Question

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

Answer 1

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

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

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

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

`x(x+1) = 12 iff x^2 + x - 12 = 0`

You now have to calculate `Delta = b^2 - 4ac`, which here equals to `49` so this quadratic equations has two real solutions, given by the quadratic formula : `(-b+sqrt(Delta))/(2a)` and `(-b-sqrt(Delta))/(2a)`. 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.