1. Find Time n = ?
Principal P = 1000, Interest Rate i = 10%, Future value FV = 1649,
for Contineous Compounding method
Solution:
`P=1000`
`i=10%=0.1` per year (Interest rate)
`FV=1649` (Future value)
`FV=P*e^(i*n)`
`:.1649=1000*e^(0.1*n)`
`:.1649/1000=e^(0.1*n)`
`e^(0.1*n)=1.65`
taking natural log on both the sides
`:.ln(e^(0.1*n))=ln(1.65)`
`:.0.1*n=0.5`
`:.n=0.5/0.1`
`:.n=5`
| Year | Principal | Yearly Interest | Total Interest | Total Balance |
| 0 | 1000 | 0 | 0 | 1000 |
| 1 | 1000 | 105.17 | 105.17 | 1105.17 |
| 2 | 1000 | 116.23 | 221.4 | 1221.4 |
| 3 | 1000 | 128.46 | 349.86 | 1349.86 |
| 4 | 1000 | 141.97 | 491.82 | 1491.82 |
| 5 | 1000 | 156.9 | 648.72 | 1648.72 |
2. Find Time n = ?
Principal P = 5000, Interest Rate i = 10%, Future value FV = 6749,
for Contineous Compounding method
Solution:
`P=5000`
`i=10%=0.1` per year (Interest rate)
`FV=6749` (Future value)
`FV=P*e^(i*n)`
`:.6749=5000*e^(0.1*n)`
`:.6749/5000=e^(0.1*n)`
`e^(0.1*n)=1.35`
taking natural log on both the sides
`:.ln(e^(0.1*n))=ln(1.35)`
`:.0.1*n=0.3`
`:.n=0.3/0.1`
`:.n=3`
| Year | Principal | Yearly Interest | Total Interest | Total Balance |
| 0 | 5000 | 0 | 0 | 5000 |
| 1 | 5000 | 525.85 | 525.85 | 5525.85 |
| 2 | 5000 | 581.16 | 1107.01 | 6107.01 |
| 3 | 5000 | 642.28 | 1749.29 | 6749.29 |
This material is intended as a summary. Use your textbook for detail explanation.
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