ALEKS - Zion Grier - Homework: [tex]$x$[/tex]

Use the model [tex]$A = Pe^{rt}$[/tex] or [tex]$A = P\left(1 + \frac{r}{n}\right)^{nt}$[/tex], where [tex]$A$[/tex] is the future value of [tex]$P$[/tex] dollars invested at interest rate [tex]$r$[/tex] compounded continuously or [tex]$n$[/tex] times per year for [tex]$t$[/tex] years.

[tex]$\$[/tex]2000[tex]$ grows to $[/tex]\[tex]$2384.88$[/tex] in 4 years under continuous compounding. Find the interest rate. Round to the nearest tenth of a percent.

The interest rate is approximately [tex]$\square$[/tex] [tex]$\%$[/tex].



Answer :

To find the interest rate [tex]\( r \)[/tex] for an investment using continuous compounding, we use the formula:

[tex]\[ A = P e^{rt} \][/tex]

Where:
- [tex]\( A \)[/tex] is the future amount,
- [tex]\( P \)[/tex] is the principal amount,
- [tex]\( r \)[/tex] is the interest rate,
- [tex]\( t \)[/tex] is the time in years,
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately 2.71828).

Given:
- [tex]\( P = 2000 \)[/tex] (the initial amount),
- [tex]\( A = 2384.88 \)[/tex] (the amount after 4 years),
- [tex]\( t = 4 \)[/tex] years.

We are tasked with finding the interest rate [tex]\( r \)[/tex].

1. Start with the given formula:
[tex]\[ A = P e^{rt} \][/tex]

2. Substitute the known values:
[tex]\[ 2384.88 = 2000 e^{4r} \][/tex]

3. Solve for [tex]\( e^{4r} \)[/tex]:
[tex]\[ \frac{2384.88}{2000} = e^{4r} \][/tex]

4. Simplify the left side:
[tex]\[ 1.19244 = e^{4r} \][/tex]

5. Take the natural logarithm (ln) of both sides to solve for [tex]\( 4r \)[/tex]:
[tex]\[ \ln(1.19244) = 4r \][/tex]

6. Compute the natural logarithm of 1.19244:
[tex]\[ \ln(1.19244) \approx 0.176 \][/tex]

7. Solve for [tex]\( r \)[/tex] by dividing both sides by 4:
[tex]\[ r = \frac{0.176}{4} \][/tex]

8. Simplify:
[tex]\[ r \approx 0.044 \][/tex]

9. Convert [tex]\( r \)[/tex] to a percentage:
[tex]\[ r \times 100 \approx 4.4\% \][/tex]

So, the interest rate is approximately 4.4%.