Problem 67 - Algebra and Trigonometry

Question

Show that \(-\ln \left(x-\sqrt{x^{2}-1}\right)=\ln \left(x+\sqrt{x^{2}-1}\right)\)

Step-by-Step Solution

Answer

The equation is correct after simplification.

Step 1: Understand the Problem

We are asked to show that \(-\ln (x-\sqrt{x^2-1}) = \ln (x+\sqrt{x^2-1})\). This involves manipulating and simplifying logarithmic expressions.

Step 2: Analyze Both Sides of the Equation

Look at the left-hand side: \(-\ln (x-\sqrt{x^2-1})\) and the right-hand side: \(\ln (x+\sqrt{x^2-1})\). The goal is to show these two expressions are equal by manipulating one side to look like the other.

Step 3: Use Logarithmic Identities

Recall the identity \(-\ln(a) = \ln\left(\frac{1}{a}\right)\). Applying it to the left side: \(-\ln (x-\sqrt{x^2-1}) = \ln \left(\frac{1}{x-\sqrt{x^2-1}}\right)\).

Step 4: Rationalize \(\frac{1}{x-\sqrt{x^2-1}}\)

Multiply the numerator and the denominator by the conjugate: \(x+\sqrt{x^2-1}\). This simplifies to:\[\frac{1}{x-\sqrt{x^2-1}} \cdot \frac{x+\sqrt{x^2-1}}{x+\sqrt{x^2-1}} = \frac{x+\sqrt{x^2-1}}{x^2-(x^2-1)} = \frac{x+\sqrt{x^2-1}}{1} = x+\sqrt{x^2-1}.\]

Step 5: Show Equality of Expressions

Thus, we have:\[-\ln(x-\sqrt{x^2-1}) = \ln\left(x+\sqrt{x^2-1}\right).\]This proves that the original equation \(-\ln (x-\sqrt{x^2-1}) = \ln (x+\sqrt{x^2-1})\) is indeed correct.

Key Concepts

Simplifying Logarithmic Expressions

Simplifying logarithmic expressions often involves using well-known logarithmic identities. One such identity is \(-\ln(a) = \ln\left(\frac{1}{a}\right)\), which allows us to transform negative logarithms into more manageable fractions.
This transformation is crucial because it makes other algebraic manipulations, such as rationalizing, more straightforward.
When simplifying expressions, remember:

• Logarithms can convert multiplication into addition: \( \log(a \times b) = \log(a) + \log(b) \).
• Division is converted into subtraction in logarithms: \( \log\frac{a}{b} = \log(a) - \log(b) \).
• Powers can be moved in front as coefficients: \( \log(a^b) = b \cdot \log(a) \).

This basic understanding of properties of logarithms is foundational when attempting to reveal underlying similarities between complex expressions or equations.

Manipulating Equations

Equation manipulation allows us to demonstrate that two seemingly different expressions are actually equivalent.
In the provided exercise, we started by using the identity \(-\ln(a) = \ln\left(\frac{1}{a}\right)\) to manipulate the left side of the equation into a more familiar format. This step unlocked further simplification.
A few key tips for manipulating equations include:

• Always perform the same operation on both sides of an equation to maintain equality.
• Seek opportunities to use identities such as the logarithm rules discussed earlier to rewrite terms in simpler forms.
• Double-check algebraic manipulations, given their complexity, especially with negatives and fractions.

Through careful manipulation and application of identities, complex equations become solvable, and their proofs evident.

Rationalizing Expressions

Rationalizing expressions is a critical step when dealing with radical terms in denominators.
By multiplying the original expression by the conjugate, we eliminate square roots from denominators without changing the expression's value.
Let's dissect the key steps:
*Identify the expression of the form \(\frac{1}{x-\sqrt{x^2-1}}\) and multiply by the conjugate \(x+\sqrt{x^2-1}\).
*Compute the result: \( \frac{1}{x-\sqrt{x^2-1}} \cdot \frac{x+\sqrt{x^2-1}}{x+\sqrt{x^2-1}} = \frac{x+\sqrt{x^2-1}}{1} = x+\sqrt{x^2-1}\).

This effective strategy significantly simplifies complex expressions. Notice how rationalizing can sometimes reveal the inherent equality between seemingly different expressions. It is an invaluable tool in algebraic manipulations that involve logarithmic expressions.