Calculus Question .... Please HELP

Let p(t) be the population of a colony of bacteria. At 11AM there are 60 bacteria and at 3PM there are 350. Assume exponential growth.
a) Find P(t) and simplify.
b) What is the size of the population at 4PM
c) When will the population reach 2000?

formula
p(t) = P*e^(k*t)
where e is approximately 2.718282 and k is the rate constant
P is the initial population

let at t = 0 at 11 AM
p(0) = 60 = P*e^(k*0)
hence
P = 60

at t = 4 for 3 PM
p(4) = 350 = 60*e^(k*4)

taking the ratio
350 = 60*e^(k*4)
taking the natural log of both sides
ln(350 / 60) = ln[e^(4k)]
ln(350 / 60) = (4k)

k = ln(35 / 6) / 4 or 0.440897148

hence

(a)
p(t) = 60*e^(0.440897148*t)

(b)
p(5) = 60*e^(0.440897148*5) = 544

(c)

2000 = 60*e^(0.440897148*t)

33.33 = e^(0.440897148*t)

ln(33.33) = 0.440897148t
t = ln(33.33) / 0.440897148 = 8 or by 7PM

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Hope this helps.