Study Notes on Exponential and Logarithmic Functions

PreCalculus: Chapter 3 Section 1 - Exponential and Logarithmic Functions

Breeding Potential for Deer Population

  • Model Assumptions:

    • No deaths due to accident, disease, hunting, or predators.

    • Over a span of 7 years, a single breeding pair can produce a population exceeding 40 deer given optimal conditions.

Operations of Exponents

  • Exercises on exponent operations:

    • 35=?3^5 = ?

    • Why? (Reasoning for operation)

    • 52=?5^2 = ?

    • Why? (Reasoning for operation)

    • (2)3=?(2)^3 = ?

    • Why? (Reasoning for operation)

Algebraic and Transcendental Functions

  • Algebraic Functions:

    • Defined as functions utilizing algebraic operations: addition, multiplication, subtraction, and division.

  • Transcendental Functions:

    • Functions whose operations transcend traditional algebraic operations.

    • Examples include exponential and logarithmic functions as well as trigonometric functions.

    • They do not conform to the same rules of addition, subtraction, multiplication, and division established for algebraic functions.

Identifying Exponential Functions

  • Examples of Exponential Functions:

    • f(x)=3xf(x) = 3^x - Yes, this is an exponential function.

    • f(x)=3xf(x) = 3^{-x} - Yes, this is also an exponential function.

    • f(x)=3f(x) = 3 - No, this does not qualify as an exponential function since the exponent is not a variable (this is a power function).

Exponential Functions Defined

  • General Form:

    • y=abxy = ab^x

    • Where:

      • xx: Any real number (variable)

      • aa: Real number constant

      • bb: Real number constant (b>0b > 0, b<br>eq1b <br>eq 1)

  • Characteristics:

    • The defining characteristic of an exponential function is that the power is the variable

Exponential Growth

  • Expression for Growth:

    • f(x)=abxf(x) = ab^x where b>1b > 1.

  • Questions for Analysis:

    • Y-intercept

    • X-intercept

    • Domain

    • Range

    • Asymptote

    • End Behavior

    • Continuity

Exponential Decay

  • Expression for Decay:

    • f(x)=abxf(x) = ab^x where 0<b<10 < b < 1 (indicates decay with fractions, decimals, or negative exponents).

  • Questions for Analysis:

    • Y-intercept

    • X-intercept

    • Domain

    • Range

    • Asymptote

    • End Behavior

    • Continuity

Graphing Exponential Functions

  • Graph Transformations:

    • Original Function: f(x)=2xf(x) = 2^x

    • Compared to transformations:

    • f(x)=2xext(reflection)f(x) = 2^{-x} ext{ (reflection)}

    • f(x)=2xext(reflectionacrossthexaxis)f(x) = -2^x ext{ (reflection across the x-axis)}

    • f(x)=2x1ext(horizontalshift)f(x) = 2^{x-1} ext{ (horizontal shift)}

The Exponential Family of Functions

  • General Form:

    • f(x)=h+kf(x) = -h + k

    • Transformed Characteristics:

      • hh: Horizontal shift (opposite direction of the sign)

      • Negative exponent influences decay curve.

      • bb: Affects the steepness of the curve and where it crosses the y-axis.

      • kk: Adjusts the vertical shift, moving upward/downward based on its sign.

One-to-One Property of Exponential Functions

  • Example:

    • For the equation: 32+5=273^{2+5} = 27

    • When expressed with a common base, the exponents must be equal:

    • 32+5=333^{2+5} = 3^{3}

    • Thus, 2+5=32 + 5 = 3.

Solving Exponential Equations

  • Try It: Solve the following:

    • 253=125625^3 = 125^6

    • Simplifying:

    • 53=565^3 = 5^6

    • 6=186 = 18

    • Which results in 33

Population Growth Example

  • Initial Forest Growth: A 500 acre forest grows at an average rate of 3% annually.

  • Yearly Calculation Sequence:

    • Year 1: 500+500imes0.03=515500 + 500 imes 0.03 = 515

    • Year 2: 515+515imes0.03=530.45515 + 515 imes 0.03 = 530.45

    • Year 3: 530.45+530.45imes0.03=546.36530.45 + 530.45 imes 0.03 = 546.36

    • Year 4: 546.36+546.36imes0.03=562.75546.36 + 546.36 imes 0.03 = 562.75

  • Efficient Calculation Method:

    • 500(1.03)12=712.88500(1.03)^{12} = 712.88

Exponential Growth & Decay Formulas

  • General Relationships:

    • For growth, the formula is:

    • N=N0(1+r)tN = N_0(1 + r)^t

      • If r>0r > 0 it indicates growth.

