Explanation:
Okay, let's try to tackle this problem. So, we need to find the number of integer values of k in the interval [-500, 500] such that the equation log(kx) = 2 log(x + 2) has exactly one real solution. Hmm, okay. Let me start by recalling some logarithm properties and think about the steps needed here.

First, since logarithms are involved, I need to remember their domains. The argument of a logarithm must be positive. So, for log(kx) to be defined, kx must be greater than 0. Similarly, for log(x + 2), the argument x + 2 must be greater than 0. Therefore, we have two conditions:

1. kx > 0
2. x + 2 > 0 ⇒ x > -2

So, the domain of x is x > -2, and also kx > 0. Depending on the sign of k, this will impose different constraints on x. Let me note that.

Now, moving on to the equation itself: log(kx) = 2 log(x + 2). Let me recall that 2 log(x + 2) can be rewritten as log((x + 2)^2). So, the equation becomes:

log(kx) = log((x + 2)^2)

Since logarithm is a one-to-one function, if log(a) = log(b), then a = b, provided that a and b are positive. Therefore, we can write:

kx = (x + 2)^2

But we also have to remember the domain restrictions. So, the solutions to this equation must satisfy kx > 0 and x > -2.

So, first, let me write the equation:

kx = (x + 2)^2 ⇒ kx = x² + 4x + 4 ⇒ x² + 4x + 4 - kx = 0 ⇒ x² + (4 - k)x + 4 = 0

So, this is a quadratic equation in terms of x: x² + (4 - k)x + 4 = 0.

Now, we need to find the number of real solutions to this equation, but considering the domain restrictions. However, the problem states that the original equation has exactly one real solution. So, even though the quadratic equation might have two solutions, we need to ensure that only one of them satisfies the domain conditions.

But first, let's analyze the quadratic equation. Let me recall that a quadratic equation ax² + bx + c = 0 has:

- Two real solutions if discriminant D = b² - 4ac > 0
- One real solution if D = 0
- No real solutions if D < 0

So, for our quadratic equation x² + (4 - k)x + 4 = 0, the discriminant D is:

D = (4 - k)² - 4 * 1 * 4 = (16 - 8k + k²) - 16 = k² - 8k

Therefore, D = k² - 8k.

So, depending on the value of k, the discriminant can be positive, zero, or negative.

But here's the thing: even if the quadratic equation has two real solutions, we need to check whether they satisfy the domain conditions. However, the problem says that the original equation has exactly one real solution, which could be due to:

1. The quadratic equation having exactly one real solution (i.e., discriminant zero), which would mean that the solution is valid (satisfies domain conditions)
2. The quadratic equation having two real solutions, but only one of them satisfies the domain conditions (i.e., one is in the domain and the other is not)

Therefore, we need to consider both cases.

So, let's break it down into two cases:

Case 1: The quadratic equation has exactly one real solution (D = 0). For this to happen, discriminant must be zero: k² - 8k = 0 ⇒ k(k - 8) = 0 ⇒ k = 0 or k = 8.

However, we need to check if these solutions satisfy the domain conditions.

First, let's check for k = 0.

If k = 0, the original equation becomes log(0 * x) = 2 log(x + 2). But log(0) is undefined, so k = 0 is not allowed. Therefore, k = 0 is invalid. Therefore, only k = 8 is a candidate here.

Wait, let's check for k = 8. If k = 8, then our quadratic equation becomes x² + (4 - 8)x + 4 = x² -4x + 4 = 0 ⇒ (x - 2)^2 = 0 ⇒ x = 2. So, x = 2. Let's check the domain conditions:

- kx = 8 * 2 = 16 > 0
- x + 2 = 4 > 0. So, both conditions are satisfied. Therefore, x = 2 is a valid solution. Therefore, for k = 8, the equation has exactly one real solution. So, k = 8 is valid.

So, Case 1 gives us k = 8.

Case 2: The quadratic equation has two real solutions, but only one of them satisfies the domain conditions.

So, in this case, we need D > 0, and exactly one of the roots satisfies the domain conditions.

So, first, D > 0 ⇒ k² - 8k > 0 ⇒ k(k - 8) > 0 ⇒ k < 0 or k > 8.

Therefore, for k < 0 or k > 8, discriminant D is positive, so quadratic has two real roots. Now, we need to check for each such k whether exactly one root is in the domain.

But before that, let me note that for k = 0, we saw that it's invalid. Similarly, for k = 8, we have discriminant zero, which we already considered.

So, now, for Case 2: k < 0 or k > 8.

Now, we need to analyze the roots of the quadratic equation and check which ones satisfy the domain conditions.

Let me recall that the quadratic equation is x² + (4 - k)x + 4 = 0.

