Problem 65 - Elementary and Intermediate Algebra

Question

Write logarithm as the sum and/or difference of logarithms of a single quantity. Then simplify, if possible. \(\log _{b} \sqrt{x y}\)

Step-by-Step Solution

Answer

Split into: \( \frac{1}{2} \log_{b}(x) + \frac{1}{2} \log_{b}(y) \).

Step 1: Apply the Square Root Logarithm Property

First, we identify the square root in the expression \( \sqrt{xy} \). The square root can be expressed as an exponent: \( (xy)^{1/2} \). Using the logarithm power property, we write this as \( \log_{b}((xy)^{1/2}) = \frac{1}{2} \log_{b}(xy) \).

Step 2: Apply the Product Logarithm Property

Since \( xy \) is the product of \( x \) and \( y \), we can use the logarithm product property: \( \log_{b}(xy) = \log_{b}(x) + \log_{b}(y) \). Substitute this into the expression from Step 1: \( \frac{1}{2} (\log_{b}(x) + \log_{b}(y)) \).

Step 3: Distribute the Coefficient

Now distribute \( \frac{1}{2} \) across the expression inside the parentheses: \( \frac{1}{2} \log_{b}(x) + \frac{1}{2} \log_{b}(y) \). This is the expression of the logarithm as the sum of logarithms.

Key Concepts

Square Root Property

When dealing with logarithms, encountering a square root might seem tricky at first. However, there's a straightforward way to handle it using properties of exponents. We know from exponent rules that taking the square root of a number is equivalent to raising it to the power of 1/2. So, for any expression under a square root, like \( \sqrt{xy} \), it can be rewritten as \((xy)^{1/2}\).
This transformation is the key step in simplifying the logarithm with a square root. Once rewritten in its exponential form, we can leverage the logarithm power property. This property tells us that \( \log_{b}(a^c) = c \cdot \log_{b}(a) \).
Thus, applying it here, we get:

• \( \log_{b}(\sqrt{xy}) = \log_{b}((xy)^{1/2}) \)
• This becomes \( \frac{1}{2} \cdot \log_{b}(xy) \)

Product Property of Logarithms

In our journey to simplify the logarithm, the next step involves understanding how to break down the product under the logarithmic sign. The product property of logarithms comes into play here.
This property states that the logarithm of a product is equal to the sum of the logarithms of the individual factors. For example, for any base \(b\) and positive numbers \(x\) and \(y\), we have:

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

Applying this property helps in splitting a complex logarithmic argument into simpler parts. Once we have \( \frac{1}{2} \cdot \log_{b}(xy) \) from the previous step, substituting the product property transforms it into \( \frac{1}{2} \cdot (\log_{b}(x) + \log_{b}(y)) \).
This step makes handling and simplifying such logarithmic expressions much more manageable.

Logarithm Simplification

Now that we've broken down the logarithmic expression into simpler parts, simplification becomes straightforward. Simplification, in this context, means distributing the coefficient across terms if applicable. Consider our expression from before: \( \frac{1}{2} \cdot (\log_{b}(x) + \log_{b}(y)) \).
To simplify further, we distribute the \( \frac{1}{2} \) to each term inside the brackets. This results in:

• \( \frac{1}{2} \cdot \log_{b}(x) + \frac{1}{2} \cdot \log_{b}(y) \)

This format is the simplest representation of the original logarithmic equation. It expresses the logarithm as the sum of simpler logarithmic terms. Such manipulation and breakdown help in better understanding and solving more complex logarithmic equations later on. Simplifying logs not only makes problem-solving easier but also enhances clarity in mathematical communication.