1. **Problem Statement:** We are given the half-life formula for Iridium-192: $$A = 100 \left(\frac{1}{2}\right)^{\frac{t}{73.83}}$$ where $A$ is the amount remaining after $t$ days. We need to find an equivalent equation in the form $A = 100 \cdot r^t$. 2. **Formula and Explanation:** The half-life formula is $$A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}$$ where $h$ is the half-life period. To rewrite this as $A = A_0 \cdot r^t$, we use the property of exponents: $$\left(\frac{1}{2}\right)^{\frac{t}{73.83}} = \left[\left(\frac{1}{2}\right)^{\frac{1}{73.83}}\right]^t$$ 3. **Intermediate Work:** Calculate the base $r$: $$r = \left(\frac{1}{2}\right)^{\frac{1}{73.83}} = e^{\ln\left(\frac{1}{2}\right) \cdot \frac{1}{73.83}} = e^{-\frac{\ln 2}{73.83}}$$ Using a calculator: $$\ln 2 \approx 0.6931$$ $$r = e^{-\frac{0.6931}{73.83}} = e^{-0.009386} \approx 0.990656$$ 4. **Final Equation:** $$A = 100 (0.990656)^t$$ 5. **Answer:** The equation that approximates the amount of Iridium-192 after $t$ days is option (3): $$A = 100 (0.990656)^t$$ This means the amount decreases by about 0.934% each day, consistent with the half-life decay.