`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...

Sunday, September 18, 2016

$\displaystyle{r}={7}%$ 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...

Formula for compounding n times per year $\displaystyle{A}={P}{{\left({1}+\frac{{r}}{{n}}\right)}}^{{{n}{t}}}$

Formula for compounding continuously $\displaystyle{A}={P}{{e}}^{{{r}{t}}}$

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)

$\displaystyle{A}={P}{{\left({1}+\frac{{r}}{{n}}\right)}}^{{{n}{t}}}$

$\displaystyle{2000}={1000}{{\left({1}+\frac{{.07}}{{1}}\right)}}^{{{1}\cdot{t}}}$

$\displaystyle{2}={{1.07}}^{{t}}$

$\displaystyle{\ln{{\left({2}\right)}}}={t}{\ln{{\left({1.07}\right)}}}$

$\displaystyle\frac{{\ln{{\left({2}\right)}}}}{{\ln{{\left({1.07}\right)}}}}={t}$

$\displaystyle{10.24}={t}$

Final answer: 10.24 years

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

$\displaystyle{A}={P}{{\left({1}+\frac{{r}}{{n}}\right)}}^{{{n}{t}}}$

$\displaystyle{2000}={1000}{{\left({1}+\frac{{.07}}{{12}}\right)}}^{{{12}\cdot{t}}}$

$\displaystyle{2}={{1.0058}}^{{{12}{t}}}$

$\displaystyle{\ln{{\left({2}\right)}}}={12}{t}{\ln{{\left({1.0058}\right)}}}$

$\displaystyle\frac{{\ln{{\left({2}\right)}}}}{{{12}{\ln{{\left({1.0058}\right)}}}}}={t}$

$\displaystyle{9.93}={t}$

Final Answer: 9.93 years

c) r=7% t=365 (daily)

$\displaystyle{A}={P}{{\left({1}+\frac{{r}}{{n}}\right)}}^{{{n}{t}}}$

$\displaystyle{2000}={1000}{{\left({1}+\frac{{.07}}{{365}}\right)}}^{{{365}\cdot{t}}}$

$\displaystyle{2}={{\left({1.00019}\right)}}^{{{365}{t}}}$

$\displaystyle{\ln{{\left({2}\right)}}}={365}{t}{\ln{{\left({1.00019}\right)}}}$

$\displaystyle\frac{{\ln{{\left({2}\right)}}}}{{{365}{\ln{{\left({1.00019}\right)}}}}}={t}$

$\displaystyle{9.90}={t}$

Final answer: 9.90 years

d) r=7% compounded continously

$\displaystyle{A}={P}{{e}}^{{{r}{t}}}$

$\displaystyle{2000}={1000}{{e}}^{{{.07}\cdot{t}}}$

$\displaystyle{2}={{e}}^{{{.07}{t}}}$

$\displaystyle{\ln{{\left({2}\right)}}}={.07}{t}{\ln{{e}}}$

$\displaystyle\frac{{\ln{{\left({2}\right)}}}}{{{.07}{\ln{{e}}}}}={t}$

$\displaystyle{9.90}={t}$

Final answer: 9.90 years

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