Answer :
To solve this problem, we need to find the number of days after which both colonies of bacteria will contain the same number of cells. Let's break down the behavior of each colony:
### Colony 1
- Initial number of cells: 32
- Doubling every day
So, the number of cells in Colony 1 after [tex]\( n \)[/tex] days is given by:
[tex]\[ \text{Cells in Colony 1} = 32 \times 2^n \][/tex]
### Colony 2
- Initial number of cells: 1024
- Doubling every 2 days
So, the number of cells in Colony 2 after [tex]\( n \)[/tex] days is given by:
[tex]\[ \text{Cells in Colony 2} = 1024 \times 2^{\left\lfloor \frac{n}{2} \right\rfloor} \][/tex]
where [tex]\( \left\lfloor \cdot \right\rfloor \)[/tex] denotes the floor function, which rounds down to the nearest whole number.
### Finding the Equilibrium Point
To find the number of days [tex]\( n \)[/tex] when the two colonies have the same number of cells, we need to solve for [tex]\( n \)[/tex] in:
[tex]\[ 32 \times 2^n = 1024 \times 2^{\left\lfloor \frac{n}{2} \right\rfloor} \][/tex]
Let's go through the steps to solve this:
1. Initial Population Growth:
- Day 0:
- Colony 1: 32 cells
- Colony 2: 1024 cells
- Day 1:
- Colony 1: [tex]\( 32 \times 2 = 64 \)[/tex] cells
- Colony 2: 1024 cells (since it doubles every 2 days)
- Day 2:
- Colony 1: [tex]\( 64 \times 2 = 128 \)[/tex] cells
- Colony 2: [tex]\( 1024 \times 2 = 2048 \)[/tex] cells (doubles as it is the end of 2 days)
- Day 3:
- Colony 1: [tex]\( 128 \times 2 = 256 \)[/tex] cells
- Colony 2: 2048 cells (next doubling on day 4)
Following this pattern, we keep calculating until the numbers match.
2. Continue the Calculation:
- Day 4:
- Colony 1: [tex]\( 256 \times 2 = 512 \)[/tex] cells
- Colony 2: [tex]\( 2048 \times 2 = 4096 \)[/tex] cells (as it's the end of 4 days)
- Day 5:
- Colony 1: [tex]\( 512 \times 2 = 1024 \)[/tex] cells
- Colony 2: 4096 cells
- Day 6:
- Colony 1: [tex]\( 1024 \times 2 = 2048 \)[/tex] cells
- Colony 2: [tex]\( 4096 \times 2 = 8192 \)[/tex] cells (doubling on day 6)
- Day 7:
- Colony 1: [tex]\( 2048 \times 2 = 4096 \)[/tex] cells
- Colony 2: 8192 cells
- Day 8:
- Colony 1: [tex]\( 4096 \times 2 = 8192 \)[/tex] cells
- Colony 2: [tex]\( 8192 \times 2 = 16384 \)[/tex] cells (doubling on day 8)
- Day 9:
- Colony 1: [tex]\( 8192 \times 2 = 16384 \)[/tex] cells
- Colony 2: 16384 cells
Finally, we see that after 9 days, both colonies will have the same number of cells: 16384.
So, the two colonies will have the same number of cells after:
9 days
### Answer
The correct answer is not among the options provided (a, b, c, or d). The actual number of days required for the two colonies to have the same number of cells is 9 days.
### Colony 1
- Initial number of cells: 32
- Doubling every day
So, the number of cells in Colony 1 after [tex]\( n \)[/tex] days is given by:
[tex]\[ \text{Cells in Colony 1} = 32 \times 2^n \][/tex]
### Colony 2
- Initial number of cells: 1024
- Doubling every 2 days
So, the number of cells in Colony 2 after [tex]\( n \)[/tex] days is given by:
[tex]\[ \text{Cells in Colony 2} = 1024 \times 2^{\left\lfloor \frac{n}{2} \right\rfloor} \][/tex]
where [tex]\( \left\lfloor \cdot \right\rfloor \)[/tex] denotes the floor function, which rounds down to the nearest whole number.
### Finding the Equilibrium Point
To find the number of days [tex]\( n \)[/tex] when the two colonies have the same number of cells, we need to solve for [tex]\( n \)[/tex] in:
[tex]\[ 32 \times 2^n = 1024 \times 2^{\left\lfloor \frac{n}{2} \right\rfloor} \][/tex]
Let's go through the steps to solve this:
1. Initial Population Growth:
- Day 0:
- Colony 1: 32 cells
- Colony 2: 1024 cells
- Day 1:
- Colony 1: [tex]\( 32 \times 2 = 64 \)[/tex] cells
- Colony 2: 1024 cells (since it doubles every 2 days)
- Day 2:
- Colony 1: [tex]\( 64 \times 2 = 128 \)[/tex] cells
- Colony 2: [tex]\( 1024 \times 2 = 2048 \)[/tex] cells (doubles as it is the end of 2 days)
- Day 3:
- Colony 1: [tex]\( 128 \times 2 = 256 \)[/tex] cells
- Colony 2: 2048 cells (next doubling on day 4)
Following this pattern, we keep calculating until the numbers match.
2. Continue the Calculation:
- Day 4:
- Colony 1: [tex]\( 256 \times 2 = 512 \)[/tex] cells
- Colony 2: [tex]\( 2048 \times 2 = 4096 \)[/tex] cells (as it's the end of 4 days)
- Day 5:
- Colony 1: [tex]\( 512 \times 2 = 1024 \)[/tex] cells
- Colony 2: 4096 cells
- Day 6:
- Colony 1: [tex]\( 1024 \times 2 = 2048 \)[/tex] cells
- Colony 2: [tex]\( 4096 \times 2 = 8192 \)[/tex] cells (doubling on day 6)
- Day 7:
- Colony 1: [tex]\( 2048 \times 2 = 4096 \)[/tex] cells
- Colony 2: 8192 cells
- Day 8:
- Colony 1: [tex]\( 4096 \times 2 = 8192 \)[/tex] cells
- Colony 2: [tex]\( 8192 \times 2 = 16384 \)[/tex] cells (doubling on day 8)
- Day 9:
- Colony 1: [tex]\( 8192 \times 2 = 16384 \)[/tex] cells
- Colony 2: 16384 cells
Finally, we see that after 9 days, both colonies will have the same number of cells: 16384.
So, the two colonies will have the same number of cells after:
9 days
### Answer
The correct answer is not among the options provided (a, b, c, or d). The actual number of days required for the two colonies to have the same number of cells is 9 days.
