1. The problem is to solve the equation $$\sqrt{48x - 5} = 3\sqrt{27x - 2}.$$\n\n2. We start by squaring both sides to eliminate the square roots. The rule is: if $a = b$, then $a^2 = b^2$.\n\n3. Squaring both sides gives: $$48x - 5 = 9(27x - 2).$$\n\n4. Expand the right side: $$48x - 5 = 243x - 18.$$\n\n5. Bring all terms to one side to isolate $x$: $$48x - 5 - 243x + 18 = 0 \implies -195x + 13 = 0.$$\n\n6. Solve for $x$: $$-195x = -13 \implies x = \frac{13}{195} = \frac{1}{15}.$$\n\n7. Check the solution in the original equation to avoid extraneous roots:\n- Left side: $$\sqrt{48 \cdot \frac{1}{15} - 5} = \sqrt{\frac{48}{15} - 5} = \sqrt{3.2 - 5} = \sqrt{-1.8},$$ which is not a real number.\n- Right side: $$3\sqrt{27 \cdot \frac{1}{15} - 2} = 3\sqrt{1.8 - 2} = 3\sqrt{-0.2},$$ also not real.\n\nSince the square roots of negative numbers are not real, $x=\frac{1}{15}$ is not a valid solution.\n\n8. Therefore, there is no real solution to the equation $$\sqrt{48x - 5} = 3\sqrt{27x - 2}.$$