How do you simplify Ln(-e) - ln(-1/e)?

3 Answers
Mar 23, 2016

This expression is not valid because ln function is only defined for positive numbers.

Explanation:

This expression is not valid because ln is not defined for negative numbers but if it was:

ln(e)-ln(1/e)

then you could use a formula: lna-lnb=ln(a/b) which would lead to:

ln(e)-ln(1/e)=ln(e/(1/e))=ln(e*e)=ln(e^2)=2*lne=2

In the last transformation I used an identity: ln(a^b)=b*lna

Aug 10, 2017

ln(-e)-ln(-1/e)=2

Explanation:

In order to simplify this, we need to use the difference law:

loga-logb-=log(a/b)

ln(-e)-ln(-1/e)=ln((-e)/(-1/e))=ln(e^2)=2lne=2

We could also use the exponent and addition laws:

bloga-=log(a^b) and loga+logb-=log(ab)

ln(-e)-ln(-1/e)=ln(-e)+ln((-1/e)^-1)
=ln(-e)+ln(-e)=ln(e^2)=2

Note that taking the log of negative numbers is only valid over the complex field.

Aug 16, 2017

If we are working in CC and using principal values of ln(z), then it simplifies to 2

Explanation:

ln(-e) = 1+pii

ln(-1/e) = -1+pii

ln(-e) - ln(-1/e) = (1+pii)- ( -1+pii) = 2+0i = 2