Problem 43 - Algebra 2 and Trigonometry

Question

Write each expression in terms of \(A\) and \(B\) if \(\log _{2} x=A\) and \(\log _{2} y=B\). \(\log _{2} x \sqrt{y}\)

Step-by-Step Solution

Answer

The expression is \( A + \frac{1}{2}B \).

Step 1: Understanding the Expression

We start with the expression \( \log_{2} x \sqrt{y} \). This can be rewritten using the property \( \log_{b}(MN) = \log_{b}M + \log_{b}N \), which says that the logarithm of a product is the sum of the logarithms.

Step 2: Applying the Logarithm Property

Using the property from the previous step, the expression becomes \( \log_{2}x + \log_{2}\sqrt{y} \). We separate the logarithm of the product into two parts.

Step 3: Simplifying \(\log_{2}\sqrt{y}\)

Since \( \sqrt{y} = y^{1/2} \), we can apply the property \( \log_{b}M^{p} = p \cdot \log_{b}M \), resulting in \( \log_{2}y^{1/2} = \frac{1}{2} \log_{2}y \).

Step 4: Substituting Given Terms

Now that we have \( \log_{2}x + \frac{1}{2}\log_{2}y \), substitute the given terms \( \log_{2}x = A \) and \( \log_{2}y = B \). This gives us \( A + \frac{1}{2}B \).

Step 5: Final Simplification

The expression \( \log_{2} x \sqrt{y} \) in terms of \(A\) and \(B\) is \( A + \frac{1}{2}B \).

Key Concepts

Properties of Logarithms

Logarithms might sound complex, but they follow simple rules that make calculations easier. One vital property is the product rule, which states \( \log_{b}(MN) = \log_{b}M + \log_{b}N \). This rule explains that when multiplying two numbers inside a logarithm, it's equivalent to adding their individual logarithms.
Another useful property is the power rule, formulated as \( \log_{b}M^{p} = p \cdot \log_{b}M \). This tells us that if we are dealing with a number raised to a power, we can "bring down" the exponent and multiply it by the logarithm of the base number.
Understanding these properties is like holding keys to unlock complex logarithmic expressions and turn them into more manageable problems. They simplify tasks that seem complicated, allowing better focus on understanding the structure of the expression itself.

Logarithmic Identities

Logarithmic identities are like the shortcuts in a maze. They guide us in simplifying expressions. For example, knowing that \( \log_{b}M = A \) and \( \log_{b}N = B \) allows us to express combinations of these logarithms elegantly.
The primary identities include:

• Product identity: \( \log_{b}(MN) = \log_{b}M + \log_{b}N \)
• Quotient identity: \( \log_{b}\left(\frac{M}{N}\right) = \log_{b}M - \log_{b}N \)
• Power identity: \( \log_{b}M^{p} = p \cdot \log_{b}M \)

Using these identities simplifies work with large expressions and helps in transforming equations into terms that we can readily evaluate and manipulate.

Algebraic Manipulation

Algebraic manipulation involves rearranging expressions to simplify or solve them. In the context of logarithms, this means using logarithmic identities and properties to rewrite equations in terms you can easily work with.
Let's look at our example problem: \( \log_{2}x \sqrt{y} \). Initially, it looks tricky, but with some manipulation:

• Apply the product rule: break it into \( \log_{2}x + \log_{2}\sqrt{y} \)
• Recognize that \( \sqrt{y} = y^{1/2} \)
• Apply the power rule to simplify further to \( \log_{2}x + \frac{1}{2}\log_{2}y \)

This transformation is only possible by grasping algebraic manipulation techniques. We then substitute known values (\( A \) for \( \log_{2}x \) and \( B \) for \( \log_{2}y \)) making it \( A + \frac{1}{2}B \).
Mastering algebraic manipulation allows us to engage with equal parts precision and creativity, turning daunting problems into crystal-clear solutions.