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.