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

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Answer 1 · 2018-03-30T02:33:51

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

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

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

Applying Rule 1 from the table above,

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

Applying Rule 6,

$ln (e^5) = 5$

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

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

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

Solving for x,

$x = (-2 +-sqrt(4 +4e^5)) / 2$

$x =( -2 + sqrt(4 + 4e^5)) / 2 " "color(green)(True) , " "(-2 -sqrt(4 + 4 e^5)) / 2 " " color(red)(false)$

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

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