Algebra 2 / Unit 5: Exponential and Logarithmic Functions
5.2 Compound Interest and the Number e
Notes
- Compound interest: $\displaystyle A = P\left(1 + \frac{r}{n}\right)^{nt}$, where $P$ is the principal, $r$ the annual rate (as a decimal), $n$ the compounds per year, $t$ the years.
- As $n$ gets huge, $\displaystyle \left(1 + \frac{1}{n}\right)^n$ approaches the number $e \approx 2.718$.
- Continuous compounding: $A = Pe^{rt}$.
- In $y = ae^{kt}$, $k > 0$ means growth and $k < 0$ means decay.
- Example: $1,000 at 5% compounded continuously for 3 years: $1000e^{0.15} \approx 1161.83$, so $1,161.83.
Practice
$1,000 is invested at 6% interest compounded monthly. What is the balance after 2 years?
$\displaystyle 1000\left(1 + \frac{0.06}{12}\right)^{24} = 1000(1.005)^{24} \approx 1127.16$.
$500 is invested at 4% compounded continuously. What is the balance after 10 years?
$500e^{0.04 \cdot 10} = 500e^{0.4} \approx 745.91$.
Which function shows continuous exponential decay?
In $ae^{kt}$, a negative $k$ means decay.
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