a culture started with 2,000 bacteria. After 4 hours, it grew to 2,200 bacteria. Predict how many bacteria will be present in 9 hours. round your answer to the nearest whole number... exponential growth formula

Guest Dec 10, 2014

#5**+10 **

Well, the starting value (N_{0} in my notation, A in yours) is 2000. This is the value at time t = 0.

At time t = 4 the value of N (my notation) or P (your notation) is 2200.

So at time t = 4 we have 2200 = 2000*e^{k*4}

Divide both sides by 2000

2200/2000 = e^{k*4}

or 1.1 = e^{k*4}

Now take logs of both sides:

ln(1.1) = ln(e^{k*4)}

or ln(1.1) = k*4 (because ln(e^{x}) = x)

Divide both sides by 4

ln(1.1)/4 = k

$${\mathtt{k}} = {\frac{{ln}{\left({\mathtt{1.1}}\right)}}{{\mathtt{4}}}} \Rightarrow {\mathtt{k}} = {\mathtt{0.023\: \!827\: \!544\: \!951\: \!081\: \!2}}$$

Now use this value of k to find N (my notation) or P (your notation) at time t = 9

P = 2000*e^{0.023827544951*9}

$${\mathtt{P}} = {\mathtt{2\,000}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left({\mathtt{0.023\: \!827\: \!544\: \!951}}{\mathtt{\,\times\,}}{\mathtt{9}}\right)} \Rightarrow {\mathtt{P}} = {\mathtt{2\,478.355\: \!127\: \!582\: \!545\: \!6}}$$

or P ≈ 2478

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Alan
Dec 10, 2014

#1**0 **

Have a look at the answer here: http://web2.0calc.com/questions/a-countrys-population-in-1995-was-56-million-in-2002-it-was-59-million-estimate-the-population-in-2016-using-exponential-growth-formula-roun and use the same method.

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The exponential growth equation is N=N_{0}e^{kt}

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Alan
Dec 10, 2014

#3**+5 **

Your expression is mathematically the same as mine Anonymous. You've just used different symbols!

Alan
Dec 10, 2014

#4**0 **

o okay but i still dont understand even after reading the other answer to the question

Guest Dec 10, 2014

#5**+10 **

Best Answer

Well, the starting value (N_{0} in my notation, A in yours) is 2000. This is the value at time t = 0.

At time t = 4 the value of N (my notation) or P (your notation) is 2200.

So at time t = 4 we have 2200 = 2000*e^{k*4}

Divide both sides by 2000

2200/2000 = e^{k*4}

or 1.1 = e^{k*4}

Now take logs of both sides:

ln(1.1) = ln(e^{k*4)}

or ln(1.1) = k*4 (because ln(e^{x}) = x)

Divide both sides by 4

ln(1.1)/4 = k

$${\mathtt{k}} = {\frac{{ln}{\left({\mathtt{1.1}}\right)}}{{\mathtt{4}}}} \Rightarrow {\mathtt{k}} = {\mathtt{0.023\: \!827\: \!544\: \!951\: \!081\: \!2}}$$

Now use this value of k to find N (my notation) or P (your notation) at time t = 9

P = 2000*e^{0.023827544951*9}

$${\mathtt{P}} = {\mathtt{2\,000}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left({\mathtt{0.023\: \!827\: \!544\: \!951}}{\mathtt{\,\times\,}}{\mathtt{9}}\right)} \Rightarrow {\mathtt{P}} = {\mathtt{2\,478.355\: \!127\: \!582\: \!545\: \!6}}$$

or P ≈ 2478

.

Alan
Dec 10, 2014