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#