# Question #cae1a

Apr 12, 2017

$x = 5$

#### Explanation:

Use the property $\log \left(a\right) + \log \left(b\right) = \log \left(a b\right)$ to simplify the equation to $\log \left(\left(x + 1\right) \cdot \left(x - 1\right)\right) = \log \left(24\right)$.

This implies that $\left(x + 1\right) \cdot \left(x - 1\right) = 24$. Then, ${x}^{2} - 1 = 24$. Add $1$ to both sides: ${x}^{2} = 25$. There are two solutions, $x = \pm 5$.

However, if we substitute $- 5$ back into the equation, we would be taking the logarithm of a negative number, which is not allowed. We can eliminate $- 5$ as an answer.

The only real solution is $5$.

Apr 12, 2017

$\textcolor{g r e e n}{x = 5}$

#### Explanation:

Remember:
[1]$\textcolor{w h i t e}{\text{XXX}} \log \left(a\right) + \log \left(b\right) = \log \left(a \cdot b\right)$
[2]$\textcolor{w h i t e}{\text{XXX}} \log \left(c\right)$ is only defined for $c > 0$

$\log \left(x + 1\right) + \log \left(x - 1\right)$
$\textcolor{w h i t e}{\text{XXX}} = \log \left({x}^{2} - 1\right) = \log \left(24\right)$

$\rightarrow {x}^{2} - 1 = 24$

$\rightarrow {x}^{2} = 25$

$\rightarrow x = \pm 5$

...but neither $\log \left(x + 1\right)$ nor $\log \left(x - 1\right)$ are defined if $x = - 5$,
so this is an extraneous solution,

leaving only
$\textcolor{w h i t e}{\text{XXX}} x = 5$