Logarithmic Difference of Logs: Quotient and Power Rules

Transformation Overview

Target expression (as given): \log_a\left(\frac{x^3}{(x-4)^2}\right)\quad\text{with } x>4\text{ (as stated in the transcript)}
Real-logarithm rule reminder: the argument of a logarithm must be positive.
The transcript intends to write the log of a quotient as a difference of logs.

In general, for base a>0, a\neq 1, the domain requires the inside of each logarithm to be positive.
Original expression requires \frac{x^3}{(x-4)^2}>0.
For the quotient, $(x-4)^2$ is positive for all x\neq 4, and x^3>0 when x>0. Hence, the simplest common domain is x>0, x\neq 4.
The transcript explicitly states x>4 to ensure the log arguments are positive in the step-by-step transformation, which aligns with the subsequent form \loga x and \loga(x-4) (both must be defined when written separately). Note: if you use the final separated form, you must have x>4 for both logs to be defined; the original log requires x>0, x\neq 4.
Base constraint: a>0, a\neq 1.

Start with the log of a quotient:
\loga\left(\frac{x^3}{(x-4)^2}\right) = \loga(x^3) - \loga((x-4)^2) (Quotient rule: \loga\left(\frac{M}{N}\right) = \loga M - \loga N.)
Express each power as a factor using the power rule:
- \loga(x^3) = 3\loga x
- \loga((x-4)^2) = 2\loga(x-4)
Substitute back:
\loga\left(\frac{x^3}{(x-4)^2}\right) = 3\loga x - 2\log_a(x-4)
Comment on the limit of further simplification: this is as far as the transcript goes, since you cannot express a difference of the arguments inside a single log using standard log rules; the expression is equivalent to the original via the quotient rule, and the separate logs reflect the product/quotient structure.

Final simplified form:
\loga\left(\frac{x^3}{(x-4)^2}\right) = 3\loga x - 2\log_a(x-4)
Domain considerations for the final form:
- To evaluate \log_a x, require x>0.
- To evaluate \log_a(x-4), require x-4>0\Rightarrow x>4.
- Therefore, the final expression (in terms of these individual logs) is defined for x>4, given the transcript’s intent.
Base condition reminder: a>0, a\neq 1.

Let x=5 and a=10:
- Original expression:
  \log{10}\left(\frac{5^3}{(5-4)^2}\right) = \log{10}\left(\frac{125}{1}\right) = \log_{10}125 \approx 2.09691
- Right-hand side after transformation:
  3\log{10}5 - 2\log{10}(5-4) = 3\cdot 0.69897 - 2\cdot \log_{10}1 = 2.09691
This numerical check confirms the algebraic equivalence under the domain conditions.

Product rule: \loga(MN) = \loga M + \log_a N
Quotient rule: \loga\left(\frac{M}{N}\right) = \loga M - \log_a N
Power rule: \loga(M^k) = k \loga M
Domain rules:
- Base: a>0, a\neq 1
- Arguments must be positive: for any log term, its argument > 0
Conceptual takeaway: Logs convert multiplication and division into addition and subtraction; bringing down exponents converts powers into multiples of logs.

Domain drift: Moving from the single-log form to a sum/difference of logs imposes the separate domain constraints of each log (e.g., x>4 for \log_a(x-4)). The original single-log form only requires x>0, x\neq 4. Always check domain after transformation.
Forgetting the power rule: Mistakes often occur by not applying \loga(x^3) = 3\loga x or \loga((x-4)^2) = 2\loga(x-4).
Misinterpreting the “cannot express a difference of the argument of a logarithm” note: you cannot represent a difference inside the log argument as a simple product/division outside the log unless you apply the standard rules (which is what’s done here).

This transformation illustrates how logarithms convert multiplicative relationships into additive ones, which is foundational in algorithms, data analysis, and modeling of power-law relationships.
In practical data work, logarithmic transformations help linearize exponential growth and stabilize variance, facilitating linear modeling and interpretation of effects.
Philosophical note: Logarithms encapsulate the idea of measuring multiplicative change through additive changes—an intuitive bridge between growth rates and cumulative effects.

Practice 1: Express \log_a\left(\frac{x^4}{(x-2)^3}\right) as a difference of logarithms and then bring down powers.
Practice 2: For a=2 and x=6, verify numerically that \log2\left(\frac{6^3}{(6-4)^2}\right) = 3\log2 6 - 2\log_2(6-4).
Practice 3: State the domain of the original log expression and compare with the domain implied by the separated logs form.

Logarithms turn products and quotients into sums and differences, and exponents into multiples; applying these rules step by step yields a clean separation of the factors inside a log, while preserving the value of the expression under the appropriate domain.