# How do you solve log_2(x+7)+ log_2 (x+8)=1?

Nov 30, 2015

I found $x = - 6$

#### Explanation:

You can use the property of the logs that tells us that:
$\log x + \log y = \log \left(x y\right)$
and write:
${\log}_{2} \left[\left(x + 7\right) \left(x + 8\right)\right] = 1$
now you apply the definition of log to change it into an exponential:
$\left[\left(x + 7\right) \left(x + 8\right)\right] = {2}^{1}$
$\left[\left(x + 7\right) \left(x + 8\right)\right] = 2$
so that:
${x}^{2} + 8 x + 7 x + 56 = 2$
${x}^{2} + 15 x + 54 = 0$
We can use the Quadratic Formula:
${x}_{1 , 2} = \frac{- 15 \pm \sqrt{225 - 216}}{2} = \frac{- 15 \pm 3}{2}$
Two solutions:
${x}_{1} = - 9$ NO it makes both argument negative.
${x}_{2} = - 6$