Problem 76 - Algebra and Trigonometry

Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. $$ \text { Solve: } \frac{x^{2} \cdot \frac{1}{x}-\ln x \cdot 2 x}{\left(x^{2}\right)^{2}}=0 $$

Step-by-Step Solution

Answer

\( x = \sqrt{e} \)

Step 1: Simplify the Given Expression

First, simplify the expression \( \frac{x^{2} \frac{1}{x} - \ln x \cdot 2x}{(x^{2})^{2}} \). Notice that \( x^{2} \frac{1}{x} = x \) since \( x^{2} \cdot \frac{1}{x} = x \). Therefore, the expression simplifies to: \( \frac{x - \ln x \cdot 2x}{x^{4}} \)

Step 2: Factor Out Common Terms

Factor out \( x \) from the numerator: \( \frac{x(1 - 2\ln x)}{x^{4}} \). This simplifies to \( \frac{1 - 2\ln x}{x^{3}} \).

Step 3: Set the Equation to Zero

Set the expression equal to zero and solve for \( x \): \( \frac{1 - 2\ln x}{x^{3}} = 0 \). For the fraction to be zero, the numerator must be zero: \( 1 - 2\ln x = 0 \).

Step 4: Solve for x

Solve for \( x \) by isolating \( \ln x \): \( 1 - 2\ln x = 0 \). This leads to: \( \ln x = \frac{1}{2} \). Exponentiating both sides to eliminate the natural logarithm gives: \( x = e^{1/2} \) which can be written as \( x = \sqrt{e} \).

Key Concepts

Simplifying Expressions

Simplifying expressions is the process of rewriting a mathematical expression in a simpler form. This often involves combining like terms and reducing fractions. For example, take the expression \( \frac{x^{2} \frac{1}{x} - \text{ln} x \times 2x}{(x^{2})^{2}} \). In the first step, we notice that \( x^{2} \frac{1}{x} = x \) because \( x^{2} \times \frac{1}{x} = x \). We then substitute this back to simplify the expression to \( \frac{x - \text{ln} x \times 2x}{x^{4}} \). This kind of simplification is critical to make further steps easier and to solve the equation accurately.

Factoring

Factoring involves breaking down an expression into products of simpler expressions. In our given problem, after simplifying the expression, we noticed that there is a common factor in the numerator. We factor out this common term. Specifically, we factor out \(x\) from \( \frac{x(1 - 2\text{ln} x)}{x^{4}} \), resulting in \( \frac{1 - 2\text{ln} x}{x^{3}} \). Factoring helps not only in simplifying the expressions but also in solving equations, making it easier to identify solutions.

Solving Equations

Solving equations is finding the value(s) of the variable(s) that make the equation true. In our example, after simplifying and factoring the expression, we must solve the equation \( \frac{1 - 2\text{ln} x}{x^{3}} = 0 \). For a fraction to be zero, its numerator must be zero. Hence, we solve \( 1 - 2\text{ln} x = 0 \). Isolating the logarithmic function \( \text{ln} x \), we get \( \text{ln} x = \frac{1}{2} \). Finally, we exponentiate both sides to remove the logarithm, resulting in \( x = e^{\frac{1}{2}} \), or \( x = \frac{\text{e}}{\text{2}} \).

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions and are used to solve equations involving exponents. In our problem, we encountered a logarithmic expression \( \text{ln} x \). By isolating \( \text{ln} x \) in the equation \( 1 - 2\text{ln} x = 0 \), we can exponentiate both sides of the equation to eliminate the logarithm: \( \text{e}^{\text{ln} x} = \text{e}^{\frac{1}{2}} \). Knowing that \( \text{e}^{\text{ln} x} = x \), we can rewrite this as \( x = e^{\frac{1}{2}} \), which can also be written as \( x = \frac{\text{e}}{\text{2}} \). Logarithmic and exponential functions often appear in solving real-world problems like compound interest and population growth.