Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.

David invested [tex]\[tex]$230[/tex] in a savings account that offers a [tex]3\%[/tex] return on the investment. The value of David's investment will be at least [tex]\$[/tex]415[/tex] after a period of [tex]\square[/tex] years.

Hint: Use the formula [tex]A = P(1 + r)^t[/tex], where [tex]A[/tex] is the amount after [tex]t[/tex] years, [tex]P[/tex] is the amount invested, [tex]r[/tex] is the rate of interest, and [tex]t[/tex] is the time period. Use a calculator to compute the answer, and round it off to the nearest year.



Answer :

To determine the number of years [tex]\( t \)[/tex] it will take for David's investment of \[tex]$230 to grow to at least \$[/tex]415 with a 3% annual interest rate, we use the formula for compound interest:

[tex]\[ A = P(1 + r)^t \][/tex]

Where:
- [tex]\( A \)[/tex] is the amount after [tex]\( t \)[/tex] years (in this case, \[tex]$415), - \( P \) is the initial investment (in this case, \$[/tex]230),
- [tex]\( r \)[/tex] is the annual interest rate (3% or 0.03),
- [tex]\( t \)[/tex] is the number of years.

First, we need to solve for [tex]\( t \)[/tex]:

[tex]\[ 415 = 230(1 + 0.03)^t \][/tex]

Next, divide both sides by 230:

[tex]\[ \frac{415}{230} = (1 + 0.03)^t \][/tex]

Simplify the fraction on the left side:

[tex]\[ 1.8043478260869565 \approx (1.03)^t \][/tex]

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

[tex]\[ \ln(1.8043478260869565) = \ln((1.03)^t) \][/tex]

Using the property of logarithms [tex]\(\ln(a^b) = b \ln(a)\)[/tex]:

[tex]\[ \ln(1.8043478260869565) = t \cdot \ln(1.03) \][/tex]

Now, solve for [tex]\( t \)[/tex]:

[tex]\[ t = \frac{\ln(1.8043478260869565)}{\ln(1.03)} \][/tex]

By using a calculator, we find:

[tex]\[ t \approx 19.96695287192616 \][/tex]

Finally, round [tex]\( t \)[/tex] to the nearest year:

[tex]\[ t \approx 20 \][/tex]

So, the value of David's investment will be at least \$415 after at least [tex]\( \boxed{20} \)[/tex] years.