Finding the error in ln(-e) = 1

Alkatran · Nov 25, 2004

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 · Nov 25, 2004

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

Integral · Nov 25, 2004

Your trouble is that

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

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

Alkatran · Nov 26, 2004

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

Finding the error in ln(-e) = 1

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