Let me denote the roots as x₁ and x₂. Then, by Vieta's formula:

x₁ + x₂ = -(4 - k) = k - 4

x₁ * x₂ = 4

So, the product of the roots is 4.

Also, the domain conditions are:

For each root x, we need:

1. x > -2 (from x + 2 > 0)
2. kx > 0 (from log(kx) being defined)

So, for each root x, we need both x > -2 and kx > 0.

So, let's think about how these conditions interact.

First, note that the product of the roots is 4. So, x₁ * x₂ = 4. Therefore, both roots are either positive or both negative? Wait, since the product is positive. Therefore, either both roots are positive or both are negative.

But let's check:

If the product is positive, then either both roots positive or both negative. However, since x > -2 is required, which is a lower bound, so if both roots are negative, but x > -2, then we need to check if they are greater than -2.

But let me think:

First, let's consider the case when k < 0. Then, for kx > 0, since k is negative, x must be negative. So, in this case, x must be negative. But also, x > -2, so x ∈ (-2, 0). However, if both roots are negative (since their product is positive) and they are in (-2, 0), then we need to check how many roots lie in (-2, 0).

Similarly, if k > 8, then k is positive. So, for kx > 0, since k is positive, x must be positive. Therefore, in this case, x > 0 (since kx > 0) and also x > -2 (which is automatically satisfied if x > 0). So, in this case, the domain is x > 0. Therefore, for k > 8, we need to check how many roots lie in x > 0.

So, let's handle k < 0 and k > 8 separately.

First, let's handle k < 0:

Case 2a: k < 0.

In this case, since product of roots is 4, both roots are negative. Also, from kx > 0, x must be negative. However, since x must also satisfy x > -2. So, the roots are negative numbers, but we need them to be greater than -2. Therefore, the roots must lie in (-2, 0). So, we need to check whether the quadratic equation has exactly one root in (-2, 0) or both roots in (-2, 0) or none.

But since the quadratic equation has two roots, and we need exactly one of them to be in (-2, 0). However, since both roots are negative (product positive) and their sum is k - 4. Since k < 0, then k - 4 is less than -4. Therefore, the sum of roots is negative. So, the sum of two negative numbers is negative, which is consistent.

But how do we know whether the roots lie in (-2, 0)?

Alternatively, perhaps we can analyze the quadratic function f(x) = x² + (4 - k)x + 4. Since the product of roots is 4, and both roots are negative.

Wait, but maybe we can use the fact that if a quadratic equation has two real roots, then to find how many roots lie in a certain interval, we can use the concept of sign changes or testing the function at the endpoints. But since we need to check whether the roots lie in (-2, 0), perhaps we can use the following approach:

For each root x, we need x ∈ (-2, 0). Also, since both roots are negative, and their product is 4. So, for example, if one root is in (-2, 0), then the other root is 4 divided by that root, which would be negative. Let's say one root is between -2 and 0, so the other root is 4 / x. Since x is between -2 and 0, 4 / x is less than -2 (since dividing a positive by a negative number gives negative, and dividing by a number between -2 and 0 gives a number less than -2). For example, if x is -1, then 4 / x is -4. If x approaches 0 from the left, 4 / x approaches -infty. If x approaches -2 from the right, 4 / x approaches -2.

Therefore, if one root is in (-2, 0), the other root is less than -2. Hence, only one root lies in (-2, 0). Therefore, for k < 0, the quadratic equation has two real roots: one in (-2, 0) and the other less than -2. Therefore, only the root in (-2, 0) satisfies the domain conditions (since the other one is less than -2, so x + 2 would be less than 0, which is invalid). Therefore, for k < 0, the equation log(kx) = 2 log(x + 2) would have exactly one real solution, as only one root is valid.

Wait, but hold on, is that always the case? Let me check with an example. Suppose k = -1. Then, quadratic equation becomes x² + (4 - (-1))x + 4 = x² + 5x + 4 = 0 ⇒ (x + 1)(x + 4) = 0 ⇒ x = -1 or x = -4. Now, check domain conditions:

For x = -1: kx = -1 * (-1) = 1 > 0, and x + 2 = -1 + 2 = 1 > 0. So valid. For x = -4: kx = -1 * (-4) = 4 > 0, but x + 2 = -4 + 2 = -2 < 0. So invalid. Hence, only one solution. So, that works.

