If $\log_6 (x-y) + \log_6 (x+y) = 2$ and $\log_y 5x = 2$, then find $x$.
I notice that the first equation has two logarithms with the same base, so I can combine them using the product rule of logarithms: $\log_6 (x-y) + \log_6 (x+y) = \log_6 [(x-y)(x+y)]$. I also notice that the right-hand side of the first equation is 2, which means that the argument of the logarithm must be equal to $6^2 = 36$. So I have $(x-y)(x+y) = 36$, which I can expand as $x^2 - y^2 = 36$. The second equation has a logarithm with base $y$, which is a bit tricky. But I can rewrite it using the change of base formula: $\log_y 5x = \frac{\log_6 5x}{\log_6 y}$. The right-hand side of the second equation is also 2, so I can cross-multiply and get $\log_6 5x = 2\log_6 y$. Using the power rule of logarithms, I can rewrite this as $\log_6 5x = \log_6 y^2$. This means that $5x = y^2$, or equivalently, $y = \sqrt{5x}$. Now I have two equations in terms of $x$ and $y$, and I want to eliminate $y$ and solve for $x$. I can substitute $y = \sqrt{5x}$ into the first equation and get $x^2 - (\sqrt{5x})^2 = 36$. Simplifying, I get $x^2 - 5x = 36$, which is a quadratic equation that I can solve by factoring or using the quadratic formula. I prefer to use the quadratic formula, so I plug in the coefficients and get $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-36)}}{2(1)}$. Simplifying, I get $x = \frac{5 \pm \sqrt{169}}{2}$, which is equivalent to $x = \frac{5 \pm 13}{2}$. So I have two possible values for $x$: $x = \frac{5 + 13}{2} = 9$ or $x = \frac{5 - 13}{2} = -4$. However, I have to check if these values are valid, because they might make the logarithms undefined or negative. If $x = 9$, then $y = \sqrt{5x} = \sqrt{45}$, and both $\log_6 (x-y)$ and $\log_6 (x+y)$ are positive, so this is a valid solution. If $x = -4$, then $y = \sqrt{5x} = \sqrt{-20}$, which is not a real number, so this is not a valid solution. Therefore, the only valid solution is $x = 9$. 