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

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

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Answer 1 · 2018-03-30T02:33:51

$x = + \sqrt{1 + e^{5}} - 1$ $color(crimson)(x =+ sqrt(1 + e^5) - 1$

Explanation:

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

$ln x + ln (x + 2) = 5$ $ln x + ln (x +2) = 5$

Applying Rule 1 from the table above,

$ln (x \cdot (x + 2)) = 5$ $ln (x * (x+2)) = 5$

Applying Rule 6,

$ln (e^{5}) = 5$ $ln (e^5) = 5$

ln(x(x+5)) = ln e^5#

$x (x + 5) = e^{5}, Removing **ln** on both sides$ $x(x+5) = e^5, " Removing **ln** on both sides"$

$x^{2} + 2 x - e^{5} = 0$ $x^2 +2x - e^5 = 0$

Solving for x,

$x = \frac{- 2 \pm \sqrt{4 + 4 e^{5}}}{2}$ $x = (-2 +-sqrt(4 +4e^5)) / 2$

$x = \frac{- 2 + \sqrt{4 + 4 e^{5}}}{2} T r u e, \frac{- 2 - \sqrt{4 + 4 e^{5}}}{2} f a l s e$ $x =( -2 + sqrt(4 + 4e^5)) / 2 " "color(green)(True) , " "(-2 -sqrt(4 + 4 e^5)) / 2 " " color(red)(false)$

$x = \sqrt{1 + e^{5}} - 1$ $color(crimson)(x = sqrt(1 + e^5) - 1$

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

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