# How do you solve  log_2(3x + 1) + log_2(x + 7) = 5 and find any extraneous solutions?

Sep 27, 2016

${\log}_{2} \left(\left(3 x + 1\right) \left(x + 7\right)\right) = 5$

$\left(3 x + 1\right) \left(x + 7\right) = {2}^{5}$

$3 {x}^{2} + x + 21 x + 7 = 32$

$3 {x}^{2} + 22 x - 25 = 0$

$3 {x}^{2} - 3 x + 25 x - 25 = 0$

$3 x \left(x - 1\right) + 25 \left(x - 1\right) = 0$

$\left(3 x + 25\right) \left(x - 1\right) = 0$

$x = - \frac{25}{3} \mathmr{and} 1$

However, $x = - \frac{25}{3}$ renders the original equation undefined, so the only solution is $x = 1$.

Hopefully this helps!