<Question>
If $\log_6 (x-y) + \log_6 (x+y) = 2$ and $\log_y 5x = 2$, then find $x$.
</Question>

<think>
<Plan_1>
Okay, so I need to solve this system of logarithmic equations. Let me write them down again to make sure I have them right. The first equation is log base 6 of (x minus y) plus log base 6 of (x plus y) equals 2. The second equation is log base y of 5x equals 2. And I need to find x. Hmm, logarithms can sometimes be tricky, especially when they're in different bases. Let me take this step by step.

First, let's recall some logarithm properties. The first equation has two logarithms with the same base being added together. I remember that log_b a + log_b c = log_b (a*c), right? So maybe I can combine those two logs into a single logarithm. Let me try that.

<prompt_1.1> Use the properties of logarithms to combine the two logarithmic equations into a single equation. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
So, combining the logs on the left side of the first equation:

log_6[(x - y)(x + y)] = 2.

Simplifying the argument inside the logarithm, (x - y)(x + y) is a difference of squares, so that becomes x² - y². So the equation becomes:

log_6(x² - y²) = 2.

Now, converting this logarithmic equation to its exponential form. Remember that log_b a = c is equivalent to b^c = a. So here, base 6 raised to the power of 2 equals x² - y². Therefore:

6² = x² - y² => 36 = x² - y².

Okay, so that's the first equation simplified to a quadratic equation: x² - y² = 36. Got that.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, moving on to the second equation: log_y (5x) = 2. This is a logarithm with base y. Again, using the definition of logarithms, this can be rewritten in exponential form. Remember that log_b a = c means that b^c = a. So here, base y raised to the power of 2 equals 5x. Therefore:

<prompt_2.1> Simplify the second logarithmic equation using the properties of logarithms and exponents. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
y² = 5x.

So now, the system of equations we have is:

1. x² - y² = 36
2. y² = 5x

So, we can substitute the second equation into the first one. Since y² is equal to 5x, we can replace y² in the first equation with 5x. Let's do that:

x² - 5x = 36.

So, bringing 36 to the left side:

x² - 5x - 36 = 0.

Now, we have a quadratic equation in terms of x. Let's try to solve this quadratic equation. We can use the quadratic formula here. The quadratic equation is ax² + bx + c = 0, so here a = 1, b = -5, c = -36. The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a). Plugging in the values:

x = [5 ± sqrt(25 + 144)] / 2.

Wait, because b² - 4ac is (-5)^2 - 4*1*(-36) = 25 + 144 = 169. So sqrt(169) is 13. Therefore:

x = [5 ± 13] / 2.

So that gives two solutions:

x = (5 + 13)/2 = 18/2 = 9,

and

x = (5 - 13)/2 = (-8)/2 = -4.

So, x = 9 or x = -4.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
But since we're dealing with logarithms, we need to check if these solutions are valid in the original equations. Because logarithms are only defined for positive arguments, so the arguments inside the logarithms must be positive. Let's check each solution.

<prompt_3.1> Verify the solutions in the context of the original logarithmic equations to ensure they are valid. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
First, x = 9.

If x = 9, then from the second equation, y² = 5x = 5*9 = 45. Therefore, y = sqrt(45) or y = -sqrt(45). But since log base y is in the second equation, y must be positive and not equal to 1. So y = sqrt(45) which is 3*sqrt(5), approximately 6.708.

Now, check the first equation's arguments: x - y and x + y. Since x = 9 and y ≈ 6.708, then x - y ≈ 9 - 6.708 ≈ 2.292, which is positive, and x + y ≈ 9 + 6.708 ≈ 15.708, which is also positive. So both arguments are positive, so that's okay. So x = 9 is a possible solution.

Now check x = -4.

If x = -4, then from the second equation, y² = 5*(-4) = -20. But y² can't be negative since squared numbers are non-negative. Therefore, y² = -20 is impossible. So x = -4 is invalid. Therefore, the only solution is x = 9.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Therefore, the answer is x = 9.

But wait, let me double-check everything because sometimes even if you think you've checked, there might be a subtlety.

First, let me verify the first equation with x = 9 and y = 3√5.

