Problem 89 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve each logarithmic equation. 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 to the logarithmic equation \(\ln (x-4)+\ln (x+1)=\ln (x-8)\) is \(x = 2\). However, this solution must be rejected because it does not fall within the domain of the original logarithmic expressions.

Step 1: Condensing the Logarithms

The first step in solving the equation \(\ln (x-4)+\ln (x+1)=\ln (x-8)\) is to condense the logarithms on the left side of the equation by using the properties of logarithms. According to the properties, the sum of two logarithms is equivalent to the logarithm of the product of the two quantities, \(\ln(a)+\ln(b)=\ln(ab)\). Hence, \(\ln (x-4)+\ln (x+1)=\ln ((x-4)*(x+1))\).

Step 2: Simplify the Equation

The multiplication within the logarithm yields \(\ln((x-4)*(x+1)) = \ln((x*x - 3*x -4)) = \ln(x^2 - 3x - 4)\). So, the equation can now be rewritten as \(\ln(x^2 - 3x - 4) = \ln(x - 8)\).

Step 3: Solve for x

Since the two sides of the equation are equal, then the expressions inside the logarithms on both sides of the equation must be equal too. Therefore, \(x^2 - 3x - 4 = x - 8\). To solve for \(x\), subtract \(x\) and add \(8\) on both sides to get \(x^2 - 4x + 4 = 0\). This is a quadratic equation \(x^2 - 2*2*x + 2^2 = 0\) which is of the form \(a^2 - 2ab + b^2\), and can therefore be rewritten as \((x - 2)^2 = 0\). Taking the square-root on both sides of the equation gives \(x - 2 = 0\), and therefore, \(x = 2\).

Step 4: Verify the Solution

The solution found, \(x = 2\), must lie in the domain of the original logarithmic expressions, \(x-4\), \(x+1\), and \(x-8\). Logarithms are defined for positive numbers, hence \(x-4\) > 0, \(x+1\) > 0 and \(x-8\) > 0. Substituting \(x = 2\) in each of these gives \(-2\), \(3\), and \(-6\) respectively. Since \(-2\) and \(-6\) are not greater than zero, \(x = 2\) is not in the domain of the original logarithmic expressions and hence, must be rejected.