In the given situation, we know that the bacteria quadruples daily, which indicates an exponential growth pattern.
The equation you referred to, \( a_1 = a_1 \cdot r^{t-1} \), can be interpreted in the context of exponential growth as follows:
- \( a_1 \) is the initial amount of bacteria.
- \( r \) is the growth factor (in this case, since the bacteria quadruples daily, \( r = 4 \)).
- \( t \) is the time in days.
Since you mentioned that there were 50 bacteria after the first day, that means:
- After 1 day, the bacteria count is 50.
- Therefore, we can denote \( a_1 \) as the initial amount of bacteria at day 0.
To find \( a_1 \), we can reverse-engineer from the observation after the first day:
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Let \( a_1 \) be the number of bacteria at day 0 (which we want to find).
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After 1 day (day 1), the number of bacteria is:
\[ a_1 \cdot r^{1-1} = a_1 \cdot r^0 = a_1 \cdot 1 = a_1 \]
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On day 2, the number of bacteria would be:
\[ a_1 \cdot r^{2-1} = a_1 \cdot r = a_1 \cdot 4 \]
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Since you're mentioning that there were 50 bacteria after the first day, that means:
\[ a_1 \cdot 4 = 50 \]
From this, we can solve for \( a_1 \):
\[ a_1 = \frac{50}{4} = 12.5 \]
So, the value that would represent \( a_1 \) (the initial number of bacteria at day 0) would be 12.5.