$2^n=n$ and similar equations Is it possible to solve equations in the form $k^n=n$ for n and if so, How? I am new to logarithms and so would be glad if someone could explain even if there is an obvious answer.
Also What about $k^{a+b}=a$ for a? Or $k^{ab}=a$?

All these equations can be standardized to the form
$$xe^x=y$$ for which the general solution has been studied in depth and is denoted as the function $x=W(y)$ known as Lambert's.


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*$k^n=n$


Write $k^n=e^{-x}$, i.e. $x=-n\log(k)$, and $e^{-x}=-\dfrac x{\log(k)}$ or $xe^x=-\log(k)$.


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*$k^{a+b}=a=k^ak^b$


Like before, $xe^x=-\log(k)k^b$, with $x=-a\log(k)$.


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*$k^{ab}=a$.


Can be rewritten $bk^{ab}=ab$, and $xe^x=-\log(k)b$, with $x=-ab\log(k)$.

In general, there is no simpler approach, but you can create equations with a known solution by working in reverse.
For instance, taking $x=-\log(2)$ so that $\log(k)=\dfrac{\log(2)}2$ yields the equation
$$(\sqrt2)^n=n$$ with the solution $n=2$.

For those not willing to take Lambert's $W$ for granted we can discuss the real roots of $xe^x=y$.
The derivative is $(x+1)e^x$ so that the function is decreasing from an horizontal asymptote $(-\infty,0)$ to the minimum at $(-1,-1/e)$, then increasing exponentially to $(\infty,\infty)$ after crossing the origin $(0,0)$.
So there are no solutions in $y$ for $x<\frac1e$, two negative solutions on both sides of $x=-1$ for $1/e<y<0$ and a single solution for $y>0$.
