Exponential and Logarithmic Functions Study Guide

Introduction to Exponential and Indicial Functions

  • Functions in which the independent variable is an index number are defined as indicial or exponential functions.
  • The general form of an exponential function is f(x)=axf (x) = a^x, where a>0a > 0 and a1a \neq 1.
  • Quantities that increase or decrease by a constant percentage over a particular time interval can be modeled using these functions.
  • Applications include:
    • Science and Medicine: Exponential growth of bacteria or exponential decay of radioactive materials.
    • Finance: Compound interest calculations and reducing balance loans.

Fundamental Index Laws

  • Index notation represents a number, aa, multiplied by itself nn times. In the expression ana^n, aa is the base and nn is the index (also referred to as the power or exponent).

  • The laws for manipulating indicial expressions are as follows:

  • Multiplication Rule: When multiplying terms with the same base, add the indices.

    • am×an=am+na^m \times a^n = a^{m+n}
    • Example: 23×24=(2×2×2)×(2×2×2×2)=272^3 \times 2^4 = (2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2) = 2^7
  • Division Rule: When dividing terms with the same base, subtract the indices.

    • am÷an=amna^m \div a^n = a^{m-n}
    • Example: 2622=2×2×2×2×2×22×2=24\frac{2^6}{2^2} = \frac{2 \times 2 \times 2 \times 2 \times 2 \times 2}{2 \times 2} = 2^4
  • Power of a Power Rule: To raise an indicial expression to another power, multiply the indices.

    • (am)n=am×n=amn(a^m)^n = a^{m \times n} = a^{mn}
    • Example: (24)3=24×24×24=24+4+4=212(2^4)^3 = 2^4 \times 2^4 \times 2^4 = 2^{4+4+4} = 2^{12}
  • Zero Power Rule: Any non-zero number raised to the power of zero equals one.

    • a0=1a^0 = 1, where a0a \neq 0
    • Proof: 23÷23=233=202^3 \div 2^3 = 2^{3-3} = 2^0. Simultaneously, 8÷8=18 \div 8 = 1. Therefore, 20=12^0 = 1.
  • Power of a Product Rule: Each factor in a product is raised to the power.

    • (ab)n=anbn(ab)^n = a^n b^n
    • Example: (2×3)4=24×34(2 \times 3)^4 = 2^4 \times 3^4
  • Power of a Quotient Rule: Both numerator and denominator are raised to the power.

    • (ab)n=anbn(\frac{a}{b})^n = \frac{a^n}{b^n}

Negative and Rational Powers

  • Negative Indices: Negative index numbers are expressed as positive indices by moving the term between the numerator and denominator.

    • an=1ana^{-n} = \frac{1}{a^n}, where a0a \neq 0
    • Rule: "Change the level, change the sign."
    • Proof: 1an=a0an=a0n=an\frac{1}{a^n} = \frac{a^0}{a^n} = a^{0-n} = a^{-n}.
  • Rational (Fractional) Indices: The index 1n\frac{1}{n} is defined as the nthn\text{th} root of the base.

    • a1n=ana^{\frac{1}{n}} = \sqrt[n]{a}
    • Example: a12=aa^{\frac{1}{2}} = \sqrt{a}, a13=a3a^{\frac{1}{3}} = \sqrt[3]{a}.
    • General Rational Form: amn=(a1n)m=(an)m=amna^{\frac{m}{n}} = (a^{\frac{1}{n}})^m = (\sqrt[n]{a})^m = \sqrt[n]{a^m}.

Indicial Equations

  • Method 1: Exact Solutions via Equating Powers

    • If the sides of an equation can be expressed using the same base, the powers can be equated.
    • Principle: If am=ana^m = a^n, then m=nm = n.
    • Example: To solve 3x=813^x = 81, rewrite as 3x=343^x = 3^4, resulting in x=4x = 4.
    • Example with Complexity: 63x1=362x36^{3x-1} = 36^{2x-3} becomes 63x1=(62)2x36^{3x-1} = (6^2)^{2x-3}. This simplifies to 3x1=4x63x-1 = 4x-6, yielding x=5x = 5.
  • Method 2: Quadratic Form Substitution

    • Equations containing terms like a2xa^{2x} and axa^x can be solved by substituting y=axy = a^x.
    • Equation: 52x4(5x)5=05^{2x} - 4(5^x) - 5 = 0 becomes y24y5=0y^2 - 4y - 5 = 0
    • Factoring: (y5)(y+1)=0(y-5)(y+1) = 0, so y=5y = 5 or y=1y = -1.
    • Re-substitution: 5x=55^x = 5 (so x=1x = 1) or 5x=15^x = -1.
    • Note: 5x=15^x = -1 has no solution because exponential functions axa^x are always positive.
  • Method 3: Trial and Error with Calculator

    • For equations where bases cannot be equated (e.g., 2x=52^x = 5), approximate solutions are found by testing values (e.g., 22.324.9932^{2.32} \approx 4.993, 22.335.0282^{2.33} \approx 5.028; therefore x2.32x \approx 2.32 to 2 decimal places).

