How do you simplify #ln 1/2#?

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Answer 1 · 2016-03-24T01:23:06

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

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#

Answer 2 · 2016-03-24T02:15:23

#0#

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

#ln1=0#, so

#ln1/2=0/2=0#