# How do you solve log_4 (x - 7) + log_4 (x + 7) = 2?

Apr 23, 2016

$x = + \sqrt{65}$

#### Explanation:

$\implies {\log}_{4} \left(\left(x - 7\right) \cdot \left(x + 7\right)\right) = 2$
$\implies {x}^{2} - {7}^{2} = {4}^{2}$
$\implies {x}^{2} = 49 + 16 = 65$
$x = \pm \sqrt{65}$
negative root not possible as log of negative number is not defined
so $x = \sqrt{65}$

Apr 23, 2016

$\sqrt{65} \approx 8.062$

#### Explanation:

${\log}_{b} m + {\log}_{b} n = {\log}_{b} \left(m n\right) .$
We can use this here and condense, assuming that x > 7..

${\log}_{4} \left(\left(x - 7\right) \left(x + 7\right)\right) = 2$

Inversely,$\left(x - 7\right) \cdot \left(x + 7\right) = {4}^{2}$.

${x}^{2} - 49 = 16$

$x = \pm \sqrt{65}$

Negative root is inadmissible, as x > 7.