1. **State the problem:** A mutant salmonella bacteria doubles every 15 minutes. Initially, there is 1 bacteria. We want to find how many bacteria there are after 1 hour. 2. **Understand the doubling time:** Since the bacteria doubles every 15 minutes, in 1 hour (which is 60 minutes), the bacteria doubles $$\frac{60}{15} = 4$$ times. 3. **Formula for exponential growth:** The number of bacteria after $n$ doubling periods is given by $$N = N_0 \times 2^n$$ where $N_0$ is the initial number of bacteria and $n$ is the number of doubling periods. 4. **Calculate bacteria after 1 hour:** Here, $N_0 = 1$ and $n = 4$, so $$N = 1 \times 2^4 = 2^4 = 16$$ 5. **Write the recursive equation:** Let $B(t)$ be the number of bacteria at time $t$ hours. Since the bacteria doubles every 15 minutes, it doubles 4 times per hour, so the recursion for each hour is $$B(t) = B\left(t - \frac{1}{4}\right) \times 2$$ with initial condition $$B(0) = 1$$ 6. **Confirm recursion for 2 hours:** To find $B(2)$, we apply the recursion 8 times (since 2 hours = 8 intervals of 15 minutes): - $B(0) = 1$ - $B(\frac{1}{4}) = 2 \times B(0) = 2 \times 1 = 2$ - $B(\frac{1}{2}) = 2 \times B(\frac{1}{4}) = 2 \times 2 = 4$ - $B(\frac{3}{4}) = 2 \times B(\frac{1}{2}) = 2 \times 4 = 8$ - $B(1) = 2 \times B(\frac{3}{4}) = 2 \times 8 = 16$ - $B(\frac{5}{4}) = 2 \times B(1) = 2 \times 16 = 32$ - $B(\frac{3}{2}) = 2 \times B(\frac{5}{4}) = 2 \times 32 = 64$ - $B(\frac{7}{4}) = 2 \times B(\frac{3}{2}) = 2 \times 64 = 128$ - $B(2) = 2 \times B(\frac{7}{4}) = 2 \times 128 = 256$ So, after 2 hours, there are 256 bacteria. **Final answers:** - a) Number of bacteria after 1 hour: $16$ - b) Recursive equation: $$B(t) = 2 \times B\left(t - \frac{1}{4}\right), \quad B(0) = 1$$ - Number of bacteria after 2 hours using recursion: $256$