To find the initial amount of bacteria, we can use the formula for exponential decay based on half-life:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
Where:
- \( N(t) \) is the remaining amount after time \( t \),
- \( N_0 \) is the initial amount,
- \( T_{1/2} \) is the half-life of the bacteria,
- \( t \) is the elapsed time.
Given:
- Half-life (\( T_{1/2} \)) = 48 days,
- Time elapsed (\( t \)) = 128 days,
- Remaining amount (\( N(t) \)) = 0.765 grams.
First, calculate the number of half-lives that have passed in 128 days:
\[ \text{Number of half-lives} = \frac{t}{T_{1/2}} = \frac{128}{48} \approx 2.6667 \]
Next, we can substitute into the decay formula:
\[ 0.765 = N_0 \left( \frac{1}{2} \right)^{2.6667} \]
Now calculate \(\left( \frac{1}{2} \right)^{2.6667}\):
\[ \left( \frac{1}{2} \right)^{2.6667} \approx 0.177 \]
Now we substitute back into the equation:
\[ 0.765 = N_0 \times 0.177 \]
To find \( N_0 \):
\[ N_0 = \frac{0.765}{0.177} \approx 4.32 \text{ grams} \]
Among the options provided, it seems there is no exact match, but the closest would be:
- 5.43 g (this is the nearest answer, but note that the calculations suggest the initial amount is approximately 4.32 grams).
So, the best available answer in your options is:
- 4.86 g (this is also close).
However, based on the calculations above, 4.32 grams is the most accurate calculated answer. If we were to rely purely on the options, the best approximation would be 4.86 g.
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