To determine the correct equation that can be solved for \( t \) when the number of cells reaches 100,000, we first need to analyze the pattern in the data provided:
- When \( t = 0 \), the number of cells is \( 8 \).
- When \( t = 1 \), the number of cells is \( 32 \).
- When \( t = 2 \), the number of cells is \( 128 \).
- When \( t = 3 \), the number of cells is \( 512 \).
The pattern shows that the number of cells doubles every hour. Specifically:
- From \( 8 \) to \( 32 \) (which is \( 8 \times 4 \))
- From \( 32 \) to \( 128 \) (which is \( 32 \times 4 \))
- From \( 128 \) to \( 512 \) (which is \( 128 \times 4 \))
This indicates that the number of cells can be described as:
\[ N(t) = 8 \times 4^t \]
where \( N(t) \) is the number of cells at time \( t \).
To find the time \( t \) when the number of cells reaches \( 100,000 \), set up the equation:
\[ 8 \times 4^t = 100,000 \]
Thus, the correct answer is:
D. \( 8 \times 4^t = 100,000 \)