1. **State the problem:** We need to find the base $x$ such that the number $110_x$ in base $x$ equals $40_5$ in base 5. 2. **Convert $40_5$ to decimal:** In base 5, $40_5 = 4 \times 5^1 + 0 \times 5^0 = 4 \times 5 + 0 = 20$. 3. **Express $110_x$ in decimal:** In base $x$, $110_x = 1 \times x^2 + 1 \times x^1 + 0 \times x^0 = x^2 + x$. 4. **Set up the equation:** Since $110_x = 40_5$, we have $$x^2 + x = 20$$ 5. **Solve the quadratic equation:** $$x^2 + x - 20 = 0$$ Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=1$, $c=-20$. Calculate the discriminant: $$\Delta = 1^2 - 4 \times 1 \times (-20) = 1 + 80 = 81$$ Calculate the roots: $$x = \frac{-1 \pm \sqrt{81}}{2} = \frac{-1 \pm 9}{2}$$ 6. **Find possible values for $x$:** - $x = \frac{-1 + 9}{2} = \frac{8}{2} = 4$ - $x = \frac{-1 - 9}{2} = \frac{-10}{2} = -5$ 7. **Check for valid base:** Base must be a positive integer greater than the largest digit in the number $110_x$, which is 1, so $x > 1$. Also, base cannot be negative. Therefore, the valid base is $x = 4$. **Final answer:** $$\boxed{4}$$