Another example: k = -8. Then quadratic equation is x² + (4 - (-8))x + 4 = x² + 12x + 4 = 0. Let me compute the roots:

Discriminant D = 144 - 16 = 128. Roots are [-12 ± sqrt(128)] / 2 = [-12 ± 8*sqrt(2)] / 2 = -6 ± 4*sqrt(2). sqrt(2) ≈ 1.414, so 4*sqrt(2) ≈ 5.656. Therefore, the roots are approximately -6 + 5.656 ≈ -0.344 and -6 - 5.656 ≈ -11.656. So, the first root is approximately -0.344 which is in (-2, 0), and the second is less than -2. So again, only one valid solution. Hence, seems like for k < 0, regardless of the value, we have exactly one valid solution. Therefore, for all k < 0, the equation has exactly one real solution.

Wait, but what if the quadratic equation has both roots in (-2, 0)? Is that possible? Let me see. Suppose that both roots are in (-2, 0). Then, since their product is 4, which is positive, but if both are between -2 and 0, then their product would be positive, but in that case, the product would be less than 4? Wait, for example, if both roots are -1, then their product is 1, which is less than 4. If both roots are approaching 0, then the product approaches 0. If both roots approach -2 from the right, then their product approaches (-2)*(-2)=4. So, the maximum product when both roots are in (-2, 0) is approaching 4. Therefore, if the product is exactly 4, which is when one root approaches -2 and the other approaches -2. But since the product is 4, if both roots are equal (i.e., discriminant zero), but discriminant zero would require k = 0 or k = 8. However, k < 0 here, so discriminant zero is not possible for k < 0. So, in the case of k < 0, the quadratic equation has two real roots, one in (-2, 0) and one less than -2. Therefore, only one solution is valid.

Wait, but let me check if there's any possible k < 0 where both roots are in (-2, 0). Suppose that both roots are in (-2, 0). Then, the sum of roots is k - 4. Let me denote the roots as α and β, both in (-2, 0). Then, α + β = k - 4, and αβ = 4.

But if α and β are both in (-2, 0), then α + β ∈ (-4, 0). Therefore, k - 4 ∈ (-4, 0) ⇒ k ∈ (0, 4). But we are considering k < 0 here. Therefore, there's a contradiction. Hence, α + β = k - 4, and for k < 0, k - 4 < -4. Therefore, the sum of roots is less than -4. But if both roots are in (-2, 0), their sum would be greater than -4 (since each root is greater than -2, so sum is greater than -4). Therefore, we have that k - 4 < -4, which is k < 0, but if both roots are in (-2, 0), then α + β > -4 ⇒ k - 4 > -4 ⇒ k > 0. Therefore, for k < 0, it's impossible for both roots to be in (-2, 0). Hence, only one root is in (-2, 0) and the other is less than -2, hence only one valid solution. Therefore, for all k < 0, the equation has exactly one real solution.

Therefore, Case 2a: k < 0 gives valid solutions (each k < 0 gives exactly one solution).

Case 2b: k > 8.

In this case, discriminant D = k² - 8k > 0. Also, k is positive, so for kx > 0, since k is positive, x must be positive. Therefore, the domain for x is x > 0. Also, we need to check whether the roots of the quadratic equation lie in x > 0.

So, the quadratic equation is x² + (4 - k)x + 4 = 0. Let's analyze its roots.

Since the product of the roots is 4 (positive) and the sum of the roots is k - 4. Since k > 8, the sum is k - 4 > 4.

So, the sum of roots is positive and greater than 4, and the product is positive (4). Therefore, both roots are positive. Because if both roots are positive, their sum and product are positive. Therefore, in this case, both roots are positive. Hence, both roots satisfy x > 0. However, we need to check if they satisfy x > -2, but since x > 0, that's automatically satisfied. Therefore, both roots are in the domain. Therefore, the original equation would have two real solutions. However, the problem requires the original equation to have exactly one real solution. Therefore, for k > 8, the equation would have two real solutions. Therefore, such k would not be acceptable.

Wait, but wait. Wait, unless one of the roots is invalid for some reason? But we have already considered that kx > 0 and x > -2. Since x > 0, both conditions are satisfied. Therefore, both roots are valid. Therefore, for k > 8, equation has two real solutions, which is not acceptable. Therefore, in Case 2b, we need to exclude k > 8. However, hold on, but maybe there is a case when one of the roots is invalid? But no, since both roots are positive and k is positive, so kx is positive. Therefore, both roots are valid. Hence, for k > 8, the equation has two solutions. Therefore, they are not acceptable.

But then, what about k = 8? For k = 8, discriminant is zero, so only one solution, which is valid as we saw. Therefore, in Case 2, only k < 0 gives valid solutions (each with exactly one solution), and k > 8 gives two solutions. Therefore, the only k's that satisfy the condition are k < 0 and k = 8.

Wait, but then wait, there might be another case where even though the quadratic equation has two solutions, one of them is invalid due to some other reason? Let me check.

