Problem 28 - Precalculus Mathematics for Calculus

Question

Evaluate the expression. (a) \(\log _{4} \sqrt{2}\) (b) \(\log _{4}\left(\frac{1}{2}\right)\) (c) \(\log _{4} 8\)

Step-by-Step Solution

Answer

(a) \(\frac{1}{4}\), (b) \(-\frac{1}{2}\), (c) \(\frac{3}{2}\).

Step 1: Simplify log expression (a)

The expression is \( \log_{4} \sqrt{2} \). We know that \( \sqrt{2} = 2^{1/2} \). By the power rule of logarithms, \( \log_{b}a^{n} = n \cdot \log_{b}a \), we can simplify as follows:\[\log_{4} \sqrt{2} = \log_{4} 2^{1/2} = \frac{1}{2} \cdot \log_{4} 2.\]

Step 2: Calculate log value for expression (a)

Next, we need to know \( \log_{4} 2 \). Since \( 4 = 2^{2} \), then \( \log_{4} 2 = \frac{1}{2} \log_{2} 2 = \frac{1}{2} \times 1 = \frac{1}{2} \). Thus:\[\log_{4} \sqrt{2} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.\]

Step 3: Simplify log expression (b)

The expression is \( \log_{4} \left( \frac{1}{2} \right) \). We can rewrite \( \frac{1}{2} \) as \( 2^{-1} \). Applying the power rule of logarithms:\[\log_{4} \left( \frac{1}{2} \right) = \log_{4} 2^{-1} = -1 \cdot \log_{4} 2.\]

Step 4: Calculate log value for expression (b)

We already found that \( \log_{4} 2 = \frac{1}{2} \). Thus:\[\log_{4} \left( \frac{1}{2} \right) = -1 \cdot \frac{1}{2} = -\frac{1}{2}.\]

Step 5: Simplify log expression (c)

The expression is \( \log_{4} 8 \). We can express 8 as \( 2^{3} \), and recognize that \( 4 = 2^{2} \). Thus, we can convert the base and rewrite:\[\log_{4} 8 = \log_{4} (2^3) = 3 \cdot \log_{4} 2.\]

Step 6: Calculate log value for expression (c)

We know \( \log_{4} 2 = \frac{1}{2} \). Therefore,\[\log_{4} 8 = 3 \cdot \frac{1}{2} = \frac{3}{2}.\]

Key Concepts

power rule of logarithms

The power rule of logarithms is an essential concept that simplifies logarithmic expressions. This rule states that when dealing with an expression of the form \( \log_{b} a^{n} \), it can be rewritten as \( n \cdot \log_{b} a \).
This transformation comes in handy, especially when dealing with exponential terms inside a logarithm.

• For example, if you have \( \log_{4} \sqrt{2} \), you recognize \( \sqrt{2} \) as \( 2^{1/2} \).
• Applying the power rule, this becomes \( \frac{1}{2} \cdot \log_{4} 2 \).

Breaking down such expressions using the power rule allows you to deal with simpler logarithmic terms individually, making calculations easier. Remember, the power \( n \) can be any real number, allowing this rule to be quite versatile.

• This rule is useful for solving logarithms involving roots, fractions, and any term with an exponent.

base conversion

Base conversion in logarithms may seem tricky at first, but with practice, it becomes an intuitive process. The idea is to express the logarithm in a different base, which is sometimes necessary for simplification.
The change-of-base formula is often useful, but in this exercise, recognizing equivalent terms in terms of a common base simplifies our work.

• Consider \( \log_{4} 2 \). Since \( 4 = 2^{2} \), you can rewrite this as \( \log_{2^{2}} 2 \).
• Logarithm properties allow you to express this as \( \frac{1}{2} \cdot \log_{2} 2 \).

Base conversion helps transform the expression into a form where simple logarithmic identities or rules apply, enabling straightforward evaluation.

• This technique maintains the conceptual understanding of the relation between bases and helps in simplifying the given logarithmic expression.

evaluating logarithms

Evaluating logarithms is the final step when you have applied the necessary rules to simplify an expression. This involves computing the actual value of the logarithm.
Once the expression is in a simplified form, like \( n \cdot \log_{b} a \), the evaluation becomes straightforward.

• For instance, to evaluate \( \log_{4} \sqrt{2} \), simplify it to \( \frac{1}{4} \) after applying the power rule and understanding \( \log_{4} 2 \).
• Similarly, evaluating \( \log_{4} \frac{1}{2} \) leads to \( -\frac{1}{2} \).
• For \( \log_{4} 8 \), the simplification using the base conversion makes it easy to find \( \frac{3}{2} \).

Logarithms may initially appear complex, but with practice in applying these techniques, one can quickly become proficient in evaluating them. Always ensure each step connects logically to the next, making understanding and error-checking easier.