Problem 72 - 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. $$\log _{4}(x+2)-\log _{4}(x-1)=1$$

Step-by-Step Solution

Answer

The exact solution for the given logarithmic equation is \(x = 2\).

Step 1: Simplify the Equation using Logarithmic Identity

By employing the properties of logarithms, namely \(\log_b a - \log_b c = \log_b \left(\frac{a}{c}\right)\), the original equation \(\log _{4}(x+2)-\log _{4}(x-1)=1\) can be rewritten as follows: \[\log _{4}\left(\frac{x+2}{x-1}\right) = 1\]

Step 2: Convert the Logarithmic Equation into Exponential Form

Convert the logarithmic equation into an equivalent equation in exponential form. Using the definition of the logarithm \(\log _{b}a = n \Longrightarrow b^n = a\), the equation can be simplified as:\[4^1 = \frac{x+2}{x-1}\]

Step 3: Solve the Resultant Quadratic Equation

This equation simplifies to \(4x - 4 = x + 2\), which further simplifies into a quadratic equation:\[3x - 6 = 0\]Solving this equation leads us to \(x = 2\).

Step 4: Check if the Solution is Within the Domain of the Original Logarithmic Expressions

For the expressions \(\log _{4}(x+2)\) and \(\log _{4}(x-1)\), the values inside the logarithm should be greater than 0. Hence check if the obtained solution is within the domain by plugging \(x = 2\) back into these expressions: \[2+2 > 0 \quad \text{and} \quad 2-1 > 0\]Both are valid for \(x = 2\). Hence, \(x = 2\) is the exact solution.