Alternatively, maybe there are some k values where one of the roots is invalid even though k is positive. Wait, for example, if one of the roots is negative, but since for k > 8, we have both roots positive. Wait, but since the product is positive and sum is positive, both roots must be positive. Therefore, there's no way for one root to be negative. Therefore, for k > 8, both roots are positive, hence both valid. Hence, two solutions. Therefore, they are not acceptable. Therefore, only k < 0 and k = 8 are acceptable.

But hold on, let me check with an example. Suppose k = 9, which is greater than 8. Then quadratic equation is x² + (4 - 9)x + 4 = x² -5x + 4 = 0 ⇒ (x - 1)(x - 4) = 0. Therefore, roots at 1 and 4. Both positive, so both satisfy x > 0 and kx positive. Hence, two solutions. So, original equation would have two solutions, which is not acceptable. Therefore, k = 9 is excluded.

Now, what about k = 8, which is the case where discriminant is zero? We saw that gives x = 2. Let me check if there are other k's where even though discriminant is positive, but one of the roots is invalid?

Wait, but for k > 8, both roots are valid. For k < 0, only one root is valid. For k = 0, invalid. For k between 0 and 8, discriminant is negative? Wait, D = k² - 8k. For k between 0 and 8, D = k(k - 8) < 0, so no real solutions. Therefore, for 0 < k < 8, equation has no real solutions. Therefore, the original equation would have no real solutions, so they are not acceptable.

Therefore, putting it all together:

The values of k for which the equation has exactly one real solution are:

- All k < 0 (since for each such k, exactly one solution)
- k = 8 (since discriminant zero and valid solution)

But wait, but hold on. Wait, let me check k = 0. If k = 0, original equation is log(0) = 2 log(x + 2). But log(0) is undefined, so k = 0 is invalid, as we saw. Similarly, for k = 8, we have one solution. So, so far, our candidates are k < 0 and k = 8.

But let me check if there are other possibilities. For example, could there be some k where the quadratic equation has one root valid and another invalid, even when discriminant is positive?

Wait, for k < 0: yes, we already considered that. For k > 8: no, both roots are valid. What about k between 0 and 8? The quadratic equation has no real roots. If k = 0, invalid. What about k = 8? discriminant zero. Therefore, seems like only k < 0 and k = 8.

But wait, hold on! Let me think again. Suppose there is some k where even though discriminant is positive (so two roots), but one of the roots is invalid. For example, suppose for some k, one of the roots is x = -2, which is on the boundary of the domain. However, x must be greater than -2, so x = -2 is not allowed. So, if one of the roots is x = -2, then that root is invalid. However, in that case, maybe there's another root that's valid?

But let's check when x = -2 is a root. Plugging x = -2 into the quadratic equation:

(-2)^2 + (4 - k)(-2) + 4 = 4 + (-8 + 2k) + 4 = 4 - 8 + 4 + 2k = 0 + 2k = 2k. Therefore, setting equal to zero: 2k = 0 ⇒ k = 0. But k = 0 is invalid because log(0) is undefined. So, only when k = 0, x = -2 is a root, but k = 0 is invalid. Therefore, no other k would have x = -2 as a root. Therefore, for other k, roots are either both valid (for k > 8), or one valid (for k < 0). Therefore, our previous conclusion holds.

But wait, there's another case. What if one of the roots is x = 0? Then, kx = 0, which is invalid. Let me check if x = 0 is a root. Plugging x = 0 into quadratic equation:

0 + (4 - k)*0 + 4 = 4 ≠ 0. Therefore, x = 0 is not a root. Therefore, no problem.

Alternatively, suppose that one of the roots is x = -2 + ε, very close to -2, but still greater than -2. However, even then, since for k < 0, we already have one root in (-2, 0) and one root less than -2, so as long as the root is in (-2, 0), it's valid. Therefore, no issues there.

Therefore, according to our analysis, the values of k are all negative integers in [-500, 500] and k = 8. But wait, the problem says "closed interval [-500, 500]". So, k is in [-500, 500], integers. Therefore, the number of integer values of k is:

Number of negative integers from -500 to -1 inclusive, plus 1 for k = 8.

But let me check if k = 8 is within the interval. Since 8 is in [-500, 500], yes.

So, first, count the number of integers k with k < 0 and in [-500, 500]. The integers from -500 up to -1 inclusive. The number of integers is 500. Because from -1 to -500, inclusive, that's 500 numbers. For example, from -1 to -n, there are n numbers. So, from -1 to -500, there are 500 numbers. Therefore, 500 values.

Then, plus k = 8, which is one more. Hence, total number of integer values is 500 + 1 = 501.

