# `r = 7%` 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 decimal

n=number of compoundings per year

t= time in years

a) r=7%  n=1 (annually)

`A=P(1+r/n)^(nt)`

`2000=1000(1+.07/1)^(1*t)`

`2=1.07^t`

`ln(2)=tln(1.07)`

`ln(2)/ln(1.07)=t`

`10.24=t`

Final answer: 10.24 years

b) r=7% n=12...

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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 decimal

n=number of compoundings per year

t= time in years

a) r=7%  n=1 (annually)

`A=P(1+r/n)^(nt)`

`2000=1000(1+.07/1)^(1*t)`

`2=1.07^t`

`ln(2)=tln(1.07)`

`ln(2)/ln(1.07)=t`

`10.24=t`

Final answer: 10.24 years

b) r=7% n=12 (monthly)

`A=P(1+r/n)^(nt)`

`2000=1000(1+.07/12)^(12*t)`

`2=1.0058^(12t)`

`ln(2)=12tln(1.0058)`

`ln(2)/[12ln(1.0058)]=t`

`9.93=t`

Final Answer: 9.93 years

c) r=7%  t=365 (daily)

`A=P(1+r/n)^(nt)`

`2000=1000(1+.07/365)^(365*t)`

`2=(1.00019)^(365t)`

`ln(2)=365tln(1.00019)`

`ln(2)/[365ln(1.00019)]=t`

`9.90=t`

Final answer: 9.90 years

d) r=7% compounded continously

`A=Pe^(rt)`

`2000=1000e^(.07*t)`

`2=e^(.07t)`

`ln(2)=.07tlne`

`ln(2)/[.07lne]=t`

`9.90=t`

Final answer: 9.90 years

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