Chapter 5 - Exponential and Logarithmic Functions - 5.6 Logarithmic and Exponential Equations - 5.6 Assess Your Understanding - Page 311: 2

$\{-2,0 \}$

Step 1. Let $u=x+3$, we have $u^2-4u+3=0 \longrightarrow (u-1)(u-3)=0 \longrightarrow u=1,3$ Step 2. For $u=1$, we have $x=u-3=-2$ Step 3. For $u=3$, we have $x=u-3=0$ Step 4. We have the solution $\{-2,0 \}$