# Question #b370e

Nov 3, 2017

x=4

#### Explanation:

${\log}_{2} x + {\log}_{2} \left(x + 4\right) = 5$

${\log}_{2} \left(x \cdot \left(x + 4\right)\right) = 5$

${\log}_{2} \left({x}^{2} + 4 x\right) = 5$

${x}^{2} + 4 x = {2}^{5} = 32$

i.e.

${x}^{2} + 4 x - 32 = 0$

$x 1 = \frac{- 4 + \sqrt{{4}^{2} - 4 \cdot 1 \cdot \left(- 32\right)}}{2 \cdot 1}$
and
$x 2 = \frac{- 4 - \sqrt{{4}^{2} - 4 \cdot 1 \cdot \left(- 32\right)}}{2 \cdot 1}$

which are

$x 1 = \frac{- 4 + \sqrt{16 + 128}}{2}$
and
$x 2 = \frac{- 4 - \sqrt{16 + 128}}{2}$

which becomes

$x 1 = - 2 + 6 = 4$
and
$x 2 = - 2 - 6 = - 8$

Check if true:

${\log}_{2} \left(4\right) + {\log}_{2} \left(4 + 4\right) = 2 + 3 = 5$, so x1 is correct,
and
${\log}_{2} \left(- 8\right) + \ldots$, nope, you cannot take the logarithm of a negative number.

So the only answer is x= 4.
Q.E.D.