##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##