To determine when the larvae will be transferred to a new environment, we first need to set up the inequality representing the exponential growth of the larvae.
Given:
- Initial length of the larva ([tex]\( L_0 \)[/tex]) = 3 mm
- Growth rate ([tex]\( r \)[/tex]) = 25% per day, which is equivalent to multiplying by 1.25 each day.
- Threshold length ([tex]\( L_t \)[/tex]) = 9.15 mm
- [tex]\( L(t) \)[/tex] represents the length of the larvae after [tex]\( t \)[/tex] days.
The growth of the larvae can be modeled with the exponential function:
[tex]\[ L(t) = L_0 \times (1.25)^t \][/tex]
We need to find the number of days [tex]\( t \)[/tex] such that the length exceeds 9.15 mm:
[tex]\[ 3 \times (1.25)^t > 9.15 \][/tex]
To solve for [tex]\( t \)[/tex], we isolate [tex]\( (1.25)^t \)[/tex]:
[tex]\[ (1.25)^t > \frac{9.15}{3} \][/tex]
[tex]\[ (1.25)^t > 3.05 \][/tex]
Next, we apply logarithms to solve for [tex]\( t \)[/tex]:
Taking the logarithm of both sides, we get:
[tex]\[ \log((1.25)^t) > \log(3.05) \][/tex]
Using the property of logarithms, [tex]\(\log(a^b) = b \cdot \log(a) \)[/tex]:
[tex]\[ t \cdot \log(1.25) > \log(3.05) \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t > \frac{\log(3.05)}{\log(1.25)} \][/tex]
Using the numerical solution we know:
[tex]\[ t > 4.99741795831278 \][/tex]
Therefore, we can conclude that [tex]\( t \)[/tex] must be greater than about 5 days.
Hence, the correct answer is:
D. [tex]\( t > 5 \)[/tex]
