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

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Answer 1 · 2016-06-13T11:25:03

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

Using that #ln(a)-ln(b) = ln(a/b)#:

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

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

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

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

#=>(x+1)/(x-1)=e#

#=>x+1=ex-e#

#=>ex-x = e+1#

#=>(e-1)x = e+1#

#:.x = (e+1)/(e-1)#