The number of eggs that Farmer Jones has in his chicken crop will grow exponentially as Farmer jones buys more chickens to increase Production. The number of eggs Farmer jones has in the coop can be modeled by the equation `y=3^x` beginning on Day 1, where x is given by x=1, and y is the number of eggs currently in the coop. IF the coop can support only 4,000 eggs ,and Farmer Jones empties the coop every day, on which day will the chickens produce too many eggs for the coop to support?

Options

A. Day 6
B. Day 7
C. Day 8
D. Day 9

Answer

To solve the problem, we need to determine on which day the number of eggs produced by Farmer Jones's chickens exceeds the capacity of the coop, which can hold a maximum of 4,000 eggs. The number of eggs produced can be modeled by the equation \( y = 3^x \), where \( x \) represents the number of days.

### Step-by-Step Solution:

1. **Understand the equation**: The number of eggs in the coop on day \( x \) is given by the equation \( y = 3^x \).

2. **Calculate the number of eggs for each day**:
- For **Day 1** (\( x = 1 \)):
\[ y = 3^1 = 3 \]
- For **Day 2** (\( x = 2 \)):
\[ y = 3^2 = 9 \]
- For **Day 3** (\( x = 3 \)):
\[ y = 3^3 = 27 \]
- For **Day 4** (\( x = 4 \)):
\[ y = 3^4 = 81 \]
- For **Day 5** (\( x = 5 \)):
\[ y = 3^5 = 243 \]
- For **Day 6** (\( x = 6 \)):
\[ y = 3^6 = 729 \]
- For **Day 7** (\( x = 7 \)):
\[ y = 3^7 = 2187 \]
- For **Day 8** (\( x = 8 \)):
\[ y = 3^8 = 6561 \]

3. **Compare with the coop capacity**: The coop can support a maximum of 4,000 eggs. We need to find the first day where \( y \) exceeds 4,000.
- On **Day 6**, \( y = 729 \) (less than 4,000)
- On **Day 7**, \( y = 2187 \) (less than 4,000)
- On **Day 8**, \( y = 6561 \) (greater than 4,000)

4. **Conclusion**: The chickens produce too many eggs for the coop to support on **Day 8**.

### Final Answer:
The answer is **Day 8**.