SOLUTION: Please Help!!! The population P of fish, in thousands, in a certain pond at time t years is modeled by the function P(t)= 1/ (1+[(1/P*)-1]e^-rt), where P* is the population at the

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Please Help!!! The population P of fish, in thousands, in a certain pond at time t years is modeled by the function P(t)= 1/ (1+[(1/P*)-1]e^-rt), where P* is the population at the      Log On


   



Question 973855: Please Help!!!
The population P of fish, in thousands, in a certain pond at time t years is modeled by the function P(t)= 1/ (1+[(1/P*)-1]e^-rt), where P* is the population at the time t=0 and r is the growth rate of the population. If P(1)=5, which of the following is equivalent to r?
A) ln(5)/P*
B) ln(5)/ln(1/P*)
C) In(4P*/5P*)
D) ln(4/5P*)
E) ln(5(P*-1)/4P*)

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
P%28t%29=+1%2F%281%2B%281%2FP%5E%22%2A%22-1%29e%5E%28-rt%29%29 so P%281%29=+1%2F%281%2B%281%2FP%5E%22%2A%22-1%29e%5E%28-r%29%29=5 .
Let's solve for r :
+1%2F%281%2B%281%2FP%5E%22%2A%22-1%29e%5E%28-r%29%29=5 --> +1%2B%281%2FP%5E%22%2A%22-1%29e%5E%28-r%29=1%2F5 --> +%281%2FP%5E%22%2A%22-1%29e%5E%28-r%29=1%2F5-1 --> %281%2FP%5E%22%2A%22-1%29e%5E%28-r%29=-4%2F5
Multiplying both sides times -1 we get
%281-1%2FP%5E%22%2A%22%29e%5E%28-r%29=4%2F5 --> %28%28P%5E%22%2A%22-1%29%2FP%5E%22%2A%22%29e%5E%28-r%29=4%2F5 --> e%5E%28-r%29=%284%2F5%29%28P%5E%22%2A%22%2F%28P%5E%22%2A%22-1%29%29 --> e%5Er=%285%2F4%29%28%28P%5E%22%2A%22-1%29%2FP%5E%22%2A%22%29 --> e%5Er=5%28P%5E%22%2A%22-1%29%2F4P%5E%22%2A%22 --> r=ln%285%28P%5E%22%2A%22-1%29%2F4P%5E%22%2A%22%29