A bacteria culture starts with 800 bacteria and grows at a rate proportional to its size. After 2 hours there will be 1600 bacteria. (a) Express the population after t hours as a function of t. population

The population is modeled by \(y=Ce^{kt},\) where C is the initial population, e is the constant 2.718..., k is a relative growth factor, t is time, and y is the population after some time t. You're given that \(C=800\), and that \(y(2)=1600\) (i.e., \(1600=800e^{2k}\)). First thing to do is to solve the equation \(1600=800e^{2k}\) for k.

so take the ln of 2 then divide that answer by 2 yes?

the function then would equal 800e^(.3465735903t) right?

Yep, \(k=\frac{\ln2}{2}\), which makes your equation \[y=800e^{\frac{\ln2}{2}t}\] (Rounded is also fine.)

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