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I was reading an article about the formation of Euler's identity and came across the following statement

Bernoulli had argued that $\ln(–1) = 0$, since $0 = \ln(1) = \ln(-1×-1) = 2\ln(-1)$.


The article goes on to say that Bernoulli is wrong in this assumption but never elaborates. From Euler's identity he is clearly mistaken but I was curious as to how and why because his argument makes sense to me.



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It manipulates log of a product as if the law $\log(ab)=\log a +\log b$, which holds for positive $a$ and $b$, continued to hold. A few steps more and he would reach $0=1$. – André Nicolas May 1 '13 at 22:37
complex log is a multi-valued function. – Jeremy May 1 '13 at 22:37
How, and why, do you think $\,\log((-1)\cdot(-1))=-2\log 1\,$ is correct? – DonAntonio May 1 '13 at 22:38
This only proves that "if the usual properties of the logarithm hold also for negative numbers, then $\log(-1)=0$". But you can't define a "well behaved" logarithm function on the negative numbers. – egreg May 1 '13 at 22:51
up vote 2 down vote accepted

Asking if $ln(x)=y$ is just a fancy way of asking does $e^y=x$. We like to write things as $y=f(x)$. However, when we have $e^y=x$, we can't "solve" for $y$. Thus, the $ln(x)=y$ notation was formed to allow use to write $y$ as a function of $x$. If $ln(-1)=0$ is a correct statement, then $e^0=-1$. However, these is clearly false. Finally, lets try to do what Bernoulli did but writing in the original form not $ln$ form.

You wrote $0=ln(1)\implies 0=ln(-1*-1)\implies 0=ln(-1).$

However, what that translates to if we don't use the $ln$ shorthand is: $1=e^0 \implies -1*-1=e^0, \implies -1=e*0$.

But if we look at the final implies that is not mathematically valid. We divided by $-1$ on the left side but not on the right side. Thus, it was not a valid operation when we did it in the $ln$ shorthand. Like the comments said, your bad assumption was assuming the laws worked for all numbers when they just work for positive ones. The reason why is what I laid out. The laws simply come from observing what happens when we are working with the exponentials not from what appears to work with the $ln$ shorthand.

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Thank you I really appreciate it , guess I was getting too caught up in the rules and less focusing on what it meant – thebmags May 2 '13 at 5:54

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