lnx+ln(x+1)=ln12

Find all real-roots

ln[x(x+1)] = ln 12

x(x+1) = 12
x^2 + x - 12 = 0
(x+4)(x-3) = 0
x = -4 or x = 3
but lnx is only defined for x > 0, so

x = 3

To solve the equation lnx + ln(x+1) = ln12, we can start by using the properties of logarithms.

1. Combine the logarithms on the left side of the equation using the property log(a) + log(b) = log(ab):

lnx + ln(x + 1) = ln12
ln(x(x + 1)) = ln12

2. Remove the natural logarithm on both sides by taking the exponential of both sides. This allows us to convert the equation into an exponential equation:

e^(ln(x(x + 1))) = e^(ln12)
x(x + 1) = 12

3. Expand the equation:

x^2 + x = 12

4. Rearrange the equation:

x^2 + x - 12 = 0

Now we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula.

5. Factoring the quadratic equation:

(x + 4)(x - 3) = 0

Setting each factor equal to zero:

x + 4 = 0 or x - 3 = 0

Solving for x:

x = -4 or x = 3

Therefore, the real roots of the equation lnx + ln(x+1) = ln12 are x = -4 and x = 3.