# How do you solve log_2(x+3)+log_2(x-3) = 4?

Apr 5, 2018

Let's use one of the log laws, which says that ${\log}_{b} \left(a\right) + {\log}_{b} \left(c\right)$ can be rewritten as ${\log}_{b} \left(a \cdot c\right)$

log_2(x+3)+log)2(x-3) = 4

${\log}_{2} \left(\left(x + 3\right) \left(x - 3\right)\right) = 4$

${\log}_{2} \left({x}^{2} - 9\right) = 4$

We can write a log , such as ${\log}_{b} \left(a\right) = c$ to be ${b}^{c} = a$ (note, the base remains the same)

${2}^{4} = {x}^{2} - 9$

$16 = {x}^{2} - 9$

$25 = {x}^{2}$

$x = 5$