Problem 32 - Algebra 2 and Trigonometry

Question

In \(27-56,\) evaluate each logarithmic expression. Show all work. $$ 8 \log _{\frac{1}{2}} 2 $$

Step-by-Step Solution

Answer

The evaluated expression is \(-8\).

Step 1: Identify the Base Change

We identify that we are dealing with a logarithm where the base is \( \frac{1}{2} \) and the argument is \(2\). This can be expressed using logarithm properties for easier calculation.

Step 2: Apply Logarithmic Property

We use the property that \( \log_b a = \frac{\log_c a}{\log_c b} \), where \( b = \frac{1}{2} \), \( a = 2 \), and choosing base \(10\) for simplification: \[ \log_{\frac{1}{2}} 2 = \frac{\log_{10} 2}{\log_{10} \left(\frac{1}{2}\right)} \]

Step 3: Simplify the Expression

Knowing that \( \log_{10} \left(\frac{1}{2}\right) = \log_{10} 1 - \log_{10} 2 = 0 - \log_{10} 2 \), we can simplify the log expression: \[ \log_{\frac{1}{2}} 2 = \frac{\log_{10} 2}{-\log_{10} 2} = -1 \]

Step 4: Calculate the Final Result

We substitute \( \log_{\frac{1}{2}} 2 = -1 \) back into the original expression: \[ 8 \log_{\frac{1}{2}} 2 = 8 \times (-1) = -8 \]

Key Concepts

Change of Base Formula

The change of base formula is a useful tool in working with logarithms that allows us to convert a logarithm to a different base, usually one that is easier to compute. This formula is expressed as follows:
\[\log_b a = \frac{\log_c a}{\log_c b}\]
Essentially, it says that the logarithm of a number with base \(b\) can be calculated by dividing the logarithm of the number with a new base \(c\) by the logarithm of the original base \(b\) with the same new base \(c\). Often, calculator-friendly bases like \(10\) or \(e\) are chosen for \(c\) because they are built into calculator functions.

• This formula is handy when dealing with bases that are difficult to compute directly.
• For example, computing \(\log_{\frac{1}{2}} 2\) can be tricky, but converting it to base \(10\) using the change of base formula simplifies the process.

Once the expression is in the common base, it's easier to evaluate using standard calculator functions, providing a pathway to solving logarithms without memorizing various base calculations.

Logarithmic Properties

Logarithmic properties are rules that simplify the manipulation and transformation of logarithmic expressions. These properties allow more complex expressions to be broken down into simpler components. Some of the essential properties include:

• Product Property: \(\log_b (xy) = \log_b x + \log_b y\)
• Quotient Property: \(\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y\)
• Power Property: \(\log_b (x^y) = y \cdot \log_b x\)

In our specific example, we apply the logarithmic property related to changing the base:
\[\log_{\frac{1}{2}} 2 = \frac{\log_{10} 2}{\log_{10} (\frac{1}{2})}\]This property transforms the original expression into a calculable form by identifying the ratio of logarithms in base \(10\).
Understanding these properties is a stepping stone to simplifying and computing complex logarithmic expressions, making seemingly daunting math tasks more approachable and logical.

Simplifying Logarithms

Simplifying logarithms involves using rules and properties to reduce a complex expression to its simplest form. This step is crucial in making further calculations manageable and accurate. In the exercise, simplifying the expression involved using logarithmic properties.
Initially, we identified the problematic log expression:

• \(\log_{10} \left(\frac{1}{2}\right)\) simplifies by the quotient property, becoming \(\log_{10} 1 - \log_{10} 2\).
• Because \(\log_{10} 1 = 0\), this further simplifies to \(-\log_{10} 2\).

With these simplifications, the calculation proceeds much more smoothly:
\[\log_{\frac{1}{2}} 2 = \frac{\log_{10} 2}{-\log_{10} 2} = -1\]Reconciling this within the original expression \(8 \log_{\frac{1}{2}} 2\), simplifies to \(-8\). This illustrates how leveraging simplification techniques enables solving exercises neat and fast, highlighting the beauty and power of algebraic manipulation in mathematics.