Problem 29 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 19-34: Solve the equation. $$ \ln x^{2}=-2 $$

Step-by-Step Solution

Answer

The solutions are \( x = e^{-1} \) and \( x = -e^{-1} \).

Step 1: Rewrite the Logarithmic Equation

We start with the logarithmic equation \( \ln x^2 = -2 \). Using the property of logarithms \( \ln a = b \) implies \( a = e^b \), we can rewrite the equation as \( x^2 = e^{-2} \).

Step 2: Solve for x by Taking the Square Root

To solve for \( x \), we take the square root of both sides of the equation \( x^2 = e^{-2} \). This gives us two possible solutions: \( x = \sqrt{e^{-2}} \) and \( x = -\sqrt{e^{-2}} \).

Step 3: Simplify the Expressions

Calculate the square root expression: \( \sqrt{e^{-2}} = e^{-1} \). Thus, the solutions can be simplified to \( x = e^{-1} \) and \( x = -e^{-1} \).

Step 4: Verify the Solutions

Substitute \( x = e^{-1} \) back into the original equation: \( \ln((e^{-1})^2) = \ln(e^{-2}) = -2 \), which is correct. Similarly, substitute \( x = -e^{-1} \): \( \ln((-e^{-1})^2) = \ln(e^{-2}) = -2 \), which is also correct.

Key Concepts

Properties of Logarithms

In mathematics, logarithms allow us to work with exponents in a manageable way. They can transform multiplication into addition, division into subtraction, and powers into multiplication. Key properties include:

• Product Property: \(\ln(ab) = \ln a + \ln b\).
• Quotient Property: \(\ln\left(\frac{a}{b}\right) = \ln a - \ln b\).
• Power Property: \(\ln a^n = n\ln a\).

In the equation \(\ln x^2 = -2\), the power property is particularly useful. It allows us to rewrite the expression \(\ln x^2\) as \(2\ln x\). This equation can be restructured by utilizing the inverse of the natural logarithm, the exponential function: if \(\ln a = b\), then \(a = e^b\). Applying these logarithmic properties helps simplify complex expressions and solve equations effectively.

Exponential Functions

Exponential functions are integral in mathematics, especially when solving logarithmic equations. The natural exponential function, denoted as \(e^x\), where \(e\) is approximately 2.71828, is the inverse function of the natural logarithm. In solving \(\ln x^2 = -2\), we employed this inverse relationship by transforming the logarithmic form into an exponential form: \(x^2 = e^{-2}\). This transformation is crucial because it converts the problem from one involving logarithms to one involving basic algebra, which is often easier to manipulate and solve.Understanding exponentials allows us to simplify expressions and equations further, especially when combined with logarithmic properties. With a grasp of how these functions interact, solving logarithmic equations becomes more straightforward.

Square Root Process

To solve the equation \(x^2 = e^{-2}\), and find the value of \(x\), the square root process is employed. Finding roots is a method to "undo" powers in an equation. Taking the square root of both sides in an equation like \(x^2 = e^{-2}\) results in \(x = \pm\sqrt{e^{-2}}\). This gives us two solutions: one positive and one negative. In solving for \(x\), this process yields two possibilities: \(x = \sqrt{e^{-2}}\) and \(x = -\sqrt{e^{-2}}\). Simplifying, \(\sqrt{e^{-2}}\) becomes \(e^{-1}\), so the solutions are \(x = e^{-1}\) and \(x = -e^{-1}\). Using the square root effectively opens up equation solutions that might otherwise remain inaccessible, providing a comprehensive set of solutions to various algebraic and mathematical problems.