# How do you solve log[(x + 3)(x - 8)] + log[(x + 3)/(x - 8)] = 2?

Sep 17, 2015

$S = - 13$

#### Explanation:

By the rules of logarithms, $\log \left(m\right) + \log \left(n\right) = \log \left(m n\right)$, so

$\log \left[\left(x + 3\right) \left(x - 8\right)\right] + \log \left[\frac{x + 3}{x - 8}\right] = \log \left[\frac{\left(x + 3\right) \left(x - 8\right) \left(x + 3\right)}{x - 8}\right]$

We cancel the (x-8) on the numerator and the denominator

$\log \left[\left(x + 3\right) \left(x - 8\right)\right] + \log \left[\frac{x + 3}{x - 8}\right] = \log \left[{\left(x + 3\right)}^{2}\right]$

The problem told us that that equaled 2, so

$\log \left[{\left(x + 3\right)}^{2}\right] = 2$

We know that if ${\log}_{b} \left(a\right) = c$ then $a = {b}^{c}$ so we have

${\left(x + 3\right)}^{2} = {10}^{2}$

We use 10 because if the log doesn't have a base, it's implicitly 10. (Also note that we couldn't have put that square to the front of the logarithm and cut it with the 2 because we'd lose a root if we did it)

Now we take the square root,

$x + 3 = \pm 10$

And solve these values
$x = 10 - 3 = 7$
$x = - 10 - 3 = - 13$

But before that, check for the values we know x can't be! Logarithms can't take null or negative arguments so we know $x \ne 3 \mathmr{and} x \ne 8$

We know that $x - 3$ and $x - 8$ can only be negative if the other is negative too. so we study the signs

------(-3)++++++++++++ $\left(x - 3\right)$
----------------------(+8)++ $\left(x - 8\right)$

++++0---------------0+++ $\left(x - 3\right) \left(x - 8\right) \mathmr{and} \frac{x - 3}{x - 8}$

We know we can't take values that are 0 or negative, so $x > 8 \mathmr{and} x < - 3$ which means there's only one possible root, -13.