Bernoulli and the natural log of negative 1 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
     $0 = \ln(1) = \ln(-1×-1) = 2\ln(-1)$.

Link
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.
Thanks,
Brian
 A: 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.  
