Notation for $\boxplus$ meaning source

So I was looking through some older questions on Math.SE when I found this question asking what a box with a plus inside meant. As it turns out,$$x\boxplus y:=x\ln y-y\ln x$$which made me wonder: Can$$x\boxplus2=(x\boxplus 2)^{-1}$$which I thought that I might be able to solve. Here is my attempt at doing so: We have that the aforementioned expression can be written as$$(x\boxplus2)^2=1$$Now we have$$(x\ln2-2\ln x)^2=1\implies\ln^2\left(\dfrac{2^x}{x^2}\right)=1\\\implies\ln\left(\dfrac{2^x}{x^2}\right)=1\implies\dfrac{2^x}{x^2}=e\implies2^x=ex^2$$and now to solve our nonlinear boi:$$2^x=ex^2\implies x\ln2=1+2\ln x\\\implies x=\dfrac{1+2\ln x}{\ln2}\implies x-\dfrac{2\ln x}{\ln2}=1\\\implies x^{-2\ln(2)}e^x=e$$However, I was having trouble solving this, so I decided to retry it:$$2^x=ex^2\implies x^{-2}2^x=e\\\implies x(2^{-0.5})^x=e^{-0.5}\implies xe^{-0.5x\ln2}=e^{-0.5}\\\implies-0.5x\ln(2)e^{-0.5x\ln2}=-0.5\ln(2)e^{-0.5}$$And now we can use Lambert's $W$ function to get$$-0.5x\ln2=W_n(-0.5\ln(2)e^{-0.5})\implies-\dfrac2{\ln2}W_n(-0.5\ln(2)e^{-0.5})=x$$which has three real solutions, being$$-(2/\ln2)W_0(\pm\ln(2)/(2\sqrt e))\text{ and }-(2/\ln2)W_{-1}(-\ln(2)/(2\sqrt e))$$and the general form for solutions to this being$$x=-(2/\ln2)W_n(\pm\ln(2)/(2\sqrt e)),W_n(\pm\ln(2)/(2\sqrt e))\ne0$$ Edit: forgot to write the case of $x\boxplus2=-1$, here it is

We also have$$x\boxplus2=-1\implies\dfrac{2^x}{x^2}=\dfrac1e\implies2^x=\dfrac{x^2}e\\\implies x^{-2}e^{x\ln2}=\dfrac1e\implies xe^{-0.5x\ln2}=\sqrt e\\\implies -0.5x\ln(2)e^{-0.5x\ln2}=-0.5\ln(2)\sqrt e\\\implies-0.5x\ln2=W_n(-0.5\ln(2)\sqrt e)\implies x=\dfrac2{\ln2}W_n(-0.5\ln(2)\sqrt e)$$

However, my question is:

Is my solution correct, or what could I do to solve my problem correctly?

  • 2
    $\begingroup$ You omitted the case of $x \boxplus 2 = -1$. $\endgroup$
    – Dan
    Nov 17 at 21:39
  • $\begingroup$ @Dan Ah, my bad. Must've forgotten about that case. $\endgroup$
    – CrSb0001
    Nov 17 at 21:49
  • 1
    $\begingroup$ Aren't you implicitly accepting that $1$ is the identity for the operation? This should be proven, if true, before proceeding further. $\endgroup$
    – MasB
    Nov 20 at 20:37
  • $\begingroup$ @MasB Would you be able to explain what you mean by that? I'm confused right now $\endgroup$
    – CrSb0001
    Nov 21 at 17:43


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