# Time to Double Your Money Calculator

Instructions: Use this calculator to get shown step-by-step the calculation of the time required to doubling certain initial amount of money $$A_0$$. Please provide annual interest rate $$r$$ and the type of compounding (yearly, semi-annually, quarterly, monthly, daily or continuously):

Interest Rate $$(r)$$ =
Compounding Period:

## Time to Double Money Calculator

This calculator will show all the steps involved in computing the amount of time needed to double an initial amount $$A_0$$) of money. Common wisdom indicates that the higher the interest rate $$r$$ you get, the shorter it will take to double your money and that is indeed the case.

It will also depend on whether the compounding occurs more frequently that once a year. Indeed, let $$k$$ be the number of times the money is compounded in a year.

For example, for yearly compounding we have $$k = 1$$, for bi-yearly compounding we have $$k = 2$$, for quarterly compounding we have $$k = 4$$, etc.

### Time to double compounded discretely

When you compound a certain amount of $$k$$ times a year, you have what is called a discrete compounding. For such type of compounding, the amount of money we will have after $$n$$ years is

$FV = A_0 \left( 1+\frac{r}{k}\right)^{ k \times n}$

So, if we wanted to double our initial amount $$A_0$$, we would need to end up with $$2 A_0$$ in the account, so that

$2 A_0 = A_0 \left( 1+\frac{r}{k}\right)^{ k \times n}$

and canceling $$A_0$$ from both sides of the equation leads to

$2 = \left( 1+\frac{r}{k}\right)^{ k \times n}$

and then applying natural log and solving for $$n$$ leads to

$n = \frac{\ln 2}{k \left( 1+\frac{r}{k}\right)}$

### Time to double compounded continously

Something interesting happens for continuous compounding. Indeed, that case is the same as considering that $$k \to \infty$$, in which case the amount of money we have after $$n$$ years is (meaning, the future value of our money):

$FV = A_0 e^{r \times n}$

So, same as in the discrete compounding case, if we wanted to double our initial amount $$A_0$$, we would need to end up with $$2 A_0$$ in the account, so that

$2 A_0 = A_0 e^{r \times n}$

and canceling again $$A_0$$ from both sides of the equation, we will get

$2 = e^{r \times n}$

and then applying natural log and solving for $$n$$ leads to

$n = \frac{\ln 2}{r)}$

Observe the very interesting fact that the number of years required to double your initial amount $$A_0$$ DOES NOT depend on the initial amount, only on the interest rate $$r$$ and the type of compounding.

In other words, doubling $1 or double$1 million will take the same amount of time, assuming the same interest rate.

### Example: Calculation of Time needed to double money

Question: How many years you need to wait until you double your money, if your bank offers you a nominal annual rate of 3.5%, which is compounded monthly?

Solution:

We need to compute the amount of time needed to double a given amount $$A_0$$

You have stated that the yearly interest rate is $$r = 0.035$$, and the compounding is done monthly.

The future value after $$n$$ years is calculated using the following formula:

$FV = \displaystyle A_0 \left( 1+\frac{r}{k}\right)^{ k \times n}$

What we want is to double the money, so we need to have $$2 A_0$$ as the future value after $$n$$ periods. So then we need $2 A_0 = \displaystyle A_0 \left( 1+\frac{r}{k}\right)^{ k \times n}$

which means that we can cancel $$A_0$$ from both sides of the equation, and we need to find $$n$$ from the following:

$2 = \displaystyle \left( 1+\frac{r}{k}\right)^{ k \times n}$

so then after applying natural log to both sides and solving for $$n$$ leads to:

$n = \displaystyle \frac{\ln 2}{12 \ln\left( 1+\frac{0.035}{12}\right)} = \displaystyle \frac{\ln 2}{12 \ln\left( 1+ 0.003\right)} = 19.83$

Finally, we plug the value we know about the interest rate and compounding period so we get:

Therefore, regardless of the initial amount $$A_0$$, the number of years required to double the initial investment for a yearly interest rate of $$r = 0.035$$, and with monthly compounding is $$n = 19.83$$ years.