Problem 61 - 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

Yes, they are equal by transforming and simplifying the expressions using logarithmic properties.

Step 1: Understanding the Problem

We need to prove that \(-\ln \left(x-\sqrt{x^{2}-1}\right)=\ln \left(x+\sqrt{x^{2}-1}\right)\). This means showing that the negative logarithm expression on the left is equal to the right logarithm expression.

Step 2: Utilize Logarithmic Properties

Recall that the property of logarithms \(-\ln(a) = \ln(\frac{1}{a})\) can be applied here. Therefore, \(-\ln \left(x-\sqrt{x^{2}-1}\right) = \ln\left(\frac{1}{x-\sqrt{x^{2}-1}}\right)\). We need to show this equals \(\ln \left(x+\sqrt{x^{2}-1}\right)\).

Step 3: Simplify the Expression

To express \(\frac{1}{x-\sqrt{x^{2}-1}}\) properly, we multiply and divide by the conjugate \(x+\sqrt{x^2-1}\). So, \[\frac{1}{x-\sqrt{x^2-1}} \times \frac{x+\sqrt{x^2-1}}{x+\sqrt{x^2-1}} = \frac{x+\sqrt{x^2-1}}{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}\].

Step 4: Utilize Difference of Squares

The product \((x-\sqrt{x^2-1})(x+\sqrt{x^2-1})\) is a difference of squares and simplifies to \[x^2 - (\sqrt{x^2-1})^2 = x^2 - (x^2 - 1) = 1\].Thus, \(\frac{x+\sqrt{x^2-1}}{1} = x+\sqrt{x^2-1}\).

Step 5: Equating and Concluding

We have shown that \(\frac{1}{x-\sqrt{x^2-1}} = x+\sqrt{x^2-1}\). Therefore, \(-\ln \left(x-\sqrt{x^{2}-1}\right) = \ln \left(x+\sqrt{x^{2}-1}\right)\)holds true.

Key Concepts

Difference of Squares

The difference of squares is a fundamental algebraic identity that states:

• For any two terms, \(a\) and \(b\), the expression \(a^2 - b^2\) can be factored into \((a-b)(a+b)\).
• This identity is widely used when simplifying expressions involving squares.

When applied to problems involving square roots, like the expression \((x - \sqrt{x^2-1})(x + \sqrt{x^2-1})\), recognizing this identity allows for simplifications that would otherwise be complex.
In our specific exercise, the goal was to manipulate the expression \(\frac{1}{x-\sqrt{x^2-1}}\). By multiplying by the conjugate \(x+\sqrt{x^2-1}\), the difference of squares comes into play.
The product simplifies to \(x^2 - (\sqrt{x^2-1})^2\), which equals 1. This step is crucial as it reduces the complexity of the expression and is a perfect example of how algebraic principles like the difference of squares streamline mathematical problems.

Logarithm Properties

Understanding the properties of logarithms is essential when working with equations involving these expressions.

• The properties include transformations, which allow us to manipulate logarithms effectively.
• One particularly useful property is dealing with negative logarithms: \(-\ln(a) = \ln\left(\frac{1}{a}\right)\).

This property was used to transform \(-\ln(x-\sqrt{x^2-1})\) into \(\ln\left(\frac{1}{x-\sqrt{x^2-1}}\right)\).
Such transformations are powerful because they allow us to convert subtraction in the logarithmic expression into division, which often simplifies the problem significantly.
Logarithm properties as a whole provide a toolkit for simplifying complex logarithmic expressions into more manageable forms. Leveraging these properties is key in proving equations or finding solutions in mathematics.

Simplifying Expressions

Simplifying expressions is a fundamental skill in algebra and calculus, allowing more complex expressions to be transformed into simpler, more understandable forms.

• Simplification involves applying mathematical identities, properties, and techniques to rewrite expressions.
• This process often makes solving and analyzing equations easier and more intuitive.

In the context of our exercise, the expression \(\frac{1}{x-\sqrt{x^2-1}}\) was simplified by multiplying by the conjugate \(x+\sqrt{x^2-1}\).
This step involved using both the difference of squares to eliminate the square root terms effectively and the logarithmic property to transform the expression’s logarithmic form.
Simplifying expressions helps in proving identities or equations by reducing equations to their core components, thereby clarifying their properties and relationships. It is an essential technique across all areas of mathematics.
Remember, each simplification step, no matter how small, is an advancement towards a more elegant and understandable solution.