Problem 40 - Algebra 2

Question

Use the properties of logarithms to evaluate each expression. \(\frac{1}{2} \log _{5} 1-2 \log _{5} 5\)

Step-by-Step Solution

Answer

-2

Step 1: Identify and Apply Logarithmic Properties

Firstly, identify the logarithmic properties in the given expression. The expression can be rewritten using the properties of logarithms as follows: \[ \frac{1}{2} \log_{5} 1 - 2 \log_{5} 5 \]This expression has two terms. The first term is \(\frac{1}{2} \log_{5} 1\), and the second term is \(2 \log_{5} 5\). For the first term, remember that the logarithm of 1 in any base is 0. Thus, \(\log_{5} 1 = 0\). For the second term, the power rule of logarithms can be used, which states that \(\log_{b}(a^n) = n \cdot \log_{b} a\). In this case, the base \(b\) is 5, the number \(a\) is 5, and the power \(n\) is 1 (since 5 is \(5^1\)). Thus, \(\log_{5} 5 = 1\). Multiply this result by the 2 on the outside of the logarithm.

Step 2: Execute the Operations

Now, apply these results back to the original equation to execute the operations.\[ \frac{1}{2} \cdot 0 - 2 \cdot 1 = 0 - 2 = -2 \]

Key Concepts

Logarithm of 1

When working with logarithmic expressions, it's important to understand that the logarithm of 1, regardless of the base, is always 0. This rule comes from the definition of a logarithm. Logarithms are the inverse of exponents. So, when asking the question, "To what power must we raise our base to achieve 1?", the answer will always be 0. For any base \( b\), \( \log_{b} 1 = 0\).

This is because any non-zero number raised to the power of 0 equals 1. Whether you have \( \log_{2} 1\), \( \log_{10} 1\), or \( \log_{5} 1\), the result is always 0. Recognizing this allows you to simplify various logarithmic expressions quickly, without needing any additional calculation.

• For example, \( \ \frac{1}{2} \cdot \log_{5} 1 = \ \frac{1}{2} \cdot 0 = 0\), which simplifies any further calculations on that term.

Power Rule of Logarithms

The power rule of logarithms is an essential property that allows you to manipulate and simplify logarithmic expressions involving powers. The rule states: \( \log_{b}(a^n) = n \cdot \log_{b} a\).

It essentially lets you "move the exponent" in front of the logarithm, turning a multiplication problem into a more manageable form. The rule can be very useful when you're faced with complex expressions or when evaluating logarithms like in our exercise.

If you have a term like \( 2 \cdot \log_{5} 5\), it can be looked at using \( \log_{5} 5\), noting that 5 is \( 5^{1}\). Thus, \( \log_{5} 5 = 1\), because the base and the input are the same, leading to the exponent 1.

• Alternatively, calculating \( n \cdot \log_{b} b = n\) gives 2 in this expression since \( 2 \cdot 1 = 2\).

Using this rule correctly lets you transform expressions into simpler forms that are easier to calculate.

Evaluating Logarithmic Expressions

Evaluating logarithmic expressions involves applying known properties, like the ones previously described, to simplify and solve expressions. It often requires recognizing how to translate parts of expressions using these rules.

For instance, consider an expression like \( \frac{1}{2} \cdot \log_{5} 1 - 2 \cdot \log_{5} 5\).

• The first step involves breaking down each term using the properties. The first part, \( \log_{5} 1\), simplifies directly to 0 as explained.
• The second part, \( 2 \cdot \log_{5} 5\), simplifies to 2, acknowledging the power rule property.

Putting these steps together, we evaluate the expression as \( 0 - 2 = -2\). By recognizing which properties apply, you can move from a complex-looking expression to a simple numerical result.

This process highlights the importance of understanding the fundamental rules of logarithms, as they offer a structured approach to tackle different values and expressions efficiently.