Problem 71 - 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. $$\log _{2}(x+2)-\log _{2}(x-5)=3$$

Step-by-Step Solution

Answer

The solution to the logarithm equation is \(x = 6\).

Step 1: Apply the properties of logarithms

Use the quotient rule of logarithms, which states that \(\log_b A - \log_b B = \log_b \frac{A}{B}\). This allows us to rewrite the equation as \(\log _{2}\frac{x+2}{x-5} = 3\).

Step 2: Remove the logs from the equation

To solve for 'x', get rid of the log by using the definition of the logarithm. An equation of the form \(\log_b A = C\) can be rewritten as \(b^C = A\). Applying this, our equation becomes \(2^3 = \frac{x+2}{x-5}\). Simplify this to obtain \(8 = \frac{x+2}{x-5}\).

Step 3: Solve the algebraic equation for 'x'

To isolate 'x', first multiply both sides of the equation by \(x-5\) giving \(8(x-5) = x+2\). Expanding and rearranging the equation gives \(8x - 40 = x + 2\), which simplifies to \(7x = 42\). So, 'x' equals \( \frac{42}{7} = 6\).

Step 4: Check the solution

To ensure the solution is valid, it should be tested in each of the original logarithmic expressions. Substituting 'x' = 6 into \(x+2\) and \(x-5\) gives 8 and 1, respectively. Since both results are positive, 'x' = 6 is a valid solution.