Problem 42 - Algebra 2 and Trigonometry

Question

In \(33-44,\) if \(\ln 2=x\) and \(\ln 3=y,\) write each of the natural logs in terms of \(x\) and \(y\) $$ \ln \frac{1}{36} $$

Step-by-Step Solution

Answer

\(-2x - 2y\)

Step 1: Understanding the Expression

You need to express \( \frac{1}{36} \) in terms of \(x\) and \(y\) where \(\ln 2 = x\) and \(\ln 3 = y\). We know that \(36 = 6^2 = (2 \cdot 3)^2\), hence \(\frac{1}{36} = (2 \cdot 3)^{-2}\).

Step 2: Use of Logarithm Properties

Utilize the logarithm property \(\ln \left(a^{-b}\right) = -b \ln a\). Thus, \(\ln \left((2 \cdot 3)^{-2}\right) = -2 \ln (2 \cdot 3)\).

Step 3: Breaking Down the Product

Use the property \(\ln(ab) = \ln a + \ln b\). Therefore, \(\ln (2 \cdot 3) = \ln 2 + \ln 3\) which becomes \(x + y\).

Step 4: Final Simplification

We substitute "\(x+y\)" for "\(\ln (2 \cdot 3)\)" into the previous expression: \(-2(x + y)\). This simplifies to \(-2x - 2y\).

Key Concepts

Logarithm Properties

Understanding properties of logarithms is essential in solving expressions involving logarithms. It's important to remember that logarithms behave in accordance with specific rules that help us simplify and manipulate these mathematical expressions:

• The first key property is the Power Rule: \( \ln(a^{-b}) = -b \ln(a) \). This tells us how to handle logarithms where the argument of the log is raised to a power. The power can be brought in front of the log and is then multiplied by the logarithm of the base.


• Another important property is the Multiplication Rule: \( \ln(ab) = \ln(a) + \ln(b) \). This means if you have the log of a product, you can split it into the sum of logs. This property is especially useful when breaking down expressions containing products.


• There's also the Division Rule: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \). This property allows you to express the log of a quotient as the difference of logs.

These properties help us to break complex logarithmic expressions into simpler parts. They are invaluable tools when performing algebraic manipulations and making difficult expressions more manageable.

Logarithmic Expressions

When working with logarithmic expressions like \( \ln\left( \frac{1}{36} \right) \), it's essential to deconstruct them using known terms. In this case, you are given that \( \ln 2 = x \) and \( \ln 3 = y \). To express \( \ln\left( \frac{1}{36} \right) \) in these terms, acknowledge that 36 is composed of factors 2 and 3.

First, notice that 36 can be written as \( 6^2 \) because 6 multiplied by itself gives 36. Further breaking down, \( 6 = 2 \times 3 \), which means \( 36 = (2 \times 3)^2 \). Hence, \( \frac{1}{36} = (2 \times 3)^{-2} \). This allows us to apply the Power Rule from logarithmic properties.

By skillfully using the properties such as the Power Rule and Multiplication Rule, complex expressions can be transformed into simpler, more manageable forms. Recognizing factorization and being able to decompose numbers to exploit their base compositions is critical in dealing with logarithmic expressions.

Algebraic Manipulation

Expressing logs in terms of given variables requires careful algebraic manipulation using the properties of logarithms. Initially, upon finding that \( \frac{1}{36} = (2 \cdot 3)^{-2} \), we apply the Power Rule: \( \ln(a^{-b}) = -b \ln(a) \). This transitions the original expression into \(-2 \ln(2 \cdot 3) \).

Next, using the Multiplication Rule, \( \ln(ab) = \ln a + \ln b \), we break down \( \ln(2 \cdot 3) \) into \( \ln 2 + \ln 3 \), which in terms of given variables \(x\) and \(y\), becomes \(x + y\).

Substituting this back into our modified expression, we get \(-2(x + y)\). Expanding this expression involves distributing the -2 to each term inside the parentheses, resulting in \(-2x - 2y\). This step-by-step manipulation showcases how algebra and properties of logarithms work in tandem to solve and simplify expressions fundamentally.