What are the solutions to $\log _6\left(x^2+8\right)=1+\log _6(x)$?
To solve the equation \(\log_6(x^2+8) = 1 + \log_6(x)\), we first identify the domain constraints. Since the logarithmic function \(\log_b(y)\) is defined only for \(y > 0\), we must have \(x > 0\) for the term \(\log_6(x)\) to be defined. Additionally, \(x^2+8 > 0\) is satisfied for all real numbers, so the primary restriction is \(x > 0\).
Next, we use properties of logarithms to simplify the equation. We know that \(1 = \log_6(6)\). Substituting this into the original equation gives:
\[\log_6(x^2+8) = \log_6(6) + \log_6(x)\]
Applying the product rule for logarithms, which states that \(\log_b(a) + \log_b(c) = \log_b(ac)\), the right side simplifies as follows:
\[\log_6(x^2+8) = \log_6(6x)\]
Since the logarithm function is one-to-one, we can equate the arguments of the logarithms:
\[x^2+8 = 6x\]
Rearranging this into a standard quadratic equation gives:
\[x^2 - 6x + 8 = 0\]
We can solve this by factoring:
\[(x - 2)(x - 4) = 0\]
This yields two potential solutions: \(x = 2\) and \(x = 4\).
Finally, we check our solutions against the initial domain constraint \(x > 0\). Both \(x = 2\) and \(x = 4\) are greater than zero, so both are valid solutions. Comparing these to the given options, we find that the correct choice is (a). Therefore, the solutions to the equation are \(x = 2\) and \(x = 4\). The final answer is \(\boxed{x=2, x=4}\).
Logarithmic equations like these are frequently used in fields such as acoustics and chemistry. For example, the pH scale used to measure the acidity of a solution is defined using a logarithmic scale, and equations similar to the one solved here are essential for determining the concentration of ions in chemical equilibrium problems. Understanding these properties allows scientists to accurately solve for unknown variables in systems where values span many orders of magnitude.