$2^n=n$ and similar equations

Question

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 $k^{ab}=a$?

Answer 1

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.

• $k^n=n$

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

• $k^{a+b}=a=k^ak^b$

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

• $k^{ab}=a$.

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

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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 $n=2$.

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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 $y>0$.

Answer 2

I think that using Lambert's $W$ is like cheating. ;-) To get a more general approach, let's study the equation $k^x=x$, with $k>0$.

Consider the function $f(x)=k^x-x$. We have $$ \lim_{x\to-\infty}f(x)=\infty $$ and $$ \lim_{x\to\infty}f(x)= \begin{cases} -\infty & \text{if $0<k\le 1$}\\[4px] \infty & \text{if $k>1$} \end{cases} $$ Moreover $$ f'(x)=k^x\log k-1 $$ If $0<k\le1$ we have $f'(x)<0$ for all $x$, so the function crosses the $x$-axis exactly once and the equation has a single solution. Since $f(0)=1$ and $f(1)=k-1<0$, we know the solution is in the interval $(0,1)$.

The more interesting case is $k>1$. The derivative vanishes for $$ k^x=\frac{1}{\log k} $$ so for $x\log k=-\log\log k$ or $$ x=-\frac{\log\log k}{\log k} $$ where the function has an absolute minimum.

Now $$ f\left(-\frac{\log\log k}{\log k}\right)= \frac{1}{\log k}+\frac{\log\log k}{\log k}= \frac{1+\log\log k}{\log k} $$ The equation has

• two solutions if $1+\log\log k<0$, that is, $\log k<e^{-1}$ or $1<k<e^{1/e}$;

• one solution if $k=e^{1/e}$;

• no solution if $k>e^{1/e}$.

The equations of the form $k^{x+b}=x$ can be studied similarly; more generally, you can study $qk^x=x$, taking into account that $k^{x+b}=k^b\cdot k^x$.

Equations like $k^{xb}=x$ are just the same as before, because they can be rewritten as $(k^b)^x=x$.