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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 ?

Michael Galuza
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Khosrotash
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    $0^0$ is generally left undefined. However, $\lim_{x\to 0^+} x^x =1$. – Argon Jul 23 '15 at 17:40
  • If you put your mouse over the point where $y = 0^0$ you will likely see a message like 'undefined' or 'no value' – Rick Jul 23 '15 at 17:42
  • $x^x$ is only defined for $x>0$ and $x$ is a real number – Dr. Sonnhard Graubner Jul 23 '15 at 17:43
  • In practically all topics $0^0=1$, in the sense of the limit and in the sense of conventions like the void product. I must add that generally all conventions are related with limits. – Masacroso Jul 23 '15 at 17:44
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    It depends on your domain. If you define $x^x$ on $[0,+\infty)$ and choose the value $1$ at $0$ to have a continuous function, then $x = 0$ is also a solution. If your domain is $(0,+\infty)$, it is of course no solution, since it's not in the domain. – Daniel Fischer Jul 23 '15 at 17:45
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    We have $$\left(\dfrac x2\right)^x=1$$ – lab bhattacharjee Jul 23 '15 at 17:45
  • @Masacroso: Surely not in context of limits. In fact, I only know it in context of polynomials. – user251257 Jul 23 '15 at 17:46
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    Just a combinatorial argument for defining $0^0=1$... $n^n$ is the number of functions from a set with $n$ elements to itself, and the number of functions from the empty set to itself is 1. – Aloizio Macedo Jul 23 '15 at 17:47
  • See the comment of Argon @user251257 – Masacroso Jul 23 '15 at 17:47
  • $0^0$ is very "dangerous" as its definition may depend on approach. As Argon mentioned, the limit of $x^x$ is $1$ (even from a negative side, the imaginary part converges to $0$). However, if you generalise to the function $f: \mathbb{R}^+ \times \mathbb{R}^+ \to \mathbb{R}: (x,y) \mapsto x^y$ you may get different results depending on the order of your limits (you will get 1 or 0 depending on $x$ or $y$ first). –  Jul 23 '15 at 17:49
  • @Ezueneok The definition of $0^0=1$ doesn't not depend on approach. One considers $0^0$ as indeterminate form (just like $\frac 00$ which is both undefined and an indeterminate form) not because it is not defined but because $x_n\to 0, y_n\to 0$ does not say much about $x_n^{y_n}$.. In other words, the power function is not continuous at $(0,0)$. – Hagen von Eitzen Jul 23 '15 at 18:06
  • I know this limit , main quote of my question is about $0^0$ . In fact I don't know to get ?$0^0=1$ or not – Khosrotash Jul 23 '15 at 18:07

3 Answers3

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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.

Khadija Mbarki
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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.

InsideOut
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  • $\log 0$ is not defined??? I though that it is $-\infty$... What is not defined is the logarithm of negative numbers on the reals. – Masacroso Jul 23 '15 at 17:52
  • This is not correct, the function $log$ is well define over $]0,+\infty[$. What are you saying is that: if $x\longrightarrow 0 \Longrightarrow log(x)\longrightarrow -\infty$. In general when you want to solve logarithms equation you suppose that the argument is strictly positive. – InsideOut Jul 23 '15 at 17:55
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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.

Hagen von Eitzen
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  • Ok, but now I have a question: my teacher of analysis at the University said to me that the exponential function is define for strictly positive arguments, in other words $a>0$. So if we watt to study that equation from the analitical point of view, we need to set $x\ge0$. So where I wrong? – InsideOut Jul 24 '15 at 07:20
  • x = 0 is not a solution, because 0^0 is indeterminate. – Olive Stemforn Jul 08 '16 at 18:29