How do you solve log(x)+log(x+1)=log(12)?

1 Answer
Dec 17, 2015

The answer is $x = 3$.

Explanation:

You first have to say where the equation is defined : it is defined if $x > - 1$ since the logarithm can't have negative numbers as argument.

Now that this is clear, you now have to use the fact that natural logarithm maps addition into multiplication, hence this :

$\ln \left(x\right) + \ln \left(x + 1\right) = \ln \left(12\right) \iff \ln \left[x \left(x + 1\right)\right] = \ln \left(12\right)$

You can now use the exponential function to get rid of the logarithms :

$\ln \left[x \left(x + 1\right)\right] = \ln \left(12\right) \iff x \left(x + 1\right) = 12$

You develop the polynomial at the left, you substract 12 at both sides, and you now have to solve a quadratic equation :

$x \left(x + 1\right) = 12 \iff {x}^{2} + x - 12 = 0$

You now have to calculate $\Delta = {b}^{2} - 4 a c$, which here equals to $49$ so this quadratic equations has two real solutions, given by the quadratic formula : $\frac{- b + \sqrt{\Delta}}{2 a}$ and $\frac{- b - \sqrt{\Delta}}{2 a}$. The two solutions here are $3$ and $- 4$. But the very 1st equation we are solving right now is only defined for $x > - 1$ so $- 4$ is not a solution of our log equation.