For a given $k>0$ constant, assuming that $x,y>0$. Also this equation has a particular name, or some mathematician associated with it?
How to solve the equation $x^y=y^x=k$?
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calculus
real-analysis
1 Answers
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WLOG suppose that $x \ge y$ and let $x = y^n$ for some $n \ge 1$. Then (ignoring $k$) the equation becomes $y^{ny} = y^{y^n}$, or $ny = y^n$, or $y^{n-1} = n$, which gives
$\displaystyle y = n^{ \frac{1}{n-1} }, x = n^{ \frac{n}{n-1} }$.
Now it only remains to find the values of $n$ such that $x^y = k$, which you should just do numerically. (I guess one must also account for the solutions where $x = y$; the graph $x^y = y^x$ has two components which intersect at $(e, e)$, which one gets from the above by taking the limit as $n \to 1$.)
I am almost positive this equation does not have a name because, in all honesty, it is not interesting.
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1Let me also mention that an alternate way to solve this equation is to write it as (log x)/x = (log y)/y. If you graph this it's not hard to see what the solutions look like. – 2010-08-07
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0Thanks for the insight, I remember now, sometime ago I have to study the behavior of the function $log(x)/x$ in relation to the question: which statement is valid $e^{\pi} > \pi^{e}$ or $e^{\pi} < \pi^{e}$. – 2010-08-07