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

Question

1364 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-13T11:25:03

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

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

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

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

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

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

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

$=>x+1=ex-e$

$=>ex-x = e+1$

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

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

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