 Sallie Banks

2021-12-30

Consider a bacterial population that grows according to the function $f\left(t\right)=500{e}^{0.05t}$ measured in minutes. After 4 hours, how many germs are present in the population? When will the population of bacteria reach 100 million? Jack Maxson

$f\left(t\right)=500x{c}^{0.05t}$
PSKt is in minutes
After 4 hours how many bacteria present
$4\text{hours}\to 4×60\to 240\text{minutes}$
$f\left(240\right)=500×e$
$=500×{e}^{12}$
$f\left(240\right)=81,317,395$ bacteria
When will be the population of bacteria each 100 million
$f\left(t\right)=500×\left({e}^{0.05t}\right)$
$\frac{f\left(t\right)}{500}={e}^{0.05×t}$
$\mathrm{ln}\left(\frac{f\left(t\right)}{500}\right)=0.05×t$
$t=\frac{1}{0.05}×\mathrm{ln}\left(\frac{f\left(t\right)}{500}\right)$
$=\frac{1}{0.05}×\mathrm{ln}\left(\frac{100,000,000}{500}\right)$
$t=244\text{minutes}12\text{second}$ Vasquez

At the starting point t = 0, there are $$f(0) = 500 e^{0.05*0} = 500$$ bacteria.
We want to find the nbr of bacteria at t = 4h. As an hour has 60 minutes, we have to calculate $$f(4 \cdot 60) = f(240) = 500 \cdot e^{0.05 \cdot 240} = 500 \cdot e^{12} = 81.377$$ Mio
When does it reach 100 Mio? It must be longer than 4h, as after 4h it is “only” 81 Mio.
$$f(t) = 500 \cdot e^{0.05t} = 100000000$$; divide both sides by 500
$$e^{0.05t} = 200000$$; take natural logarithm on both sides
$$0.05t = \ln(200000) = 12.20607$$; divide both sides by 0.05
$$t = 244.1214 = 244$$ minutes and 7.287 seconds.
It took 4h to grow to 81 mio and only 4 minutes and 7 seconds later it’s already 100 mio.

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