How do you solve #Ln (y) – Ln (y-4) = Ln (2) #?

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Answer 1 · 2016-03-10T22:03:27

#y=8#

putting everything on one side we get

#lny-ln(y-4)-ln2=0#

Use the property that #log_b (x/y)=log_b x-log_by#

#ln(y/(2(y-4)))=0#

Take the inverse log, in this case the inverse of #ln(x)# is #e^x#

#e^0=y/(2(y-4))#

#1=y/(2(y-4))#

#2y-8=y#

#2y-y=8#

#y=8#