I know it's wrong but I don't know WHY

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  • #1
Alkatran
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
anti_crank
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ln(x)^-1 = ln(-x) is incorrect. You're probably thinking of ln (1/x) = - ln (x)
 
  • #3
Integral
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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.
 
  • #4
Alkatran
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Thanks, I knew something wasn't right. I should caught on when -x = 1/x :rofl:
 

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