There is a slight problem with your solution . If you look at the graphs of y=e^x and y=x^2 , they intersect at one point . and the x value for the intersection is negative . To solve the problem , I would start with x^2=e^x . Now , take the square root of both sides . The right hand side is no problem because e^x is always positive for real x . When we take the square root of x^2 we get the absolute value of x ---> |x|=e^(x/2) . Now , we know that for our solution x is less than zero ; so , we can substitute --x for the absolute value of x-----> --x=e^(x/2) . Now , divide both sides by e^(x/2) to get --xe^(--(x/2))=1 . Now , divide both sides by 2 to get : --x/2e^(--(x/2))=1/2 . Now , take the lambert w function of both sides : W(--x/2e(--(x/2)))=W(1/2) . This gives : --x/2=W(1/2) . Multiply both sides by --2 , and you get : x=--2W(1/2) --- this gives a real solution whereas if you look up the value of the lambert W function of negative 1/2 on wolfram alpha it is complex .
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There is a slight problem with your solution . If you look at the graphs of y=e^x and y=x^2 , they intersect at one point . and the x value for the intersection is negative . To solve the problem , I would start with x^2=e^x . Now , take the square root of both sides . The right hand side is no problem because e^x is always positive for real x . When we take the square root of x^2 we get the absolute value of x ---> |x|=e^(x/2) . Now , we know that for our solution x is less than zero ; so , we can substitute --x for the absolute value of x-----> --x=e^(x/2) . Now , divide both sides by e^(x/2) to get --xe^(--(x/2))=1 . Now , divide both sides by 2 to get :
--x/2e^(--(x/2))=1/2 . Now , take the lambert w function of both sides :
W(--x/2e(--(x/2)))=W(1/2) . This gives :
--x/2=W(1/2) . Multiply both sides by --2 , and you get :
x=--2W(1/2) --- this gives a real solution whereas if you look up the value of the lambert W function of negative 1/2 on wolfram alpha it is complex .
The answer is: no solutions. W(-1÷2) is undefined