How do you solve $ln (x + 2) + ln (x - 2) = 0$ $ln (x + 2) + ln (x - 2) = 0$ ?

Question

1523 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-23T09:32:43

$x = \sqrt{5}$ $x=sqrt5$

$ln (x + 2) + ln (x - 2) = 0$ $ln(x+2)+ln(x-2)=0$ is equivalent to

$ln [(x + 2) (x - 2)] = ln 1$ $ln[(x+2)(x-2)]=ln1$

or $(x + 2) (x - 2) = 1$ $(x+2)(x-2)=1$

or $x^{2} - 4 = 1$ $x^2-4=1$

or $x^{2} - 5 = 0$ $x^2-5=0$

or $(x + \sqrt{5}) (x - \sqrt{5}) = 0$ $(x+sqrt5)(x-sqrt5)=0$

or $x = - \sqrt{5}$ $x=-sqrt5$ or $x = \sqrt{5}$ $x=sqrt5$

But as $x = - \sqrt{5}$ $x=-sqrt5$ is not in domain (as log of negative number is not possible) hence $x = \sqrt{5}$ $x=sqrt5$

How do you solve ln(x+2)+ln(x−2)=0ln (x + 2) + ln (x - 2) = 0?