Question 1140192
population growth or decay uses the continuous compounding formula.


that formula is f = p * e ^ (r * n)


f is the future value
p is the present value
e is the scientific constant of 2.718281828.....
n is the number of time periods


these are names i assigned to the variables.
other people might assign different names, but the formula is the same.


you are given that there are 1000 mosquitoes initially and there are 1400 after 1 day.


the formula becomes 1400 = 1000 * e ^ (r * 1)


divide both sides of the formula by 1000 to get:


1400 / 1000 = e ^ (r * 1)


take the natural log of both sides of the formula to get:


ln(1400 / 1000) = ln(e ^ (r * 1))


since ln(e ^ (r * 1)) is equal to r+ * 1 * ln(e) and since ln(e) is equal to 1, then the formula becomes:


ln(1400 / 1000) = r.


solve for r to get r = .3364722366


replace r in the original equation to confirm the solution is good.


1400 = 1000 * e ^ (r * 1) becomes 1400 = 1000 * e ^ (.3364722366 * 1) which becomes 1400 = 1400, confirming the solution is correct.


after 3 days, the size of the colony will be based on f = 1000 * e ^ (.3364722366 * 3) which becomes f = 2744


to find out how long until there are 70,000 mosquitoes, the formula becomes 70,000 = 1000 * e ^ (.3364722366 * n)


divide both sides of the equation by 1000 to get:


70,000 / 1000 = e ^ (.3364722366 * n)


take the natural log of both sides of the equation to get:


ln(70,000 / 1000) = ln(e ^ (.3364722366 * n))


since ln(e ^ (.3364722366 * n)) = .3364722366 * n * ln(e) and since ln(e) = 1, the equation becomes:


ln(70,000 / 1000) = .3364722366 * n


divide both sides of this equation by .3364722366 to get:


ln(70,000 / 1000) / .3364722366 = n


solve for n  to get n = 12.62658484 days.


replace n in the original equation to confirm this is true.


70,000 = 1000 * e ^ (.3364722366 * n) becomes 70,000 = 1000 * e ^ (.3364722366 * 12.62658484) which becomes 70,000 = 70,000, confirming the solution is correct.