Find the missing values assuming continuously compounded interest. What is the time to double?

Question

Initial investment $1000, Annual %rate 9.2%, Time to Double?, Amount after 10 years?

Answer 1

the underlying formula that you want to use here for continuous compounding is

A = P*e^(rt)

for doubling time you can set it up as

2P = P*e^(0.092t) and then solve for the value of t algebraically

2 = e^(0.092t)

t = ln(2) / 0.092 = 7.53, meaning that it takes about 7.53 years for the amount to double

then for the amount after 10 years you can go back to the original formula and plug in all of the values

A = 1000 * e^(0.092*10)

A = 1000(2.50929)

A = 2509.29, meaning that the accumulated amount after 10 years is $2509.29

Answer 2

Assuming the initial investment is V, Interest is R in percentage

1. After one year, the total is V (1+R)
2. After two years, the total is V(1+R)(1+R) = V(1+R)²
3. After N years, the total is V(1+R)^N

When the value is doubled,

V(1+R)^N = V*2

(1+R)^N = 2

N * ln(1+R) = ln2

In this particular case, R= 0.092, so

N*ln1.092 = ln2

N = 7.875 , which is a little under 8 years

After ten years

1000 * (1+0.092) ¹⁰ = 2411

There is also a so-called "rule of 72", which states:

The number of year it takes to double the money can be obtained by dividing 72 by the interest rate.

In this case, the interest is 9.2, so N = 7.82, which is quite close to 7.875.