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

3 Answers
Mar 23, 2016

Answer:

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

Answer:

#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

Answer:

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#