solve for x?

Question

ln(x) +ln(x+3)=1

Answer 1

Use the property of the logarithms : lnA + lnB = lnAB

ln(x) +ln(x+3)=1
ln[x(x+3)]=1

Now "antilog" the equation, meaning do the:
e^(LHS) = e^(RHS)
and simplify

x(x+3) = e¹
x² + 3x - e = 0

Use the quadratic formula

x = [-3 ± √(9 + 4e)] / 2

Type this into your calculator if you want, to get
x = [-3 + √(9 + 4e)] / 2 = 0.729
and
x = [-3 - √(9 + 4e)] / 2 = -3.729
we omit the latter solution (-3.729) because log of a negative number cannot be taken.

Hope this helps.

Answer 2

ln(x) + ln(x+3) = 1
ln(x+3) = 1 - ln(x)
e^(ln(x+3)) = e^(1 - ln(x))

x+3 = (e^(1))/(e^(ln(x)))

x+3 = e/x

(x+3)*x = e

x^2+3x -e =0

use quadratic equation

x = (-3+-sqr(9+4e)) / 2

x= (-3+sqr(9+4e)) / 2 only because the other one whould yeild a negative number any you cant do the natural log of a negative number

Answer 3

Ln[x(x+3)] = 1
x(x+3) = e
x^2 + 3x - e = 0
Solve using the genarl law
x = [-3 + sqrt ( (9 + 4e)]/2 = 0.728964295
or
x = [-3 - sqrt ( (9 + 4e)]/2 = - 3.728964295

Answer 4

ln(x) + ln(x + 3) = 1
ln[x(x - 3)] = 1
x² - 3x = e
x² - 3x - e = 0
x = [3 ± √(9 + 4e)] / 2

Since a log can't be taken of a negative number,
x = [3 + √(9 + 4e)] / 2

Answer 5

Exponentiate both sides, to get x(x+3) = e, so x*2 + 3x - 2.718 = 0, which is solvable by the quadratic formula in the usual way.

Answer 6

ln (x.(x + 3)) = 1
x.(x + 3) = 2.718
x² + 3x - 2.718 = 0
x = [- 3 ± √(9 + 10.9)] / 2
x = [- 3 ± √(19.9)] / 2
x = 0.73 , x = - 3.73

Answer 7

Help, mI use to could do this