Solve the equation $(6-x)^x=x^{6-x}$

Question

Solve the equation $(6-x)^x=x^{6-x}$

My attempt

I put $f(x)=(6-x)^x-x^{6-x}$ and $\operatorname{Dom}f=(0,6)$.

I calculated that $f(2)=f(3)=f(4)=0$

How can I prove that $2,3,4$ are the only solutions to the equation? I assume that I have something to do with the derivative: $$f^\prime(x)= (6 - x)^x \left(\log(6 - x) - \frac x{6 - x}\right) - x^{6 - x} \left(\frac{6 - x}x - \log x\right)$$

Answer 1

My new attempt is the following:

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Let $g(x)=(6-x)\log x-x\log(6-x)$ equivalent to $f(x)$. Then

$$g^\prime(x)=-\log x+\frac{6-x}x-\log(6-x)+\frac x{6-x}$$ $$g^{\prime\prime}(x)=\frac{-2(x^3-9x^2-18x+108)}{x^2(6-x)^2}$$

$g^{\prime\prime}(x)>0$ for $3<x<6$ and $g^{\prime\prime}(x)<0$ for $0<x<3$.

$3$ is a minimum for $g'$. So, for the monotoneity of $g'$ and for IVT,

$$g'(2)>0,g'(3)<0\implies \exists z_1:g'(z_1)=0$$ $$g'(3)<0,g'(4)>0\implies \exists z_2:g'(z_2)=0$$

I calculated that $g(2)=g(3)=g(4)=0$. $z_1, z_2$ are respectively a maximum and a minimum for $g$:

1. $g$ is increasing in $(0,z_1)$
2. $g$ is decreasing in $(z_1,z_2)$

3. $g$ is increasing in $(z_2,6)$

   1. $g(2)=0<g(z_1)$ because $2<z_1$ and $g$ is increasing.

   2. $g(z_1)>0>g(z_2)$ because $z_1<z_2$ and $g$ is decreasing.

There exists only one number between $z_1$ and $z_2$ s.t. $g=0$, that is $3$

3. $g(z_2)<0<g(6)$ because $z_2<6$ and $g$ is increasing.

There exists only one number between $z_2$ and $6$ s.t. $g=0$, that is $4$

Then $2,3,4$ are the only solutions of $g$ and $f$.

Answer 2

The equation is equivalent to $\dfrac{\log x}{x}=\dfrac{\log(6-x)}{6-x}$.

Now what are the variations of $f(x)=\dfrac{\log x}{x}$? The derivative is $f'(x)=\dfrac{1-\log(x)}{x^2}$, positive on $]0,e[$ and negative on $]e,+\infty[$. And $f(1)=0, f(e)=1/e$ and $f(x)\to0$ as $x\to+\infty$.

Hence, for $y\in]0,1/e[$, the equation $\dfrac{\log x}{x}=y$ has exactly two solutions $x_1,x_2$ (with $x_1<x_2)$. For $y$ decreasing from $1/e$ to $0$, $|x_1-x_2|$ increases from $0$ to $+\infty$ (because $x_1$ decreases while $x_2$ increases, and $x_2\to+\infty$). Therefore, there can be at most one case for which $|x_1-x_2|=6$ with $x_1< x_2$. Since you already found that $x_1=2,x_2=4$ is a solution, it's the only one. The other case is $x=6-x$, and then $x=3$.

Answer 3

Calling $f(x) = \frac{x}{\ln x}$ we have

$$ f(x)=f(6-x) $$

so for $x = 3$ we have a solution etc. Now analyzing the growing characteristics for $f(x)$ we can infer the existence of at least two more...

Attached the plot for $f(x)$ and $f(6-x)$