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Ln(1) = 0 ?!by alba_ei
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#1
Mar607, 03:46 PM

P: 38

supposed that we have
[tex] \ln(1) [/tex] then [tex] \frac{2}{2}\ln(1) [/tex] so [tex] \frac{1}{2}\ln(1)^2 [/tex] this is equal to [tex] \frac{1}{2}\ln(1) [/tex] and if this is equal to 0 the we can say that [tex] ln(1) = 0 [/tex] is this right , wrong, are there any explanations for this? 


#2
Mar607, 03:54 PM

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The rule of logarithms is a lnx=lnx^{a}. In this case, a=1 you cannot split it into a fraction and then only take the numerator!
If you look at the logarithm graph, you will see that the function is not defined for negative x. 


#3
Mar607, 03:54 PM

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It's wrong. ln(1) is no longer a real number, so you can't treat it like one. This is like saying sqrt(1) = (1)^{1/2} = (1)^{2/4} = ((1)^{2})^{1/4} = 1^{1/4} = 1.



#4
Mar607, 04:01 PM

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Ln(1) = 0 ?!



#6
Mar607, 04:25 PM

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Or is there something wrong with that line of logic? 


#7
Mar607, 05:22 PM

P: 26

logarithm is defined also for complex numbers.
ln(z)=ln(abs(z))+i*arg(z), where z is complex number, abs(z) is complex norm of complex number z, and arg(z) is its argument. So if 1 is treated as complex number 1+0*i, expression ln(1) gives sense, but the identity a*ln(z)=ln(z^a) is no longer true. 


#8
Mar607, 05:41 PM

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To satisfy the pedants, I shall rephrase my above answer. The natural logarithm function, whose argument is a real number and to whom we can apply the standard laws of logarithms, is not defined for negative real numbers.



#9
Mar607, 07:42 PM

P: 617

for complex z: Ln(z) = ln(z) + i*Arg(z)
so ln(1) = ln(1) + i*Arg(1) = i*pi 


#10
Mar607, 08:27 PM

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In this case, we have Ln(1) = ln(1) + iArg(z) = i*pi. 


#11
Mar607, 08:33 PM

P: 617

eh forgot the ln, fixed



#12
Mar707, 07:08 AM

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#13
Mar807, 05:31 AM

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[tex]\log_e 1 = i\pi + 2ki\pi, k\in \mathbb{Z}[/tex] Case Closed.



#14
Mar807, 07:51 AM

P: 361




#15
Mar907, 05:02 AM

P: 290

Yes, you can still change between various different bases for your logarithms in the same manner as you do for Real numbers.



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