Answer :
Certainly! To find the principal amount [tex]\( P \)[/tex] using the continuous compound interest formula, we start with the formula:
[tex]\[ A = P \times e^{(r \times t)} \][/tex]
where:
- [tex]\( A \)[/tex] is the future value,
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money),
- [tex]\( r \)[/tex] is the annual interest rate (as a decimal),
- [tex]\( t \)[/tex] is the time the money is invested for, in years, and
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately equal to 2.71828).
Given values:
- [tex]\( A = \$ 6,300 \)[/tex]
- [tex]\( r = 9.82\% = 0.0982 \)[/tex] (converted from percentage to decimal)
- [tex]\( t = 7 \)[/tex] years
We need to solve for [tex]\( P \)[/tex]. Start by rearranging the formula to solve for [tex]\( P \)[/tex]:
[tex]\[ P = \frac{A}{e^{(r \times t)}} \][/tex]
Now substitute the given values into the equation:
[tex]\[ P = \frac{6300}{e^{(0.0982 \times 7)}} \][/tex]
Thus:
[tex]\[ P = \frac{6300}{e^{0.6874}} \][/tex]
Evaluate the exponent first:
[tex]\[ e^{0.6874} \][/tex]
Using the constant [tex]\( e \)[/tex] (approximately 2.71828):
[tex]\[ e^{0.6874} \approx 1.98817 \][/tex]
Now substitute this back into the equation:
[tex]\[ P = \frac{6300}{1.98817} \][/tex]
Finally, perform the division:
[tex]\[ P \approx 3168.16 \][/tex]
Therefore, the principal amount [tex]\( P \)[/tex] is approximately:
[tex]\[ P = \$ 3168.16 \][/tex]
So, the principal amount rounded to two decimal places is [tex]\(\boxed{3168.16}\)[/tex].
[tex]\[ A = P \times e^{(r \times t)} \][/tex]
where:
- [tex]\( A \)[/tex] is the future value,
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money),
- [tex]\( r \)[/tex] is the annual interest rate (as a decimal),
- [tex]\( t \)[/tex] is the time the money is invested for, in years, and
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately equal to 2.71828).
Given values:
- [tex]\( A = \$ 6,300 \)[/tex]
- [tex]\( r = 9.82\% = 0.0982 \)[/tex] (converted from percentage to decimal)
- [tex]\( t = 7 \)[/tex] years
We need to solve for [tex]\( P \)[/tex]. Start by rearranging the formula to solve for [tex]\( P \)[/tex]:
[tex]\[ P = \frac{A}{e^{(r \times t)}} \][/tex]
Now substitute the given values into the equation:
[tex]\[ P = \frac{6300}{e^{(0.0982 \times 7)}} \][/tex]
Thus:
[tex]\[ P = \frac{6300}{e^{0.6874}} \][/tex]
Evaluate the exponent first:
[tex]\[ e^{0.6874} \][/tex]
Using the constant [tex]\( e \)[/tex] (approximately 2.71828):
[tex]\[ e^{0.6874} \approx 1.98817 \][/tex]
Now substitute this back into the equation:
[tex]\[ P = \frac{6300}{1.98817} \][/tex]
Finally, perform the division:
[tex]\[ P \approx 3168.16 \][/tex]
Therefore, the principal amount [tex]\( P \)[/tex] is approximately:
[tex]\[ P = \$ 3168.16 \][/tex]
So, the principal amount rounded to two decimal places is [tex]\(\boxed{3168.16}\)[/tex].