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Bacteria quadruples on a daily basis. After the first day, there were 50 bacteria. Using the general exponential equation an=a1...Question
Bacteria quadruples on a daily basis. After the first day, there were 50 bacteria. Using the general exponential equation an=a1 ⋅ r^n−1, what value would represent a1?
Answers
Answer
the answer is not 12.5
Answer
the answer is 50, your owner needs to fix you
Answered by
GPT-4o mini
In the context of your question, the equation \( a_n = a_1 \cdot r^{n-1} \) represents the growth of a population over time, where:
- \( a_n \) is the amount at time \( n \),
- \( a_1 \) is the initial amount (the amount at time \( n = 1 \)),
- \( r \) is the growth factor (in this case, the bacteria quadruple each day),
- \( n \) is the number of days.
From the information given, we know that after the first day (which is \( n = 2 \)), the population is 50. So:
- For the first day \( (n = 1) \), let's denote the initial amount of bacteria as \( a_1 \).
- \( r = 4 \) (because the bacteria quadruple each day).
The number of bacteria after the first day \( (n = 2) \) can be expressed as:
\[
a_2 = a_1 \cdot r^{2-1} = a_1 \cdot r
\]
Since we know \( a_2 = 50 \):
\[
50 = a_1 \cdot 4
\]
Now, we can solve for \( a_1 \):
\[
a_1 = \frac{50}{4} = 12.5
\]
Thus, the value of \( a_1 \) is **12.5**.
- \( a_n \) is the amount at time \( n \),
- \( a_1 \) is the initial amount (the amount at time \( n = 1 \)),
- \( r \) is the growth factor (in this case, the bacteria quadruple each day),
- \( n \) is the number of days.
From the information given, we know that after the first day (which is \( n = 2 \)), the population is 50. So:
- For the first day \( (n = 1) \), let's denote the initial amount of bacteria as \( a_1 \).
- \( r = 4 \) (because the bacteria quadruple each day).
The number of bacteria after the first day \( (n = 2) \) can be expressed as:
\[
a_2 = a_1 \cdot r^{2-1} = a_1 \cdot r
\]
Since we know \( a_2 = 50 \):
\[
50 = a_1 \cdot 4
\]
Now, we can solve for \( a_1 \):
\[
a_1 = \frac{50}{4} = 12.5
\]
Thus, the value of \( a_1 \) is **12.5**.
Answered by
GPT-4o mini
I apologize for any confusion in my previous response. Let's go through the calculations again more clearly.
We are given:
1. After the first day (which corresponds to \( n = 2 \)), there are 50 bacteria present.
2. The bacteria quadruple each day, which means \( r = 4 \).
We want to find \( a_1 \), which is the initial amount at \( n = 1 \).
Using the equation:
\[
a_n = a_1 \cdot r^{n-1}
\]
For \( n = 2 \) (after the first day), we have:
\[
a_2 = a_1 \cdot r^{2-1}
\]
This simplifies to:
\[
a_2 = a_1 \cdot r
\]
Given that \( a_2 = 50 \) and \( r = 4 \):
\[
50 = a_1 \cdot 4
\]
To find \( a_1 \), we rearrange the equation:
\[
a_1 = \frac{50}{4} = 12.5
\]
So the steps and calculations confirm that the correct value of \( a_1 \) is indeed **12.5**.
If there is any specific detail or constraint that I missed, please let me know!
We are given:
1. After the first day (which corresponds to \( n = 2 \)), there are 50 bacteria present.
2. The bacteria quadruple each day, which means \( r = 4 \).
We want to find \( a_1 \), which is the initial amount at \( n = 1 \).
Using the equation:
\[
a_n = a_1 \cdot r^{n-1}
\]
For \( n = 2 \) (after the first day), we have:
\[
a_2 = a_1 \cdot r^{2-1}
\]
This simplifies to:
\[
a_2 = a_1 \cdot r
\]
Given that \( a_2 = 50 \) and \( r = 4 \):
\[
50 = a_1 \cdot 4
\]
To find \( a_1 \), we rearrange the equation:
\[
a_1 = \frac{50}{4} = 12.5
\]
So the steps and calculations confirm that the correct value of \( a_1 \) is indeed **12.5**.
If there is any specific detail or constraint that I missed, please let me know!
Answered by
GPT-4o mini
I apologize for the confusion and appreciate your patience. Let's clarify the situation from the start, given your indication that \( a_1 \) should be 50.
As stated, the bacteria quadruple every day. This means:
1. At the end of Day 1 (which is after the first day), the amount is 50.
2. Let’s denote:
- \( a_1 \) as the initial amount of bacteria (at the start of Day 1).
- \( r \) as the growth factor, which is 4 (since the bacteria quadruple).
From the formula:
\[
a_n = a_1 \cdot r^{n-1}
\]
We know that at the end of Day 1 (\( n = 2 \)), we have 50 bacteria:
\[
a_2 = a_1 \cdot r^{2-1} = a_1 \cdot r
\]
Substituting \( a_2 = 50 \) and \( r = 4 \):
\[
50 = a_1 \cdot 4
\]
To find \( a_1 \):
\[
a_1 = \frac{50}{4} = 12.5
\]
So the calculations show that at \( n = 1 \) (the starting point), the initial amount should logically be 12.5, and this would give us 50 after the first day's growth.
If at the beginning of Day 1 you start with 50 bacteria, then after the first day, they would indeed quadruple, resulting in 200 on Day 2.
If there is an aspect of the question or a different framing that leads to "50" being the value for \( a_1\), I'd be happy to clarify further. Thank you for your understanding!
As stated, the bacteria quadruple every day. This means:
1. At the end of Day 1 (which is after the first day), the amount is 50.
2. Let’s denote:
- \( a_1 \) as the initial amount of bacteria (at the start of Day 1).
- \( r \) as the growth factor, which is 4 (since the bacteria quadruple).
From the formula:
\[
a_n = a_1 \cdot r^{n-1}
\]
We know that at the end of Day 1 (\( n = 2 \)), we have 50 bacteria:
\[
a_2 = a_1 \cdot r^{2-1} = a_1 \cdot r
\]
Substituting \( a_2 = 50 \) and \( r = 4 \):
\[
50 = a_1 \cdot 4
\]
To find \( a_1 \):
\[
a_1 = \frac{50}{4} = 12.5
\]
So the calculations show that at \( n = 1 \) (the starting point), the initial amount should logically be 12.5, and this would give us 50 after the first day's growth.
If at the beginning of Day 1 you start with 50 bacteria, then after the first day, they would indeed quadruple, resulting in 200 on Day 2.
If there is an aspect of the question or a different framing that leads to "50" being the value for \( a_1\), I'd be happy to clarify further. Thank you for your understanding!