How do you solve #lnx-ln(x+2)=1#?

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Answer 1 · 2016-03-19T12:44:48

#x=(2e)/(1-e)#

#ln x-ln(x+2)=1#

#ln (x/(x+2))=ln e" " "#because #ln e=1#

#x/(x+2)=e#

#x=e(x+2)#

#x=ex+2e#

#x-ex=2e#

#x(1-e)=2e#

#(x(1-e))/((1-e))=(2e)/(1-e)#

#(xcancel((1-e)))/cancel((1-e))=(2e)/(1-e)#

#x=(2e)/(1-e)#

God bless...I hope the explanation is useful.