How to solve the following equation : $2^{6-n} = n$
I have no idea of to solve it. I took logarithms on both sides. But doesn't reach at some satisfactory path. But practically I've found $n$ must be equal to $4$, for if $n<4$ then equation is not possible since LHS will become greater than $4$. Similarly $n$ can't be greater than $4$. But what is its formal attempt to find the solution? I tried lots, but never found.
The function $2^{6-n}$ is decreasing for all $n$, and the function $n$ is increasing for all $n$. Hence there is at most one solution. You found one, by trial and error, so that is the only solution. With slightly different parameters, there would be no way to find the solution exactly, only a decimal approximation, unless you use the Lambert W function, which is nonelementary.
A solution (that is elementary) to your problem is one that you have already hinted at. Show that $2^{6-n}$ is monotone decreasing. Since $n$ is monotone increasing, if the two functions intersect, they can only do so at one point. You have already found that point.
"I have no idea (how) to solve it."
On the contrary, I think you did solve it!
Try plotting the two equations $y = 2^{6-x}$ and $y=x$ to confirm your suspicions.
One function is strictly increasing, and the other is strictly decreasing. So, if they intersect at one point (which you have found) then that's the only point they intersect at.
