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.)

Question

r = 14%

(a) annually

 yr

(b) monthly

 yr

(c) daily

 yr

(d) continuously

 yr

Answer 1

2=e^.14t

ln2 = .14t

.14t =ln2

t = ln2/.14

t= about 4.95105129 years

t = about 4.95 years to double when compounded continuously

=about 4 years 11 months, 9 days 22 hours

2=(1+.14/365.25)^365.25t

2=(1.000383299)^365.25t

365.25t= log_1.0003832992 = ln2/ln1.000383299 = 1808.718555

t=about 4.952137102

t= about 4.95 years to double at daily compounding

=about 4 years 11 months, 12 days 18 hours

which is 2 days 20 hours longer than continuous compounding

but if you round off to two decimal places for years, they are the same identical doubling time

2=(1+.14/1)^1

2=(1.14)^t

log_1.142 = ln2/ln1.14 = t

t = 5.290058556

t= 5.29 years = doubling time at yearly compounding

2=(1+.14/12)^12t

2=(1.011666...)^12t

log_1.011666...2 =ln2/ln1.011666... = 12t

12t= 59.75852247

t=4.97987687

t= about 4.98 years doubling time at monthly compounding

the general formula is

A=Pe^rt for continous compounding

where P=initial investment, r=rate of interest, t=years, e=about 2.718281828...

and A= ending Amount

for doubling time A=2P

2P=Pe^rt

2=e^rt

at 14%=.14

2=e^.14t

for less than continuous compounding

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

where A,P,r,t mean the same as in continous compounding

but n= number of compounding periods per year

2=(1+14/n)^nt

plug in n=1,12 or 3.65.25 for annual, monthly or daily compounding

12 months in a year

365 1/4 days in a year

Answer 2

Compound interest formula

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

Time to double:

A/P = 2

t = ln(2)/(n•ln(1+r/n))

r = 0.14

Annually: n=1

Monthly: n=12

daily: n=365

Compound continuously

A=Pe^rt

t = ln(2)/r

Substitute values to solve for t.