@@KasyannoEZMath Now that I understand it a bit, I can say that I had a different approach. Since 2ˣ=x Then you can multiply each side by 2⁻ˣ. 2ˣ and 2⁻ˣ will cancel out and become 1 (literally). 1=x2⁻ˣ and 2=eˡⁿ⁽²⁾ so 2⁻ˣ=e⁻ˣˡⁿ⁽²⁾ so 1=xe⁻ˣˡⁿ⁽²⁾ Multiply both sides by -ln(2) so that -ln(2)=-xln(2)(e⁻ˣˡⁿ⁽²⁾) Find the Product Log of both sides. Since W(xeˣ)=x, W(-xln(2)(e⁻ˣˡⁿ⁽²⁾))=-xln(2) so W(-ln(2))=-xln2 Divide both sides by -ln(2) so that -(W(-ln(2))/ln(2))=x I got a different answer and now I'm wondering if it is correct or wrong. Please let me know.
I mean it's just a definition if you can't actually compute W(x) as a function. maybe there is a taylor series or something or useful because this feels like defining random stuff. (though it does have a wikipedia page so guess imma go read that)
@@KasyannoEZMathWhy do we decide that the equation has "no real solution" instead of "no solution"? We can't even find the meeting point of the graph of the equation.
@cinereo_argento, thanks for your time commenting. We don't decide which solution will be. This depends on what the equation is. The graph of the equation shows no intersection. This indicates that there will be no real solution, but it can have a solution in complex form. No solution means that there will be no solution at all, neither real nor complex.
@@cinereo_argentoyou're forgetting about the complex field, an equation may have no real solutions but complex, that's why we specify that, if we say it has absolutely no solutions, we're saying it has no solutions neither in the real nor the complex field.
2^x=x (e^ln(2))^x=x e^(xln(2))=x Multiply both sides with e^-xln(2) 1=xe^(-xln(2)) Multiply both sides with -ln(2) -ln(2)=-xln(2)e^(-xln(2)) Take the Lambert W function on both sides: W(-ln(2))=-xln(2) x=-W(-ln(2))/ln(2)
