Problem 70 - Precalculus

Question

Solve the logarithmic equations. Round your answers to three decimal places. $$\log (\sqrt{1-x})-\log (\sqrt{x+2})=\log x$$

Step-by-Step Solution

Answer

The solution is approximately \(x = 0.303\).

Step 1: Use Logarithmic Properties

Use the property of logarithms that states \( \log a - \log b = \log \left( \frac{a}{b} \right) \). Apply this to the equation: \[ \log \left( \frac{\sqrt{1-x}}{\sqrt{x+2}} \right) = \log x\] Simplify the left side to: \[ \log \left( \frac{\sqrt{1-x}}{\sqrt{x+2}} \right) = \log \left( \frac{1-x}{x+2} \right)\] This reduces the equation to:\[ \log \left( \frac{1-x}{x+2} \right) = \log x\]

Step 2: Remove the Logarithms

Since the logarithms are equal, their arguments must be equal:\[ \frac{1-x}{x+2} = x\] Solve this equation for \(x\).

Step 3: Solve the Algebraic Equation

Cross-multiply to eliminate the fraction:\[ 1 - x = x(x + 2)\] Simplify and arrange terms:\[ 1 - x = x^2 + 2x\] Rearrange into a standard quadratic equation:\[ x^2 + 3x - 1 = 0\]

Step 4: Solve the Quadratic Equation

Apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 3, c = -1 \):\[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-1)}}{2(1)}\] Simplify:\[ x = \frac{-3 \pm \sqrt{9 + 4}}{2}\]\[ x = \frac{-3 \pm \sqrt{13}}{2}\]Compute numerical solutions:\[ x_1 \approx \frac{-3 + 3.606}{2} = 0.303 \]\[ x_2 \approx \frac{-3 - 3.606}{2} = -3.303 \]

Step 5: Verify and Conclude

Check if potential solutions are within the domain of the original logarithmic equation, which requires \(x > 0\) and \(1-x > 0\).- \(x_1 = 0.303\) fits as it is positive and \(1-0.303 > 0\).- \(x_2 = -3.303\) is not valid as it is negative.Thus, the acceptable solution is \(x \approx 0.303\).

Key Concepts

Logarithmic Properties

Understanding the properties of logarithms is crucial when dealing with logarithmic equations. One of the most important properties is the difference of logarithms. It states that \( \log a - \log b = \log \left( \frac{a}{b} \right) \). This property is particularly useful for condensing the equation, turning it from a subtraction into a single logarithm.

By applying this property, a more complex logarithmic equation can be simplified, making it easier to solve. In our example problem, \( \log (\sqrt{1-x}) - \log (\sqrt{x+2}) = \log x \) becomes \( \log \left( \frac{\sqrt{1-x}}{\sqrt{x+2}} \right) = \log x \). This simplifies further to \( \log \left( \frac{1-x}{x+2} \right) \).

Recognizing and correctly applying these properties will help you to efficiently solve logarithmic equations and reduce them to simpler forms.

Quadratic Equation

Quadratic equations are fundamental in algebra, and they take the standard form \( ax^2 + bx + c = 0 \). In this exercise, after applying logarithmic properties and reducing the equation, we are left with \( \frac{1-x}{x+2} = x \).

Rearranging and simplifying using cross-multiplication, you get \( 1 - x = x(x + 2) \), which becomes a quadratic equation \( x^2 + 3x - 1 = 0 \). This is recognizable as a classic quadratic form.

To solve this, we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 3, c = -1 \). Solving the equation yields two potential solutions, which are examined further for validity.

Algebraic Equation Solving

Algebraic equation solving involves finding the values of variables that satisfy an equation. Here, after removing the logs by equating their arguments, the equation becomes \( \frac{1-x}{x+2} = x \).

This equation is solved by cross-multiplying to eliminate the fraction: \( 1 - x = x(x + 2) \). Cross-multiplication is an essential strategy to deal with equations involving fractions and provides a clear path to transform the equation into a standard quadratic form.

Once transformed into \( x^2 + 3x - 1 = 0 \), the equation is solved using the quadratic formula, leading us to potential solutions \( x_1 \approx 0.303 \) and \( x_2 \approx -3.303 \).

However, for logarithmic equations, you must check if the solutions lie within the domain constraints. In this case, we find that only \( x \approx 0.303 \) is valid, as it meets the condition for \( x > 0 \) and \( 1-x > 0 \).