Graphs of Exponential Functions

  • The graph shape depends on the base aa:

    • Increasing Exponential: y=axy = a^x where a>1a > 1.
    • Decreasing Exponential: y=axy = a^x where 0<a<10 < a < 1.
  • Standard Features (for basic y=axy=a^x):

    • The yy-intercept is always (0,1)(0, 1).
    • The horizontal asymptote is y=0y = 0 (xx-axis).
    • The domain is R\mathbb{R} (all real numbers).
    • The range is R+\mathbb{R}^+ (all positive real numbers).
  • Transformations:

    • Reflections: y=axy = a^{-x} reflects y=axy = a^x through the yy-axis. y=axy = -a^x reflects y=axy = a^x through the xx-axis.
    • Horizontal Translations: y=ax+by = a^{x+b} shifts the graph bb units left if b>0b > 0 and bb units right if b<0b < 0.
    • Vertical Translations: y=ax+cy = a^x + c shifts the graph up if c>0c > 0 and down if c<0c < 0. The horizontal asymptote becomes y=cy = c.
    • Example: For y=2x+35y = 2^{x+3} - 5, the asymptote is y=5y = -5, the yy-intercept is (0,3)(0, 3), the domain is R\mathbb{R}, and the range is (5,)(-5, \infty).

Logarithms: Definitions and Laws

  • A logarithm is the index (xx) in an indicial equation y=axy = a^x.

  • Equivalent Forms: ax=y    logay=xa^x = y \iff \log_a y = x.

    • Example: 32=93^2 = 9 is equivalent to log39=2\log_3 9 = 2.
    • Example: 105=100,00010^5 = 100,000 is equivalent to log10100,000=5\log_{10} 100,000 = 5.
  • Logarithm Laws (Conditions: a>0,a1,m,n>0a > 0, a \neq 1, m, n > 0):

    1. Product Rule: loga(mn)=logam+logan\log_a (mn) = \log_a m + \log_a n
    2. Quotient Rule: loga(mn)=logamlogan\log_a (\frac{m}{n}) = \log_a m - \log_a n
    3. Power Rule: logamn=nlogam\log_a m^n = n \log_a m
    4. Log of the Base: logaa=1\log_a a = 1
    5. Log of One: loga1=0\log_a 1 = 0
  • Important Domain Note: logax\log_a x is only defined for x>0x > 0. The logarithm of zero or a negative number is undefined.

Solving Logarithmic Equations

  • Logarithms to Base 10: Known as "common logarithms." These are solved using the LOG function on calculators.

  • Change of Base/Log Solution for Exponentials: If ax=ba^x = b, then x=log10blog10ax = \frac{\log_{10} b}{\log_{10} a}.

    • Example: To solve 2x=72^x = 7, x=log107log1020.84510.30102.808x = \frac{\log_{10} 7}{\log_{10} 2} \approx \frac{0.8451}{0.3010} \approx 2.808.
  • Standard Steps for Solving Equations:

    1. Simplify the equation using log laws (e.g., combining multiple logs into one).
    2. Convert the logarithmic equation into its index form equivalent.
    3. Solve the resulting algebraic equation.

Practical Applications of Exponential and Logarithmic Models

  • Paper Folding Model:
    • A sheet 0.1 mm thick folded nn times has thickness T(n)=0.1(2n)T(n) = 0.1(2^n).
    • After 10 folds: T(10)=0.1(210)=102.4mmT(10) = 0.1(2^{10}) = 102.4\,\text{mm}.
  • Gold Prices:
    • Model: P=400+50log10(5t+1)P = 400 + 50 \log_{10}(5t + 1), where tt is years since 1980.
    • In 1980 (t=0t=0), P=400+50log10(1)=$400P = 400 + 50 \log_{10}(1) = \$400.
  • Richter Scale:
    • Model: R=23log10K0.9R = \frac{2}{3} \log_{10} K - 0.9, where KK is energy in kilojoules (kJkJ).
    • Increasing the Richter value by 1 represents a significant increase in the energy released.
  • Sound Intensity (Audiology):
    • Formula: L=10log10(II0)L = 10 \log_{10}(\frac{I}{I_0}).
    • LL is sound intensity in decibels (dB).
    • II is sound intensity in watts per square metre (W/m2)\text{watts per square metre (W/m}^2).
    • I0I_0 is the reference threshold of hearing, usually 1.0×1012W/m21.0 \times 10^{-12}\,\text{W/m}^2.
  • Biology: Bacteria population growth, e.g., N=1500(20.18t)N = 1500(2^{0.18t}).
  • Physics: Cooling of a liquid, e.g., T=90(30.05t)T = 90(3^{-0.05t}) for coffee in a room.

Questions & Discussion

Career Profile: Alison Hennessy — Audiologist

  • Question 1: List two reasons why Alison finds mathematics useful in her profession.
    • Response: She uses formulas to prescribe amplification for hearing aids and calculates sound intensity levels in decibels. Additionally, math is required to interpret statistical tests in journal articles to verify manufacturer claims.
  • Question 2: Use the formula provided (L=10log10(II0)L = 10 \log_{10}(\frac{I}{I_0})) to calculate the sound intensity level for a jackhammer at a distance of 10 m, with an intensity of 1.5×103W/m21.5 \times 10^{-3}\,\text{W/m}^2.
    • Response: Using I=1.5×103I = 1.5 \times 10^{-3} and I0=1.0×1012I_0 = 1.0 \times 10^{-12}, the formula becomes L=10log10(1.5×1031.0×1012)=10log10(1.5×109)10(9.176)=91.76dBL = 10 \log_{10}(\frac{1.5 \times 10^{-3}}{1.0 \times 10^{-12}}) = 10 \log_{10}(1.5 \times 10^9) \approx 10(9.176) = 91.76\,\text{dB}.
  • Question 3: Find out what subjects are recommended to be undertaken in Year 12 to continue on to a tertiary course in this field.
    • Response: General recommendations usually include advanced sciences and mathematics such as Mathematical Methods.