Problem 78 - Algebra for College Students

Question

Express each of the following as the sum or difference of simpler logarithmic quantities. Assume that all variables represent positive real numbers. For example, $$ \begin{aligned} \log _{b} \frac{x^{3}}{y^{2}} &=\log _{b} x^{3}-\log _{b} y^{2} \\ &=3 \log _{b} x-2 \log _{b} y \end{aligned} $$ $$ \log _{b} \sqrt{x y} $$

Step-by-Step Solution

Answer

\( \log_{b} \sqrt{xy} = \frac{1}{2} \log_{b} x + \frac{1}{2} \log_{b} y \).

Step 1: Understand the problem

We need to express \( \log_{b} \sqrt{xy} \) as a sum or difference of simpler logarithmic expressions. We will use the properties of logarithms to do this.

Step 2: Apply the square root property

Recall that \( \sqrt{xy} = (xy)^{1/2} \). Therefore, \( \log_{b} \sqrt{xy} = \log_{b} (xy)^{1/2} \).

Step 3: Use the logarithm power rule

The power rule states that \( \log_{b} (a^c) = c \log_{b} a \). Using this rule, we have \( \log_{b} (xy)^{1/2} = \frac{1}{2} \log_{b} (xy) \).

Step 4: Apply logarithm product rule

The product rule for logarithms is \( \log_{b} (ab) = \log_{b} a + \log_{b} b \). Thus, \( \frac{1}{2} \log_{b} (xy) = \frac{1}{2} (\log_{b} x + \log_{b} y) \).

Step 5: Distribute the fraction

Distribute the \( \frac{1}{2} \) across the sum: \( \frac{1}{2} \log_{b} x + \frac{1}{2} \log_{b} y \).

Key Concepts

Logarithm Power Rule

When working with logarithmic expressions, one of the fundamental rules to know is the logarithm power rule. This rule is incredibly useful for simplifying expressions. It tells us that if you have a logarithm with a power, such as \( \log_{b} (a^c) \), you can bring the power down in front of the logarithm as a coefficient.

This means you transform \( \log_{b} (a^c) \) into \( c \log_{b} a \), where \( c \) is the exponent of \( a \). This rule makes expressions easier to manage and further simplifies them when combined with other properties.

• For example, \( \log_{b} (xy)^{1/2} \) becomes \( \frac{1}{2} \log_{b} (xy) \), because you bring the power \( \frac{1}{2} \) down as a coefficient.
• This rule is particularly handy in breaking down complicated expressions into simpler parts that are easier to work with.

Logarithm Product Rule

The logarithm product rule is another essential property that helps in simplifying logarithmic expressions. According to this rule, the logarithm of a product is the sum of the logarithms of its factors.

Mathematically, the product rule states: \( \log_{b} (ab) = \log_{b} a + \log_{b} b \). This rule can be applied to any two numbers or expressions \( a \) and \( b \) being multiplied together inside a logarithm.

• Taking an example from our exercise, if you have \( \frac{1}{2} \log_{b} (xy) \) you can break this down to \( \frac{1}{2} (\log_{b} x + \log_{b} y) \).
• This allows further manipulation, as each term in the equation becomes a separate, simpler logarithmic expression to work with independently.
• Using this rule often works in tandem with other rules, like the power rule, to break down complex logarithmic expressions into more manageable components.

Simplifying Logarithms

Simplifying logarithms is the process of using rules, like the logarithm power rule and the logarithm product rule, to make logarithmic expressions easier to work with. The goal is to re-express a more complex logarithmic statement in terms of simpler logarithmic quantities, often as a sum or difference of terms.

In the given exercise, through careful application of these rules, we transformed \( \log_{b} \sqrt{xy} \) into simpler components. We initially rewrote \( \sqrt{xy} \) as \( (xy)^{1/2} \), then applied the power and product rules.

• This is a multi-step process that involves:
  • Changing the format to expose powers and products.
  • Using the power rule to bring exponents forward.
  • Applying the product rule to break down the expression further.
• In our example, the final simplification yielded \( \frac{1}{2} \log_{b} x + \frac{1}{2} \log_{b} y \), which is a much cleaner, more manageable form.

By using these simplification techniques, tackling challenges involving logarithms becomes less daunting and more straightforward.