# How do you solve Log x + log (x-48) = 2?

Jan 11, 2016

$x = 50$

#### Explanation:

$\log x + \log \left(x - 48\right) = 2$

The existence condition of $\log \left(g \left(x\right)\right)$ is:

$g \left(x\right) > 0$

then

$x > 0$

in system with:

$x > 48$

$\therefore x \in \left(48 , + \infty\right)$

using the logarithm product properties

$\log \left(x \cdot \left(x - 48\right)\right) = 2$

Now we can resolve as follow:

${10}^{\log \left(x \cdot \left(x - 48\right)\right)} = {10}^{2}$

$x \left(x - 48\right) = {10}^{2}$
${x}^{2} - 48 x - 100 = 0$

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{48 \pm \sqrt{2304 + 400}}{2} =$

$= \frac{48 \pm \sqrt{2704}}{2} = \frac{48 \pm 52}{2} = 24 \pm 26$

${x}_{1} = - 2$

${x}_{2} = 50$

${x}_{1} \notin \left(48 , + \infty\right)$ is not a solution

${x}_{2} \in \left(48 , + \infty\right)$