How do you solve ln(x+1)-ln(x-2)=lnxln(x+1)ln(x2)=lnx?

2 Answers
Feb 3, 2017

(3+sqrt13)/23+132

Explanation:

ln(x+1)-ln(x-2)=lnxln(x+1)ln(x2)=lnx

x>2x>2

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

ln(x+1)-ln(x-2)=lnxln(x+1)ln(x2)=lnx

=>ln((x+1)/(x-2))=lnxln(x+1x2)=lnx

=>(x+1)/(x-2)=xx+1x2=x

(x+1)=x(x-2)(x+1)=x(x2)

x+1=x^2-2xx+1=x22x

=>x^2-3x-1=0x23x1=0

using the formula x=(-b+-sqrt(b^2-4ac))/(2a)x=b±b24ac2a

x=(3+-sqrt(3^2-4xx1xx-1))/(2xx1)x=3±324×1×12×1

x=(3+-sqrt(9+4))/(2)=(3+-sqrt13)/2x=3±9+42=3±132

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

Feb 3, 2017

x=(3+sqrt13)/2.

Explanation:

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.