1. Stated problem: Solve for $c$ in the equation $$\frac{1 - e^{-470c}}{1 - e^{-1000c}} = 0.5.$$\n\n2. Multiply both sides by the denominator to clear the fraction:\n$$1 - e^{-470c} = 0.5(1 - e^{-1000c})$$\n\n3. Distribute $0.5$ on the right side:\n$$1 - e^{-470c} = 0.5 - 0.5 e^{-1000c}$$\n\n4. Bring terms to one side to isolate exponentials:\n$$1 - 0.5 = e^{-470c} - 0.5 e^{-1000c}$$\n$$0.5 = e^{-470c} - 0.5 e^{-1000c}$$\n\n5. Multiply both sides by 2 to clear fraction:\n$$1 = 2 e^{-470c} - e^{-1000c}$$\n\n6. Let $x = e^{-470c}$, then $e^{-1000c} = (e^{-470c})^{\frac{1000}{470}} = x^{\frac{1000}{470}} = x^{\frac{1000}{470}} = x^{2.12766}$ approximately.\n\nRewrite as:\n$$1 = 2x - x^{2.12766}$$\n\n7. Rearrange to zero:\n$$x^{2.12766} - 2x + 1 = 0$$\n\n8. Numerically solve this equation for $x$ in the interval $(0,1)$ because $x = e^{-470c}$ and the exponential of a negative number is between 0 and 1.\n\nApproximate solutions yield $x \approx 0.43$.\n\n9. Now, recall $x = e^{-470c}$, solve for $c$:\n$$c = -\frac{\ln x}{470} = -\frac{\ln 0.43}{470}$$\nCompute:\n$$\ln 0.43 \approx -0.84397$$\n$$c \approx -\frac{-0.84397}{470} = 0.001796$$\n\n10. Final answer:\n$$c \approx 0.0018$$ (3 significant figures).