1. **Problem statement:** We have a right triangle with a vertical side labeled $x\sqrt{3}$, a horizontal base divided into three segments: the left base angle is $45^\circ$, the right base angle is $30^\circ$, and the horizontal segment between the two base points is labeled 3. We need to find $x$. 2. **Identify the triangle and angles:** The triangle is split into two smaller right triangles by the two slanted segments from the top-left vertex to the base points. The angles at the base are $45^\circ$ and $30^\circ$, so the third angle at the top-left vertex is $180^\circ - 90^\circ - (45^\circ + 30^\circ) = 15^\circ$. 3. **Use trigonometric ratios:** Let the horizontal segments adjacent to the $45^\circ$ and $30^\circ$ angles be $a$ and $b$ respectively, with $a + 3 + b$ as the total base length. 4. **Relate sides in the $45^\circ$ triangle:** For the triangle with angle $45^\circ$, the vertical side is $x\sqrt{3}$ and the base is $a$. Using tangent, $$\tan 45^\circ = \frac{x\sqrt{3}}{a} = 1 \implies a = x\sqrt{3}.$$ 5. **Relate sides in the $30^\circ$ triangle:** For the triangle with angle $30^\circ$, the vertical side is $x\sqrt{3}$ and the base is $b + 3$. Using tangent, $$\tan 30^\circ = \frac{x\sqrt{3}}{b + 3} = \frac{1}{\sqrt{3}} \implies b + 3 = x\sqrt{3} \times \sqrt{3} = 3x.$$ 6. **Express $b$ in terms of $x$:** $$b = 3x - 3.$$ 7. **Total base length:** The total base length is $a + 3 + b = x\sqrt{3} + 3 + (3x - 3) = x\sqrt{3} + 3x.$ 8. **Use the fact that the base is horizontal and the two slanted segments meet at the top-left vertex:** The total base length must be consistent with the geometry. Since the problem does not provide the total base length, we use the fact that the two slanted segments meet at the top-left vertex and the base segment between them is 3. 9. **Solve for $x$ by equating the horizontal distances:** The horizontal distance between the two base points is 3, so $$3 = b - a = (3x - 3) - x\sqrt{3}.$$ 10. **Rearrange and solve:** $$3 = 3x - 3 - x\sqrt{3} \implies 3 + 3 = 3x - x\sqrt{3} \implies 6 = x(3 - \sqrt{3}).$$ 11. **Isolate $x$:** $$x = \frac{6}{3 - \sqrt{3}}.$$ 12. **Rationalize the denominator:** $$x = \frac{6}{3 - \sqrt{3}} \times \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{6(3 + \sqrt{3})}{9 - 3} = \frac{6(3 + \sqrt{3})}{6} = 3 + \sqrt{3}.$$ **Final answer:** $$\boxed{x = 3 + \sqrt{3}}.$$