Algebra and Trigonometry: Properties of Logarithms

Properties of Logarithms and Core Identities

  • Foundational Definitions and Domains:   - For all properties established in this section, MM and aa must be positive real numbers.   - The base aa must satisfy the condition a1a \neq 1.   - The variable rr represents any real number.

  • Theorem: Basic Properties of Logarithms (1 of 4):   - The Exponentiation Property: The number loga(M)\log_a(M) is the specific exponent to which the base aa must be raised to obtain the value MM.     - Mathematically expressed as: aloga(M)=Ma^{\log_a(M)} = M   - The Inverse Power Property: The logarithm with base aa of the base aa raised to a specific power is equal to that power.     - Mathematically expressed as: loga(ar)=r\log_a(a^r) = r

  • Establishing Fundamental Logarithmic Values (Example 1):   - Proving loga(1)=0\log_a(1) = 0:     - This identity was originally established through graphing, but can be proven algebraically.     - Let y=loga(1)y = \log_a(1).     - Convert to exponential form: ay=1a^y = 1.     - Since any base raised to the power of 00 is 11, we solve for yy to find y=0y = 0.     - Conclusion: loga(1)=0\log_a(1) = 0   - Proving loga(a)=1\log_a(a) = 1:     - Let y=loga(a)y = \log_a(a).     - Convert to exponential form: ay=aa^y = a.     - Since a1=aa^1 = a, we solve for yy to find y=1y = 1.     - Conclusion: loga(a)=1\log_a(a) = 1

  • Examples of Basic Property Application (Example 2):   - Scenario A: log0.2(0.2)2\log_{0.2}(0.2)^{-\sqrt{2}} results in 2-\sqrt{2} by applying loga(ar)=r\log_a(a^r) = r.   - Scenario B: ln(ekt)\ln(e^{kt}) results in ktkt (noting that the base of the natural log is ee).   - Scenario C: 2log2(π)2^{\log_2(\pi)} results in π\pi by applying aloga(M)=Ma^{\log_a(M)} = M.

Expansion and Condensation of Logarithmic Expressions

  • Theorem: Operational Properties of Logarithms (2 of 4):   - The Product Rule: The log of a product is equal to the sum of the logs of its factors.     - loga(MN)=loga(M)+loga(N)\log_a(MN) = \log_a(M) + \log_a(N)   - The Quotient Rule: The log of a quotient is equal to the difference of the logs of the dividend and divisor.     - loga(MN)=loga(M)loga(N)\log_a\left(\frac{M}{N}\right) = \log_a(M) - \log_a(N)

  • Theorem: Power Property of Logarithms (3 of 4):   - The Power Rule: The log of a value raised to a power is equal to the product of that power and the log of the value.     - loga(Mr)=rloga(M)\log_a(M^r) = r\log_a(M)

  • Expansion Examples (Writing as a Sum or Difference):   - Example 3: Expanding loga(xx2+1)\log_a(x\sqrt{x^2+1}) as a sum of logarithms.     - By the Product Rule: loga(x)+loga(x2+1)\log_a(x) + \log_a(\sqrt{x^2+1})     - Expressing the radical as a power: loga(x)+loga((x2+1)12)\log_a(x) + \log_a((x^2+1)^{\frac{1}{2}})     - Using the Power Rule to express exponents as factors: loga(x)+12loga(x2+1)\log_a(x) + \frac{1}{2}\log_a(x^2+1)   - Example 4: Expanding a quotient expression as a difference of logarithms while expressing all powers as factors.   - Example 5: Expanding a complex expression involving products, quotients, and square roots into a sum and difference of logarithms, ensuring all powers are moved to the front as factors.

  • Condensation Examples (Example 6: Writing as a Single Logarithm):   - Scenario A: loga(7)+loga(3)=loga(7×3)=loga(21)\log_a(7) + \log_a(3) = \log_a(7 \times 3) = \log_a(21).   - Scenario B: 2loga(x)+12loga(9)2\log_a(x) + \frac{1}{2}\log_a(9).     - Apply the Power Rule in reverse: loga(x2)+loga(912)\log_a(x^2) + \log_a(9^{\frac{1}{2}}).     - Simplify the constant: loga(x2)+loga(3)\log_a(x^2) + \log_a(3).     - Apply the Product Rule: loga(3x2)\log_a(3x^2).   - Scenario C: loga(x+1)loga(x)=loga(x+1x)\log_a(x+1) - \log_a(x) = \log_a\left(\frac{x+1}{x}\right).

Equality and Approximation Techniques

  • Theorem: Logarithmic Equality (4 of 4):   - If M=NM = N, then loga(M)=loga(N)\log_a(M) = \log_a(N).   - Conversely, if loga(M)=loga(N)\log_a(M) = \log_a(N), then M=NM = N.   - This assumes MM, NN, and aa are positive real numbers and a1a \neq 1.

  • Approximation without Common Bases (Example 7):   - Task: Approximate log4(17)\log_4(17) and round to four decimal places.   - Conceptual Logic: The expression log4(17)\log_4(17) asks: "4 raised to what exponent equals 17?"   - Let y=log4(17)y = \log_4(17).   - Estimation:     - 42=164^2 = 16     - 43=644^3 = 64     - Therefore, the value of yy must be between 22 and 33, and significantly closer to 22 than to 33.   - Numerical Result: Rounded to four decimal places, the value is approximately 2.04372.0437.

The Change-of-Base Formula

  • Theorem: Change-of-Base Formula:   - This formula allows for the conversion of a logarithm with any valid base aa into a ratio of logarithms with a new base bb.   - If a1a \neq 1, b1b \neq 1, and MM are positive real numbers, then: loga(M)=logb(M)logb(a)\log_a(M) = \frac{\log_b(M)}{\log_b(a)}   - In practice, the formula is most useful when converting to base 1010 (common logarithm) or base ee (natural logarithm) because these are standard on calculators:     - Base 10: loga(M)=log(M)log(a)\log_a(M) = \frac{\log(M)}{\log(a)}     - Base e: loga(M)=ln(M)ln(a)\log_a(M) = \frac{\ln(M)}{\ln(a)}

  • Application of Change-of-Base (Example 8):   - Scenario A: Approximate log5(89)\log_5(89).     - Calculation: ln(89)ln(5)2.7889\frac{\ln(89)}{\ln(5)} \approx 2.7889   - Scenario B: Approximate log2(5)\log_{\sqrt{2}}(\sqrt{5}).     - Calculation: log(5)log(2)2.3219\frac{\log(\sqrt{5})}{\log(\sqrt{2})} \approx 2.3219

Graphing with Non-Standard Bases (Example 9)

  • Procedure for Graphing Utility:   - Graphing utilities typically only have keys for log10\log_{10} (log\log) and loge\log_e (ln\ln).   - To graph a function like y=log2(x)y = \log_2(x), one must use the Change-of-Base Formula.   - Step 1: Express the function in terms of base 1010 or base ee.     - Conversion: y=log(x)log(2)y = \frac{\log(x)}{\log(2)} or y=ln(x)ln(2)y = \frac{\ln(x)}{\ln(2)}.   - Step 2: Input either of these quotient expressions into the graphing utility to render the graph of y=log2(x)y = \log_2(x).