Problem 42 - Algebra and Trigonometry

Question

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

\( \log_{5}(x + 1) \)

Step 1: Identify Logarithm Difference

Begin by recognizing that the expression involves a difference of two logarithms: \( \log_{5}(x^2 - 1) - \log_{5}(x - 1) \). According to the laws of logarithms, specifically the difference of logarithms, you can combine these into a single logarithm by dividing the arguments.

Step 2: Apply Logarithm Quotient Rule

Use the quotient rule of logarithms: \( \log_{b}(M) - \log_{b}(N) = \log_{b}\left(\frac{M}{N}\right) \). Here, set \( M = x^2 - 1 \) and \( N = x - 1 \), so you get: \( \log_{5}\left(\frac{x^2 - 1}{x - 1}\right) \).

Step 3: Simplify the Argument

The argument \( \frac{x^2 - 1}{x - 1} \) can be simplified by recognizing that \( x^2 - 1 \) is a difference of squares: \( x^2 - 1 = (x - 1)(x + 1) \). Substitute this back into the expression so it's \( \frac{(x - 1)(x + 1)}{x - 1} \).

Step 4: Cancel Common Terms

In \( \frac{(x - 1)(x + 1)}{x - 1} \), cancel out the common term \( x - 1 \), giving you \( x + 1 \). However, remember \( x eq 1 \) since the denominator cannot be zero.

Step 5: Write Final Logarithmic Expression

Write the simplified combined expression: \( \log_{5}(x + 1) \). This is the expression resulting from the original problem after applying the logarithmic laws.

Key Concepts

Logarithm Quotient Rule

The Logarithm Quotient Rule is a fundamental concept in logarithms that helps simplify expressions involving the difference of two logarithmic terms. When you see an expression like \( \log_{b}(M) - \log_{b}(N) \), you can apply this rule to combine it into one logarithmic term, \( \log_{b}\left(\frac{M}{N}\right) \). This rule is very useful when you're faced with complex expressions because it allows you to condense and work with just one logarithmic term instead of two.

In our exercise, this rule applies directly to the expression \( \log_{5}(x^2 - 1) - \log_{5}(x - 1) \). According to the Quotient Rule, this is equivalent to \( \log_{5}\left(\frac{x^2 - 1}{x - 1}\right) \). The base, which is 5 in this case, remains the same while the arguments are divided.

This rule helps by simplifying expressions to an easier-to-manage form, paving the way for additional simplifications to be made, such as in the next steps where the argument \( \frac{x^2 - 1}{x - 1} \) gets further reduced.

Difference of Squares

Recognizing a difference of squares is a crucial step in simplifying mathematical expressions. The difference of squares formula states that \( a^2 - b^2 = (a - b)(a + b) \). This is often encountered in algebra, and identifying it can significantly simplify an expression.

For our example, the argument \( x^2 - 1 \) is a clear case of a difference of squares because it can be rewritten as \( (x)^2 - (1)^2 \). Applying the formula, you get \( (x - 1)(x + 1) \). This understanding allows the expression \( \frac{x^2 - 1}{x - 1} \) to be simplified as \( \frac{(x - 1)(x + 1)}{x - 1} \).

• Helps to break down expressions
• Reduces complexity for further simplification

By simplifying using the difference of squares, common terms like \( x-1 \) can be canceled out in the given fraction, as seen in the solution. However, it's important to remember that \( x eq 1 \) because it would make the original denominator zero.

Simplifying Expressions

Simplifying expressions is about making them as straightforward and understandable as possible, often by reducing them to their simplest form. In the context of the problem, after applying the Quotient Rule and recognizing the difference of squares, we need to simplify \( \frac{(x - 1)(x + 1)}{x - 1} \).

The simplification process often involves canceling out terms that appear in both the numerator and the denominator. In this expression, \( x - 1 \) is a common term that appears in both, which can be canceled out under the prerequisite condition that \( x eq 1 \). This gives us \( x + 1 \) as the final simplified term.

After cancellation, rewriting it in terms of a logarithm gives us \( \log_{5}(x + 1) \). Simplifying expressions like this is crucial in mathematics, as it reduces errors and makes problems easier to solve and interpret. Always take care to check for any restrictions like undefined values that may arise through this process.