Answer
$x=-1$
Work Step by Step
$\log_6(x+3)+\log_6(x+4)=1$
Using the product rule of logarithms:
$\log_6((x+3)\cdot(x+4))=1$
Simplify:
$x^2+7x+12=6^1$
$x^2+7x+6=0$
$(x+6)(x+1)=0$
$x_1+6=0$
$x_1=-6$
$x_2+1=0$
$x_2=-1$
Verify answers:
$\log_6(-6+3)+\log_6(-6+4)=1$
$\log_6(-3)+\log_6(-2)=1$
It isn't possible to take a logarithm of a negative answer; therefore x=-6 isn't an answer.
$\log_6(-1+3)+\log_6(-1+4)=1$
$\log_6(2)+\log_6(3)=1$
$\log_6(2\cdot3)=1$
$6=6^1\checkmark$
The only answer is x=-1.
