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How do you simplify #ln(1/2) - ln(1) #?

1 Answer

Feb 9, 2016

#-ln2#

Explanation:

From laws of logs, #ln(a/b)=lna-lnb#

and #ln(1)=0# since #e^0=1#.

Therefore #ln(1/2)-ln(1)=(ln1-ln2)-ln1#

#=0-ln2-0#

#=-ln2#

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