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

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Answer 1 · 2015-12-20T13:19:51

#x=1/(e-1)#

Given: #ln(x+1)-ln(x)=1#

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

#e^(ln((x+1)/x))=e^1#

#(x+1)/x=e#

#x+1 = x*e#

#x-x*e = -1#

#x*(1-e)=-1#

#x=1/(e-1)#