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

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Answer 1 · 2015-09-09T18:11:17

Use properties of $ln$ to find $ln(x^2-4) = ln(5)$ and hence find solution: $x = 3$

$ln(x^2-4) = ln((x-2)(x+2)) = ln(x-2)+ln(x+2)$

$= ln(5)$

So $x^2-4 = 5$

Add $4$ to both sides to get: $x^2 = 9$

So $x = +-3$

If $x = -3$ , then $x - 2 < 0$ and $x + 2 < 0$ , so neither $ln(x-2)$ nor $ln(x+2)$ have Real values.

If $x = 3$ then:

$ln(x-2) + ln(x+2) = ln(1) + ln(5) = 0 + ln(5) = ln(5)$

So the only solution is $x = 3$

Answer 2 · 2015-09-09T18:12:50

Well you can write it as follows

$ln(x-2)+ln(x+2)=ln5=>ln(x^2-4)=ln5=>x^2-4=5=>x^2=9=>x=-3,x=+3$

But x-2>0 so the solution is x=3

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