '22 at 16:04
  • See https://math.stackexchange.com/q/2910250 (the equation is a bit different, but the same method applies) – L. F. Jul 09 '22 at 16:10
  • You can solve it with the Lambert W function, but it isn't pretty. https://en.wikipedia.org/wiki/Lambert_W_function – Thomas Andrews Jul 09 '22 at 16:11
  • @EthanBolke and @ SlipEternal So is it just like finding square root of 2, in which we can't find it by just using some mathematical operations and properties (distributive, addition, factorisation, etc.), but we need some ALGORITHM (which I call non-mathematical procedure) to find closest value (but never exact, the degree of exactness depends number of looping wwe do). I'm I right? –  Jul 09 '22 at 16:11
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    @ByteManager it is not clear what you mean. Most of mathematics is algorithmic. If you refer to a vast portion of mathematics as a "non-mathematical procedure", then we are not speaking using the same terminology. If we have different definitions, there is no way to communicate effectively. – SlipEternal Jul 09 '22 at 16:17
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    @ByteManager There is no way to solve this equation using what's generally characterized as "high school algebra". You have to do it numerically, or invent a special function (the Lambert W function) that magically (that is, by definition) does the job. – Ethan Bolker Jul 09 '22 at 16:20
  • In addition to the two solutions identified by SlipEternal, there will be one more between $40$ and $44$ (not too bad to show by hand). With a calculator, it can be shown to be between $43$ and $44$. – paw88789 Jul 09 '22 at 16:21
  • The Lambert-W-function needs to be determined numerically as well , so there is no workaround. – Peter Jul 09 '22 at 16:35
  • @SlipEternal in your earlier comment, you mentioned the term 'numerical analysis', after googling for it's meaning, I think that's what I mean when I mentioned 'ALGORITHM': by doing some guesses, I predict the interval in which my solution will be there, then by repeated (this is what I meant by LOOPING) magnifying (further minimising the interval size), I will know the value of my solution upto some number decimals. –  Jul 09 '22 at 16:57
  • @ByteManager long division is performed using an algorithm that loops. Using the algorithm, you get one significant figure per loop. Do you place long division in the same category? Similarly, square roots can be solved using a similar algorithm to long division, which you already said uses a "looping" algorithm. I disagree with your assessment that these algorithms are "non-mathematical" – SlipEternal Jul 09 '22 at 17:03
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    Yes even I disagree my classification (now) of these Algorithms as Non-mathematical, I just didn't knew any other term to this, hence I simply used 'Algorithm' and looping.. –  Jul 09 '22 at 17:06
  • You mentioned $\sqrt{2}$ as being something that "cannot be calculated". What about $\dfrac{5}{2}=2.5$? Is that calculable since it has a finite decimal expansion? What about $\dfrac{1}{3}=0.3333...$? That has a repeating decimal expansion. I suspect you are either asking one of the following: do any of the solutions have finite decimal expansions? Or are any solutions rational numbers? The second question assumes $1/3$ is "ok". – SlipEternal Jul 09 '22 at 17:15
  • @SlipEternal I wanted to confirm whether the only way to find the solution to the equation 2^n = n^8 is the Numerical analysis or not. I mentioned finding approximated root-2 value method to describe 'numerical analysis' (as at that time I didnt knew this term). I wasn't asking about the type of decimals the roots have (repeating, non-repeating, etc). –  Jul 09 '22 at 17:59
  • @ByteManager your question I becoming contradictory. I thought I was on the right track where I thought you were asking if the solution was rational or irrational. Numerical analysis is not the only solution. There are algebraic techniques that may or may not yield results. There are infinite series that might yield a solution. So, your question might be too broad. As you said, you don't know what you don't know. You can study Calculus, real analysis, and in particular Taylor expansions to see if that could yield a solution without numerical analysis. – SlipEternal Jul 09 '22 at 19:49

3 Answers3

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The closed form for the solutions can be expressed in terms of the Lambert W function, a non-elementary function.

You have: $$e^{n\log 2/8}=\pm n$$

Multiplying both sides by $\mp\frac{\log 2}{8}e^{-n\log 2/8}$ and letting $y=-\frac{\log 2}8 n$ you get:

$$ye^y = \mp\frac{\log 2}{8}$$ or $$y=W\left(\mp\frac{\log 2}8\right)$$ Solving for for $n,$ you get $$n=\frac{-8W\left(\pm\frac{\log 2}8\right)}{\log2},$$ where we've replace $\mp$ with $\pm$ in this final formula.

This gives three solutions, since $W$ has a branch for negative values. There are two values for $W(-\log 2/8)$ but only one value of $W(\log 2/8).$

But this isn't very edifying.

Thomas Andrews
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If you're happy to work with some approximation, you can find the root using something like Newton-Raphson's method for example. It is tedious but it works "by hand".

In any case, there seems to exist 3 roots to the equation for $n\in\mathbb R$. Numerically, these are around $n\sim-0.923131786$, $n\sim1.099997030$, and $n\sim43.55926044$, if that's helpful.

PC1
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All done by hand

Suppose that we look for the zeros of function $$f(x)=2^x-x^8$$ A very quick expansion shows that there are roots close to $\pm 1$.

Using Taylor expansion $$f(x)=-\frac{1}{2}+(x+1) \left(8+\frac{\log (2)}{2}\right)+O\left((x+1)^2\right)$$ gives as an estimate $$x_1=-1+\frac{1}{16+\log (2)}=-0.940$$ while the exact solution is $x_1=-0.923$

Similarly $$f(x)=1+(x-1) (2 \log (2)-8)+O\left((x-1)^2\right)$$ gives as an estimate $$x_2=1+\frac{1}{8-2\log (2)}=1.151$$ while the exact solution is $x_2=1.100$

For the other root, considering that $f(x)$ varies very fast, consider instead $$g(x)=\log \left(2^x\right)-\log \left(x^8\right)=x\log(2)-8\log(x)$$ $$g'(x)=\log (2)-\frac{8}{x} \qquad \text{and} \qquad g''(x)=\frac{8}{x^2}~>~0$$

The first derivative cancels at $$x_*=\frac{8}{\log (2)}$$ $$f(x)=\left(8-8 \log \left(\frac{8}{\log (2)}\right)\right)+\frac{1}{16} \log ^2(2) \left(x-\frac{8}{\log (2)}\right)^2+O\left(\left(x-\frac{8}{\log (2)}\right)^3\right)$$ giving $$x_3=\frac{8}{\log (2)}+\frac{8}{\log (2)}\sqrt{2 \left(\log \left(\frac{8}{\log (2)}\right)-1\right)}=31.17$$ which is far from the solution $(x_3=43.56)$ but which gives an order of magnitude.

Claude Leibovici
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