How do you solve $ln (x – 2) + ln (x + 2) = ln 5$ ?

Question

1542 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-09T11:55:19

x=-3 or x=3

Using the property that says :

$ln(a) +ln(b) = ln(a*b)$
We have:

$ln(x-2)+ln(x+2) = ln5$

$ln((x-2)*(x+2)) = ln5$

Rasing exponential both sides we will have:

$(x-2)*(x+2) =5$

Applying polynomial property on the equation above that says:

$a^2 - b^2 = (a-b)*(a+b)$

We have: $(x-2)*(x+2) = x^2-4$

So,
$x^2 - 4 = 5$
$x^2 - 4 -5 =0$
$x^2 - 9 =0$
$(x-3)*(x+3)=0$

So,
$x-3=0$ thus $x=3$

Or,

$x+3=0$ thus $x=-3$

How do you solve ln (x – 2) + ln (x + 2) = ln 5 ln (x – 2) + ln (x + 2) = ln 5?