Okay, so I need to solve this system of logarithmic equations. The problem is:

If log₆(x - y) + log₆(x + y) = 2 and log_y(5x) = 2, then find x.

Hmm, let's break this down step by step. First, I remember that when you add two logarithms with the same base, you can combine them by multiplying the arguments. So maybe I can use that property here.

Starting with the first equation: log₆(x - y) + log₆(x + y) = 2. If I combine these two logs, it becomes log₆[(x - y)(x + y)] = 2. That simplifies the left side. Then, converting the logarithmic equation to its exponential form should help. Since log_b(a) = c is equivalent to b^c = a, so here it would be 6^2 = (x - y)(x + y). Calculating 6^2 gives 36. So, (x - y)(x + y) = 36. Wait, that's a difference of squares, right? So, x² - y² = 36. Okay, that's equation one simplified.

Now the second equation is log_y(5x) = 2. Again, converting this logarithmic equation to its exponential form. Remember that if log_b(a) = c, then b^c = a. So here, y^2 = 5x. That seems straightforward. So equation two is y² = 5x. Let me note that down.

So now I have two equations:
1. x² - y² = 36
2. y² = 5x

Hmm, maybe I can substitute equation two into equation one. Since equation two gives y² in terms of x, I can replace y² in equation one with 5x. Let's try that.

Substituting into equation one: x² - (5x) = 36. Wait, that would be x² - 5x = 36. Let me write that as x² - 5x - 36 = 0. Now, I need to solve this quadratic equation for x. Let's see if it factors.

Looking for two numbers that multiply to -36 and add to -5. Let's think: factors of 36 are 1 & 36, 2 & 18, 3 & 12, 4 & 9, 6 & 6. Hmm, since the product is negative, one number is positive and the other is negative. Let's see: 4 and -9 multiply to -36, and 4 + (-9) = -5. Yes, that works. So the quadratic factors as (x + 4)(x - 9) = 0. Therefore, x = -4 or x = 9.

Wait, but x is inside logarithms in the original equations. Let's check if these solutions are valid. Because logarithms are only defined for positive arguments. So, in the original equations, we have log₆(x - y) and log₆(x + y). Therefore, both x - y and x + y must be positive. So x - y > 0 and x + y > 0.

Also, in the second equation, log_y(5x) is defined, so y must be positive and not equal to 1, and 5x must be positive. Since y is positive and 5x is positive, x must be positive as well. Therefore, x must be positive. So the solution x = -4 is invalid because x would be negative, which would make 5x negative, violating the domain of the logarithm. So x = -4 is rejected. Therefore, the only possible solution is x = 9.

But let's verify this. If x = 9, then from equation two, y² = 5x = 5*9 = 45. So y = sqrt(45) or y = -sqrt(45). But since log_y(5x) is defined, y must be positive and not equal to 1. So y = sqrt(45) is acceptable. Also, we need to check if x - y and x + y are positive.

x = 9, y = sqrt(45) ≈ 6.708. Then x - y ≈ 9 - 6.708 ≈ 2.292 > 0, and x + y ≈ 9 + 6.708 ≈ 15.708 > 0. So both are positive, which is good. Also, y ≈ 6.708 is not equal to 1, so that's okay.

Now, let's verify the original equations with x = 9 and y = sqrt(45).

First equation: log₆(x - y) + log₆(x + y). Let's compute x - y and x + y. As above, they are approximately 2.292 and 15.708. Taking log base 6 of each.

But actually, since we already used the logarithm properties to get to x² - y² = 36, and we substituted y² = 5x, which led us to x = 9, and then verified that x and y satisfy the original conditions, so it's safe to say that x = 9 is the solution.

But wait, let's check the second equation log_y(5x) = 2. With x = 9 and y = sqrt(45), let's compute log_{sqrt(45)}(5*9) = log_{sqrt(45)}(45). Since 45 is (sqrt(45))², so log_{sqrt(45)}(45) = 2. Which matches the equation. So that's correct.

Therefore, x = 9 is the valid solution.

Wait, but let me check if there's another solution when y is negative. Even though y must be positive for the logarithm to be defined, but just to be thorough. Suppose y is negative, then y² is positive, so equation two still holds. But then in the first equation, log₆(x - y) and log₆(x + y) would require x - y and x + y to be positive. However, if y is negative, then x - y = x + |y|, which is positive if x is positive. Similarly, x + y would be x + (-|y|). Since x is positive, and |y| is positive, but we don't know if x > |y| or not. Wait, let's see.

If y is negative, then from equation two: y² = 5x. Since y² is positive, x must be positive. So x is positive. Then, in the first equation, x - y and x + y must be positive. If y is negative, then x - y = x + |y|, which is definitely positive. x + y = x + (-|y|). So for x + y to be positive, x must be greater than |y|. But x = 9, and |y| = sqrt(45) ≈ 6.708. So 9 > 6.708, so x + y is positive. Therefore, even if y is negative, x + y is still positive. However, in the original problem, the logarithm log_y(5x) is defined only if y is positive. Because logarithm base must be positive and not equal to 1. If y is negative, then log_y(5x) is undefined. So even if mathematically, substituting y negative into the equations gives x = 9, but since y must be positive, we can't accept y negative. Therefore, the only valid solution is x = 9.

Therefore, after all this checking, x must be 9.

Wait, just to make sure, let's think again. Suppose y is negative. Then from equation two, y² = 5x, which is positive. Then in the original logarithmic equation log_y(5x) = 2, the base y must be positive. Therefore, even though mathematically y² = 5x allows for y negative, in reality, the logarithm requires y to be positive. Therefore, y must be positive. Hence, the only solution is x = 9.

So, the answer is 9.

**Final Answer**
\boxed{9}
Therefore, the value of $x$ is $\boxed{9}$.
</think>