# A child is now 3 years old - dad opens an account with $10,000 it earns 4.5% annual interest. a)...

## Question:

A child is now 3 years old - dad opens an account with $10,000 it earns 4.5% annual interest.

a) construct formula {eq}(A(t)=A_o(a)^t) {/eq}

b) how much money will be in the account when he is 10 years old

c) if dad wants an account to grow to 100,000 when he is 18 then what should original amount to be

## Exponential Growth: Investment Predictions

An exponential growth model can be used in calculating the total amount of money earned when interest is compounded continuously.

Suppose that:

- An original amount {eq}A_o {/eq}

- With an interest rate {eq}p\% {/eq}

- Is invested for {eq}t {/eq} years and is compounded yearly

- The amount earned in {eq}t {/eq} years is then {eq}A_t = A_o(1 + \frac{p}{100})^t {/eq}

## Answer and Explanation:

a. Consider a general formula:

{eq}A(t) = A_oa^t {/eq}

Given the initial amount:

{eq}A_o = 10,000 {/eq}

We know that:

{eq}a = 1 + \frac{p}{100} {/eq}

And the interest rate is:

{eq}p = 4.5 {/eq}

Our formula becomes:

{eq}A(t) = 10,000(1.045)^t {/eq}

b. A child will be 10 years old in 7 years, meaning we are going to solve our equation for {eq}t = 7 {/eq}:

{eq}A(7) = 10,000(1.045)^7\approx 13,608.62 {/eq}

c. The child will be 18 in 15 years. We wish to ind the initial amount now expecting:

{eq}A(15) = 100,000 {/eq}

Solving for the initial amount:

{eq}A_o = \dfrac{A(t)}{1.045^t} = \dfrac{100,000}{1.045^{15}}\approx 51,672.04 {/eq}

Thus:

a. The equation for this model is {eq}\boxed{A(t) = 10,000(1.045)^t} {/eq}

b. When the child is 10 years old, there will be about {eq}\boxed{\$13,608.62} {/eq} in the account

c. The original amount must be about {eq}\boxed{\$51,672.04} {/eq} in order to grow to $100,000 by the age of 18

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from High School Algebra I: Help and Review

Chapter 6 / Lesson 10