Question

A person invested \$6,000$6,000 in an account growing at a rate allowing the money to double every 11 years. How much money would be in the account after 19 years, to the nearest dollar?

Answer 1

Answer: $19,866

Step-by-step explanation:

We know that the initial investment is $6,000. And it doubles every 11 years.

Then after 11 years, the amount of money will be 2*$6,000

After another 11 years (a total of 22) the amount of money will be 2*(2*$6,000)

or $6,000*(2)^2

We already can see that this is an exponential growth equation.

After n times 11 years, the amount of money will be;

M = $6,000*(2)^n

But n represents groups of 11 years, then if y represents the number of single years, we can write the equation as:

M = $6,000*(2)^(y/11)

You can see that y = 11 is equivalent to n = 1

y = 22 is equivalent to n = 2

and so on.

Then the equation:

M = $6,000*(2)^(y/11)

Tell us how much money will be in the account after y years.

Now we only need to replace y by 19, to get:

M = $6,000*(2)^(19/11) = $19,866

Where i rounded the amount of money to the nearest dollar.