Answer
$x = -3$
Work Step by Step
$\log_4 (x+4) + \log_4 (x+7) = 1$
$\log_4 (x+4)(x+7) = 1$
$(x+4)(x+7) = 4^{1}$
$(x+4)(x+7) = 4$
$x(x+7)+4(x+7) = 4$
$x^{2} + 7x + 4x+28 = 4$
$x^{2} + 11x + 28 - 4 = 0$
$x^{2} + 11x + 24 = 0$
$x^{2} + 8x + 3x + 24 = 0$
$x(x+8) + 3(x+8) = 0$
$(x+3)(x+8) = 0$
$x = -3, -8$
$x = -3$
Check:
When $x= -3$
$\log_4 (x+4) + \log_4 (x+7) \overset{?}{=} 1$
$\log_4 ((-3)+4) + \log_4 ((-3)+7) \overset{?}{=} 1$
$\log_4 (1) + \log_4 (4) \overset{?}{=} 1$
$0 + 1 \overset{?}{=} 1$
$1 = 1$
When $x = -8$
$\log_4 (x+4) + \log_4 (x+7) \overset{?}{=} 1$
$\log_4 (-8+4) + \log_4 (-8+7) \overset{?}{=} 1$
$\log_4 (-4) + \log_4 (-1) \overset{?}{=} 1$
Does not exist, since you cannot have a negative logarithm.