Logarithms: The population, of algae...how long to double

[wind]

Hi, I need help with this problem.

The population, P, of algae in a fish tank can be modelled by a function of the form \(\displaystyle \L\ P=P_{o}a^{t}\), where \(\displaystyle \L\ P_{o}\) is the initial population and t is time, in hours. At t=o, the algea population is measured to be 200. At t=3, the population is 800.

a) determine the values of \(\displaystyle \L\ P_{o}\) and a
b) How long will it take the population to double?
c) Determine the rate of change of the algae population after each time.
i) 1h
ii) 6h
d) Will this model hold true for all time?
---
a)

At t=o, P=200
At t=3, P=800

\(\displaystyle \L\ 200=P_{o}a^{0}\)
\(\displaystyle \L\ 200=P_{o}\)

\(\displaystyle \L\ 800=200a^{3}\)
\(\displaystyle \L\ 4=a^{3}\)
\(\displaystyle \L\sqrt[3]{4}=a\)

b) if population was 2

\(\displaystyle \L\ 4=2(4^{\frac{1}{3}})^{t}\)

\(\displaystyle \L\ ln4=ln 2(4^{\frac{1}{3}})^{t}\)

\(\displaystyle \L\ ln4=t ln 2(4^{\frac{1}{3}})\)

.... :?

Thanks

 

[arthur ohlsten]

P=Po a^t

at t=0 P=200
at t=3 P=800

a)
eq1) 200=Po a^0
eq1) 200=Po answer substitute

b)
when does P=2Po?

P=200 a^t but at t=3 P=800 substitute
eq2) 800=200 a^3
eq2) 400=a^3
eq2) a=[400^1/3]
eq2) a= 2[50]^1/3

P=200 [400^1/3]^t
400=200[400^1/3]^t
2={400^1/3}^t take log
log 2 = t log[400]^1/3
log2=[1/3]log 400 t
3log2=2log4 t
3 log 2 =4 log 2 t
t=3/4 hours
t=45 minutes answer

c)
P=Po a^t take derivative
dP/dt= Poa^t lna where Po=200 a=400^1/3
i. substitute t=1 hour
ii. substitute t=6 hours

d) no.
When the tank is saturated the process stops

Arthur