Question
A scientist started witha sample of 8 cells. The sample increased as shown in the tableTime (Hours)Number of Cells081 322 1283 512Assume that the pattern in the table continues. Which equation can be solved for t, the time in hours when the number of cells will reach 100,000?(1 point)Responses4×8t=100,0004 times 8 to the t th power is equal to 100 comma 0004×t8=100,0004×t8=100,0008×4t=100,0008 times 4 to the t th power is equal to 100 comma 0008×t4=100,000
Answers
GPT-4o mini
To analyze the growth of the cell sample given in the table, we can see that the number of cells is doubling every hour.
Starting with 8 cells:
- At 0 hours: \(8\)
- At 1 hour: \(16\) (which is \(8 \times 2^1\) or \(8 \times 4^0\))
- At 2 hours: \(32\) (which is \(8 \times 2^2\) or \(8 \times 4^1\))
- At 3 hours: \(64\) (which is \(8 \times 2^3\) or \(8 \times 4^2\))
- At 4 hours: \(128\) (which is \(8 \times 2^4\) or \(8 \times 4^3\))
- At 5 hours: \(256\) (which is \(8 \times 2^5\) or \(8 \times 4^4\))
From this pattern, we observe that the number of cells can be represented by the equation:
\[
\text{Number of Cells} = 8 \times 2^t
\]
Where \(t\) is the time in hours.
Now to find when the number of cells will reach 100,000, we set up the equation:
\[
8 \times 2^t = 100,000
\]
To match this with the responses provided, we can rewrite \(8\) as \(4^1\) and express \(2\) as \(4^{1/2}\), leading us to:
\[
8 \times (4^{1/2})^{2t} = 100,000 \quad \text{which simplifies to}\quad 4^{1} \times 4^{t} = 100,000
\]
This gives us:
\[
4^{t+1} = 100,000
\]
However, the answer that closely relates to a pattern of doubling from \(4\) can also be presented logarithmically:
The correct equation from the options you've provided is:
\[
4 \times 8^t = 100,000
\]
Thus, the equation you need to solve for \(t\) is:
\[
8 \times 4^t = 100,000
\]
So, the best response from the provided options would be:
**8×4t=100,000** (8 times 4 to the t-th power is equal to 100,000).
Starting with 8 cells:
- At 0 hours: \(8\)
- At 1 hour: \(16\) (which is \(8 \times 2^1\) or \(8 \times 4^0\))
- At 2 hours: \(32\) (which is \(8 \times 2^2\) or \(8 \times 4^1\))
- At 3 hours: \(64\) (which is \(8 \times 2^3\) or \(8 \times 4^2\))
- At 4 hours: \(128\) (which is \(8 \times 2^4\) or \(8 \times 4^3\))
- At 5 hours: \(256\) (which is \(8 \times 2^5\) or \(8 \times 4^4\))
From this pattern, we observe that the number of cells can be represented by the equation:
\[
\text{Number of Cells} = 8 \times 2^t
\]
Where \(t\) is the time in hours.
Now to find when the number of cells will reach 100,000, we set up the equation:
\[
8 \times 2^t = 100,000
\]
To match this with the responses provided, we can rewrite \(8\) as \(4^1\) and express \(2\) as \(4^{1/2}\), leading us to:
\[
8 \times (4^{1/2})^{2t} = 100,000 \quad \text{which simplifies to}\quad 4^{1} \times 4^{t} = 100,000
\]
This gives us:
\[
4^{t+1} = 100,000
\]
However, the answer that closely relates to a pattern of doubling from \(4\) can also be presented logarithmically:
The correct equation from the options you've provided is:
\[
4 \times 8^t = 100,000
\]
Thus, the equation you need to solve for \(t\) is:
\[
8 \times 4^t = 100,000
\]
So, the best response from the provided options would be:
**8×4t=100,000** (8 times 4 to the t-th power is equal to 100,000).