Integrated Math 3 / Unit 3: Exponential and Logarithmic Functions
3.6 Modeling with Exponentials and Logarithms
Notes
- Compound interest: $\displaystyle A = P\left(1 + \frac{r}{n}\right)^{nt}$.
- Continuous growth: $A = Pe^{rt}$.
- Half-life: amount $\displaystyle = A_0 \left(\frac{1}{2}\right)^{t/h}$.
- Use logs to solve for time: $\displaystyle 1.05^t = 2 \Rightarrow t = \frac{\ln 2}{\ln 1.05}$.
Practice
$1,000 is invested at 5% interest compounded annually. What is the balance after 3 years?
$1000(1.05)^3 = 1157.625$. Compound interest earns interest on interest, so it beats the simple-interest $1,150.
A sample has 80 grams of a substance with a half-life of 5 days. How much is left after 15 days?
15 days is 3 half-lives: $\displaystyle 80 \cdot \left(\frac{1}{2}\right)^3 = 10$.
$2,000 is invested at 3% compounded continuously. About how much is there after 10 years?
$2000e^{0.03 \cdot 10} = 2000e^{0.3} \approx 2000(1.3499) = 2699.72$.
Ready for more? Take the Unit 3 test.