Question

Assume a general exponential growth/decay model. A bacteria culture starts with 500 bacteria and doubles every 3.5 hours. Find an expression for the number of bacteria after t hours.

Answer 1

550e^35t
500=Qoe^(ln2)/(3.5)(t)
500e^(ln2)/(3.5)(t)
2e^3.5t

Answer 2

once simple way to do it is to write
\[500\times 2^{\frac{t}{3.5}}\] but i guess that is not an option

Answer 3

you want to solve for \(r\) in the formula
\[500e^{rt}\] and you know if \(t=3.5\) then you double to \(1000\)
set
\[1000=500e^{3.5r}\] and solve for \(r\) via
\[2=e^{3.5r}\]
\[\ln(2)=3.5r\]
\[r=\frac{\ln(2)}{3.5}\]

Answer 4

therefore you formula is the same as choice C
it is always C