SOLUTION: A bird species in danger of extinction has a population that is decreasing exponentially \left(A = A_0 e^{kt}\right). 6 years ago the population was at 1710 and today only 950 o

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: A bird species in danger of extinction has a population that is decreasing exponentially \left(A = A_0 e^{kt}\right). 6 years ago the population was at 1710 and today only 950 o      Log On


   

Question 1077241: A bird species in danger of extinction has a population that is decreasing exponentially \left(A = A_0 e^{kt}\right).
6 years ago the population was at 1710 and today only 950 of the birds are alive. Once the population drops below 150, the situation will be irreversible. When will this happen?
Found 2 solutions by josgarithmetic, jorel1380:
Answer by josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
From 6 years ago to now:
950=1710%2Ae%5E%28kt%29, k will be found negative in value.
e%5E%28kt%29=950%2F1710
e%5E%28kt%29=95%2F171
e%5E%28kt%29=5%2F9
ln%28e%5E%28kt%29%29=ln%285%2F9%29
kt=ln%285%2F9%29, and t quantity used was t=6;
k=%281%2F6%29ln%285%2F9%29
k=-0.09796
MODEL: highlight_green%28A=A%5Bo%5De%5E%28-0.09796t%29%29

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If A=150 and A%5Bo%5D=950, what is t?
You do this on your own?

Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
From the question, we get:
950=1710*e^6k
So:
950/1710=e^6k
0.55555555555555555555555555555556=e^6k
ln 0.55555555555555555555555555555556= ln e^6k=6k*ln e
k=-0.09796444415035316803162185676981
Then:
150/950=e^t*-0.09796444415035316803162185676981
0.15789473684210526315789473684211=e^t*-0.09796444415035316803162185676981
ln 0.15789473684210526315789473684211=ln e^t*-0.09796444415035316803162185676981
t=18.84180 years from now. ☺☺☺☺