# How do you solve log_3 (x + 3) + log_3 (x + 5) = 1?

Dec 6, 2015

x=-2

#### Explanation:

$\log \left(b a s e 3\right) \left(x + 3\right) + \log \left(b a s e 3\right) \left(x + 5\right) = 1$-> use product rule of logarithm

log(base3)((x+3)(x+5))=1 write in exponential form

${3}^{1} = \left(x + 3\right) \left(x + 5\right)$
${x}^{2} + 8 x + 15 = 3$
${x}^{2} + 8 x + 12 = 0$
$\left(x + 6\right) \left(x + 2\right) = 0$

$x + 6 = 0 \mathmr{and} x + 2 = 0$

x=-6 or x=-2

x=-6 is extraneous . An extraneous solution is root of transformed but it is not a root of original equation.

so x=-2 is the solution.