Question

1. a certain culture of bacteria, triple every 25h. the initial count of shows 1000 bacteria present.

a) write an exponential function that models the given conditions.
po = 1000
for every 25 h
p(t) = 1000(3)^{t/25}
b) approximately how many bacteria will there be in 4 days?
4 days = 96 hours
p = 1000(3)^{96/25}
= 67943 bacteria
c) how many bacterial were there 3 days prior to the count?
p = 1000(3)^{-73/25} = 40 bacteria
d) when will there be 10000 bacteria?
10000 = 1000(3)^{t/25}
t = 25\frac{10}{3}
t = 52.4 hours

Explanation:

Step1: Write the exponential function

The general form of an exponential growth function is \(P(t)=P_0\cdot a^{\frac{t}{k}}\), where \(P_0\) is the initial amount, \(a\) is the growth factor, \(t\) is the time, and \(k\) is the time it takes for the quantity to change by the factor \(a\). Here, \(P_0 = 1000\), \(a = 3\) (since the bacteria triple), and \(k=25\) (triple every 25 hours). So the function is \(P(t)=1000\cdot3^{\frac{t}{25}}\).

Step2: Calculate the number of bacteria in 4 days

First, convert 4 days to hours. Since 1 day has 24 hours, 4 days have \(t = 4\times24=96\) hours. Substitute \(t = 96\) into the function \(P(t)=1000\cdot3^{\frac{t}{25}}\), we get \(P(96)=1000\cdot3^{\frac{96}{25}}\).

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Step3: Calculate the number of bacteria 3 days prior

Convert 3 days to hours, \(t=- 3\times24=-72\) (negative because it is prior). Substitute \(t = - 72\) into the function \(P(t)=1000\cdot3^{\frac{t}{25}}\), we get \(P(-72)=1000\cdot3^{\frac{-72}{25}}\).

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Step4: Find the time when there are 10000 bacteria

Set \(P(t)=10000\), so \(10000 = 1000\cdot3^{\frac{t}{25}}\). Divide both sides by 1000: \(10=3^{\frac{t}{25}}\). Take the natural logarithm of both sides: \(\ln(10)=\ln(3^{\frac{t}{25}})\). Using the property \(\ln(a^b)=b\ln(a)\), we have \(\ln(10)=\frac{t}{25}\ln(3)\). Then \(t = 25\times\frac{\ln(10)}{\ln(3)}\).

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Answer:

a) \(P(t)=1000\cdot3^{\frac{t}{25}}\)
b) \(67943\) bacteria
c) \(40\) bacteria
d) \(t\approx52.4\) hours