Problem 38 - Algebra 2 and Trigonometry

Question

Write each expression in terms of \(A\) and \(B\) if \(\log _{2} x=A\) and \(\log _{2} y=B\). \(\log _{2}(x y)^{3}\)

Step-by-Step Solution

Answer

The expression \(\log_2(xy)^3\) in terms of \(A\) and \(B\) is \(3A + 3B\).

Step 1: Recognize the Logarithm Property

We start by recognizing that the logarithm property of multiplication states that \(\log_b(MN) = \log_b M + \log_b N\). Therefore, in our case where we have \(\log_2(xy)^3\), we will use this property inside another property that deals with exponents.

Step 2: Apply the Exponentiation Property

The expression \(\log_2(xy)^3\) can be simplified using the exponential property of logarithms \(\log_b(M^n) = n\log_b M\). This means that \(\log_2(xy)^3 = 3 \cdot \log_2(xy)\).

Step 3: Apply the Multiplication Property of Logarithms

Next, apply the multiplication property of logarithms, which states \(\log_2(xy) = \log_2 x + \log_2 y\). So, \(\log_2(xy) = A + B\) because we were given \(\log_2 x = A\) and \(\log_2 y = B\).

Step 4: Combine All Parts

Substitute \(\log_2(xy) = A + B\) into \(3\log_2(xy)\) to get \(3(A + B)\). Thus, \(\log_2(xy)^3 = 3(A + B)\).

Step 5: Distribute the 3

Finally, distribute the 3 across the terms within the parenthesis: \(3(A + B) = 3A + 3B\). Therefore, the expression is written in terms of \(A\) and \(B\) as \(3A + 3B\).

Key Concepts

Understanding the Multiplication Property of Logarithms

The multiplication property of logarithms is a fundamental tool when dealing with logarithmic expressions. This property makes it easier to simplify logarithmic functions that involve the product of two numbers.

• Essentially, this property states: \(\log_b(MN) = \log_b M + \log_b N\).
• In practical terms, if you're given the logarithms of two separate numbers, you can find the logarithm of their product by simply adding the two logarithms together.

To see this in action, consider the expression \(\log_2(xy)\) from our exercise. Here, by applying the multiplication property, we separate the variables into \(\log_2 x + \log_2 y\).
This allows us to rewrite the complex expression in a simpler form using known values, like \(A = \log_2 x\) and \(B = \log_2 y\).
The multiplication property not only helps simplify expressions but also demonstrates how logarithmic calculations easily relate to basic addition.

Applying the Exponentiation Property of Logarithms

The exponentiation property of logarithms is another important tool that simplifies expressions by breaking them down into understandable parts. This property is incredibly useful when logarithmic expressions involve powers or exponents.

• The property can be written as: \(\log_b(M^n) = n\log_b M\).
• This means that the logarithm of a power is just the exponent multiplied by the logarithm of the base itself.

In our specific case, the expression \(\log_2(xy)^3\) uses this property to transform into \(3\log_2(xy)\). This is achieved by recognizing that the exponent 3 in \((xy)^3\) can be taken out front, thus simplifying your calculation significantly.
The exponentiation property further clarifies how complex powers in logarithms can be distilled into simpler, linear expressions, thus making calculations much quicker and more intuitive.

Exploring Logarithmic Expressions in Terms of Known Values

Logarithmic expressions can often be simplified or rewritten in terms of known values, which makes solving equations more straightforward. In our exercise, we started with \(\log_2(xy)^3\) and ultimately transformed it into a cleaner expression using properties of logarithms.First, we applied the exponentiation property to factor out the exponent, converting it into \(3\log_2(xy)\). Then, the multiplication property helped rewrite \(\log_2(xy)\) as \(\log_2 x + \log_2 y\).
This meant we could express it in terms of the given variables \(A\) and \(B\), known as \(A = \log_2 x\) and \(B = \log_2 y\), resulting in \(3(A + B)\). By distributing the 3, we end up with the final expression \(3A + 3B\).
This approach helps you break down and conquer challenging logarithmic expressions step by step, dramatically simplifying the process of solving logarithmic equations when you have defined components like \(A\) and \(B\). It highlights how the restructuring of terms leads to clearer and more manageable mathematical expressions.