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

Mar 7, 2016

$S = \left\{20\right\}$

#### Explanation:

Use the log property that $\log \left(a\right) + \log \left(b\right) = \log \left(a b\right)$, so

$\log \left(x - 15\right) + \log \left(x\right) = 2$
$\log \left(x \left(x - 15\right)\right) = 2$

Since no base was specified we assume the base of the logarithm was $10$, take the base-10 exponential of both sides, i.e.:

${10}^{\log \left(x \left(x - 15\right)\right)} = {10}^{2}$

We know that ${b}^{{\log}_{b} \left(a\right)} = a$, and that ${10}^{2} = 100$ so

$\left(x \left(x - 15\right)\right) = 100$

Expand the product and take that 100 to the LHS

${x}^{2} - 15 x - 100 = 0$

$x = \frac{15 \pm \sqrt{{15}^{2} - 4 \cdot 1 \cdot \left(- 100\right)}}{2 \cdot 1} = \frac{15 \pm \sqrt{225 + 400}}{2}$
$x = \frac{15 \pm \sqrt{625}}{2} = \frac{15 \pm 25}{2}$
${x}^{'} = \frac{15 + 25}{2} = \frac{40}{2} = 20$
${x}^{'} = \frac{15 - 25}{2} = \frac{- 10}{2} = - 5$

However, recall that this was originally a logarithmic formula; we can have anything equal to 0 or a negative number in one!

So we know that

$x - 15 > 0$
$x > 0$

$20 - 15 = 5 > 0$ and $20 > 0$ so it's a valid solution, but $- 5 < 0$ so it isn't one. So the set of solutions is $S = \left\{20\right\}$