Question

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).

Answer 1

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

Answer 2

is it given that e is the base e or is it given that e = 9?

Answer 3

e is refering to the base of ln (I'm pretty sure since I'm doing the calculus log unit)

Answer 4

So no I guess

Answer 5

okay, then at t = 4 , y = 7200
so 7200 = 800 e^4k
or e^4k = 9. simplify that to get k

Answer 6

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 :)

Answer 7

e^4k = 9
taking ln on both sides 4k = ln 9
k = (ln9)/4
so y = 800 e^ t/4(ln 9)

Answer 8

since e^ln 9 = 9,
y = 800*9^t/4

Answer 9

t/4 ln9 = ln 9^t/4
note that a ln b = ln b^a

Answer 10

Why exactly does e^ln9 = 9?