The initial size of a bacterial culture is 800. After 4h there are 7200 bactera. By starting with y=Ce^kt, show that the simplified equation is y=800(9)^(t/4).

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The initial size of a bacterial culture is 800. After 4h there are 7200 bactera. By starting with y=Ce^kt, show that the simplified equation is y=800(9)^(t/4).

Mathematics
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y = Ce^kt, where y is the bacteria number, t is the time and C and K are constants. at t = 0, y = 800 (initial bacteria culture, y = 800) so 800 = Ce^(0*k) or C = 800, since e^0 = 1
is it given that e is the base e or is it given that e = 9?
e is refering to the base of ln (I'm pretty sure since I'm doing the calculus log unit)

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So no I guess
okay, then at t = 4 , y = 7200 so 7200 = 800 e^4k or e^4k = 9. simplify that to get k
Ok so I got ln9/4 = k, put that back into the equation to have y=800e^((ln9/4)t). I looked at my notes and found the same thing where the number beside the ln would become the value for "e". So in this case, 9 would replace e giving me the right answer. I don't know how that works though. I'd have the right answer I just don't know how that works :( Thanks for the help too :)
e^4k = 9 taking ln on both sides 4k = ln 9 k = (ln9)/4 so y = 800 e^ t/4(ln 9)
since e^ln 9 = 9, y = 800*9^t/4
t/4 ln9 = ln 9^t/4 note that a ln b = ln b^a
Why exactly does e^ln9 = 9?

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