Problem 57 - Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

State the exact solutions before approximation. If more than one valid solution exists, list those. Do remember to add final values in decimal form rounding to three decimal places

Step 1: Apply Logarithm Properties

Apply the property of logarithm \(\ln(a) - \ln(b) = \ln(a/b)\) to simplify the equation to:\(\ln(x+5) = \ln \left(\frac{x-1}{x+1}\right)\)

Step 2: Apply the Definition of Logarithm

Since \(\ln(a) = \ln(b)\) implies \(a=b\) for positive real numbers, we equate the arguments of the logarithm terms so \(x+5 = \frac{x-1}{x+1}\).

Step 3: Solve Equation

Cross-multiply and simplify to solve for x. This will yield a quadratic equation. Collect like terms and solve the quadratic equation by the quadratic formula.

Step 4: Check Solutions

Check each solution within the domain of the original equation. Remember the argument of a logarithm (in this case, \(x+5\), \(x-1\) and \(x+1\)) must be greater than 0. Make sure to validate the solutions within this restriction.

Step 5: Approximate the Result

Once all solutions are validated, present the solution as an approximate value up to three decimal places.

Key Concepts

Logarithmic Properties

Understanding the properties of logarithms is crucial for solving logarithmic equations effectively. One of the most important properties is a difference of logs property: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). This property allows us to condense a subtraction of two natural logs into a single logarithm. For example, in the equation \( \ln(x+5) = \ln(x-1) - \ln(x+1) \), this property becomes very handy. Performing the steps results in \( \ln(x+5) = \ln\left(\frac{x-1}{x+1}\right) \).
This simplification step is key to transforming the problem into a simpler form that can be solved more easily. Other logarithmic properties include the product property \( \ln(a \times b) = \ln(a) + \ln(b) \) and the power property \( \ln(a^b) = b \cdot \ln(a) \). Mastering these allows you to manipulate equations to your advantage.

Solving Quadratic Equations

After simplifying the logarithmic equation and equating the arguments, you sometimes end up with a quadratic equation.
For instance, after cross-multiplying \( x+5 = \frac{x-1}{x+1} \), you obtain a quadratic form. Solving such equations requires rearranging terms to fit the standard quadratic format, \( ax^2 + bx + c = 0 \).
To solve, you can use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

• Ensure you identify the coefficients \( a \), \( b \), and \( c \) correctly.
• Calculate the discriminant \( b^2 - 4ac \) to check the nature of solutions: real, distinct, or complex.
• Apply the formula to find \( x \) values.

Always verify solutions in the context of the original equation, as the logarithmic constraints may disallow some roots by falling out of acceptable domains.

Domain Restrictions

In logarithmic functions, domain restrictions play a significant role. The argument of any logarithm must be positive.
For instance, in the equation \( \ln(x+5) = \ln(x-1) - \ln(x+1) \), each expression within a log function must respect this rule:

• \( x+5 > 0 \)
• \( x-1 > 0 \)
• \( x+1 > 0 \)

From this, we infer the feasible range for \( x \). So, \( x > -5 \), \( x > 1 \), and \( x > -1 \), hence, the tightest restriction is \( x > 1 \). Ensuring all solutions respect this range is essential to maintain the validity of the logarithms in the equation and ultimately to ensure a correct result.

Cross-Multiplication

Cross-multiplication is a useful algebraic technique, especially when dealing with fractions in equations. This method helps eliminate denominators to simplify equations.
Consider the equation \( x+5 = \frac{x-1}{x+1} \). By cross-multiplying, we transform it into \( (x+5)(x+1) = x-1 \). This yields a polynomial equation that is often easier to solve.

• First, simplify both sides of the equation after cross-multiplying.
• Expand the products and collect like terms to form a solvable linear or quadratic expression.

This method makes it easier to handle equations involving rational expressions and is a powerful tool in various mathematical situations. It’s crucial for simplifying complex equations and setting the stage for other solving techniques like factoring or applying the quadratic formula.