<prompt_4.1> Re-verify the solution by substituting the values back into the original equations. </prompt_4.1>

But wait, let me check if there are other possible values for y. When I solved y² = 45, y could be positive or negative.

<prompt_4.2> Consider other possible values for y and confirm the validity of the solution. </prompt_4.2>

Is there another possible value for y? For example, if y is negative, but log base y is defined?

<prompt_4.3> Explore the implications of y being negative and confirm the solution. </prompt_4.3>

Let me check once again.

Original equations:

<prompt_4.4> Re-evaluate the solution by substituting the values back into the original equations. </prompt_4.4>

Is there a possibility that when we combined the logs, we missed something?

<prompt_4.5> Verify the approach used to combine the logarithms and confirm the solution. </prompt_4.5>

Another thing: when dealing with logarithms, the base must be positive and not equal to 1. So in the second equation, log base y, so y must be positive and not 1.

<prompt_4.6> Confirm the constraints on the variables and finalize the solution. </prompt_4.6>
</Plan_4>

<Execution_4>
<execution_4.1>
Compute (x - y)(x + y). That's 9² - (3√5)^2 = 81 - 9*5 = 81 - 45 = 36. Which matches the first equation, which after conversion was x² - y² = 36. So that's correct.

Also, check the second equation: log_y (5x) = log_{3√5} (45). Let's compute that. Since 45 is 9*5, and 9 is (3)^2, and 3√5 is the base. Let me see: (3√5)^2 = 9*5 = 45. So (3√5)^2 = 45, so log_{3√5} (45) = 2. Which matches the second equation. So that's correct.

Therefore, x = 9 is a valid solution.
</execution_4.1>

<execution_4.2>
But logarithm base y must be positive and not equal to 1. So if y is negative, log base y is undefined. Therefore, y must be positive, so y = 3√5 is the only valid solution. So that's correct.

Therefore, x = 9 is the only valid solution.
</execution_4.2>

<execution_4.3>
Let me check. The logarithm log_y (5x) is defined only if y > 0 and y ≠ 1, and the argument 5x > 0. Since x is 9 in this case, 5x is 45, which is positive. So y must be positive and not 1. So y = 3√5 is approximately 6.708, which is positive and not 1, so that's okay. If we took y = -3√5, then y is negative, which would make the base of the logarithm invalid. So we can discard the negative y. Therefore, only y = 3√5 is valid.

Therefore, x = 9 is the only solution.
</execution_4.3>

<execution_4.4>
log_y(5x) = 2. So if y is negative, log base negative number is undefined in real numbers. So y must be positive. Therefore, y = 3√5 is the only possibility. So x = 9 is the only valid solution.

Wait, but let me verify the first equation again. The first equation is log_6(x - y) + log_6(x + y) = 2. So if x = 9 and y = 3√5, then x - y = 9 - 3√5. Let's compute that numerically. 3√5 is approximately 3*2.236 = 6.708. So 9 - 6.708 ≈ 2.292, which is positive, as we said before. Similarly, x + y ≈ 15.708. So both are positive, so the logs are defined. Then, the product (x - y)(x + y) = x² - y² = 36, which is 6², so log base 6 of 36 is 2, which matches the first equation. So that's correct.

For the second equation, log_y(5x) = 2. If y = 3√5 and x = 9, then 5x = 45. So log base y (45) = log_{3√5}(45). Let's check if that's equal to 2. Let's compute:

If log_{3√5}(45) = 2, then (3√5)^2 = 45. Let's compute (3√5)^2: 9*5 = 45. Yes, that's correct. So that's valid.

So everything checks out. Therefore, x = 9 is the solution.

Wait, but let me check if there are any other possible solutions. For example, when solving the quadratic equation, we got x = 9 and x = -4. But x = -4 was rejected because y² would be negative. Are there any other possibilities?

