Chapter 5 - Exponential and Logarithmic Functions - 5.6 Logarithmic and Exponential Equations - 5.6 Assess Your Understanding - Page 311: 24

$2$

Step 1. Rewrite the equation as $log_5[(x+3)(x-1)]=1 \longrightarrow log_5(x^2+2x-3)=1$ Step 2. Thus $x^2+2x-3=5\longrightarrow x^2+2x-8=0\longrightarrow (x+4)(x-2)=0\longrightarrow x=-4, 2$. Step 3. Check answers, only $x=2$ fits the original equation.