SOLUTION: An initial population of 30 fish is introduced into a lake. This fish population grows according to a continuous exponential growth model. There are 69 fish in the lake after 12 ye
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-> SOLUTION: An initial population of 30 fish is introduced into a lake. This fish population grows according to a continuous exponential growth model. There are 69 fish in the lake after 12 ye
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Question 1189542: An initial population of 30 fish is introduced into a lake. This fish population grows according to a continuous exponential growth model. There are 69 fish in the lake after 12 years.
(a) Let âtâ be the time (in years) since the initial population is introduced, and let y be the number of fish at time t. Write a formula relating y to t.
Use exact expressions to fill in the missing parts of the formula. Do not use approximations.
(b) How many fish are there 14 years after the initial population is
introduced?
Do not round any intermediate computations, and round your
answer to the nearest whole number. Found 2 solutions by Theo, ikleyn:Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! the continuous exponential growth formula is:
f = p * e ^ (r * t)
f is the future value.
p is the present value.
e is the scientific constant e which is equal to 2.718281828.....
r is the interest rate per time period.
t is the number of time periods.
for this problem, the time periods are in years.
the initial population is 30.
12 years later, the population is 69.
f = 69
p = 30
t = 12
r equals what you want to find.
the formula becomes:
69 = 30 * e ^ (r * 12)
divide both sides of the equation by 30 to get:
69/30 = e ^ (r * 12
take the natural log of both sides of the equation to get:
ln(69/30) = ln(e ^ (r * 12)
since ln(e ^ x) = x * ln(e) and since ln(e) = 1, this becomes:
ln(69/30) = 12 * r
divive both sides of this equation by 12 to get:
ln(69/30) / 12 = r
solve for r to get:
r = .0694090936.
confirm by replacing r with that in the original equationand solving for f to get:
f = 30 * e ^ (.0694090936 * 12) = 69.
this confirms the value of r is correct.
to determine how many fish are there after 14 years, the equation becomes:
f = 30 * e ^ (.0694090936 * 14) = 79.27514835.
You can put this solution on YOUR website! .
An initial population of 30 fish is introduced into a lake. This fish population grows according to
a continuous exponential growth model.
There are 69 fish in the lake after 12 years.
(a) Let âtâ be the time (in years) since the initial population is introduced,
and let y be the number of fish at time t. Write a formula relating y to t.
Use exact expressions to fill in the missing parts of the formula. Do not use approximations.
(b) How many fish are there 14 years after the initial population is
introduced?
Do not round any intermediate computations, and round your
answer to the nearest whole number.
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Write an exponential function connecting y and t, as you read the problem
y = .
You use the factor 30, since 30 fish is the initial population (at t= 0).
The base "b" is unknown now; determine it from the given data at t= 12 years
69 = .
Divide both sides by 30
= , or 2.3 = .
Take logarithm base 10 from both sides
log(2.3) = 12*log(b)
log(b) = = 0.030144
b = = 1.071875.
Thus, your exponential function is fully determined: it is y = .
Check at t= 12: y = = 69 ! PRECISELY CORRECT !
Part (a) is completed.
For part (b), simply substitute t= 14 into the formula and calculate
y = = 79.2 = 79 fish (rounded).
Solved.
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If you want to see many other similar and different solved problems on population growth, look into the lesson
- Population growth problems
in this site.
