Problem 20 - Intermediate Algebra

Question

For Problems \(15-22\), solve each logarithmic equation. $$ \log (x+1)-\log (x+2)=\log \frac{1}{x} $$

Step-by-Step Solution

Answer

The solution is \(x = \sqrt{2}\).

Step 1: Combine Logarithms on the Left Side

We notice that the left-hand side of the equation contains two logarithmic functions, \ \(\log(x+1) - \log(x+2)\). We can utilize the property of logarithms \( \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \) to combine them: \[\log\left(\frac{x+1}{x+2}\right) = \log\left(\frac{1}{x}\right).\]

Step 2: Set the Arguments Equal

Since the logarithms are equal and assuming the bases are the same, we can equate their arguments:\[\frac{x+1}{x+2} = \frac{1}{x}.\]

Step 3: Cross-Multiply to Eliminate Fractions

Cross-multiply to solve the equation for \(x\):\[(x+1)x = (x+2)\cdot1.\] This simplifies to:\[x^2 + x = x + 2.\]

Step 4: Simplify and Rearrange the Equation

Move all terms to one side to set the quadratic equation to zero:\[x^2 + x - x - 2 = 0 \]Simplifying gives us:\[x^2 - 2 = 0.\]

Step 5: Solve the Quadratic Equation

Add 2 to both sides to isolate \(x^2\):\[x^2 = 2.\]Take the square root of both sides to solve for \(x\):\[x = \pm \sqrt{2}.\]

Step 6: Check for Extraneous Solutions

Consider each solution by plugging it back into the original logarithmic equation to ensure no invalid results occur:- If \(x = \sqrt{2}\), ensure the log arguments are positive: \(x+1 > 0\) and \(x+2 > 0\).- If \(x = -\sqrt{2}\), \(x+1 = 1-\sqrt{2}\) and \(x+2 = 2-\sqrt{2}\), both must be greater than zero, which are not, making \(-\sqrt{2}\) an extraneous solution.Thus, \(x = \sqrt{2}\) is valid and \(x = -\sqrt{2}\) is not valid.

Key Concepts

Properties of Logarithms

Logarithms are powerful tools that are often used to solve equations involving exponential relationships. One of the essential properties of logarithms used in this exercise is the rule that allows you to combine or separate logarithmic expressions:

• \(\log(a) - \log(b) = \log\left(\frac{a}{b}\right)\)

This property is particularly helpful for simplifying expressions by combining logs into a single log function. Using this property simplifies the equation on the left side:\(\log(x+1) - \log(x+2) = \log\left(\frac{x+1}{x+2}\right)\).
By employing this property, you can more easily isolate the variable and solve the equation. It is handy in many scenarios, especially when dealing with subtraction or summation of log terms.
Remember, the logarithm’s base must be consistent when applying these rules, as they are dependent on the logarithmic identity and properties.

Quadratic Equations

Quadratic equations are a type of polynomial equation that can be written in the standard form:\(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are real numbers, and \(a eq 0\). The solutions can often be found using factoring, completing the square, or using the quadratic formula. In this problem, after merging logarithms and setting their arguments equal:\[\frac{x+1}{x+2} = \frac{1}{x}\],you end up with a quadratic equation in step 4 after cross-multiplying:\[x^2 - 2 = 0\].
Solving quadratic equations involves rearranging terms to match the standard form and then applying appropriate methods to find the variable's value.

• Quadratic formula provides the roots: \(x = \pm \sqrt{2}\).

Quadratic roots need to be verified in the context of the original problem to ensure they are valid solutions in the scenario given.

Extraneous Solutions

Extraneous solutions are results that satisfy the transformed equation but not the original equation. This often happens in equations that involve logarithms or radicals.
When checking for extraneous solutions, after calculations, each solution needs to be substituted back into the original equation to validate the result. In the problem:

• The solution \(x = \sqrt{2}\) works when substituted back, addressing the requirement that the arguments of the original logarithms must be positive: \(x + 1 > 0\) and \(x + 2 > 0\).
• However, the solution \(x = -\sqrt{2}\) actually results in non-positive values for the logarithmic arguments: \(1 - \sqrt{2} < 0\) and \(2 - \sqrt{2} < 0\).

These invalid conditions dismiss \(-\sqrt{2}\) as an extraneous solution.
Identifying extraneous solutions is crucial to ensuring that only valid solutions are considered, maintaining the integrity of mathematical solutions.