Problem 31 - Algebra and Trigonometry

Question

In Exercises 21–42, evaluate each expression without using a calculator. $$ \cdot \log _{2} \frac{1}{\sqrt{2}} $$

Step-by-Step Solution

Answer

The value of the expression \(\log _{2} \frac{1}{\sqrt{2}}\) is -0.5.

Step 1: Rewrite the fraction

First, we need to note that \(\frac{1}{\sqrt{2}}\) is equivalent to \(2^{-\frac{1}{2}}\). This is because the denominator \(\sqrt{2}\) equals \(2^{\frac{1}{2}}\), and when we have a number in the denominator, we can rewrite it in the numerator with a negative exponent. Hence \(\frac{1}{\sqrt{2}}\) can be written as \(2^{-\frac{1}{2}}\). So, the logarithmic expression becomes \(\log _{2} 2^{-\frac{1}{2}}\).

Step 2: Use the properties of logarithms

The base of the logarithm and the base of the exponent in the argument are the same, which allows us to use the property of logarithms that states \(\log _{b} b^{a} = a\). This means that the exponent can come down in front of the logarithm, as the logarithm to a certain base of a number raised to that base equals to the exponent of the number. Thus, \(\log _{2} 2^{-\frac{1}{2}} = -\frac{1}{2}\).

Key Concepts

Logarithmic Expressions

Understanding logarithmic expressions is a cornerstone of higher mathematics. While they may seem challenging at first, they follow specific rules that make them easier to handle.
A logarithm is essentially an exponent in disguise. It’s an operation that asks the question: to what power do we raise a given base number to get another number?
An expression like \(\log_b a\) is read as 'the logarithm of \(a\) to the base \(b\)'. The answer is the power you raise \(b\) to in order to get \(a\). In our exercise, we had \(\log_2 \frac{1}{\sqrt{2}}\), and this represents the power you need to raise 2 to obtain \(\frac{1}{\sqrt{2}}\). Understanding this concept is vital for working with these kinds of expressions without reaching for a calculator.

Properties of Logarithms

Logarithms aren't just mystical entities floating around in your math textbook – they follow very specific rules called properties. These properties are fantastic tools that can help you simplify and evaluate logarithmic expressions.

• Product Property: \(\log_b(m \cdot n) = \log_b m + \log_b n\)
• Quotient Property: \(\log_b(\frac{m}{n}) = \log_b m - \log_b n\)
• Power Property: \(\log_b(m^n) = n \cdot \log_b m\)

These rules allow us to break down complex logarithmic expressions into simpler forms that are easier to work with. The exercise you’ve encountered uses the Power Property to extract the exponent from a logarithm, greatly simplifying the evaluation process.

Evaluating Expressions Without Calculator

In the age of technology, it might be tempting to solve every math problem with a calculator. However, understanding the principles behind mathematical expressions allows for deeper learning and often faster solutions.
When evaluating logarithmic expressions, such as in the given exercise, we apply our knowledge of properties of logarithms and rewrite expressions into a form that we can solve mentally. For instance, recognizing that a number in the denominator with a square root can be written as a negative exponent is crucial. Then applying logarithm properties to simplify the expression further enables us to find the answer without the need for a calculator.
This not only improves your mental mathematical skills but also ensures you have a robust conceptual understanding, which is essential in more advanced topics.

Exponents and Radicals

Exponents and radicals are two sides of the same coin. An exponent refers to the number of times a number (the base) is multiplied by itself. A radical, on the other hand, is a root, such as a square root or cube root.
Understanding the relationship between exponents and radicals is essential when working with logarithms. Any radical can be expressed as a fraction in an exponent. For example, the square root of a number is the same as raising that number to the \(\frac{1}{2}\) power. The cube root would be the \(\frac{1}{3}\) power, and so on. This concept helps to demystify the process of dealing with square roots and higher-level roots in logarithmic expressions, leading to a clearer pathway to solving problems.