How Can You Explore Solutions for ln() with Negative Values?

[Alkatran]

1/ln(x) = ln(x)^-1 = ln(-x)
ln(-e) = 1/ln(e) = 1/1 = 1

ln() is only defined over positive values, but you can find solutions like so... where's the error.

 

[anti_crank]

ln(x)^-1 = ln(-x) is incorrect. You're probably thinking of ln (1/x) = - ln (x)

 

[Integral]

Your trouble is that

\ln (-x) \ne (\ln(x))^{-1})
This is the correct way to do it.
\ln (-x) = \ln(-1 * x) = \ln(-1) + \ln(x)


There will be complex solutions to this but none on the Real line.

 

[Alkatran]

Thanks, I knew something wasn't right. I should caught on when -x = 1/x