Question

this exercise uses the population growth model.
a certain culture of the bacterium rhodobacter sphaeroides initially has 45 bacteria and is observed to double every 4 hours.
(a) find an exponential model $n(t) = n_02^{t/a}$ for the number of bacteria in the culture after $t$ hours.
$n(t) = \square$

(b) estimate the number of bacteria after 17 hours. (round your answer to the nearest whole number.)
\underline{\quad\quad\quad\quad} bacteria

(c) after how many hours will the bacteria count reach 1 million? (round your answer to one decimal place.)
$t = \underline{\quad\quad\quad\quad}$ hr

Explanation:

Part (a)

Step 1: Identify \( n_0 \) and \( a \)

The initial number of bacteria \( n_0 \) is given as 45. The bacteria double every 4 hours, so in the formula \( n(t)=n_02^{t/a} \), \( a \) is the time it takes to double, which is 4.

Step 2: Write the model

Substitute \( n_0 = 45 \) and \( a = 4 \) into the formula: \( n(t)=45\cdot2^{t/4} \)

Part (b)

Step 1: Substitute \( t = 17 \)

We use the model from part (a): \( n(17)=45\cdot2^{17/4} \)

Step 2: Calculate the exponent

First, calculate \( \frac{17}{4}=4.25 \)

Step 3: Calculate \( 2^{4.25} \)

\( 2^{4.25}=2^4\cdot2^{0.25}=16\cdot\sqrt[4]{2}\approx16\cdot1.1892\approx19.0272 \)

Step 4: Multiply by 45

\( n(17)=45\cdot19.0272\approx856.224 \), round to the nearest whole number: 856

Part (c)

Step 1: Set up the equation

We want to find \( t \) when \( n(t) = 1000000 \). So, \( 45\cdot2^{t/4}=1000000 \)

Step 2: Solve for \( 2^{t/4} \)

Divide both sides by 45: \( 2^{t/4}=\frac{1000000}{45}\approx22222.2222 \)

Step 3: Take the logarithm

Take the natural logarithm of both sides: \( \ln(2^{t/4})=\ln(22222.2222) \)
Using the logarithm power rule \( \ln(a^b)=b\ln(a) \), we get \( \frac{t}{4}\ln(2)=\ln(22222.2222) \)

Step 4: Solve for \( t \)

Multiply both sides by \( \frac{4}{\ln(2)} \): \( t = \frac{4\ln(22222.2222)}{\ln(2)} \)
Calculate \( \ln(22222.2222)\approx10.0107 \) and \( \ln(2)\approx0.6931 \)
\( t=\frac{4\cdot10.0107}{0.6931}\approx\frac{40.0428}{0.6931}\approx57.8 \)

Answer:

s:
(a) \( n(t)=\boldsymbol{45 \cdot 2^{t/4}} \)

(b) \(\boldsymbol{856}\) bacteria

(c) \(\boldsymbol{57.8}\) hr