Talk:Imaginary logarithm

Proof

I don't care for the first step:

$\log _{b}(-x)=\log _{b}(x)+\log _{b}(-1).\quad (1)$

This happens to be true for positive x and for the usual branch of the logarithm, but it's a little too glib. Melchoir 14:29, 20 June 2006 (UTC)

I see what you're saying, but I still think it would work. For example, if x=3 and b=5, then

${5^{3}}\times 5^{-1}=5^{(3-1)}=5^{2}$

meaning that the property still holds since

$\log _{b}(xy)=\log _{b}(x)+\log _{b}(y)\!\,$ because $b^{x}\cdot b^{y}=b^{x+y}$ Carifio24 16:19, 21 June 2006 (UTC)

Well there doesn't seem to be any more dispute, so I'll remove the tag. -- He Who Is^{[ Talk ]} 04:34, 6 July 2006 (UTC)

I see that this is considered for deletion, so if it's deleted then this is pointless, but I can agree now that the title is wrong, perhaps 'logarithm of a negative number' would be a better choice? --Carifio24 16:53, 29 July 2006 (UTC)