How do you solve $ln (x + 2) + ln (x - 2) = 3$ $ln(x+2)+ln(x-2)=3$ ?

Question

How do you solve $ln (x + 2) + ln (x - 2) = 3$ $ln(x+2)+ln(x-2)=3$ ?

1 Answer

Impact of this question

1453 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-05T19:03:30

You can write it as:
$ln [(x + 2) \times (x - 2)] = 3$ $ln[(x+2)×(x-2)]=3$
And:
$(x + 2) \times (x - 2) = e^{3} = 20$ $(x+2)×(x-2)=e^3=20$
So you have:
$x^{2} - 4 = 20$ $x^2-4=20$
$x^{2} = 24$ $x^2=24$
$x = \pm \sqrt{24}$ $x=+-sqrt(24)$
$x = \pm 2 \sqrt{6}$ $x=+-2sqrt(6)$
We use the positive answer in our original expression.

Science

Math

Humanities

... and beyond

How do you solve $ln (x + 2) + ln (x - 2) = 3$ $ln(x+2)+ln(x-2)=3$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve ln(x+2)+ln(x-2)=3ln(x+2)+ln(x−2)=3ln(x+2)+ln(x-2)=3?

1 Answer

Related questions

Impact of this question

How do you solve $ln (x + 2) + ln (x - 2) = 3$ $ln(x+2)+ln(x-2)=3$ ?