# How do you calculate to the nearest hundredth of a year how long it takes for an amount of money to double if interest is compounded continuously at 3.8%?

Mar 29, 2015

This is an example of an exponential function.

These are allways of the form $N = B \cdot {g}^{t}$
Where $g =$growth factor

With an interest rate of 3.8%, your growth factor will be:

$\frac{100 + 3.8}{100} = 1.038$

To double your money the ratio of $N$ and $B$ will be $2$

So ${1.038}^{t} = 2$

Two ways to solve this:

By logs:
$\log {1.038}^{t} = t \cdot \log 1.038 = \log 2$
$\to t = \log \frac{2}{\log} 1.038 = 18.585 \approx 18.59$ years

By GC:
$Y 1 = {1.038}^{X}$ and $Y 2 = 2$ and use intersect (same result).

Extra :
You could even say it would be on day 215 of the 19th year, but that was not the question.