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.
[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.
