How do you solve $ln (x + 5) + ln (x - 1) = 2$ $Ln(x + 5) + ln(x - 1) = 2$ ?

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Answer 1 · 2016-06-21T12:56:47

$x = \sqrt{9 + e^{2}} - 2$ $x = sqrt(9+e^2) -2$ .

e = 2.7183, nearly.

$x > 1$ $x>1$ to make $log (x - 1)$ $log (x-1)$ real.

Use ${ln e}^{n} = n and ln a + ln b = ln a b$ $ln e^n=n and ln a + ln b = ln ab$ ..

The given equation is

$ln ((x + 5) (x - 1)) = {ln e}^{2}$ $ln((x+5)(x-1))= ln e^2$ .

So, $(x + 5) (x - 1) = e^{2}$ $(x+5)(x-1)=e^2$ . Solving,

$x = - 2 \pm \sqrt{9 + e^{2}}$ $x = -2+-sqrt(9+e^2)$

The negative root is inadmissible.

So, $x = \sqrt{9 + e^{2}} - 2$ $x = sqrt(9+e^2)-2$ .

e = 2.7183, nearly...

How do you solve ln(x+5)+ln(x−1)=2Ln(x + 5) + ln(x - 1) = 2?