How do you simplify #ln 1/2#?

2 Answers
Mar 24, 2016

Answer:

I found: #-ln(2)=-0.69315# when the original question stated #ln(1/2)#...!

Explanation:

I would use a property of the logs where you have:
#logx-logy=log(x/y)#
To write:
#ln(1/2)=ln(1)-ln(2)=0-ln(2)=-ln(2)=-0.69315#

Mar 24, 2016

Answer:

#0#

Explanation:

Assuming you meant #ln1/2# and not #ln(1/2)#:

#ln1=0#, so

#ln1/2=0/2=0#