Wait, but hold on. Wait, the problem says "closed interval [-500, 500]". So, integers from -500 to 500 inclusive. So, the negative integers from -500 to -1 inclusive: that's 500 numbers. Then, k = 8 is one more. But wait, we need to check if k = 8 is the only value for Case 1 (discriminant zero). We had k = 8 and k = 0. But k = 0 is invalid, so only k = 8. So, total number of valid k is 500 + 1 = 501. But let me check again.

Wait, but wait. What about k = 8 is included in the interval [-500, 500], which it is. So, the answer would be 500 + 1 = 501. However, let me think again. Let me make sure that there are no other values of k where the equation has exactly one solution. For example, is there a possibility when the quadratic equation has two solutions, but one of them is invalid? But for k > 8, both are valid. For k < 0, one is valid. For 0 < k < 8, no solutions. For k = 0, invalid. For k = 8, one solution. So, yes, only those.

Wait, but hold on! There is another case. What if the quadratic equation has two solutions, but one of them is invalid due to kx > 0? Wait, but for k > 8, we established both roots are positive, so kx positive, so valid. For k < 0, one root is negative (but x > -2) and the other is less than -2 (so invalid). So, only one valid. But is there a case where for some k < 0, the quadratic equation's roots are such that both are in (-2, 0), hence both valid? Earlier, we showed that's impossible. Because for k < 0, the sum of roots is k - 4, which is less than -4. However, if both roots are in (-2, 0), the sum would be greater than -4, so conflicting. Therefore, not possible. Therefore, our conclusion holds.

Therefore, the answer is 500 (for k < 0) + 1 (for k = 8) = 501. However, wait, but hold on. Let me check again the count of negative integers. The interval is from -500 to 500 inclusive. How many integers are there from -500 to -1 inclusive? Well, from -1 to -500, that's 500 numbers. Because if you take 500 - (-1) = 501? Wait, no. Wait, the number of integers from a to b inclusive is b - a + 1. So, if we have from -500 to -1 inclusive, that is, a = -500, b = -1. Then number of integers is (-1) - (-500) + 1 = 499 + 1 = 500. Yes. So that's correct. So 500 negative integers. So total is 500 + 1 = 501. So is 501 the answer?

But wait, wait. Wait, hold on. Wait, k = 8. But what if k = 8 is not in the interval? Wait, no, the interval is [-500, 500], so 8 is within it. Therefore, answer is 501.

But let me check again with another example. Let me check if k = -500 is included. Yes, since the interval is closed. So k = -500 is included. So, number of negative integers from -500 to -1 inclusive is 500 numbers. Then, adding k = 8. Therefore, total 501.

But wait, but maybe there is a mistake here. Because for k < 0, we need to check that the equation actually has exactly one solution. But we already established that for each k < 0, the equation has exactly one solution. Therefore, 500 values. Plus k = 8. Therefore, 501.

But wait, but in the problem statement, it says "closed interval [-500, 500]". So, k can be from -500 to 500 inclusive. But let me check if k = -500 is allowed. For k = -500, the equation is log(-500x) = 2 log(x + 2). Then, domain requires -500x > 0, so x < 0, and x > -2. So x ∈ (-2, 0). Then, solving the quadratic equation:

x² + (4 - (-500))x + 4 = x² + 504x + 4 = 0. The roots are [-504 ± sqrt(504² - 16)] / 2. Which are real numbers. As we saw, one root is in (-2, 0) and another is less than -2, so only one valid solution. So, k = -500 is acceptable.

Similarly, k = -1 is acceptable. So, yes, all negative integers from -500 to -1 inclusive are acceptable, which is 500 values. Then k = 8 is another value, so total of 501.

But wait, wait. Wait, hold on. Wait, when k = 8, we have exactly one solution. But what about k = 8? But what if there is another k where the equation has exactly one solution? For example, suppose that for some k > 8, one of the roots is invalid? Wait, we had considered that for k > 8, both roots are positive. Suppose that one of the roots is x = -2? Wait, but we saw that only when k = 0, x = -2 is a root, but k = 0 is invalid. Therefore, no. Therefore, for k > 8, both roots are positive. Hence, both valid. Therefore, they give two solutions. Hence, they are not counted.

But here's a thought. What if the quadratic equation has one root which is x = -2? But we saw that only happens when k = 0. But k = 0 is invalid. Therefore, no.

Alternatively, maybe there's some k where one of the roots is invalid even for k < 0? For example, suppose for some k < 0, the root in (-2, 0) actually makes log(kx) undefined? Wait, no. Because for k < 0, and x in (-2, 0), then kx is positive? Let me check. If k is negative and x is negative, then kx is positive. Yes. So, log(kx) is defined. Also, x + 2 > 0 because x is in (-2, 0). Therefore, both sides are defined. Therefore, yes, valid. Therefore, the solution is valid. Therefore, all k < 0 are okay.

