Problem 45 - Algebra 2 and Trigonometry

Question

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

Step-by-Step Solution

Answer

\( \frac{1}{2}A - \frac{1}{2}B \).

Step 1: Express as Exponents

The expression \( \log_{2} \sqrt{\frac{x}{y}} \) involves a square root. Recall that a square root is equivalent to raising a number to the power of \( \frac{1}{2} \). Thus, we can rewrite the expression as \( \log_{2} \left( \frac{x}{y} \right)^{\frac{1}{2}} \).

Step 2: Apply the Power Rule

Using the logarithmic power rule \( \log_{b}(m^n) = n \cdot \log_{b}(m) \), apply it to the expression \( \log_{2} \left( \frac{x}{y} \right)^{\frac{1}{2}} \). This gives us \( \frac{1}{2} \cdot \log_{2} \left( \frac{x}{y} \right) \).

Step 3: Use the Quotient Rule

Apply the logarithmic quotient rule \( \log_{b} \frac{m}{n} = \log_{b}(m) - \log_{b}(n) \) to the expression \( \log_{2} \left( \frac{x}{y} \right) \). This gives us \( \log_{2}(x) - \log_{2}(y) \).

Step 4: Substitute Known Values

Substitute the given values \( \log_{2}(x) = A \) and \( \log_{2}(y) = B \) into the expression \( \log_{2}(x) - \log_{2}(y) \). This results in \( A - B \).

Step 5: Simplify the Expression

Substitute \( A - B \) back into the expression from Step 2: \( \frac{1}{2} \cdot (A - B) \). Simplifying this results in \( \frac{1}{2}A - \frac{1}{2}B \).

Key Concepts

Logarithmic Power Rule

The logarithmic power rule is a useful property when dealing with exponents inside a logarithmic expression. It states that for any real number base, such as base 2 in this case, and any real numbers \(m\) and \(n\), the rule is expressed as:

• \(\log_{b}(m^n) = n \cdot \log_{b}(m)\)

This means you can "bring down" the exponent as a coefficient in front of the logarithm. Let's apply this rule to the expression \(\log_{2} \left( \frac{x}{y} \right)^{\frac{1}{2}}\). Here, the exponent \(\frac{1}{2}\) from the square root can be moved in front of the log, resulting in:

• \(\frac{1}{2} \cdot \log_{2} \left( \frac{x}{y} \right)\)

This step is crucial as it simplifies the application of the next rules by reducing complexity inside the log expression.

Logarithmic Quotient Rule

The logarithmic quotient rule is another important property that helps simplify expressions where division is involved inside a logarithm. This rule is represented by:

• \(\log_{b} \frac{m}{n} = \log_{b}(m) - \log_{b}(n)\)

By applying this rule, you can split the logarithm of a division into the difference of two logarithms, simplifying calculations. In our scenario, we apply it to \(\log_{2} \left( \frac{x}{y} \right)\). By doing this, we get:

• \(\log_{2}(x) - \log_{2}(y)\)

Replacing these with the given expressions for \(\log_{2}(x) = A\) and \(\log_{2}(y) = B\), the expression simplifies to:

• \(A - B\)

This simplification sets the stage for easier manipulation of the log terms.

Expressions in Terms of Variables

Expressions in terms of variables involve substituting known values for variables or expressions to simplify an equation. In the exercise, substituting given values into the logarithmic equation is essential to finding a solution.
In our example, the problem provides two variables: \( \log_{2} x = A \) and \( \log_{2} y = B \). After applying the power rule, and the quotient rule, we arrive at the simplified expression \(A - B\).
The final step involves substituting these known variables into the previously complex expression. By doing so, it simplifies to \( \frac{1}{2}(A - B) \), then further simplifies to:

• \(\frac{1}{2}A - \frac{1}{2}B\)

This approach enables us to convey results of complex logarithmic equations in terms of basic variables like \(A\) and \(B\), providing an intuitive understanding.

Properties of Exponents

Properties of exponents are the foundational rules that dictate how exponents operate within algebraic expressions. A critical aspect of this math problem involves understanding that a square root \(\sqrt{m}\) can be expressed as an exponent of \(\frac{1}{2}\). Hence,

• \(\sqrt{m} = m^{\frac{1}{2}}\)

This basic property allows us to transform roots into exponent form, which can then be easily manipulated using the logarithmic power rule. In our given exercise, recognizing that \(\sqrt{\frac{x}{y}}\) can be rewritten as \((\frac{x}{y})^{\frac{1}{2}}\) allowed further application of logarithmic rules.

• Knowing these properties facilitates a deeper understanding and handling of more complex logarithmic and algebraic expressions effectively.