SOLUTION: the frog population in a small pond grows exponentially. The current population is 85 frogs, and the relative growth rate is 18% per year. a. find function that models the popu

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Question 81409This question is from textbook Precalculus
: the frog population in a small pond grows exponentially. The current population is 85 frogs, and the relative growth rate is 18% per year.
a. find function that models the population after t years.
b. find the projected population after 3 years.
c. find the number of years required for the frog population to reach 600.
This question is from textbook Precalculus

Found 2 solutions by stanbon, SMM52091:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
the frog population in a small pond grows exponentially. The current population is 85 frogs, and the relative growth rate is 18% per year.
a. find function that models the population after t years.
A(t) = A(0)(1.18)^t
A(t) = 85(1.18)^t
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b. find the projected population after 3 years.
A(3) = 85(1.18)^3
A(3)= 139.65 frogs
Rounding down: 139 frogs
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c. find the number of years required for the frog population to reach 600.
600 = 85(1.18)^t
7.0588=1.18^t
Take the log of both sides to ge:
t=[log7.0588]/[log1.18]
t=11.81 years
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Cheers,
Stan H.

Answer by SMM52091(3) About Me  (Show Source):
You can put this solution on YOUR website!
Forumula for this type of problem would be P=er%5E%28t%29 where P is the new amount, e is the original, r is the rate, and t is the time. You can create that formula by simply thinking about what the question is asking. If one year goes by, then the new amount of frogs is 18% greater, or 85+%2A+118%25 OR 85+%2A+1.18. If two years go by, then you'd get 85+%2A+1.18 after year one, and then multiply that by 1.18 AGAIN for the second year, so technically 85+%2A+1.18+%2A+1.18 Also known as +85+%2A+1.18%5E2 So....
a) the formula as a function would be... f%28t%29=er%5Et
b) plug in to find out the population after 3 years (t)
f%283%29+=+%2885%29%281.18%5E3%29
f%283%29+=+%2885%29%281.643032%29
f%283%29+=+139.65772 Don't forget to round off, because you can't have .65772 of a frog, thats ridiculous. So 140 frogs after 3 years
c)plug in to find out the amount of f(t) = 600
600+=+%2885%29%281.18%5Et%29
7.058823529+=+1.18%5Et
log+7.058823529+=+t+log+1.18
%28log+7.058823529%29%2F%28log+1.18%29+=+t
11.80729861+=+t
It would take 11.80729861 years to reach 600 frogs.