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Two friends want to buy a house and need to save a down payment of [tex]$12,000[/tex]. They currently have [tex]$7,000[/tex] in a bank account earning [tex]3.25\%[/tex] interest, compounded continuously.

Which function models the situation and will help determine how long they will need to save in order to have their down payment?

[tex]
\begin{array}{l}
7000 = 12000(1.0325)^t \\
12000 = 7000 e^{0.325 t} \\
12000 = 7000 e^{0.0325 t} \\
7000 = 12000 e^{0.0325 t}
\end{array}
[/tex]

Answer :

To solve this problem, we need to determine how long it will take for the initial amount of [tex]$7,000, which is earning interest at a rate of 3.25% compounded continuously, to grow to the needed down payment amount of $[/tex]12,000.

The formula for continuous compounding is given by:

[tex]\[ A = P \times e^{rt} \][/tex]

Where:
- [tex]\( A \)[/tex] is the future amount of money to be achieved ([tex]$12,000 in this case).
- \( P \) is the principal (initial amount, $[/tex]7,000 here).
- [tex]\( r \)[/tex] is the interest rate (expressed as a decimal, so 0.0325 for 3.25%).
- [tex]\( t \)[/tex] is the time in years.
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately 2.71828).

The task is to solve for [tex]\( t \)[/tex]. The equation becomes:

[tex]\[ 12000 = 7000 \times e^{0.0325t} \][/tex]

To isolate [tex]\( t \)[/tex], follow these steps:

1. Divide both sides by 7000 to get:

[tex]\[ \frac{12000}{7000} = e^{0.0325t} \][/tex]

2. Simplify the left side:

[tex]\[ \frac{12000}{7000} = \frac{12}{7} \approx 1.7143 \][/tex]

3. To solve for [tex]\( t \)[/tex], take the natural logarithm (ln) of both sides:

[tex]\[ \ln\left(1.7143\right) = 0.0325t \][/tex]

4. Solve for [tex]\( t \)[/tex] by dividing both sides by 0.0325:

[tex]\[ t = \frac{\ln(1.7143)}{0.0325} \][/tex]

After performing the calculation, you find that:

[tex]\( t \approx 16.58 \)[/tex]

So, it will take approximately 16.58 years for the friends to save enough for the down payment at this interest rate with continuous compounding.

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