SOLUTION: Suppose $2,900 is invested in an account at an annual interest rate of 3.5% compounded continuously. How long (to the nearest tenth of a year) will it take the investment to double

Question 1136388: Suppose $2,900 is invested in an account at an annual interest rate of 3.5% compounded continuously. How long (to the nearest tenth of a year) will it take the investment to double in size?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the continuous compounding formula is f = p * e ^ (r * n)

f if the future value
p is the present value
e is the scientific constant of 2.718281828.....
r is the rate per time period.
n is the number of time periods.

in your problem:

p = 2900
f = 2 * 2900
r = 3.5% / 100 = .035 per year.
n = the number of years you want to find.

the formula becomes:

2 * 2900 = 2900 * e ^ (.035 * n)

simplify this to get:

5800 = 2900 * e ^ (.035 * n)

divide both sides of this equation by 2900 to get:

2 = e ^ (.035 * n)

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

ln(2) = ln(e ^ (.035 * n)

since ln(a^b) = b*ln(a), this equation becomes:

ln(2) = .035 * n * ln(e).

since ln(e) = 1, this equation becomes:

ln(2) = .035 * n

divide both sides of this equation by .035 to get:

ln(2) / .035 = n

solve for n to get:

n = ln(2) / .035 = 19.80420516.

confirm by replacing n in the original equation of 5800 = 2900 * e ^ (.035 * n) to get:

5800 = 2900 * e ^ (.035 * 19.80420516) which becomes:

5800 = 5800

this confirms the solution is correct.

the solution is that investment will double in 19.8 years rounded to the nearest tenth of a year.

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