Question

The number of bacteria triples every 5 days if the bacteria population starts with 10 bacteria then the equation a of t equals 10 of t 0.2 t can represent the number of bacteria found the value of a of 15

Answer 1

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.