3
$\begingroup$

So, I was adviced to ask a new question on my problem (as the first one wasn't very precise), that is to solve the system of equations: $$\begin{cases} x\cdot y=6 \\ x^y+y^x=17 \end{cases}$$ where:

  • solutions are considered up to symmetry,
  • $x,y\in \{z\in\Bbb R\mid z\geq 0\}$
    I'm interested in methods different from the Newton's method or the listing that if $xy=6$, then $x=2, y=3$ or $x=1,y=6$ (and so on) and then manually checking whether such pair satisfies $x^y+y^x=17$.
$\endgroup$
  • $\begingroup$ You could use a graphing tool and graph both equations. Seems like the combination (2,3) and (3,2) are the only ones that work $\endgroup$ – imranfat Aug 25 '15 at 20:10
3
$\begingroup$

From the first equation we have $$y=\frac6x.$$ Plugging this into the second equation yields $$x^{\tfrac6x}+\left(\frac6x\right)^x=17.$$ Call this expression on the left hand side $f(x)$. We want to solve the equation $f(x)=17$. It is obvious that $x=2$ and $x=3$ solve this equation. It remains to show that there are no other solutions.

To see this, first rewrite $f$ as $$f(x)=e^{\frac6x\log x}+e^{x\log\frac6x}.$$ Computing the derivatives of $$g(x)=\frac6x\log x\qquad\text{and}\qquad h(x)=x\log\frac6x$$ immediately shows that $g$ is strictly increasing on $(0,e]$ and strictly decreasing on $[e,\infty)$. Similarly, $h$ is strictly increasing on $(0,\frac6e]$ and strictly decreasing on $[\frac6e,\infty)$. Since the exponential function is increasing, we may conclude that $f$ is increasing on $(0,\frac6e]$ and decreasing on $[e,\infty)$.

Therefore $f(x)<17$ for $x\in(0,2)\cup(3,\infty)$ and $f(x)>17$ for $x\in(2,\frac6e]\cup[e,3)$.

We now only have to verify that $f(x)>17$ for $x\in[\frac6e,e]$. Note that $g$ is increasing on $[\frac6e,e]$, so $g(x)\geq g(\frac6e)=e\log(\frac6e)$ for $x\in[\frac6e,e]$. Similarly, $h$ is decreasing there, so $h(x)\geq h(e)=e\log(\frac6e)$ for $x\in[\frac6e,e]$. Therefore, for $x\in[\frac6e,e]$ we have $$h(x)\geq e^{e\log(\frac6e)}+e^{e\log(\frac6e)}=2\left(\frac6e\right)^e>17.$$

Therefore the only solution is $\{x,y\}=\{2,3\}$.

$\endgroup$
  • $\begingroup$ Great! Thank you so much! $\endgroup$ – VanDerWarden Aug 25 '15 at 20:19
  • $\begingroup$ Are you sure that $h(x)\geq h(e)=e\log(\frac6e)$? This function is decreasing, for $ x\in[\frac6e,e]$. Shouldn't it be like $h(x)<\ldots$? $\endgroup$ – VanDerWarden Aug 26 '15 at 10:10
  • 1
    $\begingroup$ @hetajr: Yes, because $h$ is decreasing, $x\leq e$ implies $h(x)\geq h(e)=e\log\frac{6}{e}$. $\endgroup$ – Dejan Govc Aug 26 '15 at 10:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.