Alternatively, could there be another value of y when x = -4? But y² = 5x would require y² = -20, which is impossible. So no. Therefore, x = -4 is invalid. So only x = 9 is the solution.
</execution_4.4>

<execution_4.5>
So, log_6(x - y) + log_6(x + y) = 2. For the logs to be defined, both x - y and x + y must be positive. So x - y > 0 and x + y > 0. So x > y and x > -y. Since x is positive (as we found x = 9), and y is positive (since y = 3√5 ≈6.708), so x > y and x > -y is automatically satisfied because x is positive and -y is negative, so x > -y is always true. But x - y must be positive, so x > y. In our case, x = 9 and y ≈6.708, so 9 > 6.708, which is true. Similarly, x + y is positive. So that's okay.

So, in the case of x = -4, y² would be negative, which is impossible, so x = -4 is invalid. So x must be 9.

But let me check the first equation again. Let's compute log_6(x - y) + log_6(x + y). If x = 9 and y = 3√5, then x - y = 9 - 3√5, and x + y = 9 + 3√5. Let's compute their product:

(9 - 3√5)(9 + 3√5) = 9² - (3√5)^2 = 81 - 9*5 = 81 - 45 = 36. So the product is 36. Then log_6(36) = log_6(6^2) = 2, which matches the first equation. So that's correct.

Second equation: log_y(5x) = log_{3√5}(5*9) = log_{3√5}(45). Let's verify if that equals 2. Let me compute log_{3√5}(45). By definition, this is the exponent you need to raise 3√5 to get 45. Let's write 45 as 45 = 3√5 raised to some power.

First, let's express 3√5 in exponential form. 3√5 is 5^(1/2) * 3^1. Hmm, maybe another approach. Let me compute (3√5)^2. That is 9*5 = 45. So (3√5)^2 = 45. Therefore, log_{3√5}(45) = 2. So that's correct. Therefore, the second equation is satisfied.

So both equations are satisfied with x = 9 and y = 3√5. Therefore, x = 9 is the solution.

But just to make sure, let's recap all steps:

1. Combine the two logarithms in the first equation to get a single logarithm, leading to x² - y² = 36.

2. From the second equation, express y² in terms of x: y² = 5x.

3. Substitute y² into the first equation, leading to x² - 5x - 36 = 0.

4. Solve the quadratic equation, getting x = 9 or x = -4.

5. Check validity: x = -4 leads to y² = -20, which is impossible, so discard. x = 9 leads to y² = 45, which is valid, and checking back in original equations confirms correctness.

Therefore, the answer is x = 9. I don't see any mistakes in this reasoning.
</execution_4.5>

<execution_4.6>
So when we found y² = 45, y must be positive, so y = sqrt(45). So negative y is invalid. So that's correct.

Therefore, x = 9 is the only solution.

But let me just verify the first equation again.

Original first equation: log_6(x - y) + log_6(x + y) = 2.

Substituting x = 9 and y = 3√5:

First, compute x - y: 9 - 3√5. Since √5 is approximately 2.236, so 3√5 ≈ 6.708. So 9 - 6.708 ≈ 2.292, which is positive, so log_6 is defined.

x + y: 9 + 3√5 ≈ 15.708, which is positive, so log_6 is defined.

Then, log_6(2.292) + log_6(15.708). Let's compute these values.

First, log_6(2.292). Let's change the base to natural log: ln(2.292)/ln(6). Let me approximate.

ln(2.292) ≈ 0.832, ln(6) ≈ 1.792. So 0.832 / 1.792 ≈ 0.464.

Similarly, ln(15.708) ≈ 2.755, divided by ln(6) ≈ 1.792, so ≈ 1.538.

Adding 0.464 + 1.538 ≈ 2.002, which is approximately 2. So that checks out.

Similarly, the second equation: log_y(5x) = 2. With y = 3√5 and x =9, 5x = 45. So log_{3√5}(45) = 2.

Let me verify that. Let me write 3√5 as 5^(1/2) * 3, but maybe another approach.

Express 45 in terms of 3√5. Let me compute (3√5)^2 = 9*5 = 45. Therefore, (3√5)^2 = 45, so log_{3√5}(45) = 2. Which is exactly the second equation. So that's correct.

Therefore, both equations are satisfied with x =9 and y = 3√5. So x =9 is the solution. x = -4 is invalid because it leads to y² being negative. So the answer is 9.

Therefore, after thorough checking, x = 9 is the valid solution.
</execution_4.6>
</Execution_4>

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