Thursday, January 16, 2014

r = 5% Find the time necessary for $1000 to double when it is invested at a rate of r compounded (a) anually, (b) monthly, (c) daily, and (d) continuously

 
 
Formula for compounding n times per year: A=P(1+r/n)^(nt)
Formula for compounding continuously: A=Pe^(rt)
A=Final Amount
P=Initial Amount
r=rate of investment expressed as a percent
n=number of compoundings per year
t=time in years
 
a) r=5% n=1 (annually)
A=P(1+r/n)^(nt)
2000=1000(1+.05/1)^(1*t)
2=1.05^t
ln(2)=tln(1.05)
ln(2)/ln(1.05)=t
14.21=t
Final Answer: 14.21 years
 
b) r=5% n=12 (monthly)
A=P(1+r/n)^(nt)
2000=1000(1+.05/12)^(12*t)
2=(1.00416)^(12t)
ln(2)=12tln(1.00416)
ln(2)/[12ln(1.00416)]=t
13.89=t
Final Answer: 13.89 years
 
c) r=5% n=365 (daily)
A=P(1+r/n)^(nt)
2000=1000(1+.05/365)^(365*t)
2=(1.000136)^(365t)
ln(2)=365tln(1.00136)
ln(2)/[365ln(1.00136)]=t
13.86=t
Final Answer: 13.86 years
 
d)A=Pe^(rt)
2000=1000e^(.05*t)
2=e^(.05t)
ln(2)=.05tlne
ln(2)/[.05lne]=t
13.86=t
Final Answer: 13.86 years
 

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