Problem 89 - College Algebra

Question

Solve each logarithmic equation in Exercises \(49-92 .\) Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \ln (x-4)+\ln (x+1)=\ln (x-8) $$

Step-by-Step Solution

Answer

The solution of the logarithmic equation \( \ln (x-4)+\ln (x+1)=\ln (x-8) \) is \( x = 2 \).

Step 1: Combine the logarithms on left-hand side

Use the properties of logarithms to combine the two logarithms on the left side of the equation. The product rule of logarithms states that \( \ln a + \ln b = \ln ab \). Therefore, you can rewrite the equation as \( \ln [(x-4)(x+1)] = \ln (x-8) \).

Step 2: Cancel out the logarithms

Remember, if \( \ln a = \ln b \), then \( a = b \). Use this rule to simplify the equation into a normal quadratic equation: \( (x-4)(x+1) = x-8 \).

Step 3: Expand and simplify

First, expand the left-hand side of the equation: \( x^2 - 3x - 4 = x - 8 \). Then, simplify the equation by subtracting x and adding 8 to both sides, which results in the equation \( x^2 - 4x + 4 = 0 \).

Step 4: Solve the quadratic equation

This is now a simple quadratic equation and can be solved by factoring: It factors to \( (x - 2)^2 = 0 \). Thus, \( x = 2 \).

Step 5: Check the solution

The solution must lie within the domain of the original logarithmic expressions. Both \( \ln (x-4) \) and \( \ln (x+1) \) are undefined for values of x which make the expressions inside the logarithms less than or equal to 0. However, x = 2 is valid in all expressions in the original equation \( \ln (x-4)+\ln (x+1)=\ln (x-8) \). Hence, x = 2 is the valid solution.

Key Concepts

Properties of Logarithms

Logarithms are mathematical expressions that help to find the exponent to which a base must be raised to obtain a given number. There are several important properties of logarithms that allow us to manipulate and simplify expressions involving them. One of the most commonly used properties is the 'product rule of logarithms'. This rule states that:

• If you have two logarithms with the same base that are added together, you can combine them into a single logarithm by multiplying the arguments. Mathematically, this is expressed as: \( \ln(a) + \ln(b) = \ln(ab) \).

For example, in our original equation \( \ln(x-4) + \ln(x+1) = \ln(x-8) \), the left-hand side can be condensed using the product rule, resulting in \( \ln[(x-4)(x+1)] \). This simplification is crucial, as it allows us to equate the arguments of the logarithms directly, bypassing the complex logarithmic expressions and leading us to a simpler equation to solve. Another useful property is the 'power rule', which can be used if any terms have exponents. Understanding and applying these rules enables one to simplify and solve logarithmic equations more effectively.

Quadratic Equations

A quadratic equation is a polynomial equation of the form: \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). The solutions to quadratic equations can be found using several methods such as factoring, completing the square, or applying the quadratic formula.

• Factoring involves expressing the equation as a product of its linear factors. If viable, this method is often the quickest.
• The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) and works for any quadratic equation.

In the solution of the logarithmic equation given, after cancelling out the logarithms, we are left with a quadratic equation: \( (x-4)(x+1) = x-8 \). By expanding and rearranging this equation (\(x^2 - 4x + 4 = 0 \)), we simplify it into an easily factorable form: \((x - 2)^2 = 0\), which tells us that \(x = 2\) is the sole solution. Recognizing quadratic forms and knowing how to solve them is essential, especially as they often appear as intermediate steps in solving logarithmic equations.

Domain of Logarithmic Functions

Understanding the domain of logarithmic functions is vital because logarithms are only defined for positive arguments. A logarithm such as \( \ln(x) \) is undefined for non-positive \( x \). This means that when solving logarithmic equations, you must always check that your solutions lie within the valid domain. In the original exercise, the expressions \( \ln(x-4) \) and \( \ln(x+1) \) need \( x-4 > 0 \) and \( x+1 > 0 \), respectively. Further simplifying gives us \( x > 4 \) and \( x > -1 \), of which the most restrictive condition is \( x > 4 \). Additionally, for \( \ln(x-8) \) to be defined, \( x-8 > 0 \) thus \( x > 8 \) is needed. Therefore, the domain that satisfies all these conditions is \( x > 8 \). However, our solution \( x = 2 \) does not fall within this range. This means that although \( x = 2 \) satisfies the quadratic equation derived from the logarithmic equation, it has to be rejected based on the domain conditions. Thus, understanding and applying the domain restrictions of logarithmic functions is crucial to correctly solving and validating solutions.