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)