Problem 68 - 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. $$\log _{6}(x+5)+\log _{6} x=2$$

Step-by-Step Solution

Answer

The solution to the logarithmic equation \(\log _{6}(x+5)+\log _{6} x=2\) is \(x = 4\)

Step 1: Combine Logs with the Same Base

Using the product rule for logarithms, \(\log_b(MN) = \log_bM + \log_bN\), which allows the logs to be combined when the bases are the same. Combine the two logarithms into one: \(\log _{6}((x+5)x) = 2\)

Step 2: Convert Logarithm to Exponential Form

Convert the equation from logarithmic form to exponential form, using the rule \(b^y = x\) corresponds to \(\log_b x = y\). This turns into: \(6^2 = (x+5)x\)

Step 3: Simplify and Solve the Quadratic Equation

First simplify \(6^2 = (x+5)x\) into \(36 = x^2 + 5x\). Rearrange the equation to get a quadratic equation equal to zero: \(0 = x^2 + 5x - 36\). Now solve the quadratic equation. You get \(x = -9\) and \(x = 4\) as solutions.

Step 4: Check the Solutions

Ensure that the solutions fit into the domain of the original logarithmic expressions. Plugging in -9 for \(x\) results in the expression \(\log_6(-4)\), not defined for real numbers, hence, is discarded as an extraneous solution. However, when 4 is plugged in, all expressions are positive and in domain, hence, 4 is a valid solution.

Step 5: Obtain Decimal Approximation

The exact answer is 4. Although there is no need to find a decimal approximation in this case, because the exact answer is already a decimal. If necessary, use a calculator to calculate approximations.