The number N of bacteria in a culture is given by N = 200e^{kt} . If N=300 when t = 4 \enspace hours , find k and then determine approximately how long it will take for the number of bacteria
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find k, we make it the subject of the formula, that is, {eq}300 = 200 e^{4k} \text{ so that } 1.5 = e^{4k} {/eq} (dividing both sides by 200).
Next, we take the logarithm of both sides of the equation to give {eq}\ln{e^{4k}} = \ln{1.5} \text{ so that } 4k = \ln{1.5} \text{ and } k = \dfrac{\ln{1.5}}{4} = 0.1014 {/eq} (to 4 decimal places).
How long will it take for the bacteria to triple in size? Here, we will make time the subject of the formula. Thus, {eq}N = 200 e^{0.1014t} \text{ and } 900 = 200 e^{0.1014t} {/eq} (triple the initial value of N). This gives us {eq}4.5 = e^{0.1014t} \text{ and } t = \dfrac{\ln{4.5}}{0.1014} = 14.833 {/eq} (to 3 decimal places)
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Next, we take the logarithm of both sides of the equation to give {eq}\ln{e^{4k}} = \ln{1.5} \text{ so that } 4k = \ln{1.5} \text{ and } k = \dfrac{\ln{1.5}}{4} = 0.1014 {/eq} (to 4 decimal places).
How long will it take for the bacteria to triple in size? Here, we will make time the subject of the formula. Thus, {eq}N = 200 e^{0.1014t} \text{ and } 900 = 200 e^{0.1014t} {/eq} (triple the initial value of N). This gives us {eq}4.5 = e^{0.1014t} \text{ and } t = \dfrac{\ln{4.5}}{0.1014} = 14.833 {/eq} (to 3 decimal places)
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