Question

the change in a sample of bacteria, p, is represented by the function $p(t) = 80(3)^{\frac{t}{6}}$, where t is the number of hours. which of the following statements best describes the initial number of bacteria and the change in the number of bacteria over time? (1 point) \bigcirc the initial number of bacteria is 80. the number of bacteria triples 6 times per hour. \bigcirc the initial number of bacteria is 240. the number of bacteria triples 6 times per hour. \bigcirc the initial number of bacteria is 80. the number of bacteria triples every hour. \bigcirc the initial number of bacteria is 80. the number of bacteria triples every 6 hours.

Explanation:

Step1: Analyze the initial number

The general form of an exponential function is \( P(t)=a(b)^t \), where \( a \) is the initial amount. In the function \( P(t) = 80(3)^{\frac{t}{6}} \), when \( t = 0 \), \( P(0)=80(3)^0=80\times1 = 80 \). So the initial number of bacteria is 80.

Step2: Analyze the growth rate

Rewrite the exponent: \( (3)^{\frac{t}{6}}=3^{\frac{1}{6}\times t}=(3^{\frac{1}{6}})^t \)? No, wait, actually, for exponential growth, if we have \( P(t)=a(b)^{\frac{t}{k}} \), it means the quantity triples (when \( b = 3 \)) every \( k \) hours. Here, the exponent is \( \frac{t}{6} \), so when \( t \) increases by 6, the exponent increases by 1, so \( 3^{\frac{t + 6}{6}}=3^{\frac{t}{6}+1}=3^{\frac{t}{6}}\times3^1 \), which means the number of bacteria triples every 6 hours.

Answer:

The initial number of bacteria is 80. The number of bacteria triples every 6 hours.