At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.