1. The problem states that the number of bacteria $B$ is given by the formula $$B = 50 \cdot 10^{d/2}$$ where $d$ is the number of days. 2. We want to find the number of days $d$ when the bacteria count reaches 800,000. We set up the equation: $$800000 = 50 \cdot 10^{d/2}$$ 3. Divide both sides by 50: $$\frac{800000}{50} = 10^{d/2} \implies 16000 = 10^{d/2}$$ 4. Take the base-10 logarithm of both sides to solve for $d/2$: $$\log_{10}(16000) = \log_{10}\left(10^{d/2}\right)$$ 5. Using the logarithm property $\log_{10}(a^b) = b \log_{10}(a)$ and $\log_{10}(10) = 1$: $$\log_{10}(16000) = \frac{d}{2}$$ 6. Multiply both sides by 2 to isolate $d$: $$d = 2 \log_{10}(16000)$$ 7. This is the exact answer in terms of base-ten logarithms. **Final answer:** $$d = 2 \log_{10}(16000)$$