Therefore, the answer should be 501.

But wait, hold on. But the problem says "closed interval [-500, 500]". Therefore, the integers k from -500 to 500 inclusive. But k = 8 is included, so total of 500 negative integers, plus 1 for k = 8. So, total 501. But wait, maybe there is a mistake here. Let me check the count again.

Number of integers from -500 to -1 inclusive: Let me think of it as the number of integers from 1 to 500, which is 500. Similarly, from -500 to -1 is also 500 numbers. Therefore, yes. Then, adding k = 8. So 501.

However, let me check once again if there are other values of k where the equation has exactly one solution. For example, if discriminant is zero. We had discriminant zero when k = 0 or k = 8. But k = 0 is invalid, so only k = 8. Therefore, only one such value.

Alternatively, is there another case when the equation has exactly one solution? For example, when one root is valid and another is invalid, but discriminant is positive. But for k < 0, as we saw, one is valid and one is invalid. For k > 8, both are valid. For other k, either no solutions or one solution (k=8). Therefore, yes, only the cases considered.

Thus, answer is 501. However, wait, wait! Wait, the problem says "exactly one real solution". However, for k < 0, we have exactly one real solution. For k = 8, exactly one real solution. So total 501.

But hold on, let me check if there are any overlaps or something. For example, is k = 8 included in the k < 0? No, since 8 is positive. Therefore, no overlap. Therefore, the total number is 500 + 1 = 501. Therefore, the answer is 501.

But now, wait. Wait, the problem says "closed interval [-500, 500]". So, k can be -500, -499,..., -1, 0, 1,..., 500. But we are to count the number of integer values of k in this interval for which the equation has exactly one real solution. So, as per our analysis, all negative integers (from -500 to -1, inclusive) and k = 8. So, the total number is 500 + 1 = 501.

But wait, let me check the problem statement again. It says "exactly one real solution". Wait, but for k < 0, even though the quadratic equation has two real roots, only one of them is valid. Therefore, the equation has exactly one real solution. For k = 8, the quadratic equation has one real solution. Therefore, both cases satisfy the condition. Hence, 501.

However, I'm a bit unsure because sometimes there can be edge cases. Let me check for k = -2. Let me take k = -2. Then quadratic equation is x² + (4 - (-2))x + 4 = x² + 6x + 4 = 0. Discriminant D = 36 - 16 = 20 > 0. So roots are [-6 ± sqrt(20)] / 2 = [-6 ± 2*sqrt(5)] / 2 = -3 ± sqrt(5). sqrt(5) ≈ 2.236. Therefore, roots are approximately -3 + 2.236 ≈ -0.764 and -3 - 2.236 ≈ -5.236. Therefore, x ≈ -0.764 is in (-2, 0), valid. The other root is less than -2, invalid. So, exactly one solution. Correct. Similarly for k = -1. So, seems okay.

Wait, but what about k = -4? Let me try k = -4. Then quadratic equation is x² + (4 - (-4))x + 4 = x² + 8x + 4 = 0. Discriminant D = 64 - 16 = 48 > 0. Roots are [-8 ± sqrt(48)] / 2 = [-8 ± 4*sqrt(3)] / 2 = -4 ± 2*sqrt(3). sqrt(3) ≈ 1.732. So, roots ≈ -4 + 3.464 ≈ -0.536 and -4 - 3.464 ≈ -7.464. So, again, one valid solution. Correct.

Another check for k = 8: equation is log(8x) = 2 log(x + 2). Which reduces to 8x = (x + 2)^2. Which is x² + 4x + 4 - 8x = x² -4x +4 = (x - 2)^2 = 0. So x = 2. Check domain: 8*2 = 16 > 0, and x + 2 = 4 > 0. Valid. So, one solution. Correct.

So, seems like our reasoning is correct.

Therefore, the answer is 501. Therefore, the number of integer values of k is 501.

Wait, but wait! Wait, the problem says "closed interval [-500, 500]". So, k can be -500, -499,..., 0,..., 500. So, the number of integers from -500 to 500 inclusive is 1001 numbers. But we are to count how many of them satisfy the condition. As per our analysis, it's 500 negative numbers plus 1 (for k = 8). So 501. But let me check once more if k = 8 is the only case where discriminant is zero.

Yes, discriminant D = k² - 8k. Setting D = 0: k² - 8k = 0 ⇒ k(k - 8) = 0 ⇒ k = 0 or k = 8. As before, k = 0 invalid, so only k = 8. Therefore, 501 total.

But the problem is in the original equation log(kx) = 2 log(x + 2). Wait, does the equation have any restrictions on k? For example, k can't be zero, as log(0) is undefined. But k is allowed to be negative, as long as kx is positive. Which is considered in our analysis.

