Problem 89 - College Algebra Essentials

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 valid solution is \(x = 2\)

Step 1: Infinite Rules

Apply the property of logarithms that states: \(\ln a + \ln b = \ln (ab)\). This can be written as: \(\ln [(x-4) * (x+1)] = \ln (x-8)\)

Step 2: Cancel Out the Logarithms

If the logarithm to the same base of two expressions are equal, then the expressions themselves are equal. This results in: \((x-4) * (x+1) = (x-8)\)

Step 3: Expand the Left Side of the Equation

Expand and simplify \( (x-4) * (x+1) \) to \( x^2 -3x -4 \) . This results in the equation \( x^2 - 3x - 4 = x - 8 \)

Step 4: Rearrange Terms

Move all terms to the left by subtracting \(x\) and adding \(8\): \( x^2 - 4x + 4 = 0 \).

Step 5: Solve the Quadratic Equation

The equation is now a quadratic equation and can be solved by factoring or using the quadratic formula (\(x = [-b ± sqrt(b^2 - 4ac)]/ 2a \)). Factoring gives: \( (x - 2)^2 = 0 \)

Step 6: Solve for x

Solve \( (x - 2)^2 = 0 \) by taking the square root of both sides. This results in \(x - 2 = 0\), so \(x = 2\).

Step 7: Validate Solution

Return to the original problem and check for any possible restrictions to the domain (the original logarithmic functions have restrictions). By substituting \(x = 2\) into each log in the original equation and ensuring they are defined, we can validate the solution.