We know that negative number to the power of any

wingtipti 2022-01-23 Answered
We know that negative number to the power of any integers or some fractions will always have a solution. Is it possible for us to solve (2)13 or (2)e, by modifying/extending our current number system?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

sineurosi0f
Answered 2022-01-24 Author has 14 answers
Consider the value of the "number" (2)e; this is equivalent to considering
x=(2)e
lnx=eln(2)
lnxe=ln(2)
elnxe=2
12elnxe=1
elnxe+ln12=1
Given the equation ea+bi=ea(cosb+isinb), we only need to find x such that lnxe=iπ+2niπln12 or lnx=eiπ+2iπ+eln2 which is
x=eeiπ+2iπ+eln2=2e(cos(eπ+2π)+isin(eπ+2π)
which is an infinite set of unique complex values in a circle of radius 2e
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

New questions