C6

Exponential and Logarithmic Functions

Domain, Asymptotes, and Graphs of Logarithmic Functions

Logarithmic functions are the inverses of exponential functions and have important properties regarding their domains and asymptotes. Understanding these properties is essential for graphing and solving equations involving logarithms.

• Domain: The domain of a logarithmic function $f(x) = ext{log}_b(x)$ is $x > 0$.

• Vertical Asymptote: The graph of $f(x) = ext{log}_b(x)$ has a vertical asymptote at $x = 0$.

• Transformations: Shifting, stretching, and reflecting logarithmic graphs can be described by changes to the function's formula. For example, $f(x) = -5 ext{log}_3(x) - 4$ involves reflection, vertical stretch, and vertical shift.

Example: For $f(x) = ext{log}_2(x-3)$, the domain is $x > 3$ and the vertical asymptote is $x = 3$.

Finding Intercepts of Logarithmic Functions

Intercepts are found by setting $x$ or $y$ to zero and solving for the other variable.

• x-intercept: Set $y = 0$ and solve for $x$.

• Example: For $y = ext{log}_5(x)$, set $0 = ext{log}_5(x)$, so $x = 1$.

Compound Interest and Exponential Growth

Compound Interest Formulas

Compound interest is calculated on both the initial principal and the accumulated interest from previous periods. The frequency of compounding affects the total amount accrued.

• General Formula: $P(t) = P_0 (1 + r)^t$

• Compound Interest (n times per year): $P(t) = P_0 \left(1 + \frac{r}{n}\right)^{nt}$

• Continuous Compounding: $P(t) = P_0 e^{rt}$

• Variables:

  • $P_0$ = initial principal

  • $r$ = annual interest rate (as a decimal)

  • $n$ = number of compounding periods per year

  • $t$ = number of years

Example: $1000$invested at $6\%$ annual interest for $10$ years, compounded annually:

• $P(10) = 1000(1.06)^{10} \approx 1790.85$

Periodic and Continuous Compounding

Interest can be compounded at different intervals: annually, quarterly, monthly, or continuously. More frequent compounding results in slightly higher returns.

• Quarterly: $P(t) = P_0 \left(1 + \frac{r}{4}\right)^{4t}$

• Monthly: $P(t) = P_0 \left(1 + \frac{r}{12}\right)^{12t}$

• Continuously: $P(t) = P_0 e^{rt}$

Example: $1000$at $6\%$ for $10$ years, compounded monthly:

• $P(10) = 1000 \left(1 + \frac{0.06}{12}\right)^{120} \approx 1822.12$

Doubling Time and Interest Rate Calculations

To find the time required for an investment to double, set $P(t) = 2P_0$ and solve for $t$.

• Continuous Compounding: $2P_0 = P_0 e^{rt} \implies t = \frac{\ln 2}{r}$

• Finding Interest Rate: If you know the doubling time, solve $r = \frac{\ln 2}{t}$

Example: At $8\%$ interest compounded continuously, doubling time is $t = \frac{\ln 2}{0.08} \approx 8.66$ years.

Exponential Growth and Decay Models

General Exponential Function

Exponential functions model growth and decay in many real-world contexts, such as finance, population, and radioactive decay.

• General Form: $f(t) = ab^t$

• Growth: $b > 1$

• Decay: $0 < b < 1$

Example: If $a = 400$, $b = 0.86$, then $f(t) = 400(0.86)^t$ models exponential decay.

Radioactive Decay and Half-Life

Radioactive decay is a common example of exponential decay, where the amount of substance decreases by a fixed percentage over equal time intervals.

• Half-life: The time required for half the substance to decay.

• Formula: $A = A_0 \left(\frac{1}{2}\right)^{t/h}$, where $h$ is the half-life.

Example: If $400$grams decay to $250$ grams in $3$years, solve $250 = 400e^{-0.154t}$ for $t$.

Population Growth

Exponential growth models are used for populations that grow by a constant percentage over equal time intervals.

• General Form: $P(t) = P_0 b^t$

• Doubling Time: The time it takes for the population to double.

Example: A population of bacteria doubles every $3$hours. If starting with $100$ bacteria, after $t$ hours: $P(t) = 100 \cdot 2^{t/3}$.

Practice Problems and Applications

Application of Exponential and Logarithmic Models

Exponential and logarithmic models are widely used in finance, science, and everyday life. Practice problems help reinforce understanding of these concepts.

• Calculating accumulated value with different compounding intervals

• Modeling inflation and population growth

• Finding half-life and decay rates

• Solving for time or rate in doubling scenarios

Summary Table: Compound Interest Formulas

Compounding Type	Formula	Example Calculation
Annually	$P(t) = P_0 (1 + r)^t$	$1000(1.06)^{10} \approx 1790.85$
Quarterly	$P(t) = P_0 \left(1 + \frac{r}{4}\right)^{4t}$	$1000(1.015)^{40} \approx 1814.02$
Monthly	$P(t) = P_0 \left(1 + \frac{r}{12}\right)^{12t}$	$1000(1.005)^{120} \approx 1822.12$
Continuously	$P(t) = P_0 e^{rt}$	$1000e^{0.6} \approx 1822.12$

Summary Table: Exponential Growth and Decay

Model Type	Formula	Key Parameter
Exponential Growth	$f(t) = ab^t$	$b > 1$
Exponential Decay	$f(t) = ab^t$	$0 < b < 1$
Half-life	$A = A_0 \left(\frac{1}{2}\right)^{t/h}$	$h$ = half-life

Additional info: These notes include expanded explanations and context for formulas and applications, as well as tables summarizing key models and formulas for exam preparation.