The number of bacteria in the early stages of a growing colony of bacteria can be approximated using the function: $$N(t) = 450 e^{0.065t}$$ where $t$ is the time, measured in hours, and $N(t)$ is the number of bacteria in the colony at time $t$ - Leaving Cert Mathematics - Question 9 - 2020

To find the time when $N(t) = 790$:

1. Set the equation:

   $790 = 450 e^{0.065t}$

2. Solve for $t$:

   $\ln(790) = \ln(450) + 0.065t$

3. Rearranging gives:

   $t = \frac{\ln(790) - \ln(450)}{0.065}$

   Calculating this gives: $t \approx 7.7$

4. Therefore, the time taken is approximately 7.7 hours.

To find the average number of bacteria:

1. Use the average formula:

   $\text{Average} = \frac{1}{12-3}\int_{3}^{12} N(t) dt$

2. Substituting $N(t)$:

   $\int_{3}^{12} 450 e^{0.065t} dt$

3. The integral evaluates to:

   $\left[ \frac{450}{0.065} e^{0.065t} \right]_{3}^{12}$

4. Calculate:

   $\frac{450}{0.065}(e^{0.78} - e^{0.195})$

5. Average is approximately 743.

To find the rate of change:

1. Compute the derivative:

   $N'(t) = 450 e^{0.065t} \cdot 0.065$

2. Evaluate at $t = 12$:

   $N'(12) = 450 e^{0.78} \cdot 0.065 \approx 63.8$

3. Thus, the rate of change is approximately 63.8 bacteria per hour.

To solve for $N'(t)$ at 90:

1. Set up the equation:

   $29.25 e^{0.065t} > 90$

2. Solve for $t$:

   $e^{0.065t} > \frac{90}{29.25} \implies 0.065t > \ln\left(\frac{90}{29.25}\right)$

3. This gives:

   $t > \frac{\ln\left(\frac{90}{29.25}\right)}{0.065} \approx 18$

4. Therefore, the least value of $k$ is 18.

Set the equations equal:

1. We have:

   $450 e^{0.065t} = 220 e^{0.17t}$

2. Simplifying gives:

   $\frac{450}{220} = e^{0.17t - 0.065t}$

3. Taking logarithms:

   $\ln\left(\frac{450}{220}\right) = (0.17 - 0.065)t$

4. Solve for $t$:

   $t = \frac{\ln\left(\frac{450}{220}\right)}{0.105} \approx 7$

5. Therefore, the time at which the number of bacteria in both colonies will be equal is approximately 7 hours.