About the solution of $2^x-x^x=0$

Question

Obviously $x=2$ is root of this equation $$2^x=x^x$$ if you plot it by some graphing software ,you will see x=0 is another root.

and now,my question:

is it true that $x=0$ is the solution of equation ?

Answer 1

You can write $2^{x}-x^{x}=\exp(x\log(2))-\exp(x\log(x))$ and since the function $x\log(x)$ has an extension by continuity then we can consider that $0^{0}=\lim_{x\to 0^{+}}x\log(x)$ so if $x=0$ then $\exp(x\log(2))-\exp(x\log(x))=0.$ Also, $2$ is a trivial solution.

Answer 2

Consider the following fact

$$2^x=x^x=2^{x\log_2x} $$

so you can solve $x=x\log_2x$. Suppose $x\neq 0$: then $1=\log_2x\Longrightarrow x=0$ So this solution can not be accept.

In other words you can not accept $x=0$ since the equality above needs to be true, and $\log_20$ is not defined.

Answer 3

It may be worth reconsidering for a moment what $a^b$ is in the first place. Namely, it is not primarily defined as $a^b:=\exp(b\ln a)$. Instead, the fundamental definition is of course: repeated multiplication. This way $a^b$ is defined for $b\in N_0$ and arbitrary $a$. The first natural extension is to the case $b\in\mathbb Z$, $a\ne 0$, and then to rational $b$ with $a>0$, and then perhaps to rational $b$ with odd denominator and $a<0$. Then finally, we use log and exp to define $a^b$ for $a>0$ $b\in\mathbb R$ (not to mention complex arguments). At any rate, when we talk about $0^0$ (and not about $\lim_{h\to 0}f(h)^{g(h)}$ given $\lim_{h\to 0}f(h)=\lim_{h\to 0}g(h)=0$) its value is $1$ because we have an empty product, the value of which does not depend on the value of the nonexisting factors.

With this in mind, we note that $2^0=1$ and $0^0=1$ so that $x=0$ is a solution to the equation in question. Clearly $x=2$ is also a soluton. For $x>2$ we have $x^x>2^x$, for $0<x<2$ we have $x^x<2^x$, so we have fond the only two solutions with $x\in [0,\infty)$. If we want to consider negative $x$ at all, we may want to restrict to the case of ratonal $x$ with odd denominator (else we get problems with defining $x^x$ unambigously), in which case $x^x<0<2^x$, so no addition solution is found there.