How do you solve #ln(x-3)-ln(x-5)=ln5#?

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Answer 1 · 2016-07-22T06:24:01

#x=11/2#

The solution must be:

#x-3>0 and x-5>0#

that's #x>5#

Since #loga-logb=log(a/b)#

the given equation is equivalent to:

#ln((x-3)/(x-5))=ln5#

#(x-3)/(x-5)=5#

#x-3=5(x-5)#

#x-3=5x-25#

#4x=22#

#x=11/2#

that's >5