Problem 37 - Algebra 2 and Trigonometry

Question

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

Step-by-Step Solution

Answer

\(\log_{2} x^{2}y = 2A + B\).

Step 1: Express the Problem in Logarithmic Form

The given expression is \( \log_{2}(x^2 y) \). We need to express this in terms of \( A \) and \( B \), given \( \log_{2} x = A \) and \( \log_{2} y = B \).

Step 2: Apply Logarithm Power Rule

The logarithm power rule states that \( \log_{b}(m^n) = n \log_{b} m \). Apply this rule to \( x^2 \) in our expression: \[ \log_{2}(x^2) = 2 \log_{2} x = 2A. \]

Step 3: Separate the Logarithm of a Product

We use the property of logarithms that \( \log_{b}(mn) = \log_{b}m + \log_{b}n \). Apply this to the expression \( \log_{2}(x^2 y) \): \[ \log_{2}(x^2 y) = \log_{2}(x^2) + \log_{2} y. \]

Step 4: Substitute Known Logarithm Values

Substitute the previously calculated \( \log_{2}(x^2) = 2A \) and \( \log_{2} y = B \) into the expression: \[ \log_{2}(x^2 y) = 2A + B. \]

Key Concepts

Power Rule for Logarithms

Let's start with understanding the power rule for logarithms. This is an essential tool when dealing with exponential expressions in logarithmic form. The rule states that if you have a logarithm \(\log_b(m^n)\), it can be simplified to \(n \cdot \log_b(m)\). In simple terms, you can "bring down" the exponent as a multiplier in front of the logarithm.

In the context of our exercise, where the expression is \(\log_{2}(x^2 y)\), the power rule allows us to take the 2 out of the exponent in \(x^2\), transforming it into \(2 \cdot \log_{2}(x)\). This gives us \(2A\), because we know \(\log_{2}(x) = A\).

The power rule simplifies complex logarithmic expressions and is especially useful when dealing with variable exponents. Always remember, the exponent becomes a coefficient! This conversion is pivotal as it makes calculations more straightforward, turning complex operations into simpler addition or subtraction tasks.

Product Rule for Logarithms

Following the first step, let's move on to the product rule for logarithms. If you have a product inside a logarithm, such as \(\log_b(mn)\), this rule states that it can be split into a sum of two separate logarithms: \(\log_b(m) + \log_b(n)\).

In our given expression \(\log_{2}(x^2 y)\), after applying the power rule to \(x^2\), you still have the product \(x^2 \cdot y\). By using the product rule, you can split this into \(\log_{2}(x^2) + \log_{2}(y)\).

• The first term \(\log_{2}(x^2)\) becomes \(2A\) thanks to the power rule.
• The second term stays as \(\log_{2}(y)\), which is simply \(B\) since \(\log_{2}(y) = B\).

By applying the product rule, we break down complicated multiplication within a logarithm into simpler additions, making it much easier to handle and interpret.

Substitution in Logarithms

Once you've applied both the power and product rules in our logarithmic expression, the final step is substitution. Substitution is a method of replacing a given value within an expression based on known equalities.

In the step-by-step solution, after applying the power and product rules, we reached an important stage. We had the following transformed expression: \(\log_{2}(x^2) + \log_{2}(y)\). With our given values, \(\log_{2}(x) = A\) and \(\log_{2}(y) = B\), you simply substitute these into the expression.

• Instead of \(\log_{2}(x^2)\), use \(2A\) since \(\log_{2}(x) = A\) and power rule gives us \(2 \cdot \log_{2}(x)\).
• Instead of \(\log_{2}(y)\) just use \(B\).

So now, the final expression \(\log_{2}(x^2 y)\) simplifies neatly into \(2A + B\). Substitution helps resolve equations into known quantities, making your solution clear and concise.