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The change of base formula is an essential tool in mathematics, especially when dealing with complex logarithmic expressions. At its core, this formula allows us to convert logarithms from one base to another, simplifying the calculation process.
Suppose we want to convert the logarithm of a number to a different base. The change of base formula is expressed as:

• \( \log_{b} x = \frac{\log_{k} x}{\log_{k} b} \)

Here, \( k \) is a new base, often chosen to be 10 (common logarithms) or \( e \) (natural logarithms), which computer and calculator engineers find convenient. This conversion allows for easier computation using a standard set of base values.
In the given exercise, we employed this formula to convert \( \log_{3b} 2 \) and \( \log_{3b} 15 \) into expressions involving known values: \( A, B, \) and \( C \). By re-expressing these logs with a base \( b \), we can leverage already defined logarithmic operations for simplification.

Simplifying logarithmic expressions requires a solid understanding of logarithm properties. These include product, quotient, and power rules, as well as recognizing specific logarithmic identities.
For example, the expression \( \log_{b}(xy) \) can be simplified using the product rule to \( \log_{b}(x) + \log_{b}(y) \). Similarly, the quotient rule states that \( \log_{b}(\frac{x}{y}) = \log_{b}(x) - \log_{b}(y) \).
In our exercise, to handle expressions like \( \log_{b} 3b \), we apply these identities. We split the log into \( \log_{b} 3 + \log_{b} b \). Recognizing that \( \log_{b} b = 1 \) is crucial, as it simplifies to add 1 to the logarithm of 3, resulting in \( B + 1 \).
These simplifications not only make calculations manageable but are vital for converting complex expressions into simplified terms that relate directly to known quantities like \( A, B, \) and \( C \).

Precalculus encompasses a wide range of mathematical concepts vital for calculus readiness, including logarithmic expressions. Tackling logarithm problems involves understanding both complex derivations and simplifications.
Take the task of evaluating \( \log_{3b} 15 \). Initially, 15 is expressed as the product of two factors, 3 and 5. Using the product rule in logarithm properties, we represent this as \( \log_{3b} (3 \times 5) = \log_{3b} 3 + \log_{3b} 5 \).
Additionally, using the change of base formula allows breaking this complex expression into simpler base \( b \) components, such as \( \frac{B}{B+1} \) for \( \log_{3b} 3 \) and \( \frac{C}{B+1} \) for \( \log_{3b} 5 \). Completing this requires summing these sub-expressions to arrive at a simplified form \( \frac{B+C}{B+1} \).
Familiarity with these techniques and simplifications empowers students to handle precalculus problems confidently, setting the stage for future calculations involving more advanced concepts.