Notes
Exponentials and logarithms
- $f(x) = a \cdot b^x$ grows when $b > 1$ and decays when $0 < b < 1$. It has a horizontal asymptote and never reaches it.
- $\log_b x = y$ means $b^y = x$. Logs are inverses of exponentials, with domain $x > 0$.
- Useful values: $\log_b 1 = 0$, $\log_b b = 1$, $\ln e^k = k$.
Properties
- $\log_b (MN) = \log_b M + \log_b N$, $\displaystyle \log_b \frac{M}{N} = \log_b M - \log_b N$, $\log_b M^p = p\log_b M$.
- Change of base: $\displaystyle \log_b x = \frac{\ln x}{\ln b}$.
Solving and modeling
- Match bases or take logs to solve exponentials. Convert to exponential form to solve logs, and check for extraneous solutions.
- Continuous model $A = Pe^{rt}$. Doubling and half-life problems multiply by 2 or $\displaystyle \frac{1}{2}$ for each period.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$5^3 = 125$.
$81 = 3^4$, so $x - 1 = 4$ and $x = 5$.
Quotient rule: $\log_2 8 = 3$.
Product rule and power rule.
$x(x - 8) = 9$ gives $x = 9$ or $-1$, and $-1$ is extraneous.
The starting amount is 200 and it multiplies by 3 each day.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A bacteria population is modeled by $N(t) = 500e^{0.4t}$, where $t$ is in hours.
(a) State the initial population and whether the model shows growth or decay.
Show the answer and scoring guide
$N(0) = 500e^0 = 500$. Since $0.4 > 0$ the model shows growth.
Check off the points you earned:
(b) Write an exact expression for the population after 5 hours.
Show the answer and scoring guide
$N(5) = 500e^{0.4 \cdot 5} = 500e^2$.
Check off the points you earned:
(c) Find when the population reaches 2,000. Give an exact answer and a decimal rounded to two places.
Show the answer and scoring guide
$\displaystyle 500e^{0.4t} = 2000 \Rightarrow e^{0.4t} = 4 \Rightarrow 0.4t = \ln 4 \Rightarrow t = \frac{\ln 4}{0.4} = 2.5\ln 4 \approx 3.47$ hours.
Check off the points you earned: