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What are the complex solutions to e^x = 2x?
I know that this is related to the Lambert W() function, which is only defined on the range x >= -1/e, and that x = -W(-1/2) here.
http://www.wolframalpha.com/input/?i=e%5Ex+%3D+2x%...
However it seems that there is a system of equations that provides a complex solution to this equation. Does this only have a couple of solutions, or infinitely many? What do they look like?
2 Answers
- 5 years ago
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- Dragon.JadeLv 77 years ago
Hello,
Some inquiries about the Lambert function in Wikipedia would have yielded the answer:
x = -W₀(-½)
is the only real solution.
This is the use of the W₀ branch of the Lambert function.
The W₋₁ branch will yield a fully complex solution.
So you will get at most two complex solutions, one of them being real.
Regards,
Dragon.Jade :-)
