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I thought up of a problem that i cant find the answer to. It is:

$$ n/ 2^n=2 $$

Can this be solved? or is it impossible to?

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$\forall n \geq 1, \; n < 2^{n}$ –  jibounet Mar 12 at 10:57
    
$n=2^{1+n}$ is a normal form of this equation. You can easily establish a bound on $n$. –  AlexR Mar 12 at 10:57
    
There are no solutions even if $n \in \mathbb R$ –  dani_s Mar 12 at 10:58
    
From a less technical viewpoint, it is easy to see from trial and error that the function has a maximum value of about 0.53 where n is about 0.5 –  Alan Gee Mar 12 at 11:09

3 Answers 3

up vote 4 down vote accepted

Considering the problem from an algebraic point of view, the equation $$ x/ 2^x=2 $$ has no simple solution except in terms of Lambert function. The explicit solution is given by $$x=-\frac{W(-2 \log (2))}{\log (2)}$$ which is a complex number ($x=0.12792 -2.18169 i$). So, even if $\mathbb R$, there is no solution (as mentioned by dani_s).

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There're no solutions $n\in\Bbb R$. Clearly, $n$ can't be negative. In zero we have

$$n<2^{1+n}.$$

Left hand side has a derivative with respect to $n$ equal to $1$.

Right hand side has a derivative with respect to $n$ equal to $2^{1+n}\ln 2$ which is obviously greater than $1$ for all $n\ge 0$. Therefore, $\forall n\in \Bbb R$ we have $$n<2^{1+n}.$$

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Another (easy) way to see that there are no real solutions to this problem is plotting the two functions $$\left\{\begin{aligned} y&=n\\ y&=2^{n+1} \end{aligned}\right.$$ which will eventually meet where your equation has solutions.

You can do it in many ways. Probably the most easy way to do that is using WolframAlpha and asking literally for it

plot x and 2^(x+1), x between -10 and 10

Or use this link directly.

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