how do you find x when Lnx + Ln(x+3)=1 im confuesd i only got up to Lnx(x+3)= 1 thanksss
2007-11-20 15:39:20 · 3 answers · asked by meow 2 in Science & Mathematics ➔ Mathematics
but if Lnx(x+3)=1 => x(x+3)=e => ??? would you do like a quadratic equation or like how would you factor it? like if it equaled x2+3x-e=0
2007-11-20 15:51:55 · update #1
ln(x) + ln(x+3) = 1
ln(x*(x+3) = 1 (since ln a + ln b = ln(ab))
=>ln(x^2+3x) = ln(e) (since ln e = 1)
taking out logs
x^2 + 3x = e
x^2 + 3x = 2.72 (since e ≈ 2.72)
x^2 + 3x - 2.72 = 0
x = -3 +/- sqrt(9 + 10.87)/2
x = -3 +/- sqrt(19.87)/2
x = (-3+/- 4.46)/2
x = 1.46/2 = 0.73 (ignoring negative value)
2007-11-20 15:57:44 · answer #1 · answered by mohanrao d 7 · 0⤊ 0⤋
Ln (x -x + 3) = 1
Ln 3 = 1
2007-11-20 23:46:04 · answer #2 · answered by rnygelle87 2 · 0⤊ 0⤋
Assuming that Ln is natural logs, you can take the "antilog" of each side to get x(x+3)= e. This gives you a quadratic to solve.
2007-11-20 23:47:55 · answer #3 · answered by cattbarf 7 · 0⤊ 0⤋