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A strain of E. coli SC18del-recA718 is placed into a nutrient broth at 30° Celsius and allowed to grow. The data below are collected. Theory states that the number of bacteria in the petri dish will initially grow according to the law of uninhibited growth. The population is measured using an optical device in which the amount of light that passes through the petri dish is measured.

(a) Draw a scatter diagram treating time as the independent variable.

(b) Using a graphing utility, build an exponential model from the data.

(c) Express the function found in part (b) in the form N(t) = N0 e^kt.

(d) Graph the exponential function found in part (b) or (c) on the scatter diagram.

(e) Use the exponential function from part (b) or (c) to predict the population at x = 6 hours.

(f) Use the exponential function from part (b) or (c) to predict when the population will reach 2.1.

Step by step solution

01

Given information 

Given a table with time x and population y

Use this table to frame an exponential function

02

Part(a): Step 1 :Plot the table 

Plot all the points

03

Part(b)  step 1: 

A graphing utility fits the data in Table to an exponential model of the form y = ab^x using the Exponential regression option.

By using graphing utility

$y=0.03386{(1.947)}^x$

04

Part(c): Step 1 : 

$$\begin{gathered} Toexpressy=a{b}^{x}intheformN=N_0{e}^{kt} \\ a{b}^x=N_0{e}^{kt} \\ a=N_0{e}^{kt} \\ b=e^k \\ y=0.03386{(1.947)}^x{A}_0=a=0.03386 \\ b=e^k \\ 1.947=e^k \\ k=\ln (1.947)=0.66628 \\ Asaresult,N=0.033861e^{0.66628t} \end{gathered}$$

05

Part(d) : Step 1 : Graph the exponential function 

The graph of the exponential equation is

06

Part(e) : step 1: population 

We need to find the population at x= 6 hours

$$\begin{gathered} y=0.03386{(1.947)}^{x}substitute,x=6 \\ y=0.03386{(1.947)}^6 \\ y=1.84451 \end{gathered}$$

07

Part(f) : step 1: find out the time 

given y=2.1

$$\begin{gathered} y=0.03386{(1.947)}^{x}2.1=0.03386{(1.947)}^x \\ \frac{2.1}{0.03386}={(1.947)}^x \\ \ln \frac{2.1}{0.03386}=x\ln (1.947) \\ \frac{\ln \frac{2.1}{0.03386}}{\ln (1.947)}=x \\ x=6.19468 \end{gathered}$$

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