Question$\log (x+3)+\log (x)=1$

Studdy Solution
Solve the quadratic equation using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 3$, and $c = -10$:
$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1}$ $x = \frac{-3 \pm \sqrt{9 + 40}}{2}$ $x = \frac{-3 \pm \sqrt{49}}{2}$ $x = \frac{-3 \pm 7}{2}$
Calculate the two potential solutions:
$x = \frac{-3 + 7}{2} = 2$ $x = \frac{-3 - 7}{2} = -5$
Since $x$ must be positive (as assumed), we discard $x = -5$.
The value of $x$ is:
$$\boxed{2}$$