Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

1 Answer

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.