1. Find Interest Rate i = ?
Principal P = 1000, Time n = 5 Year, Future value FV = 1649,
for Contineous Compounding method
Solution:
`P=1000`
`n=5` years (Number of periods)
`FV=1649` (Future value)
`FV=P*e^(i*n)`
`:.1649=1000*e^(i*5)`
`:.1649/1000=e^(i*5)`
`e^(i*5)=1.65`
taking natural log on both the sides
`:.ln(e^(i*5))=ln(1.65)`
`:.i*5=0.5`
`:.i=0.5/5`
`:.i=0.1`
`:.i=10 %` per year
| Year | Principal | Yearly Interest | Total Interest | Total Balance |
| 0 | 1000 | 0 | 0 | 1000 |
| 1 | 1000 | 105.21 | 105.21 | 1105.21 |
| 2 | 1000 | 116.28 | 221.49 | 1221.49 |
| 3 | 1000 | 128.51 | 350 | 1350 |
| 4 | 1000 | 142.03 | 492.03 | 1492.03 |
| 5 | 1000 | 156.97 | 649 | 1649 |
2. Find Interest Rate i = ?
Principal P = 5000, Time n = 3 Year, Future value FV = 6749,
for Contineous Compounding method
Solution:
`P=5000`
`n=3` years (Number of periods)
`FV=6749` (Future value)
`FV=P*e^(i*n)`
`:.6749=5000*e^(i*3)`
`:.6749/5000=e^(i*3)`
`e^(i*3)=1.35`
taking natural log on both the sides
`:.ln(e^(i*3))=ln(1.35)`
`:.i*3=0.3`
`:.i=0.3/3`
`:.i=0.1`
`:.i=10 %` per year
| Year | Principal | Yearly Interest | Total Interest | Total Balance |
| 0 | 5000 | 0 | 0 | 5000 |
| 1 | 5000 | 525.77 | 525.77 | 5525.77 |
| 2 | 5000 | 581.06 | 1106.84 | 6106.84 |
| 3 | 5000 | 642.16 | 1749 | 6749 |
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
