Question

Juan's parents put $ 10,000 into a college education savings account at the rate of 6% compounded annually. The chart below shows the value of the original investment at the end of years 1 and 2. If no further deposits or withdrawals are made, what will the value of the original investment be at the end of year 4? Round your answer to the nearest dollar.

Answer 1

The value of the original investment at the end of year 4 is $12,625.

To find the value of the original investment at the end of year 4, we will use the compound interest formula, which is given by:


`\[ A = P \left(1 + (r)/(n)\right)^(nt) \]`

where:
- A is the amount of money accumulated after n years, including interest.
- P is the principal amount (the initial amount of money).
- r is the annual interest rate (in decimal form).
- n is the number of times that interest is compounded per year.
- t is the time the money is invested for, in years.

In this case, we have the following information:
- Principal amount, P, is $10,000.
- Annual interest rate, r , is 6% or 0.06 in decimal form.
- Interest is compounded annually, so n is 1.
- We want to find the amount at the end of year 4, so t is 4.

Now, we just need to plug these values into the compound interest formula:

`\[ A = 10000 \left(1 + (0.06)/(1)\right)^(1 \cdot 4) \]`

`\[ A = 10000 \left(1 + 0.06\right)^4 \]`

`\[ A = 10000 \left(1.06\right)^4 \]`

Now, let's calculate
`\( (1.06)^4 \)`:

`\[ (1.06)^4 = 1.06 * 1.06 * 1.06 * 1.06 \]`

`\[ (1.06)^4 \approx 1.2625 \]` (rounded to four decimal places)

Multiplying this by the principal:

`\[ A = 10000 * 1.2625 \]`
A = 12625

Answer 2

`\begin{gathered} I\text{ will be solving the question in the image attached below.} \\ p\text{ = \$300, r = 0.06\%, A = Thr}ee\text{ times, 3 }*300\text{ = \$900} \\ A\text{ = P}e^(rt) \\ 900\text{ = 300 }* e^(0.06t) \\ (900)/(300)\text{ = }e^(0.06t) \\ 3\text{ = }e^(0.06t) \\ \text{Taking Log of boths sides.} \\ \ln 3\text{ = }\ln e^(0.06t) \\ 1.098\text{ = 0.06t} \\ t\text{ = }(1.098)/(0.06)\text{ = 18.31years} \end{gathered}`