The following can't be true, but I can't discover my error. Can anyone see it?
$\displaystyle
\frac{i \pi n}{x}=\ln(-1^{\frac{n}{x}})=\ln(-1)+\ln(1^{\frac{n}{x}})=i\pi
$
(n is a positive integer and x is a positive real number)
Thanks
The following can't be true, but I can't discover my error. Can anyone see it?
$\displaystyle
\frac{i \pi n}{x}=\ln(-1^{\frac{n}{x}})=\ln(-1)+\ln(1^{\frac{n}{x}})=i\pi
$
(n is a positive integer and x is a positive real number)
Thanks
$\displaystyle (-1)^{\frac{n}{x}}=(-1)(-1)....(-1)$ for $\displaystyle \frac{n}{x}$ times
or $\displaystyle \left(\sqrt{-1}\right)^{\frac{2n}{x}}$
Using the laws of logs.. $\displaystyle log(ab)=loga+logb$
but you cannot split $\displaystyle (-1)^{\frac{n}{x}$ into $\displaystyle (-1)(1)^{\frac{n}{x}}$