# How do you solve log_2 (x-2)+log_2 (x+4)=4?

Feb 12, 2015

If log_2(x−2)+log_2(x+4)=4
then $x = 4$
as can be seen by replacing $x$ with $4$ in the sum from the original equation.
${\log}_{2} \left(4 - 2\right) + {\log}_{2} \left(4 + 4\right)$
= ${\log}_{2} \left(2\right) + {\log}_{2} \left(8\right)$

Since ${2}^{1} = 2$ and ${2}^{3} = 8$
(and therefore ${\log}_{2} \left(2\right) = 1$ and ${\log}_{2} \left(8\right) = 3$

when x = 4,
log_2(x−2)+log_2(x+4)= 1+3 = 4

How to get there :
Let ${2}^{j} = x - 2$ (i.e. $j = {\log}_{2} \left(x - 2\right)$)
and
let ${x}^{k} = x + 4$ (i.e. $k = {\log}_{2} \left(x + 4\right)$)

From the given equation
$j + k = 4$

$\left({2}^{j}\right) \cdot \left({2}^{k}\right) = \left(x - 2\right) \left(x + 4\right)$
combining, we get:
${2}^{j + k} = {x}^{2} + 2 x - 8$
but, since $j + k = 4$
${2}^{4} = 16 = {x}^{2} + 2 x - 8$
therefore
${x}^{2} + 2 x - 24 = 0$

Simple factoring then gives
$\left(x - 4\right) \left(x + 6\right) = 0$
so
$x = 4$ or $x = - 6$

If $x = - 6$
then
log_2(x−2) and ${\log}_{2} \left(x + 4\right) = 4$
would be ${\log}_{2}$'s of negative values (impossible).

Therefore
$x = 4$