How do you solve #ln(3x+1)-ln(5+x)=ln2#?

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Answer 1 · 2015-10-01T08:48:02

#x=9#

#ln(3x+1)-ln(5+x)=ln(2)=># using laws of logs:

#ln[(3x+1)/(x+5)]=ln(2)=># if #ln(A)=ln(B)hArrA=B#:

#(3x+1)/(x+5)=2=># multiply by #(x+5)#:

#3x+1=2(x+5)=># expand right side:

#3x+1=2x+10=># subtract#-2x# and 1 from both sides:

#3x-2x=10-1=># simplify:

#x=9#