But wait, in our analysis, k can be any negative integer from -500 to -1 inclusive. And k = 8. So, total of 500 + 1 = 501 integers.

But I'm a bit uncertain because I recall sometimes problems like this might have more cases. Wait, let me think again. Maybe there's a case when the quadratic equation has two solutions, both valid, but one of them is extraneous? For example, when solving log(kx) = 2 log(x + 2), we assumed that kx > 0 and x + 2 > 0, but perhaps there's another extraneous solution when we did the logarithm manipulation?

Wait, when we started with log(kx) = 2 log(x + 2), we converted it to kx = (x + 2)^2. However, this step is valid only if both sides are defined, i.e., log(kx) and log(x + 2) are defined. So, we have already considered that kx > 0 and x + 2 > 0. Therefore, all solutions to kx = (x + 2)^2 that are in the domain are valid. Therefore, there are no extraneous solutions. Therefore, our analysis holds.

Therefore, I think the answer is 501. However, let me check an example where k is positive but less than 8, but maybe some edge case?

Wait, for example, take k = 4. Then quadratic equation is x² + (4 - 4)x + 4 = x² + 0x + 4 = x² + 4 = 0 ⇒ x² = -4, which has no real solutions. So, no solution. Correct. For k = 5, quadratic equation x² + (4 - 5)x + 4 = x² - x + 4 = 0. Discriminant D = 1 - 16 = -15 < 0. No solution. So, correct.

Another example, k = 8: one solution. Correct.

So, yes, seems okay.

Therefore, I think the answer is 501. However, to make sure, let me check if k = -500 is allowed. We said that yes, since x must be in (-2, 0) for k = -500. For k = -500, x ≈ -0.764 is valid. Therefore, yes.

But let me check if there's a mistake in the count. The interval from -500 to 500 inclusive has 1001 integers. The number of integers k with k < 0 is 500 (from -500 to -1 inclusive). Then, k = 8 is another value, so total 501. Therefore, answer is 501. Therefore, I think that is correct.

But wait, just to check if k = -500 is included. Yes, since -500 is in [-500, 500]. Similarly, -1 is included. So, numbers from -500 to -1 inclusive is 500 numbers. Then k = 8. So, 501.

But let me check if there are any other values of k where the equation has exactly one solution. For example, suppose k = something where the quadratic equation has two solutions, but one of them is invalid? But for k > 8, both solutions are valid. For k < 0, one is valid, one invalid. For k between 0 and 8, no solution. For k = 0, invalid. For k = 8, one solution. Therefore, yes, only the above cases. Therefore, 501.

But wait, just another thought. What if k = -2? Wait, we checked that already. So, yes, seems okay.

Therefore, I think the answer is 501.

Wait, but let me check again the count of negative integers. Let me think of the number of integers between -500 and -1 inclusive. The number is equal to 500. Because from -1 to -500, that's 500 numbers. For example, from -1 to -n, there are n numbers. Therefore, from -1 to -500, there are 500 numbers. Therefore, correct.

Therefore, the answer is 501. But to cross-verify, let me check how many integers are there in total. From -500 to 500 inclusive, it's 500 - (-500) + 1 = 1001 numbers. If we have 500 negative, 1 zero, and 500 positive numbers. But we are only counting the negative and k = 8. Therefore, 500 + 1 = 501. Yes. Therefore, answer is 501.

But hold on! Wait, in the interval [-500, 500], the number of negative integers is 500 (from -1 to -500), and then k = 8 is one specific positive integer. Therefore, total of 501. Therefore, I think that is correct.

But, wait, the problem says "closed interval [-500, 500]". So, 500 is included. Therefore, k = 8 is included. Therefore, answer is 501. So, I think the answer is \boxed{501}.

But wait, I remember sometimes in these problems, there's a possibility that when k = 8, maybe there's another solution or something. But no, we checked that k = 8 gives exactly one solution. Therefore, I think 501 is correct.

Alternatively, maybe I made a mistake in assuming that for all k < 0, exactly one solution exists. Let me think if there is any k < 0 where the quadratic equation has two solutions in (-2, 0). But as per earlier reasoning, since the sum of roots is k - 4, and k < 0, so sum is less than -4. But if two roots are in (-2, 0), then their sum would be greater than -4. Contradiction. Hence, no. Therefore, only one solution. Hence, correct.

Therefore, I think the answer is 501.

**Final Answer**
\boxed{501}

To solve the given problem, we are asked to find how many integer values of $ k $ in the interval $[-500, 500]$ satisfy the condition that the equation

$$
\log(kx) = 2\log(x + 2)
$$

has **exactly one real solution**.

