SOLUTION: Jason invested $3000 at 9% interest rate compounded continuously. What is the minimum number of years Jason must wait to have at least $4000?
Group of answer choices
3 years. A
Question 1150532: Jason invested $3000 at 9% interest rate compounded continuously. What is the minimum number of years Jason must wait to have at least $4000?
Group of answer choices
3 years. A
2 years B
5years C
4years D Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! formula for continuous compounding is f = p * e ^ (r * t)
f is the future value
p is the present value
r is the interest rate per time period.
t is the number of time periods.
f = 4000
p = 3000
r = .09 per year
t will be in years.]
e is the scientific constant of 2.718281828...
formula becomes 4000 = 3000 * e ^ ( .09 * t).
divide both sides of this formula by 3000 to get 4000 / 3000 = e ^ (.09 * t).
take the natural log of both sides of this equation to get ln(4000/3000) = ln(e ^ (.09 * t)).
since ln(e ^ (.09 * t)) is equal to .09 * t * ln(e) and since ln(e) is equal to 1, the formula becomes ln(4000 / 3000) = .09 * t.
divide both sides of this formula by .09 to get ln(4000 / 3000) / .09 = t
solve for t to get t = 3.196467472.
that's greater than 3 years but less than 4 years.
the minimum number of years is 3 and the exact point in time when 4000 is earned is somewhere within the third year before getting to the fourth year.
f = p * e ^ (r * t) becomes f = 3000 * e ^ (.09 * 3.196467472) = 4000.
