How long does it take an investment of $1000 to double if the rate is 5.5% and interest is

Question

a computed as simple interest

b compounded yearly

c what happens if the original principal is $10,000

Answer 1

(a) Simple interest is just

I = Prt

P = 1000 (original principal)

r = 0.055 (interest rate per year)

t = time in years

We need to accumulate $1000 in interest to double the investment.

1000 = (1000)(0.055)(t)

1 = 0.055t

t=18.18 years

If the interest is paid at the end of each year, you won't actually have doubled the investment until the 19th year interest is added.  This would require you to round up to 19 years.

 

(b) Compound interest applies the rate to an increased balance at the end of each year.

A = P(1 + r)^t

A = 2000 (the final balance, including interest)

P = 1000 (original principal)

r= 0.055 (interest rate per year)

t = time in years

2000 = (1000)(1.055)^t

2 = 1.055^t

This is an exponential equation.  Use logarithms to solve it.

log(2) = [log(1.055)](t)

t = [log(2)] / [log(1.055)]

t = 12.95 years

(NOTE: you can use common logs or natural logs.  Both will give the same result.)

As above, you may need to round this up to 13 years, depending on how picky they are about actually doubling the investment.

 

(c) If the original amount is 10,000 (or any other amount), the time to double will be the same!

I = Prt

P = Prt

1=rt       The original principal divides out of the equation.

 

A = P(1 + r)^t

2P = P(1 + r)^t

2 = (1 + r)^t      The original principal divides out of the equation

 

NOTE: For monthly compounding, you would need to divide the annual rate by 12 to get the monthly rate (0.0045833...).  Then the time t will be in months.  Try it.  It will double faster.  151.57 months = 12.63 years.