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Sagot :
Sure! Let's solve the problem step-by-step to find out the amount [tex]\( P \)[/tex] that must be invested at an interest rate of 12% compounded continuously to obtain a balance of [tex]$130,000 in 10 years.
1. Identify the Given Values:
- Future value \( A = \$[/tex]130,000 \)
- Interest rate [tex]\( r = 0.12 \)[/tex] (which is 12% expressed as a decimal)
- Time period [tex]\( t = 10 \)[/tex] years
2. Formula for Continuous Compounding:
The formula for the future value with continuous compounding interest is given by:
[tex]\[ A = P \times e^{rt} \][/tex]
Here:
- [tex]\( A \)[/tex] is the future value
- [tex]\( P \)[/tex] is the principal amount (the amount to be invested)
- [tex]\( e \)[/tex] is the base of natural logarithms, approximately equal to 2.71828
- [tex]\( r \)[/tex] is the annual interest rate (as a decimal)
- [tex]\( t \)[/tex] is the time in years
3. Rearranging the Formula to Solve for [tex]\( P \)[/tex]:
To find the present value [tex]\( P \)[/tex], we need to rearrange the formula:
[tex]\[ P = \frac{A}{e^{rt}} \][/tex]
4. Substitution with Given Values:
Substitute [tex]\( A = 130000 \)[/tex], [tex]\( r = 0.12 \)[/tex], and [tex]\( t = 10 \)[/tex]:
[tex]\[ P = \frac{130000}{e^{0.12 \times 10}} \][/tex]
5. Calculation:
- Calculate the exponent:
[tex]\[ 0.12 \times 10 = 1.2 \][/tex]
- Calculate [tex]\( e^{1.2} \)[/tex]:
[tex]\[ e^{1.2} \approx 3.320117 \][/tex]
- Divide the future value by the calculated exponent:
[tex]\[ P = \frac{130000}{3.320117} \approx 39155.25 \][/tex]
Thus, the amount [tex]\( P \)[/tex] that must be invested today at an interest rate of 12% compounded continuously to obtain a balance of [tex]$130,000 in 10 years is approximately \(\$[/tex]39,155.25\).
Summary Table:
| Future Value ([tex]\(A\)[/tex]) | Interest Rate ([tex]\(r\)[/tex]) | Time ([tex]\(t\)[/tex]) | Present Value ([tex]\(P\)[/tex]) |
|----------------------|-----------------------|-----------------|------------------------|
| \[tex]$130,000 | 12% | 10 years | \$[/tex]39,155.25 |
Rounded to the nearest cent, the amount to be invested is [tex]\(\$39,155.25\)[/tex].
- Interest rate [tex]\( r = 0.12 \)[/tex] (which is 12% expressed as a decimal)
- Time period [tex]\( t = 10 \)[/tex] years
2. Formula for Continuous Compounding:
The formula for the future value with continuous compounding interest is given by:
[tex]\[ A = P \times e^{rt} \][/tex]
Here:
- [tex]\( A \)[/tex] is the future value
- [tex]\( P \)[/tex] is the principal amount (the amount to be invested)
- [tex]\( e \)[/tex] is the base of natural logarithms, approximately equal to 2.71828
- [tex]\( r \)[/tex] is the annual interest rate (as a decimal)
- [tex]\( t \)[/tex] is the time in years
3. Rearranging the Formula to Solve for [tex]\( P \)[/tex]:
To find the present value [tex]\( P \)[/tex], we need to rearrange the formula:
[tex]\[ P = \frac{A}{e^{rt}} \][/tex]
4. Substitution with Given Values:
Substitute [tex]\( A = 130000 \)[/tex], [tex]\( r = 0.12 \)[/tex], and [tex]\( t = 10 \)[/tex]:
[tex]\[ P = \frac{130000}{e^{0.12 \times 10}} \][/tex]
5. Calculation:
- Calculate the exponent:
[tex]\[ 0.12 \times 10 = 1.2 \][/tex]
- Calculate [tex]\( e^{1.2} \)[/tex]:
[tex]\[ e^{1.2} \approx 3.320117 \][/tex]
- Divide the future value by the calculated exponent:
[tex]\[ P = \frac{130000}{3.320117} \approx 39155.25 \][/tex]
Thus, the amount [tex]\( P \)[/tex] that must be invested today at an interest rate of 12% compounded continuously to obtain a balance of [tex]$130,000 in 10 years is approximately \(\$[/tex]39,155.25\).
Summary Table:
| Future Value ([tex]\(A\)[/tex]) | Interest Rate ([tex]\(r\)[/tex]) | Time ([tex]\(t\)[/tex]) | Present Value ([tex]\(P\)[/tex]) |
|----------------------|-----------------------|-----------------|------------------------|
| \[tex]$130,000 | 12% | 10 years | \$[/tex]39,155.25 |
Rounded to the nearest cent, the amount to be invested is [tex]\(\$39,155.25\)[/tex].
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