Notes
Arithmetic and geometric sequences
- Arithmetic: add the same common difference $d$ each step. $a_n = a_0 + dn$. These match linear functions.
- Geometric: multiply by the same common ratio $r$ each step. $g_n = g_0 r^n$. These match exponential functions.
Exponential functions
$f(x) = ab^x$ with $a \ne 0$, $b > 0$, $b \ne 1$.
- If $a > 0$: growth when $b > 1$, decay when $0 < b < 1$. The graph is always concave up and has a horizontal asymptote at $y = 0$.
- Over equal-length input intervals, outputs change by equal factors (the ratio is constant). That's how you spot exponential data.
- Percent change: growth by $r\%$ per period gives $\displaystyle b = 1 + \frac{r}{100}$; decay gives $\displaystyle b = 1 - \frac{r}{100}$.
- Equivalent forms: $f(x) = ab^x$ can be rewritten, for example $b^{x+k} = b^k b^x$ (a horizontal shift is the same as a vertical stretch). Continuous growth uses $e$: $f(t) = ae^{kt}$.
Logarithms
$\log_b c = a$ means $b^a = c$. A log answers "what power?"
- $\log_b x$ is the inverse of $b^x$: domain $x > 0$, vertical asymptote at $x = 0$.
- Rules: $\log_b(xy) = \log_b x + \log_b y$, $\displaystyle \log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y$, $\log_b(x^n) = n\log_b x$.
- Change of base: $\displaystyle \log_b x = \frac{\ln x}{\ln b}$.
- Over equal-length output intervals of a log function, inputs change by equal factors. Logs grow, but slower and slower (concave down when $b > 1$).
Solving equations
- Exponential equation: take a log of both sides. $5 \cdot 2^x = 40 \Rightarrow 2^x = 8 \Rightarrow x = 3$.
- Log equation: rewrite in exponential form, then check that every log in the original has a positive input.
Semi-log plots and modeling
On a semi-log plot (the $y$-axis is logarithmic), exponential data appears as a line. If $y = ab^x$, then $\log y = \log a + x\log b$.
Topics in this unit
Each topic has its own short notes and practice.
- 2.1Change in Arithmetic and Geometric Sequences
- 2.2Change in Linear and Exponential Functions
- 2.3Exponential Functions
- 2.4Exponential Function Manipulation
- 2.5Exponential Function Context and Data Modeling
- 2.6Competing Function Model Validation
- 2.7Composition of Functions
- 2.8Inverse Functions
- 2.9Logarithmic Expressions
- 2.10Inverses of Exponential Functions
- 2.11Logarithmic Functions
- 2.12Logarithmic Function Manipulation
- 2.13Exponential and Logarithmic Equations and Inequalities
- 2.14Logarithmic Function Context and Data Modeling
- 2.15Semi-log Plots
Practice
Each term is 3 times the previous one, so the ratio is constant.
Decreasing by 4% leaves 96% each year, so the growth factor is $0.96$.
$2^5 = 32$, so the log is 5.
$\log_3 9 + \log_3 x^4 = 2 + 4\log_3 x$.
Divide by 3: $4^x = 16$, so $x = 2$.
$\log y$ is linear in $x$ exactly when $y = ab^x$.
The input of the log must be positive: $x - 2 > 0$.
Swap and solve: $\displaystyle x = e^{2y} \Rightarrow 2y = \ln x \Rightarrow y = \frac{1}{2}\ln x$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A cup of coffee cools in a room. Let $D(t)$ be how many degrees Fahrenheit the coffee is above room temperature $t$ minutes after it is poured.
(a) At $t = 0$ minutes the difference is 60 °F, and at $t = 10$ minutes it is 30 °F. Write an exponential model $D(t) = ab^t$.
Show the answer and scoring guide
$a = 60$. $60b^{10} = 30$ gives $b^{10} = 0.5$, so $b = 0.5^{1/10} \approx 0.933$. $D(t) = 60(0.5)^{t/10}$.
Check off the points you earned:
(b) Use your model to find when the difference reaches 15 °F.
Show the answer and scoring guide
$60(0.5)^{t/10} = 15 \Rightarrow (0.5)^{t/10} = 0.25 = (0.5)^2$, so $t = 20$ minutes.
Check off the points you earned:
(c) Explain why an exponential model fits better than a linear model here.
Show the answer and scoring guide
The difference is cut in half every 10 minutes, so it changes by a constant factor over equal time intervals instead of a constant amount.
Check off the points you earned: