Expanded Logarithms: Rules, Strategy, and Examples (Based on Transcript)

Expanded Logarithms: Rules, Strategy, and Examples

Step-by-step Strategy for Expanding Logarithms

• Goal: expand a single logarithm into a sum of logarithms, one term per factor inside the argument.
• Start by converting all radicals to rational exponents (so no radicals remain).
• If there is a multiplication inside the log, use the product rule to turn it into addition of logs.
• If there is a division inside the log, use the quotient rule to turn it into subtraction of logs.
• Apply the power rule:
  • 
    \logb(A^k) = k \cdot \logb(A)
  • Bring powers to the front as coefficients for each term inside the log.
• Important sign rule when expanding:
  • Terms on the top (numerator) become positive (added).
  • Terms in the bottom (denominator) become negative (subtracted).
• After expansion, simplify terms where possible (e.g., logb b = 1, logb 1 = 0).
• If there are common bases or powers, consider rewriting to simplify the expression further.

Converting Radicals to Rational Exponents

• Any radical: (\sqrt[n]{A}) becomes (A^{1/n}).
• For a product under a radical, apply the exponent to each factor:
  • Example: (\sqrt[3]{x^2 y^2 z^3} = x^{\tfrac{2}{3}} y^{\tfrac{2}{3}} z^{1}).
• This step is necessary before applying logarithm rules.

Product inside Log → Sum of Logs

• If the inside is a product: (\logb( A \cdot B \cdot C ) = \logb A + \logb B + \logb C).
• After factoring out powers (see below), each term contributes a coefficient in front of its logarithm.

Quotient inside Log → Difference of Logs

• If the inside is a quotient: (\logb\left( \frac{A}{B} \right) = \logb A - \log_b B).
• This yields negative terms for each factor in the denominator.

Handling Powers Within Logs

• When a factor is raised to a power, move the exponent to the front:
  • ( \logb (A^k) = k \logb A ).
• Practically: bring all exponents to the front of their respective logarithms as coefficients.
• Do not bring exponents past to other terms; apply to the correct term only.

Worked Examples

Example 1: log base 2 of the cube root of x^2 y^2 z^3

• Expression: \log_2 \sqrt[3]{x^2 y^2 z^3}
• Step 1 (radicals to rational exponents):
  \sqrt[3]{x^2 y^2 z^3} = x^{\tfrac{2}{3}} y^{\tfrac{2}{3}} z^{1}
• Step 2 (product → sum):
  \log2 \left( x^{\tfrac{2}{3}} y^{\tfrac{2}{3}} z^{1} \right) = \log2 \left( x^{\tfrac{2}{3}} \right) + \log2 \left( y^{\tfrac{2}{3}} \right) + \log2 \left( z^{1} \right)
• Step 3 (powers become coefficients):
  = \tfrac{2}{3} \log2 x + \tfrac{2}{3} \log2 y + 1 \cdot \log_2 z
• Final expanded form:
  \log2 \sqrt[3]{x^2 y^2 z^3} = \frac{2}{3}\log2 x + \frac{2}{3}\log2 y + \log2 z
• Simplification note: if any argument equals the base, use (\log_b b = 1) or other identities to simplify further.

Example 2: log base 3 of (3^{3/4} m^{1/2} n^{a} p^{b})

• Expression: \log_3\left( 3^{\tfrac{3}{4}} m^{\tfrac{1}{2}} n^{a} p^{b} \right)
• Step 1 (product inside log):
  = \log3(3^{\tfrac{3}{4}}) + \log3(m^{\tfrac{1}{2}}) + \log3(n^{a}) + \log3(p^{b})
• Step 2 (powers bring down):
  = \tfrac{3}{4} \log3 3 + \tfrac{1}{2} \log3 m + a \log3 n + b \log3 p
• Step 3 (simplify (\log3 3 = 1)): = \frac{3}{4} + \frac{1}{2} \log3 m + a \log3 n + b \log3 p
• Note: if any of the inner bases equal the log base (e.g., (m=3)), the evaluation can yield a constant term (e.g., (\log_3 3 = 1)).
• Quick observation: you can sometimes simplify further if specific values are given (e.g., if a = 0 or if n or p equals a power of 3).

Example 3: log base 2 of (\dfrac{32 x^3 y^2}{z^2})

• Expression: \log_2 \left( \dfrac{32 x^3 y^2}{z^2} \right)
• Step 1 (lower to lowest base): (32 = 2^5)
• Step 2 (expand as product and quotient):
  = \log2(2^5) + \log2(x^3) + \log2(y^2) - \log2(z^2)
• Step 3 (powers brought down):
  = 5 \log2 2 + 3 \log2 x + 2 \log2 y - 2 \log2 z
• Step 4 (simplify (\log2 2 = 1)): = 5 + 3 \log2 x + 2 \log2 y - 2 \log2 z
• Final expanded form: \log2\left(\dfrac{32 x^3 y^2}{z^2}\right) = 5 + 3\log2 x + 2\log2 y - 2\log2 z

Example 4: log base 5 of (\dfrac{25 x^{2/3} y^{2/3}}{w^3 c})

• Expression: \log_5\left( \dfrac{25 x^{\tfrac{2}{3}} y^{\tfrac{2}{3}}}{w^3 c} \right)
• Step 1 (rewrite 25 as (5^2)):
  = \log_5\left( \dfrac{5^2 x^{\tfrac{2}{3}} y^{\tfrac{2}{3}}}{w^3 c} \right)
• Step 2 (split across product/quotient):
  = \log5(5^2) + \log5\left( x^{\tfrac{2}{3}} \right) + \log5\left( y^{\tfrac{2}{3}} \right) - \log5(w^3) - \log_5 c
• Step 3 (powers bring down):
  = 2 \log5 5 + \tfrac{2}{3} \log5 x + \tfrac{2}{3} \log5 y - 3 \log5 w - \log_5 c
• Step 4 (simplify (\log5 5 = 1)): = 2 + \frac{2}{3} \log5 x + \frac{2}{3} \log5 y - 3 \log5 w - \log_5 c
• Final expanded form: \log5\left( \frac{25 x^{2/3} y^{2/3}}{w^3 c} \right) = 2 + \frac{2}{3}\log5 x + \frac{2}{3}\log5 y - 3\log5 w - \log_5 c

Natural-Log (ln) Version

• The natural log version uses the same rules with base e:
• Example (same structure as Example 1):
  \ln\left( \sqrt[3]{x^2 y^2 z^3} \right) = \frac{2}{3} \ln x + \frac{2}{3} \ln y + \ln z
• If the inside includes a factor like (e^{2/3}):
  \ln\left( e^{\tfrac{2}{3}} x^{1} y^{\tfrac{2}{3}} \right) = \frac{2}{3}\ln e + \ln x + \frac{2}{3}\ln y = \frac{2}{3} + \ln x + \frac{2}{3}\ln y
• Key reminder: ln is log base e, and the inverse relationships between exponentials and logs carry over with base e.

Inverse Functions and Compositions (fog in golf)

• Core idea: exponentials and logarithms are inverse functions.
• Practice: find f inverse for basic functions and verify by composition.

Basic example: Find the inverse of f(x) = 2x - 3

• Solve for x: y = 2x - 3 => x = (y + 3)/2
• Inverse function: f^{-1}(x) = \frac{x + 3}{2}
• Notation for inverse: you write f^{-1} in place of x inside the outer function when checking composition:
  • Check: f\left(f^{-1}(x)\right) = x and f^{-1}\left(f(x)\right) = x
• Concept: the composition of a function with its inverse yields the identity on the appropriate domain (the argument becomes the input variable).
• Domain considerations: the identity result reflects that the functions undo each other on the domain where both are defined.

Exponential and Log Inverses

• For same base b (> 0, b ≠ 1):
  • \log_b ( b^x ) = x
  • b^{\log_b x} = x
• These identities are used to simplify expanded logs and to switch between exponential and logarithmic forms.
• Important constraint: the inverses require the same base when applying these identities.

Practice Tips and Common Pitfalls (from the transcript)

• Always convert radicals to rational exponents first, then apply log rules.
• When expanding, convert to the lowest common base when dealing with numeric factors (e.g., 32 = 2^5, 9 = 3^2).
• If there is a fraction inside the log, treat the denominator as a subtraction: log(A/B) = log(A) - log(B).
• Do not forget to bring exponents down to the front as coefficients for each corresponding log term.
• After expansion, simplify constants when possible (e.g., logb b = 1, logb (b^k) = k).
• If you have multiple terms inside the log, you will usually end with multiple log terms with signs corresponding to numerator/denominator.
• For ln versions, use the same rules with base e; ln e = 1, so terms like ln(e^{2/3}) simplify to 2/3.
• Composer hint (fog in golf): remember f and g inverses satisfy f(g(x)) = x and g(f(x)) = x; this helps in recognizing inverse relationships between logs and exponentials.
• When selecting a base for conversion, aim for the lowest base possible to simplify coefficients (e.g., convert 32 to 2^5 rather than keeping 32).

Quick Reference: Key Formulas to Memorize

• Product rule: \logb(AB) = \logb A + \log_b B
• Quotient rule: \logb\left( \frac{A}{B} \right) = \logb A - \log_b B
• Power rule: \logb(A^k) = k \logb A \quad\text{and}\quad A^k\; \logb A^k = k \logb A
• Change of base (conceptual): exponentials and logarithms are inverses; use same base in a composition.
• Base restrictions: for log_b(x) to be defined, x > 0 and b > 0, b ≠ 1.

End of Notes