Problem 50 - Elementary and Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

\(-2 - \log_6 r\)

Step 1: Apply Logarithm of a Quotient Rule

Using the logarithm quotient rule, which states that \( \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N \), we can separate the logarithm as follows:\( \log_6 \frac{1}{36r} = \log_6 1 - \log_6 (36r) \).

Step 2: Simplify the Logarithm of 1

The logarithm of 1 to any base is always 0, so:\( \log_6 1 = 0 \). Therefore, the expression simplifies to:\( 0 - \log_6 (36r) \).

Step 3: Distribute the Negative Sign

Distribute the negative sign to get:\( -\log_6 36 - \log_6 r \).

Step 4: Simplify Using Logarithm Properties

Breaking down \( \log_6 36 \), note that \( 36 = 6^2 \). So, \( \log_6 36 = \log_6 6^2 = 2 \log_6 6 \). Further simplification gives:Since \( \log_6 6 = 1 \), thus \( 2 \log_6 6 = 2 \times 1 = 2 \).Therefore, the expression becomes:\( -2 - \log_6 r \).

Key Concepts

Logarithm of a Quotient

Understanding the logarithm of a quotient is an essential part of simplifying logarithmic expressions. When you have a fraction inside a logarithm, this is where the quotient rule for logarithms comes into play. The rule states that the logarithm of a quotient is equal to the difference of the logarithms:

• For instance, \( \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N \).

To apply this, consider our exercise case:

• Inside the logarithm, we have \( \frac{1}{36r} \),leading to:\( \log_6 \left( \frac{1}{36r} \right) = \log_6 1 - \log_6 (36r) \).

By breaking down complex fractions this way, we can start resolving and simplifying the individual parts.

Logarithm of a Product

Sometimes, within a logarithmic term, you come across a multiplication. Knowing the logarithm of a product rule can be incredibly handy here. This rule states that the logarithm of a product is the sum of the logarithms of the factors:

• So for a product, like,\( \log_b (M \cdot N) = \log_b M + \log_b N \).
• In our solved exercise involving \( \log_6(36r) \),we note back in our problem:\( \log_6 (36) + \log_6 (r) \).

Using the product rule, we can decompose multiplication into sum terms, simplifying bit by bit.

Logarithm Simplification

Logarithm simplification is key to solving logarithmic expressions efficiently. Simplifying logarithms often involves substituting known values or simplifying using known logarithmic bases:

• The first noticeable step here is worth considering \( \log_6 1 = 0 \) because a logarithm of any non-zero base of 1 is zero.
• Then we delve into: \( \log_6 36 \).Breaking it down utilizes recognizing powers of the base, like:\( 36 = 6^2 \), making it \( 2 \log_6 6 \).

Since \( \log_6 6 = 1 \),it effectively simplifies to the value 2. The cumulative result when rewriting takes into account all these little shifts.

Algebraic Expressions

Algebraic expressions, when coupled with logarithms, present interesting challenges. Different steps might be needed for simplification depending on the variables involved.

• In our example, this is seen in\( \log_6 r \).Here the focus is on keeping the variable expressions simple while logging its behavior.
• These steps illustrate how algebraic terms can be organized or split up, depending on whole numbers or known logarithm identities.
• Combining these helps in making the expression of the problem as straightforward as possible: \( -2 - \log_6 r \).

Simplifying doesn't end in formulaic changes. It also means recognizing where algebraic values fit into the logarithm solution path.