Problem 24 - Algebra and Trigonometry

Question

\(15-24\) Evaluate the expression. $$ \begin{array}{llll}{\text { (a) } \log _{4} \sqrt{2}} & {\text { (b) } \log _{4}\left(\frac{1}{2}\right)} & {} & {\text { (c) } \log _{4} 8}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Simplifying Part (a)

Evaluate \( \log_{4} \sqrt{2} \). The square root of 2 can be expressed as 2 raised to the power of 1/2, thus \( \sqrt{2} = 2^{1/2} \). Applying the logarithm power rule \( \log_b x^n = n \log_b x \), we have:\[\log_{4}(2^{1/2}) = \frac{1}{2} \log_{4}(2)\]To solve \( \log_{4}(2) \), remember that \( 4 = 2^2 \), so the change of base formula gives us:\[\log_{4}(2) = \frac{\log_{2}(2)}{\log_{2}(4)} = \frac{1}{2}\]This results in:\[\log_{4}(\sqrt{2}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\]

Step 2: Simplifying Part (b)

Evaluate \( \log_{4}\left(\frac{1}{2}\right) \). Let's use the logarithm quotient rule, \( \log_b \frac{x}{y} = \log_b x - \log_b y \):\[\log_{4}\left(\frac{1}{2}\right) = \log_{4}(1) - \log_{4}(2) \]We know that \( \log_{4}(1) = 0 \) because any number raised to the power of 0 is 1, so:\[\log_{4}\left(\frac{1}{2}\right) = 0 - \log_{4}(2) = -\frac{1}{2}\]

Step 3: Simplifying Part (c)

Evaluate \( \log_{4}(8) \). Express 8 as a power of 2: \( 8 = 2^3 \). Using the power rule of logarithms:\[\log_{4}(2^3) = 3 \cdot \log_{4}(2)\]From Part (a), we know \( \log_{4}(2) = \frac{1}{2} \), so:\[3 \times \frac{1}{2} = \frac{3}{2}\]Thus, \( \log_{4}(8) = \frac{3}{2} \).

Key Concepts

Logarithm Power Rule

The logarithm power rule is a very helpful rule when dealing with logarithmic expressions involving powers. It states that for any logarithm of the form \( \log_b{x^n} \), it can be simplified to \( n \cdot \log_b{x} \). This means you can bring the exponent in front of the logarithm as a multiplier.

• Let's say you have \( \log_4{2^{1/2}} \). According to the power rule, this can be rewritten as \( \frac{1}{2} \cdot \log_4{2} \).
• This is particularly useful for simplifying expressions and makes it easier to solve logarithmic problems with integer or fractional exponents.
• Another example: if you have \( \log_5{125} \), and knowing \( 125 = 5^3 \), it becomes \( 3 \cdot \log_5{5} \), further simplifying to \( 3 \), because any number log to its own base is 1.

The logarithm power rule is thus an essential tool in evaluating logarithms with exponential terms.

Logarithm Quotient Rule

When you encounter a logarithmic expression with a division, like \( \log_b{\left( \frac{x}{y} \right)} \), the quotient rule can simplify it. This rule states that \( \log_b{\left( \frac{x}{y} \right)} = \log_b{x} - \log_b{y} \).

• For example, take \( \log_4{\left( \frac{1}{2} \right)} \). This can be broken down into \( \log_4{1} - \log_4{2} \).
• We know \( \log_4{1} = 0 \) because any base raised to 0 equals 1, so this simplifies to \(-\log_4{2}\) which you've already calculated as \(-\frac{1}{2} \).
• Using the quotient rule makes handling divisions in logarithms intuitive and manageable.

The logarithm quotient rule provides a straightforward way to break down complex logarithmic expressions, especially when division is involved.

Change of Base Formula

Sometimes, you need to find a logarithm that isn’t easily solvable with its current base. That’s where the change of base formula becomes incredibly useful. It states that for any logarithm \( \log_b{x} \), you can convert it to another base, usually a more convenient one. The formula is: \[\log_b{x} = \frac{\log_k{x}}{\log_k{b}}\] Here, \( k \) can be any positive base useful for calculation, such as 10 or the natural log base \( e \).

• For example, to evaluate \( \log_4{2} \), using \( k = 2 \), you can rewrite it as \( \frac{\log_2{2}}{\log_2{4}} \).
• Since \( \log_2{2} = 1 \) and \( \log_2{4} = 2 \), this simplifies to \( \frac{1}{2} \).
• This method can simplify expressions and make solving them much easier, especially when dealing with uncommon bases.

The change of base formula is a powerful tool in mathematics, providing flexibility and ease in evaluating logarithms.

Evaluating Logarithmic Expressions

Evaluating logarithmic expressions involves breaking down the problem into simpler parts using various logarithmic rules. It usually requires a good grasp of other mathematical operations like exponentiation and manipulation using base conversion.

• For example, evaluating \( \log_4{8} \) involves expressing 8 in terms of base 2, making it \( 2^3 \).
• Using the logarithm power rule, it becomes \( 3 \cdot \log_4{2} \), which simplifies to \( \frac{3}{2} \).
• Practically, every logarithmic evaluation uses a combination of power rule, quotient rule, and sometimes the change of base formula, depending on the complexity and structure of the expression.

Understanding these steps and rules allows you to tackle complex logarithmic expressions efficiently and accurately.