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

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Answer 1 · 2017-01-16T14:45:48

#x=(sqrt(1+4e)-1)/2=1.223#

#lnx+ln(x+1)=1#

Hence #x(x+1)=e#

i.e. #x^2+x-e=0#

and using quadratic formula

#x=(-1+-sqrt(1^2+4e))/2#

#=(-1+-sqrt(1+4e))/2#

#=(-1+-3.446)/2#

#=(2.446)/2# or #-4.446/2#

#=1.223# or #-2.223#

But as #x=-2.223# is not in domain, we have #x=(sqrt(1+4e)-1)/2=1.223#
graph{lnx+ln(x+1)-1 [-5, 5, -2.5, 2.5]}