To find the number of bacteria at time \( t = 15 \) days given that the initial population is 10 and it triples every 5 days, we can use the equation you provided. However, let's clarify the equation you mentioned.
If the bacteria population triples every 5 days, the equation that describes the population over time (in days) can be expressed as:
\[ N(t) = N_0 \cdot 3^{t/5} \]
Where:
- \( N(t) \) is the number of bacteria at time \( t \),
- \( N_0 \) is the initial amount of bacteria (10 bacteria in this case),
- \( t \) is time in days,
- The factor \( 3^{t/5} \) accounts for the tripling every 5 days.
Starting with \( N_0 = 10 \), the equation becomes:
\[ N(t) = 10 \cdot 3^{t/5} \]
Now, we want to find \( N(15) \):
\[ N(15) = 10 \cdot 3^{15/5} \] \[ N(15) = 10 \cdot 3^{3} \] \[ N(15) = 10 \cdot 27 \] \[ N(15) = 270 \]
Thus, the number of bacteria after 15 days is 270.
Please note that the equation you mentioned \( a(t) = 10 \cdot t^{0.2t} \) doesn't accurately represent the situation of tripling every 5 days. Instead, the exponential growth equation used above is the correct form for this problem.
4 answers