# How do you solve log_2(x+2) + log_2(x+6)=5?

Apr 17, 2016

$x = 2$

#### Explanation:

Use $\log a + \log b = \log a b$ and, if a^x=b, log_a(a^x)=x,

${\log}_{2} \left(x + 2\right) + {\log}_{2} \left(x + 6\right) = {\log}_{2} \left(\left(x + 2\right) \left(x + 6\right)\right) = 5$, with $x + 2 > 0 \mathmr{and} x + 5 > 0$. It is sufficient that $x > - 2$.

So, $\left(x + 2\right) \left(x + 6\right) = {2}^{5} = 32$
${x}^{2} + 8 x - 20 = 0$.
The roots are $x = 2$ and $x = - 10$.
As $x > - 2$, the root $- 10$ is inadmissible. So, $x = 2$.