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

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Answer 1 · 2017-02-03T11:55:48

#(3+sqrt13)/2#

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

#x>2#

using the law of logs # " "lnA-lnB=ln(A/B)#

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

#=>ln((x+1)/(x-2))=lnx#

#=>(x+1)/(x-2)=x#

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

#x+1=x^2-2x#

#=>x^2-3x-1=0#

using the formula #x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(3+-sqrt(3^2-4xx1xx-1))/(2xx1)#

#x=(3+-sqrt(9+4))/(2)=(3+-sqrt13)/2#

#because x>2" required solution "(3+sqrt13)/2#

Answer 2 · 2017-02-03T12:10:48

# x=(3+sqrt13)/2#.

We use the following Rules of Log Fun.:

#(1): log a+logb=log(ab); (2): loga=logb rArr a=b.#

Rewriting the given eqn. as, #ln(x+1)=ln(x-2)+lnx#

#rArr ln(x+1)=ln{x(x-2)}=ln(x^2-2x)#

#rArr x+1=x^2-2x#

#rArr x^2-3x-1=0#

Using the Quadratic Formula to solve this, we have,

#x=[-(-3)+-sqrt{(-3)^2-4(1)(-1)}]/{2(1)}, i.e.,#

#x=(3+-sqrt13)/2#

#"As, "x=(3-sqrt13)/2<0, ln x" undefined. :. "x=(3+sqrt13)/2#.