Problema Solution

Determine the time necessary for P dollars to double when it is invested at interest rate r compounded annually, monthly, daily, and continuously. (Round your answers to two decimal places.) The rate is 10%

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Let


P = is the principal


r = 0.10 the annual interest rate of 10%


t = the time in years


A = 2P future value


- Annually:


A = P(1 + i)^n


m = 1 is the number of compounding periods per year


n = m*t = t is the number of compounding periods


i = r/m = 0.10/1 = 0.10 is the interest rate per period


2P = P(1 + 0.10)^t


1.10^t = 2

........


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........

t = 7.27 years



- Monthly:


A = P(1 + i)^n


m = 12 is the number of compounding periods per year


n = m*t = 12t is the number of compounding periods


i = r/m = 0.10/12 = 0.1/12 is the interest rate per period


2P = P(1 + 0.1/12)^(12t)


(1 + 0.1/12)^(12t) = 2


12t*log(1.00833) = log(2)

........


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........

t = 6.96 years



(d) Daily:


A = P(1 + i)^n


m = 365 is the number of compounding periods per year


n = m*t = 365*t is the number of compounding periods


i = r/m = 0.10/365 = 0.1/365 is the interest rate per period


2P = P(1 + 0.1/365)^(365t)


(1 + 0.1/365)^(365t) = 2


365t*log(1.00027 = log(2)

........


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........

t = 7.03 years


(e) Continuously:


A = P*e^(r*t), where e = 2.71828 is Napier's constant


2P = P*e^(0.10*t)


e^(0.10*t) = 2


0.10t = ln(2)

........


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........

t = 6.93 years