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Solve the given set of equations:

$x^y=y^x$ and $x^2=y^3$ where $x,y \in \mathbb{R}$

Would any other solution exist other that $x=y=1$ because I think $x^2=y^3$ will only be true for $x=y=1$ or $x=y=0$

Neil
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Akira
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    $x^2=y^3$ has the solution $x=8$, $y=4$ and many (related) others. – André Nicolas May 23 '16 at 19:44
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    Since you are in the real numbers, $x^2=y^3$ has uncountable solutions, including one for each value of $x$ and each positive value of $y$. – pancini May 23 '16 at 19:45
  • $0^0$ is ill-defined so $x=y=0$ is not a valid solution for both equations. For $x^y=y^x$, this is essentially equivalent to $x^{1/x}=y^{1/y}$. If there is a solution on the interval $(1,e)$, there will be one on the interval $(e,\infty)$ – Neil May 23 '16 at 19:46

5 Answers5

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Note that if $x^2=y^3$, then $y=x^{2/3}$. Therefore, we find that

$$x^{x^{2/3}}=\left(x^{2/3}\right)^x \tag 1$$

Taking the logarithm of both sides of $(1)$ yields

$$x^{2/3}\log(x)=\frac23 x\log(x) \tag2$$

Solutions to $(2)$ are $x=1$ and $x=27/8$. For $x=1$, $y=1$ and for $x=27/8$, $y=9/4$.

Mark Viola
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Write $x=y^{3/2}$ and substitute in the first equation to get $y^{3y/2}=y^{y^{3/2}}$; if $y \neq 0,1$, then the exponents must be equal so that $3y/2=y^{3/2}$, which gives $y=\dfrac{9}{4}$ and $x=\dfrac{27}{8}$.

Aravind
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Hint Since $x^2 \geq 0$, we can write any solution to the second equation as $(x, y) = (t^3, t^2)$ for some $t \geq 0$. Substituting in the first equation gives $$t^{3 t^2} = t^{2 t^3},$$ and taking the logarithm of both sides gives $$3 t^2 \log t = 2 t^3 \log t .$$ This leads to two solutions (three if you take $0^0$ to be defined).

Travis Willse
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Following my answer here, we can deduce from the first equation that for some $a \neq 1$, we have $$ x = a^{1/(a-1)}, \quad y = a^{a/(a-1)} $$ Thus, the second equation becomes $$ a^{2/(a-1)} = a^{3a/(a-1)} \implies\\ a = a^{3a/2} \implies\\ 1 = a^{3a/2 -1} \implies\\ \left(\frac 32 a - 1\right) \log a = 0 $$ Thus, we get a solution at $a = 2/3$, which is to say $$ x = (2/3)^{-3} = \frac{27}{8}, \qquad y = (2/3)^{-2} = \frac{9}{4} $$

Community
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Ben Grossmann
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Equation $x^2=y^3$ has a clear parametrization $(x,y)=(t^3,t^2)$.

On the other hand $$x^y=y^x\iff \frac{\log x}{x}=\frac{\log y}{y}$$ which is easily seen to have the solutions $x=y$ for $0\lt x\le1$ and $x=e$. Besides for all $x$ in the open interval $(1,e)$ there is a unique $y\in (e,\infty)$ satisfying the equation. (All this from the well known graphic of the function $f(x)=\frac{\log x}{x}$)

Piquito
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