SOLUTION: The population of fish in a lake is decreasing. There are currently 24,000 fish in the lake. The population is decreasing 6% each year. In how many years will there be 1/4 of the c

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: The population of fish in a lake is decreasing. There are currently 24,000 fish in the lake. The population is decreasing 6% each year. In how many years will there be 1/4 of the c      Log On


   

Question 585462: The population of fish in a lake is decreasing. There are currently 24,000 fish in the lake. The population is decreasing 6% each year. In how many years will there be 1/4 of the current number of the fish in the lake? Use formula T(n)=a(1+r)^n to solve problem?
-please someone help.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
T(n)=a(1+r)^n

T(n)=24000(1+(-0.06))^n

T(n)=24000(1-0.06)^n

T(n)=24000(0.94)^n

So the formula is T(n)=24000(0.94)^n


Now plug in T(n) = 24000/4 = 6000 and solve for n

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T(n)=24000(0.94)^n

6000=24000(0.94)^n

6000/24000=(0.94)^n

1/4=(0.94)^n

ln(1/4) = ln((0.94)^n)

ln(1/4) = n*ln(0.94)

ln(1/4)/ln(0.94) = n

n = ln(1/4)/ln(0.94)

n = 22.4046

So in roughly 22.4046 years, the population will be 1/4 of the current population.
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