Problem 40 - 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 \(y\) $$ \ln \frac{1}{2} $$

Step-by-Step Solution

Answer

\(\ln \frac{1}{2} = -x\)

Step 1: Recall log property for reciprocal

The logarithm of a reciprocal can be expressed using the property \(\ln \left( \frac{1}{a} \right) = -\ln a\). Apply this property to find \(\ln \frac{1}{2}\).

Step 2: Substitute in given values

Given \(\ln 2 = x\), we substitute into the property: \(-\ln 2 = -x\). Thus, \(\ln \frac{1}{2} = -x\).

Key Concepts

Logarithmic Properties

Logarithmic properties play an essential role in simplifying and solving logarithmic expressions. These properties are similar to the rules we use for basic arithmetic operations. They help us break down complex logarithmic expressions into simpler forms, making it easier to understand and work with them effectively.

Some important logarithmic properties include:

• Product Property: The logarithm of a product is the sum of the logarithms. This is expressed as \(\log_b (mn) = \log_b m + \log_b n\).
• Quotient Property: The logarithm of a quotient is the difference of the logarithms. It is written as \(\log_b \left( \frac{m}{n} \right) = \log_b m - \log_b n\).
• Power Property: The logarithm of a number raised to an exponent is the exponent times the logarithm of the number. This can be expressed as \(\log_b (m^n) = n \cdot \log_b m\).
• Change of Base Formula: This allows the conversion of logarithms from one base to another, given by \(\log_b m = \frac{\log_k m}{\log_k b}\).

These properties are fundamental tools when dealing with logarithmic functions and equations. By transforming and simplifying expressions using these properties, complex problems become easier to solve.

Reciprocal Logarithms

A reciprocal logarithm refers to the logarithm of a reciprocal. This concept uses the logarithmic property \(\ln \left( \frac{1}{a} \right) = -\ln a\).

This property indicates that the logarithm of the reciprocal of a number is simply the negative of the logarithm of the original number. As demonstrated in the exercise, when you have a reciprocal like \(\frac{1}{2}\), you can apply this property:

• Start with \(\ln \left( \frac{1}{a} \right) = -\ln a\).
• For \(\ln \frac{1}{2}\), applying the property gives \(-\ln 2\).
• Given that \(\ln 2 = x\), substituting it results in \(-x\).

This method of dealing with reciprocal logarithms allows for straightforward simplification in these cases. Understanding this property allows students to transform and simplify logarithms of reciprocals easily.

Logarithm Transformation

Logarithm transformation involves changing logarithmic expressions to simplify them or solve equations. This transformation is often necessary for handling complex logarithmic terms or when solving exponential equations.

A common type of transformation is using logarithmic properties to express complex logs in terms of simpler, known values. For example, using given values like \(\ln 2 = x\) and \(\ln 3 = y\), we can express other logarithms in terms of \(x\) and \(y\) by applying various logarithmic properties.

• Through the reciprocal property used in the original exercise, \(\ln \frac{1}{2}\) becomes \(-x\) because \(\ln 2 = x\).
• Such transformations depend on a fundamental understanding of simplifying expressions using known values.

Transforming logarithmic expressions using known logarithmic identities and properties allows you to tackle more advanced problems effectively. This transformation is key in using logs to analyze and interpret data in fields like mathematics and the sciences.