# `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%...

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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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