1. **State the problem:** We want to find $k$ such that $$1-(1-\text{sigmoid}(x_1+k))(1-\text{sigmoid}(x_2+k)) = a \times \left(1-(1-p_1)(1-p_2)\right)$$ where $p_1 = \text{sigmoid}(x_1)$ and $p_2 = \text{sigmoid}(x_2)$. 2. **Recall the sigmoid function:** $$\text{sigmoid}(x) = \frac{1}{1+e^{-x}}$$ Important property: $1 - \text{sigmoid}(x) = \text{sigmoid}(-x)$. 3. **Rewrite the left side:** $$1-(1-\text{sigmoid}(x_1+k))(1-\text{sigmoid}(x_2+k)) = 1 - \text{sigmoid}(-x_1 - k) \times \text{sigmoid}(-x_2 - k)$$ 4. **Rewrite the right side:** $$a \times \left(1-(1-p_1)(1-p_2)\right) = a \times \left(1 - \text{sigmoid}(-x_1) \times \text{sigmoid}(-x_2)\right)$$ 5. **Set the equation:** $$1 - \text{sigmoid}(-x_1 - k) \times \text{sigmoid}(-x_2 - k) = a \times \left(1 - \text{sigmoid}(-x_1) \times \text{sigmoid}(-x_2)\right)$$ 6. **Isolate the product term:** $$\text{sigmoid}(-x_1 - k) \times \text{sigmoid}(-x_2 - k) = 1 - a + a \times \text{sigmoid}(-x_1) \times \text{sigmoid}(-x_2)$$ 7. **Express sigmoid in exponential form:** $$\text{sigmoid}(-x) = \frac{1}{1+e^{x}}$$ So, $$\text{sigmoid}(-x_1 - k) = \frac{1}{1+e^{x_1 + k}}, \quad \text{sigmoid}(-x_2 - k) = \frac{1}{1+e^{x_2 + k}}$$ 8. **Write the product:** $$\frac{1}{(1+e^{x_1 + k})(1+e^{x_2 + k})} = 1 - a + a \times \frac{1}{(1+e^{x_1})(1+e^{x_2})}$$ 9. **Invert both sides:** $$ (1+e^{x_1 + k})(1+e^{x_2 + k}) = \frac{1}{1 - a + a \times \frac{1}{(1+e^{x_1})(1+e^{x_2})}}$$ 10. **Simplify the denominator on the right:** $$1 - a + \frac{a}{(1+e^{x_1})(1+e^{x_2})} = \frac{(1 - a)(1+e^{x_1})(1+e^{x_2}) + a}{(1+e^{x_1})(1+e^{x_2})}$$ 11. **Therefore:** $$ (1+e^{x_1 + k})(1+e^{x_2 + k}) = \frac{(1+e^{x_1})(1+e^{x_2})}{(1 - a)(1+e^{x_1})(1+e^{x_2}) + a}$$ 12. **Expand the left side:** $$1 + e^{x_1 + k} + e^{x_2 + k} + e^{x_1 + x_2 + 2k} = \frac{(1+e^{x_1})(1+e^{x_2})}{(1 - a)(1+e^{x_1})(1+e^{x_2}) + a}$$ 13. **Let $A = e^{k}$, $B = e^{x_1}$, and $C = e^{x_2}$:** $$1 + A B + A C + A^2 B C = \frac{(1+B)(1+C)}{(1 - a)(1+B)(1+C) + a}$$ 14. **Rewrite as a quadratic in $A$:** $$A^2 B C + A (B + C) + 1 = \frac{(1+B)(1+C)}{(1 - a)(1+B)(1+C) + a}$$ 15. **Set:** $$D = \frac{(1+B)(1+C)}{(1 - a)(1+B)(1+C) + a}$$ 16. **Quadratic equation:** $$A^2 B C + A (B + C) + (1 - D) = 0$$ 17. **Solve for $A$ using quadratic formula:** $$A = \frac{-(B + C) \pm \sqrt{(B + C)^2 - 4 B C (1 - D)}}{2 B C}$$ 18. **Finally, find $k$:** $$k = \ln(A)$$ **Summary:** - Compute $B = e^{x_1}$, $C = e^{x_2}$. - Compute $D = \frac{(1+B)(1+C)}{(1 - a)(1+B)(1+C) + a}$. - Solve quadratic for $A$. - Take $k = \ln(A)$. This gives the value of $k$ satisfying the original equation.