Chapter 5 - Exponential and Logarithmic Functions - 5.6 Logarithmic and Exponential Equations - 5.6 Assess Your Understanding - Page 311: 39

$\frac{1}{3}, 729$

Step 1. Let $u=log_3x$, rewrite the equation as $u^2-5u-6=0 \longrightarrow (u+1)(u-6)=0 \longrightarrow u=-1, 6$ Step 2. For $u=-1$, we have $log_3x=-1 \longrightarrow x=\frac{1}{3}$. Step 3. For $u=6$, we have $log_3x=6 \longrightarrow x=3^6=729$. Step 4. Check answers, $x=\frac{1}{3}, 729$ fit the original equation.