Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\ln (x+5)=\ln (x-1)-\ln (x+1)\)

Short answer

The solution for the given logarithmic equation, approximated to three decimal places is \(x = -3.000\)

Step-by-step solution

Simplify the equation using logarithmic properties

The right side of the equation can first be simplified using the quotient rule of the logarithms \(\ln (a)-\ln (b)=\ln (\frac{a}{b})\). The equation is then written as follows: \(\ln (x+5) = \ln ((x-1) / (x+1))\).

Equate the arguments of the logarithms

Since the logarithms bases are equal, then their arguments can be equated. The equation becomes \(x+5 = (x-1) / (x+1)\)

Simplify the equation into a quadratic equation by clearing the fraction

Multiply each side by \(x+1\) to clear the fraction. The equation then becomes \((x+5)(x+1) = x-1\). Upon simplifying, we obtain a quadratic equation: \(x^2 + 6x + 5 = x - 1\).

Equate the equation to zero and solve for 'x'

To find the roots of the quadratic equation, equate the equation to zero, in this case we Subtract \(x\) and add \(1\) to both sides of the equation, i.e., \(x^2 + 5x + 6 = 0\). Then solve for 'x'. The roots of this equation are \(x = -3\) and \(x = -2\). We find solutions to the equation by using the quadratic formula or factoring

Validate the solutions

A logarithm of a negative number is undefined, therefore from the solutions, \(x = -2\) results in undefined solution in the original equation. So the only solution here is \(x = -3\), which is valid since it does not result in the logarithm of a negative number in the original equation.

Key concepts

Properties of Logarithms

Logarithms are a powerful mathematical tool used to address problems involving exponential relationships. They come with several important properties, which can simplify complex equations, like the logarithmic equation in our exercise. One key property is the **quotient rule**, which is extremely useful in solving our current problem. The quotient rule states that

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

This property transforms a subtraction between two logarithms on the right side of the equation into division within a single logarithm. This simplification allowed us to write our original equation as \( \ln(x+5) = \ln \left( \frac{x-1}{x+1} \right) \).
Logarithmic equations often benefit from this kind of transformation, turning potential stumbling blocks into manageable steps. Understanding these properties ensures a smooth path from a confusing logarithmic expression to a clear equation that can be equated and solved.

Quadratic Equations

At the heart of many algebraic problems lie quadratic equations. In this particular problem, transforming a logarithmic equation into a quadratic one simplifies the trouble around the logs. Once we have equated the arguments by setting \( x+5 = \frac{x-1}{x+1} \), multiplying out terms converts the problem to a familiar form: a quadratic equation.
Quadratic equations have the general form

• \( ax^2 + bx + c = 0 \)

with specific parameters for our equation: \( x^2 + 5x + 6 = 0 \). Solving this form involves various methods such as factoring, using the quadratic formula, or completing the square. These methods give us solutions, or "roots," that represent values for \(x\) where the equation equals zero. Often, in these exercises, you might find more than one root, but it's critical to evaluate their validity in the context of the original problem.

Solving Equations Algebraically

Solving equations algebraically involves using mathematical operations and properties to find the value of a variable. Each step in the calculation logically follows from the one before, forming a clear path to the solution.
In this exercise, we maintain balance and accuracy by performing the same operation on both sides of our equations. For example, we multiplied each side by \(x+1\) to eliminate the fraction. Once transformed into a quadratic equation, the exercise required careful algebraic manipulation, where we:

• Adjusted terms to equate everything to zero \(x^2 + 5x + 6 = 0\)
• Used the quadratic formula or factoring to solve \((x+3)(x+2)=0\)
• Identified roots \(x = -3\) and \(x = -2\)\

Finally, we ensured the validity of the roots by checking them against the initial conditions of the problem, especially since logarithms have undefined scenarios for negative arguments. This step-by-step logical process helps in finding the correct solutions algebraically.