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# 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

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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
+27223
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Well, the starting value (N0 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*ek*4

Divide both sides by 2000

2200/2000 = ek*4

or 1.1 = ek*4

Now take logs of both sides:

ln(1.1) = ln(ek*4)

or ln(1.1) = k*4    (because ln(ex) = 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*e0.023827544951*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
+27223
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Alan  Dec 10, 2014
#2
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no its P=Ae^kt

Guest Dec 10, 2014
#3
+27223
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Your expression is mathematically the same as mine Anonymous.  You've just used different symbols!

Alan  Dec 10, 2014
#4
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o okay but i still dont understand even after reading the other answer to the question

Guest Dec 10, 2014
#5
+27223
+10

Well, the starting value (N0 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*ek*4

Divide both sides by 2000

2200/2000 = ek*4

or 1.1 = ek*4

Now take logs of both sides:

ln(1.1) = ln(ek*4)

or ln(1.1) = k*4    (because ln(ex) = 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*e0.023827544951*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