Question

what are the potential solutions of \\(\log_{4} x + \log_{4}(x + 6) = 2\\)?
\\(\bigcirc\\) \\(x = -2\\) and \\(x = -8\\)
\\(\bigcirc\\) \\(x = -2\\) and \\(x = 8\\)
\\(\bigcirc\\) \\(x = 2\\) and \\(x = -8\\)
\\(\bigcirc\\) \\(x = 2\\) and \\(x = 8\\)

Explanation:

Step1: Use log property

Using the property \(\log_a M + \log_a N=\log_a(MN)\), we get \(\log_4 [x(x + 6)]=2\).

Step2: Convert to exponential form

By the definition of logarithms, if \(\log_a b = c\), then \(b=a^c\). So \(x(x + 6)=4^2 = 16\).

Step3: Solve the quadratic equation

Expand to get \(x^2+6x - 16 = 0\). Factor it as \((x + 8)(x - 2)=0\). So \(x=-8\) or \(x = 2\).

Step4: Check domain of log

For \(\log_4 x\) and \(\log_4(x + 6)\), \(x>0\) and \(x+6>0\) (i.e., \(x>- 6\)). So \(x=-8\) is invalid. The potential solution (before checking domain fully, but from quadratic) is \(x = 2\) and \(x=-8\) (but we note domain restricts later, but the question is about potential solutions from the equation before domain check fully, or the quadratic result).

Answer:

C. \(x = 2\) and \(x=-8\)