SOLUTION: The population of insects on an island has grown by 50% over an 5 month period. For simplicity, we will assume that the percentage growth per month has been constant, and compunde

Question 1025348: The population of insects on an island has grown by 50% over an 5 month period. For simplicity, we
will assume that the percentage growth per month has been constant, and compunded each month.
(After all, the increase in the population should depend on the current population, not what it was
a month, or 5 months, ago.)
(a) Without calculation, do you excpect the monthly percentage increase to be 10%, more than
10%, or less than 10%?
(b) Find the monthly percentage increase, both exactly, and also to 3 significant figures.
(c) By a process of guessing and checking, find out, to the nearest half month�
, how long it will
take for the population of insects to double. It may be useful to note that if a quantity doubles,
it has increased by 100%.

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
with compounding you would expect the monthly percent increase to be less than 10%.

it would be exactly 10% if it was straight line growth.

with compounding 10% a month for 5 months would be greater than 50% growth.

you would therefore expect the monthly growth rate to be less than 10% when you use compounding.

to find the monthly percent increase, use the following general formula.

f = p * (1 + r) ^ n

r is the interest rate per month.
n is the number of months.

p is equal to 1
f is equal to 1.5 because the number of flies increased by 50% in 5 months.

formula becomes 1.5 = 1 * (1 + r) ^ 5

simplify to get 1.5 = (1 + r) ^ 5.

raise each side of the equation to the power of 1/5 to get:

1.5 ^ (1/5) = ((1 + r) ^ 5) ^ (1/5

you will get 1.5 ^ (1/5) = 1 + r

subtract 1 from both sides of this equation to get:

1.5 ^ (1/5) - 1 = r

solve for r to get:

r = .0844717712

round to 3 significant digits to get r = .0845.

you want to know how long it will take for the population of insects to double.

since f = p * (1 + r) ^ n, and since:

f = 2
p = 1
1 + r = 1.0844717712, then:

formula becomes 2 = 1.0844717712 ^ n

you want to solve for n.

you can use logs or you can do guess and check.

we'll do guess and check, since that's what they asked you to do.

you know n is going to be greater than 5, because f = 1.5 when n = 5.

try n = 10.

store 1.0844717712 into memory of your calculator to make subsequent calculations easy.

1.0844717712 ^ 10 = 2.25

make n = 8.

you get 1.0844717712 ^ 8 = 1.91.....

make n = 9.

you get 1.0844717712 ^ 9 = 2.074.....

the number of insects will double between the 8th month and the 9th month.

you can try n = 8.5 to get 1.0844717712 ^ 8.5 = 1.992....

you can try n = 8.6 to get 1.0844717712 ^ 8.6 = 2.008.

the number of insects will double sometime between the 8.5th and 8.6th month.

to the nearest half month, it would take 8.5 months for the population to double.

if i were to solve exactly, i would use logs.

i would start with 2 = 1.0844717712 ^ n

i would take the log of both sides of the equation to get:

log(2) = log(1.0844717712 ^ n)

by the laws of logarithms, this would becomes log(2) = n * log(1.0844717712).

i would solve for n to get n = log(2) / log(1.0844717712) = 8.547556457 months.

round that to the nearest 10th and you get 8.5 months.

the guess and check method and the log method agree with each other.

if you didn't take logs yet, don't worry about it.

you will, eventually.

a note on significant figures, which i determined to mean significant digits.

that's different than rounding to s set number of decimal places.

.0844717712 rounded to 3 decimal places would give you .084

1.0844717712 round to 3 significant digits would give you 1.0845.

this is based on the following reference which discusses significant digits.

http://www.purplemath.com/modules/rounding2.htm

the bad news is that it can get a little complex to figure it out.

the good news is that you won't have to do it that often.

most of the time you will just round to the number of decimal places desired.

nearest tenth, nearest one hundredth, etc.

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