Question

Solve the equation:

$x^{x+1}=x$

Answer 1

0, 1

Answer 2

We are asked to solve the equation $x^{x+1}=x$.

We can rewrite the equation as $x^{x+1} - x = 0$.
Factor out $x$: $x(x^x - 1) = 0$.

This gives two possibilities:

1. $x = 0$
2. $x^x - 1 = 0$, which means $x^x = 1$.

Let's check the first possibility, $x=0$, in the original equation $x^{x+1}=x$.
Substitute $x=0$: $0^{0+1} = 0^1$.
By definition, $0^1 = 0$.
The right side of the equation is $x=0$.
So, $0^1 = 0$, which is $0=0$. This is true.
Thus, $x=0$ is a solution.

Now let's consider the second possibility, $x^x = 1$.
We need to find the values of $x$ that satisfy $x^x=1$.

Case 1: The base is 1.
If $x=1$, then $1^1 = 1$. This is true.
So $x=1$ is a solution to $x^x=1$.
Let's check $x=1$ in the original equation $x^{x+1}=x$.
Substitute $x=1$: $1^{1+1} = 1^2 = 1$.
The right side is $x=1$.
So, $1^2 = 1$, which is $1=1$. This is true.
Thus, $x=1$ is a solution to the original equation.

Case 2: The exponent is 0.
The equation is $x^x=1$. The exponent is $x$. If $x=0$, the exponent is 0.
However, if $x=0$, the base is also 0, leading to $0^0$.
The expression $0^0$ is typically considered indeterminate in calculus, but it is often defined as 1 in combinatorics and other areas.
If we were to define $0^0=1$, then $x=0$ would be a solution to $x^x=1$.
But we already considered $x=0$ directly in the original equation and found it is a solution.
The step $x(x^x - 1) = 0$ assumes $x^x$ is defined when $x \neq 0$.
The equation $x^x=1$ is derived from $x^{x+1}=x$ by dividing by $x$, which is valid only if $x \neq 0$.
So, when solving $x^x=1$, we are looking for solutions where $x \neq 0$.
For $x \neq 0$, $x^x=1$ means the exponent $x$ must be 0, but this contradicts $x \neq 0$.

Case 3: The base is -1 and the exponent is an even integer.
Let $x=-1$. The equation is $(-1)^{-1} = 1$.
$(-1)^{-1} = \frac{1}{(-1)^1} = \frac{1}{-1} = -1$.
So, $-1 = 1$, which is false.
Thus, $x=-1$ is not a solution to $x^x=1$.
Let's check $x=-1$ in the original equation $x^{x+1}=x$.
Substitute $x=-1$: $(-1)^{-1+1} = (-1)^0$.
If we define $a^0=1$ for $a \neq 0$, then $(-1)^0=1$.
The right side is $x=-1$.
So, $1 = -1$, which is false.
Thus, $x=-1$ is not a solution to the original equation.

Let's consider the domain of the expression $x^{x+1}$.
The expression $a^b$ is generally defined as $e^{b \ln a}$. This requires the base $a$ to be positive, i.e., $x>0$.
If we restrict $x>0$, the equation $x^{x+1}=x$ becomes $x^{x+1}=x^1$.
Since the base $x$ is positive and not equal to 1 (if $x=1$, it is a solution as checked), we can equate the exponents:
$x+1 = 1$
$x = 0$.
This contradicts the assumption $x>0$.
So, under the restriction $x>0$, the only solution is $x=1$.

However, the problem does not state any restriction on $x$. We need to consider other possible values of $x$ for which $x^{x+1}$ is defined.
The expression $x^{x+1}$ is defined for:

• $x>0$ (standard definition $e^{(x+1)\ln x}$)
• $x=0$ (as $0^1=0$)
• $x<0$ if $x+1$ is a rational number $p/q$ where $q$ is odd.

Let's revisit the solutions we found: $x=0$ and $x=1$.
For $x=0$, $0^{0+1} = 0^1 = 0$. The right side is 0. $0=0$. $x=0$ is a solution.
For $x=1$, $1^{1+1} = 1^2 = 1$. The right side is 1. $1=1$. $x=1$ is a solution.

