1. **State the problem:** We start with 550 bacteria that double every 30 hours. We want to find the time $t$ when the number of bacteria reaches 810. 2. **Formula used:** The growth of bacteria doubling every $T$ hours is modeled by the exponential growth formula: $$ N(t) = N_0 \times 2^{\frac{t}{T}} $$ where $N_0$ is the initial amount, $T$ is the doubling time, and $t$ is the time elapsed. 3. **Plug in known values:** $$ 810 = 550 \times 2^{\frac{t}{30}} $$ 4. **Isolate the exponential term:** $$ \frac{810}{550} = 2^{\frac{t}{30}} $$ 5. **Simplify the fraction:** $$ \frac{\cancel{810}}{\cancel{550}} = \frac{81}{55} $$ 6. **Take the logarithm base 2 of both sides:** $$ \log_2\left(\frac{81}{55}\right) = \frac{t}{30} $$ 7. **Solve for $t$:** $$ t = 30 \times \log_2\left(\frac{81}{55}\right) $$ 8. **Calculate the logarithm:** Using change of base formula, $$ \log_2\left(\frac{81}{55}\right) = \frac{\ln\left(\frac{81}{55}\right)}{\ln(2)} $$ 9. **Evaluate numerically:** $$ \ln\left(\frac{81}{55}\right) \approx \ln(1.4727) \approx 0.387 $$ $$ \ln(2) \approx 0.693 $$ 10. **Calculate $t$:** $$ t = 30 \times \frac{0.387}{0.693} \approx 30 \times 0.558 = 16.7 $$ **Final answer:** It will take approximately **16.7 hours** for the bacteria to grow from 550 to 810 under the given conditions.