Problem 87 - College Algebra Essentials

Question

Let \(\log _{b} 2=A\) and \(\log _{b} 3=C .\) Write each expression in terms of \(A\) and \(C\). \(\log _{b} \sqrt{\frac{2}{27}}\)

Step-by-Step Solution

Answer

\(\frac{1}{2}(A - 3C)\)

Step 1: Rewrite the expression

Firstly, rewrite the given expression using the property of logarithms that states the logarithm of a square root can be written as half of the original logarithm. This yields: \(\log _{b}\sqrt{\frac{2}{27}} = \frac{1}{2}\log_{b}\frac{2}{27}\)

Step 2: Apply the logarithm of a fraction rule

Next, apply another property of logarithms, the logarithm of a fraction rule, to split the log fraction into subtraction of two logs: \(\frac{1}{2}\log_{b}\frac{2}{27} = \frac{1}{2}(\log _{b} 2 - \log _{b} 27)\)

Step 3: Break down \(\log _{b} 27\)

Break down \(\log _{b} 27\) further. As 27 can be written as \(3^{3}\), this step yields: \(\frac{1}{2}(\log _{b} 2 - \log _{b} 27) = \frac{1}{2}(\log _{b} 2 - \log _{b} 3^{3})\)

Step 4: Apply the power rule of logarithms

Use the power rule, which states \(\log_b{a^n} = n \log_b{a}\), to process the \(\log _{b} 3^{3}\). This step gives: \(\frac{1}{2}(\log _{b} 2 - \log _{b} 3^{3}) = \frac{1}{2}(\log _{b} 2 - 3 \log _{b} 3)\)

Step 5: Substitute given values

Substitute \(\log _{b} 2\) as \(A\) and \(\log _{b} 3\) as \(C\) according to the problem. This results in: \(\frac{1}{2}(\log _{b} 2 - 3 \log _{b} 3) = \frac{1}{2}(A - 3C)\)

Step 6: Simplify the expression

Simplify the expression to find the final answer: \(\frac{1}{2}(A - 3C)\)

Key Concepts

Logarithm Properties

Understanding the properties of logarithms is essential for solving equations that involve these interesting mathematical concepts. Three fundamental properties can help simplify or manipulate expressions:

• Product Property: The logarithm of a product is the sum of the logarithms of its individual factors. Mathematically, this is expressed as \( \log_b (xy) = \log_b x + \log_b y \).
• Quotient Property: The logarithm of a quotient is the difference between the logarithms of the numerator and denominator: \( \log_b \left( \frac{x}{y} \right) = \log_b x - \log_b y \).
• Power Property: The logarithm of a number raised to an exponent can be simplified by bringing the exponent in front of the log: \( \log_b (x^n) = n \log_b x \).

These properties make it easier to manipulate and solve complex logarithmic expressions. For example, in simplifying \( \log_b \sqrt{\frac{2}{27}} \), the expression benefits from these properties to arrive at a form that uses simpler terms \( A \) and \( C \).

Logarithm Rules

The rules of logarithms might seem intimidating at first glance, but they are instrumental in breaking down complex expressions. Each rule serves a specific purpose:

• Change of Base Rule: It allows you to transform logs into different bases, useful for computation with non-standard bases. It states:\( \log_b a = \frac{\log_c a}{\log_c b} \).
• Log of One Rule: The log of one, regardless of base, is always zero: \( \log_b 1 = 0 \).
• Base Rule: Where the base equals the value, the result is one: \( \log_b b = 1 \).

The application of these rules is crucial, particularly with multi-step problems such as simplifying logarithmic fractions. For example, when confronted with \( \frac{2}{27} \), the logarithmic fraction rule allows you to express it in terms of simple subtraction of two logs rather than a single complex log expression.

Power Rule of Logarithms

Among the most powerful tools in logarithmic manipulation is the power rule of logarithms. This rule is especially useful when dealing with expressions where a number inside the logarithm is raised to a power. The power rule states:

When \( x^n \) is inside a logarithm, you can "take out" the exponent, placing it in front: \( \log_b (x^n) = n \log_b x \).
For example, in the expression \( \log_b 3^3 \), applying the power rule simplifies it to \( 3 \log_b 3 \). This simplification is directly used in exercises to make them more manageable and understandable.

Employing the power rule allows for straightforward substitutions, such as exchanging \( \log_b 3 \) for \( C \), and reducing heavy logarithmic expressions into manageable terms like \( \frac{1}{2}(A - 3C) \). That's why it's one of the first rules you should try out when dealing with logs raised to powers.