Problem 48 - Algebra and Trigonometry

Question

\(45-54\) . Use the Laws of Logarithms to combine the expression. $$ \log _{5}\left(x^{2}-1\right)-\log _{5}(x-1) $$

Step-by-Step Solution

Answer

The expression simplifies to \( \log_{5}(x+1) \).

Step 1: Identify the Logarithm Difference Rule

The expression \( \log_{b}(A) - \log_{b}(B) \) can be simplified using the logarithm difference rule, which states that \( \log_{b}\left(\frac{A}{B}\right) = \log_{b}(A) - \log_{b}(B) \). This allows us to combine the two separate logarithmic terms into a single logarithm.

Step 2: Simplify using the Logarithm Difference Rule

Apply the difference rule to the given expression. Here we identify \( A = x^2 - 1 \) and \( B = x - 1 \). Therefore, the expression \( \log_{5}(x^2 - 1) - \log_{5}(x - 1) \) becomes \( \log_{5}\left( \frac{x^2 - 1}{x - 1} \right) \).

Step 3: Simplify the Fraction

The expression \( \frac{x^2 - 1}{x - 1} \) can be simplified further by recognizing that \( x^2 - 1 \) is a difference of squares. It can be factored into \((x+1)(x-1)\). Thus, \( \frac{x^2 - 1}{x - 1} = \frac{(x+1)(x-1)}{x-1} = x+1 \) because \( x-1 \) cancels out in the numerator and denominator.

Step 4: Final Simplified Logarithmic Expression

Substituting the simplified fraction \( x+1 \) back into the logarithmic expression from Step 2, we get \( \log_{5}(x+1) \) as the final simplified form of the original expression.

Key Concepts

Logarithm Difference Rule

The logarithm difference rule is an essential tool in simplifying expressions involving logarithms by combining them into a single term. This rule states that if you have two logarithms with the same base that are subtracted, such as \( \log_{b}(A) - \log_{b}(B) \), you can simplify them into one logarithm: \( \log_{b}\left(\frac{A}{B}\right) \). To apply this rule:

• Ensure the bases of the logarithms are the same.
• Subtract the logarithms.
• Place the result as a fraction inside a single logarithm with the same base.

For example, in the expression \( \log_{5}(x^2 - 1) - \log_{5}(x - 1) \), we can use this rule to combine them into \( \log_{5}\left(\frac{x^2 - 1}{x - 1}\right) \). This turns two separate logarithmic terms into a single, more manageable expression.

Combining Logarithms

Combining logarithms is a useful strategy for simplifying mathematical expressions and solving equations that involve multiple logarithms. When we combine logarithms by using their rules, we reduce the complexity of calculations.Here’s how you can combine logarithms effectively:

• Use the difference rule to convert subtractions into fractions within a Logarithm.
• The other main rule is the Logarithm of a product, which states \( \log_{b}(XY) = \log_{b}(X) + \log_{b}(Y) \).
• Use these rules to turn complex logarithmic expressions into simpler ones.

By applying these techniques, expressions like \( \log_{5}\left(\frac{x^2 - 1}{x - 1}\right) \) can be achieved from \( \log_{5}(x^2 - 1) - \log_{5}(x - 1) \). This simplification leads to easier further calculations or derivations within math problems.

Difference of Squares

The difference of squares is a common algebraic pattern that helps simplify expressions. When an expression fits the pattern \( a^2 - b^2 \), it can be factored into \((a+b)(a-b)\).To see how this works, consider the expression \( x^2 - 1 \):

• This can be rewritten as \( (x)^2 - (1)^2 \), where \(a = x\) and \(b = 1\).
• Using the difference of squares formula, it factors into \( (x+1)(x-1) \).

In the context of the original problem, recognizing \( x^2 - 1 \) as a difference of squares allows us to simplify the entire expression by canceling out identical terms in the numerator and denominator, transforming \( \frac{(x+1)(x-1)}{x-1} \) to \( x+1 \).Mastering this concept allows for quick simplifications and problem-solving in both algebra and calculus.