Problem 30 - Algebra and Trigonometry

Question

Use the Laws of Logarithms to expand the expression. $$ \log _{5} \sqrt{\frac{x-1}{x+1}} $$

Step-by-Step Solution

Answer

\( \frac{1}{2} \log_{5}(x-1) - \frac{1}{2} \log_{5}(x+1) \)

Step 1: Convert square root to exponent

Start by converting the square root into an exponent. The square root of an expression is equivalent to raising the expression to the 1/2 power. Thus, we can rewrite the expression as: \( \log_{5} \left( \frac{x-1}{x+1} \right)^{1/2} \).

Step 2: Use power rule for logarithms

Use the power rule of logarithms which states \( \log_{b}(a^n) = n \cdot \log_{b}(a) \). Apply this to the expression to move the exponent outside the logarithm: \( \frac{1}{2} \cdot \log_{5} \left( \frac{x-1}{x+1} \right) \).

Step 3: Apply quotient rule for logarithms

Use the quotient rule for logarithms which states \( \log_{b}(\frac{a}{c}) = \log_{b}(a) - \log_{b}(c) \). Apply it to \( \log_{5} \left( \frac{x-1}{x+1} \right) \): \( \log_{5}(x-1) - \log_{5}(x+1) \).

Step 4: Distribute the factor

Now distribute the \( \frac{1}{2} \) across the entire expression from Step 3: \( \frac{1}{2} \cdot \left( \log_{5}(x-1) - \log_{5}(x+1) \right) \), which results in \( \frac{1}{2} \log_{5}(x-1) - \frac{1}{2} \log_{5}(x+1) \).

Key Concepts

Power Rule of Logarithms

The power rule of logarithms is a helpful tool when working with expressions where a term is raised to a power. Essentially, it allows you to "bring down" the exponent in front of the logarithm, making the expression easier to manage.
Here’s why it works: when you have an expression like \( \log_{b}(a^n) \), the power rule rewrites this as \( n \cdot \log_{b}(a) \). This adjustment simplifies calculations<,br> especially when dealing with complex logarithms that involve exponents.
In our example, once we rewrote the square root as an exponent, the expression became \( \log_{5} \left( \frac{x-1}{x+1} \right)^{1/2} \). Using the power rule, we moved the \( \frac{1}{2} \) outside the logarithm: \( \frac{1}{2} \cdot \log_{5} \left( \frac{x-1}{x+1} \right) \).
This process is important because it clears the way for applying other rules like the quotient rule more effectively.

Quotient Rule of Logarithms

When dealing with division inside a logarithm, the quotient rule of logarithms is your best friend. This rule transforms a division inside the logarithm into a subtraction outside, which is usually easier to handle.
The rule states \( \log_{b}\left( \frac{a}{c} \right) = \log_{b}(a) - \log_{b}(c) \). In simple terms, the division turns into a subtraction of two separate logarithms.
For the exercise expression \( \log_{5} \left( \frac{x-1}{x+1} \right) \), applying the quotient rule means rewriting it as \( \log_{5}(x-1) - \log_{5}(x+1) \).
Using this rule not only simplifies the expression but also sets up for easy application of any coefficients or constants that are multiplied later, like the \( \frac{1}{2} \) factor from the power rule.

Logarithmic Expressions

Understanding logarithmic expressions is essential for solving various math problems. A logarithmic expression involves the concept of a logarithm, which is essentially the inverse operation to exponentiation.
Think of it as answering the question: "To what power must the base be raised, to achieve a certain number?" Logarithms function as tools that simplify complex calculations, whether evaluating, expanding, or condensing expressions.
In the example provided, we start with a complex logarithmic expression involving a square root and division. The task is to expand this expression using the laws of logarithms, such as the power and quotient rules. By expanding, each term becomes separately dealt with, making the expression easier to evaluate or further manipulate if needed.
This approach is particularly useful in algebra and higher math courses, as it provides clarity and simplifies calculations while reinforcing the ability to manipulate mathematical expressions.