Answer
$e^{\frac{ln3ln5}{ln15}}\approx1.921$
Work Step by Step
Step 1. Rewrite the equation to get $\frac{ln\ x}{ln5}+\frac{ln\ x}{ln3}=1 \longrightarrow (\frac{1}{ln5}+\frac{1}{ln3})ln\ x=1 \longrightarrow ln\ x=\frac{ln3ln5}{ln3+ln5}=\frac{ln3ln5}{ln15} \longrightarrow x=e^{\frac{ln3ln5}{ln15}}\approx1.921$
Step 2. Check, $x=e^{\frac{ln3ln5}{ln15}}\approx1.921$ fits the original equation.