Problem 50 - Intermediate Algebra

Question

Write each logarithm as the sum and/or difference of logarithms of a single quantity. Then simplify, if possible. See Example 4. $$ \log _{6} \frac{1}{36 r} $$

Step-by-Step Solution

Answer

\(\log_{6} \frac{1}{36 r} = -\log_{6}(36) - \log_{6}(r)\)

Step 1: Identify the Structure of the Logarithm

We begin by examining the expression \(\log_{6} \frac{1}{36r}\) which is a logarithm of a quotient. In the step ahead, we'll apply the quotient rule to break this down.

Step 2: Apply the Quotient Rule

The quotient rule for logarithms states that \(\log_{b}(\frac{x}{y}) = \log_{b}(x) - \log_{b}(y)\). Using this, we write: \(\log_{6}(1) - \log_{6}(36r)\).

Step 3: Simplify the Logarithm of a Constant

Since \(\log_{6}(1) = 0\) (because any log with a base of any number for 1 is 0), we can simplify the expression from Step 2 to \(-\log_{6}(36r)\).

Step 4: Expand Using the Product Rule

Finally, apply the product rule \(\log_{b}(xy) = \log_{b}(x) + \log_{b}(y)\) on \(\log_{6}(36r)\). This becomes \(- (\log_{6}(36) + \log_{6}(r))\), which simplifies to \(-\log_{6}(36) - \log_{6}(r)\).

Key Concepts

Quotient Rule

When you encounter a logarithm that involves a quotient, like \( \log_{b} \frac{x}{y} \), the quotient rule comes into play. This rule helps simplify the expression by breaking the logarithm of a fraction into a difference of logarithms. The rule is expressed as:

• \( \log_{b}(\frac{x}{y}) = \log_{b}(x) - \log_{b}(y) \)

To understand why this works, remember that division corresponds to subtraction in the world of exponents and logarithms. Therefore, breaking it down as a difference makes it simpler to interpret. Consider the example from our exercise:

We begin with \( \log_{6} \frac{1}{36r} \). By identifying the expression as a quotient, we apply the quotient rule:

• \( \log_{6}(1) - \log_{6}(36r) \)

This first step allows us to see how the rule transforms a complex-looking fraction into a more manageable form. It's important to note that \( \log_{6}(1) \) becomes 0, since any base logarithm of 1 equals 0.

Product Rule

After simplifying our initial query using the quotient rule, we come across a product within a logarithmic expression. The product rule for logarithms becomes useful here. It helps simplify expressions where the logarithm of a product needs to be expanded or simplified. The rule states:

• \( \log_{b}(xy) = \log_{b}(x) + \log_{b}(y) \)

This rule is grounded in multiplication concepts from exponential expressions. Just as multiplication corresponds to addition in exponents, it corresponds to addition in logarithms. Looking back at our example, after applying the quotient rule, we encounter:

• \(-\log_{6}(36r) \)

We further simplify it using the product rule:

• \(-(\log_{6}(36) + \log_{6}(r)) \)

By doing so, we are separating the factors within the product, which offers a way to simplify our expressions further. Breaking it down this way helps students better manage more complex logarithmic expressions.

Logarithmic Simplification

Logarithmic simplification involves breaking down complex logarithmic expressions into simpler, equivalent forms. By applying rules like the quotient and product rules, these expressions become easier to manipulate and understand.

• The goal of simplification is to make operations on logarithms straightforward and intuitive.

In our example, we began with \( \log_{6} \frac{1}{36r} \). After applying both the quotient and product rules, we ended with:

• \(-\log_{6}(36) - \log_{6}(r) \)

Each step, from applying the rules to simplifying constants, removes complexity from the expression through careful manipulation. Simplifying \( \log_{6}(1) = 0 \) is one such example of simplifying constants.
The method behind simplification helps in increasing one's fluency with logarithms and builds up confidence for more complex algebraic or calculus problems. By constantly practicing these steps, students become adept at identifying and applying the most efficient strategies for simplifying logarithms.