Question

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

Answer 1

`(3+sqrt13)/2`

Explanation:

`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

`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`.