Is there a way to solve this problem 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? 
 A: 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).
A: 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 $$n<2^{1+n}.$$
A: 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.
