If $1300 is invested in an account at the annual interest rate of 4.4% compounded continuously. how long will it take the investment to double in size. ( round to the nearest tenth of a year)

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# 2 Answers

Use the equation A = (P)(e^rt) in which P is the principle, r is the rate and t is the time.

Since we are looking for the time it takes for P to double we can rewrite the above as

2P = (P)(e^rt)

Divide each side by P

2 = e^rt

The rate is 0.044

2 = e^(0.044)t

Take the natural logarithm of each side

ln(2) = (0.044)t

t = ln(2)/0.044 = (0.69315)/(0.044) = 15.753 years = 15.8 years (rounded to nearest tenth of a year).

A=Pe^(rt) where A is the account value, P is the initial principal, r is the annual interest rate, and t is the time in years. A=2600, P = 1300, r=0.044, solve for t.

2600 = 1300e^(.044t)

2=e^(.044t)

ln 2 = .044t

t = ln 2 /.044≈15.8