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

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

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Answer 1 · 2016-03-08T18:21:58

$x \approx 4.9$ $x~~4.9$

Explanation:

$ln (x + 2) + ln (x - 2) = 3$ $ln(x+2)+ln(x-2)=3$
$ln [(x + 2) (x - 2)] = 3$ $ln[(x+2)(x-2)]=3$ -> use property ${log}_{b} x + {log}_{b} y = {log}_{b} (x y)$ $log_bx+log_by=log_b(xy)$
$ln (x^{2} - 4) = 3$ $ln (x^2-4) =3$
$e^{3} = x^{2} - 4$ $e^3=x^2-4$ -> use definition ${log}_{b} x = n \Leftrightarrow b^{n} = x$ $log_bx=n iffb^n=x$
$e^{3} + 4 = x^{2}$ $e^3+4=x^2$ -> solve for x
$x = \pm \sqrt{e^{3} + 4}$ $x=+-sqrt(e^3+4)$
$x \approx 4.9, x \approx$ $x~~4.9, x~~$ -4.9

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

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