# How do you solve log(x-15)=2-logx?

Sep 18, 2015

$x = 20$

#### Explanation:

Put everything that's a log on the same side
$\log \left(x - 15\right) + \log \left(x\right) = 2$

Remember that $\log \left(m\right) + \log \left(n\right) = \log \left(m n\right)$
$\log \left(x \left(x - 15\right)\right) = 2$

If ${\log}_{a} \left(b\right) = c$, then $b = {a}^{c}$
$x \left(x - 15\right) = {10}^{2}$

Expand and solve the quadratic equation

${x}^{2} - 15 x = 100 \rightarrow {x}^{2} - 15 x - 100 = 0$
$x = \frac{15 \pm \sqrt{225 - 4 \cdot 1 \left(- 100\right)}}{2} = \frac{15 \pm \sqrt{225 + 400}}{2}$
$x = \frac{15 \pm \sqrt{625}}{2} = \frac{15 \pm 25}{2}$

${x}_{1} = \frac{15 + 25}{2} = \frac{40}{2} = 20$
${x}_{2} = \frac{15 - 25}{2} = - \frac{10}{2} = - 5$

Remember that since we were dealing with logarithms, we can't have null or negative arguments, so
$x - 15 > 0 \rightarrow x > 15$
$x > 0$

We conclude that any answers must follow $x > 15$, which only of the two answers do, thus, the answer is $x = 20$