# We know that negative number to the power of any

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 ${\left(-2\right)}^{\frac{1}{3}}$ or ${\left(-2\right)}^{e}$, by modifying/extending our current number system?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

sineurosi0f
Consider the value of the "number" ${\left(-2\right)}^{e}$; this is equivalent to considering
$x={\left(-2\right)}^{e}$
$\mathrm{ln}x=e\mathrm{ln}\left(-2\right)$
$\frac{\mathrm{ln}x}{e}=\mathrm{ln}\left(-2\right)$
${e}^{\frac{\mathrm{ln}x}{e}=-2}$
${\frac{12}{e}}^{\frac{\mathrm{ln}x}{e}=-1}$
${e}^{\frac{\mathrm{ln}x}{e}+\mathrm{ln}\frac{1}{2}=-1}$
Given the equation ${e}^{a+bi}={e}^{a}\left(\mathrm{cos}b+i\mathrm{sin}b\right)$, we only need to find x such that $\frac{\mathrm{ln}x}{e}=i\pi +2ni\pi -\mathrm{ln}\frac{1}{2}$ or $\mathrm{ln}x=ei\pi +2\ne i\pi +e\mathrm{ln}2$ which is
$x={e}^{ei\pi +2\ne i\pi +e\mathrm{ln}2}={2}^{e}\left(\mathrm{cos}\left(e\pi +2\ne \pi \right)+i\mathrm{sin}\left(e\pi +2\ne \pi \right)$
which is an infinite set of unique complex values in a circle of radius ${2}^{e}$