Question

A person invested $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:

P(19)=$19,852

To the nearest dollar.

Step-by-step explanation:

Exponential Growth

The natural growth of some magnitudes can be modeled by the equation:

$P(t)=P_o(1+r)^t$

Where P is the actual amount of the magnitude, Po is its initial amount, r is the growth rate and t is the time.

We are given the condition that an investment of Po=$6,000 in an account doubles every 11 years. The final value of the investment in t=11 is P(11)=$12,000.

Substituting into the general equation:

$12,000=6,000(1+r)^{11}$

Dividing by 6,000 and swapping sides:

$(1+r)^{11}=2$

Taking the 11th root:

$1+r=\sqrt[11]{2}$

$1+r=1.065$

Substituting into the formula:

$P(t)=6,000(1.065)^t$

Now we need to find the money in the account after t=19 years:

$P(19)= 6,000(1.065)^{19}$

P(19)=$19,852

To the nearest dollar.