To find the missing values assuming continuously compounded interest, we use the continuously compounded interest formula and some logarithmic properties.
Let's start with the given information:
- Annual interest rate: [tex]\( r = 3.8\% = 0.038 \)[/tex]
- Amount after 10 years: [tex]\( A = \$17,000 \)[/tex]
- Continuously compounded interest formula: [tex]\( A = P \cdot e^{rt} \)[/tex], where:
- [tex]\( P \)[/tex] is the initial investment
- [tex]\( r \)[/tex] is the annual interest rate
- [tex]\( t \)[/tex] is the time in years
### Step 1: Find the Initial Investment
Given:
[tex]\[ A = 17000 \][/tex]
[tex]\[ r = 0.038 \][/tex]
[tex]\[ t = 10 \][/tex]
We use the formula to find [tex]\( P \)[/tex]:
[tex]\[ 17000 = P \cdot e^{(0.038 \cdot 10)} \][/tex]
[tex]\[ 17000 = P \cdot e^{0.38} \][/tex]
Solving for [tex]\( P \)[/tex]:
[tex]\[ P = \frac{17000}{e^{0.38}} \][/tex]
Calculating the value of [tex]\( P \)[/tex]:
[tex]\[ P = 11625.64 \][/tex]
### Step 2: Find the Time to Double the Investment
To double the investment, the amount [tex]\( A \)[/tex] will be [tex]\( 2P \)[/tex]:
[tex]\[ 2P = P \cdot e^{(0.038 \cdot t_{\text{double}})} \][/tex]
Solving for the time [tex]\( t_{\text{double}} \)[/tex]:
[tex]\[ 2 = e^{(0.038 \cdot t_{\text{double}})} \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ \ln(2) = 0.038 \cdot t_{\text{double}} \][/tex]
Solving for [tex]\( t_{\text{double}} \)[/tex]:
[tex]\[ t_{\text{double}} = \frac{\ln(2)}{0.038} \][/tex]
Calculating the value of [tex]\( t_{\text{double}} \)[/tex]:
[tex]\[ t_{\text{double}} = 18.24 \][/tex]
### Summary
- Initial Investment: \[tex]$11625.64
- Annual % Rate: 3.8%
- Time to Double: 18.24 years
- Amount After 10 Years: \$[/tex]17000
