Chapter 5 - Exponential and Logarithmic Functions - 5.6 Logarithmic and Exponential Equations - 5.6 Assess Your Understanding - Page 311: 22

$-1$

Step 1. Rewrite the equation as $log_6[(x+4)(x+3)]=1 \longrightarrow log_6(x^2+7x+12)=1$ Step 2. Thus $x^2+7x+12=6\longrightarrow x^2+7x+6=0\longrightarrow (x+1)(x+6)=0\longrightarrow x=-1,-6$. Step 3. Check answers, only $x=-1$ fits the original equation.