Are there any other solutions?
We derived the equation $x^x=1$ by assuming $x \neq 0$.
Let's consider $x<0$. Let $x=-y$ where $y>0$.
The equation $x^x=1$ becomes $(-y)^{-y}=1$.
$\frac{1}{(-y)^y} = 1$.
$(-y)^y = 1$.
Let $y = p/q$ where $p, q$ are positive integers with $\gcd(p,q)=1$.
$(-p/q)^{p/q} = 1$.
For $(-p/q)^{p/q}$ to be a real number, the denominator $q$ must be odd.
Then $(-p/q)^{p/q} = (\root q \of {-p/q})^p = (-\root q \of {p/q})^p$.
$(-\root q \of {p/q})^p = 1$.
If $p$ is even, $p=2k$ for some integer $k \ge 1$ (since $y>0$, $p>0$).
$(-\root q \of {p/q})^{2k} = ((\root q \of {p/q})^2)^k = (\root q \of {(p/q)^2})^k = 1$.
Since $p/q > 0$, $\root q \of {(p/q)^2} > 0$.
So $(\root q \of {(p/q)^2})^k = 1$ implies $\root q \of {(p/q)^2} = 1$ (if $k \ge 1$).
$(p/q)^2 = 1^q = 1$.
$p^2 = q^2$. Since $p, q$ are positive, $p=q$.
Since $\gcd(p,q)=1$, this implies $p=1$ and $q=1$.
So $y = p/q = 1/1 = 1$.
This gives $x = -y = -1$.
We already checked $x=-1$ in the original equation and found it is not a solution.

If $p$ is odd, $p=2k+1$ for some integer $k \ge 0$.
$(-\root q \of {p/q})^{2k+1} = -(\root q \of {p/q})^{2k+1} = 1$.
$(\root q \of {p/q})^{2k+1} = -1$.
Since $p/q > 0$, $\root q \of {p/q} > 0$.
A positive number raised to any power is positive.
So $(\root q \of {p/q})^{2k+1} > 0$.
This means $(\root q \of {p/q})^{2k+1} = -1$ has no solution.

So, the equation $x^x=1$ has no solutions for $x<0$.
The equation $x^x=1$ has only one solution $x=1$ for $x \neq 0$.

The solutions to $x(x^x - 1) = 0$ are $x=0$ or $x=1$.

Let's verify if there are any other cases for $x^{x+1}=x$ for $x<0$.
Let $x=-y$ where $y>0$.
$(-y)^{-y+1} = -y$.
$(-y)^{-(y-1)} = -y$.
$\frac{1}{(-y)^{y-1}} = -y$.
$1 = -y (-y)^{y-1}$.
$1 = -y (-1)^{y-1} y^{y-1}$.
$1 = -(-1)^{y-1} y^y$.
$1 = (-1)^y y^y$.

If $y$ is an integer, $y=n$ for $n \in \{1, 2, 3, \dots\}$.
$1 = (-1)^n n^n$.
If $n$ is even, $n=2k$ for $k \ge 1$. $1 = (-1)^{2k} (2k)^{2k} = (2k)^{2k}$. Since $2k \ge 2$, $(2k)^{2k} \ge 2^2=4$. No solution.
If $n$ is odd, $n=2k+1$ for $k \ge 0$. $1 = (-1)^{2k+1} (2k+1)^{2k+1} = -(2k+1)^{2k+1}$. $(2k+1)^{2k+1}=-1$. Since $2k+1 \ge 1$, $(2k+1)^{2k+1} \ge 1$. No solution.

If $y=p/q$ with $q$ odd.
$1 = (-1)^{p/q} (p/q)^{p/q}$.
$(-1)^{p/q} = (\root q \of {-1})^p = (-1)^p$.
$1 = (-1)^p (p/q)^{p/q}$.
If $p$ is even, $1 = (p/q)^{p/q}$. This implies $p/q = 1$ (as $z^z=1$ for $z>0$ implies $z=1$). So $y=1$, $x=-1$. We checked $x=-1$, not a solution.
If $p$ is odd, $1 = -(p/q)^{p/q}$. $(p/q)^{p/q}=-1$. No solution as $p/q>0$.

The only solutions are $x=0$ and $x=1$.