      • If r<0r < 0 it indicates decay.

      • N0N_0: Initial amount, rr: Rate, tt: Time.

Compound Interest Formula

  • General Formula:

    • A=P(1+racrn)ntA = P(1 + rac{r}{n})^{nt}

    • Where:

      • AA: Amount after time t

      • PP: Principal amount (initial investment)

      • nn: Number of times the interest is compounded in a year

      • tt: Number of years

      • rr: Interest rate as a decimal.

Example of Compound Interest Calculation

  • Scenario: Investment of $20,000 for 3 years at 6% compounded daily.

  • Formula Application:

    • A=20000(1+rac0.06365)365imes3A = 20000(1 + rac{0.06}{365})^{365 imes 3}

    • Result: A23943.99A ≈ 23943.99

Effects of Compounding Frequency

  • Comparison of Amounts Based on Compounding Frequency:

    • Invest $100 at 5% for 1 year:

    • Compounded Annually: 105.00105.00

    • Compounded Semiannually: 105.06105.06

    • Compounded Quarterly: 105.09105.09

    • Compounded Monthly: 105.12105.12

    • Compounded Daily: 105.13105.13

    • Compounded Hourly: 105.13105.13

Continuous Compounding Formula

  • Formula:

    • A=PertA = Pe^{rt}

    • Natural Base:

    • Form derived from: extlimnoext(1+rac1n)n<br>ightarrowe2.7182ext{lim}_{n o ext{∞}} (1 + rac{1}{n})^n <br>ightarrow e ≈ 2.7182…

    • This base (e) is preferred in real-world applications of exponential growth.

Carbon-14 Dating Example

  • Decay Rate:

    • Carbon-14 isotope decay rate: -0.012%.

  • Initial Amount:

    • If a bone contains 3.4 grams of Carbon-14, the amount remaining after 450 years is calculated as:

    • N=3.4(10.00012)450<br>ightarrowN3.22extgramsN = 3.4(1 - 0.00012)^{450} <br>ightarrow N ≈ 3.22 ext{ grams}

Half-Life Calculation Example

  • Radium Decay:

    • Half-life: 1620 years.

  • Calculation:

    • Starting mass: 10 lbs.

    • Quantity left after 850 years:

    • N=10(1/2)rac8501620<br>ightarrowN6.95extlbs.N = 10(1/2)^{ rac{850}{1620}} <br>ightarrow N ≈ 6.95 ext{ lbs.}

Population Prediction Example

  • Initial Population: 7,500, with a continuous growth rate of 2.3%.

  • Population after 8 years calculation:

    • N=7500e0.023imes8<br>ightarrowN9015N = 7500e^{0.023 imes 8} <br>ightarrow N ≈ 9015

Turkey Consumption Projection Example

  • Initial Year: t = 0 corresponds to 1939.

  • Model for Consumption:

    • f(t)=2.3(3)0.033tf(t) = 2.3(3)^{0.033t}

  • Prediction for Year 2010:

    • f(71)30.2extlbs.f(71) ≈ 30.2 ext{ lbs.}

Distinction Between Growth and Decay in Exponential Functions

  • For Exponential (not continuous):

    • N=N0(1+r)tN = N_0(1 + r)^t

  • For Continuous Exponential:

    • N=N0ertN = N_0e^{rt}

    • Where r>0r > 0 indicates growth, and r<0r < 0 indicates decay.

Exponential Rate of Increase Example

  • Initial Data: 125 deer in 2000, increasing to 264 deer in 2010.

  • Modeling Method:

    • Use the equation N(t)=N0(1+r)tN(t) = N_0(1 + r)^t

    • Find rr in the context of data provided for accurate growth modeling.

Another Turkey Consumption Example

  • Growth Model Established: Similar to previous example with different initial conditions leading to:

    • f(t)=2.3(3)0.033tf(t) = 2.3(3)^{0.033t}

  • Amount consumed in 2010:

    • Resulting in f(t)=73f(t) = 73.

Assignment Instructions

  • Complete exercises P-226 #3-21 odd, 29-33 odd, 47-67 odd.

Final Notes on Operations and Graphs of Exponentials

  • Additional exercises answering questions regarding transformations and comparisons of different functions accurate to properties of exponential graphs.