Question

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%?

Answer 1

This is an example of an exponential function.

These are allways of the form `N=B*g^t`
Where `g=`growth factor

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

`(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*log1.038=log2`
`->t=log2/log1.038=18.585~~18.59` years

By GC:
`Y1=1.038^X` and `Y2=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.