Chapter 5 - Exponential and Logarithmic Functions - 5.6 Logarithmic and Exponential Equations - 5.6 Assess Your Understanding - Page 312: 94

$e^{\frac{ln2ln6}{ln12} }\approx1.648$

Step 1. Rewrite the equation to get $\frac{ln\ x}{ln2}+\frac{ln\ x}{ln6}=1 \longrightarrow (\frac{1}{ln2}+\frac{1}{ln6})ln\ x=1 \longrightarrow ln\ x=\frac{ln2ln6}{ln2+ln6}=\frac{ln2ln6}{ln12} \longrightarrow x=e^{\frac{ln2ln6}{ln12} }\approx1.648$ Step 2. Check, $x=e^{\frac{ln2ln6}{ln12} }\approx1.648$ fits the original equation.