---

### Step 1: Domain Analysis

The logarithmic equation is defined only when:

- $ kx > 0 $, because $ \log(kx) $ is defined only if its argument is positive.
- $ x + 2 > 0 $, because $ \log(x + 2) $ is defined only if the argument is positive.

So the domain of $ x $ is:

$$
x > -2 \quad \text{and} \quad kx > 0
$$

This gives us a restriction on how $ x $ behaves relative to $ k $:

- If $ k > 0 $, then $ x > 0 $
- If $ k < 0 $, then $ x < 0 $, but we also need $ x > -2 $, so $ x \in (-2, 0) $

If $ k = 0 $, $ kx = 0 $, which makes $ \log(0) $ undefined, so $ k = 0 $ is **not** allowed.

---

### Step 2: Solve the Equation

Given:
$$
\log(kx) = 2\log(x + 2)
$$

Using logarithmic identities:

$$
\log(kx) = \log((x + 2)^2)
\Rightarrow kx = (x + 2)^2
$$

Expanding the right-hand side:

$$
(x + 2)^2 = x^2 + 4x + 4
\Rightarrow x^2 + (4 - k)x + 4 = 0
$$

So the quadratic equation is:
$$
x^2 + (4 - k)x + 4 = 0
$$

Let’s denote this as $ f(x) = x^2 + (4 - k)x + 4 $

---

### Step 3: Nature of the Roots

We analyze the number of real roots of the quadratic equation.

**Discriminant** $ D $ is:
$$
D = (4 - k)^2 - 16 = k^2 - 8k
$$

So:

- $ D > 0 $ when $ k^2 - 8k > 0 $ ⟹ $ k < 0 $ or $ k > 8 $
- $ D = 0 $ when $ k = 0 $ or $ k = 8 $
- $ D < 0 $ when $ 0 < k < 8 $

---

### Step 4: Determine Validity of Roots for Each Case

Let’s consider the three cases:

#### Case 1: $ k = 0 $ → Invalid
As noted earlier, $ k = 0 $ leads to an undefined equation, so this case is **excluded**.

#### Case 2: $ k = 8 $
Then $ D = 0 $, so the quadratic has one real solution (double root):

$$
x = \frac{-(4 - 8)}{2} = \frac{4}{2} = 2
$$

Check domain:
- $ kx = 8 \cdot 2 = 16 > 0 $
- $ x + 2 = 4 > 0 $

✅ Valid. So **this value of $ k = 8 $ is acceptable**.

#### Case 3: $ k < 0 $ (i.e., $ k \in [-500, -1] $)
- The discriminant is $ D = k^2 - 8k > 0 $
- The quadratic has two real roots
- Since $ k < 0 $, for $ kx > 0 $, $ x < 0 $

This implies $ x $ must be negative, and from domain: $ x > -2 $. So $ x \in (-2, 0) $

However, the product of the roots is 4 (from the constant term of the quadratic), so both roots must have the same sign. Since $ k < 0 $, both roots are negative.

Let’s denote the roots as $ x_1 $ and $ x_2 $. Since their product is 4 and both are negative, and their sum is $ k - 4 < -4 $, the sum of two negative numbers must be less than -4.

This forces one root to be in $ (-2, 0) $ and the other to be **less than -2**.

Only the root in $ (-2, 0) $ satisfies the domain, so for all $ k < 0 $, the equation has exactly **one valid solution**.

#### Case 4: $ 0 < k < 8 $
Then $ D < 0 $, so the quadratic has **no real roots**.
So, the equation has **no real solutions** → not acceptable.

#### Case 5: $ k > 8 $
Then $ D > 0 $, and the quadratic has two real roots.

- Product of roots = 4 ⇒ both roots are positive
- Sum of roots = $ k - 4 $, which is greater than 4

Hence, **both roots are positive**, and since $ x > -2 $ is already satisfied for them, **both are valid**.

Thus, this leads to **two real solutions**, which is invalid.

---

### Step 5: Count the Valid Integer Values of $ k $

From the above, we have two sets of valid values:

1. All $ k < 0 $ in the interval $[-500, 500]$ → these are the integers from **-500 to -1**, inclusive.
   - Total: $ 500 $ values

2. $ k = 8 $ → **1 additional** value

So, the total number of valid values is:

$$
500 + 1 = \boxed{501}
$$

---

### ✅ Final Answer

$$
\boxed{501}
$$

[Assistant]
Explanation:
Okay, let's try to tackle this problem. So, we need to find the number of integer values of k in the interval [-500, 500] such that the equation log(kx) = 2 log(x + 2) has exactly one real solution. Hmm, okay. Let me start by recalling some logarithm properties and think about the steps needed here.

First, since logarithms are involved, I need to remember their domains. The argument of a logarithm must be positive. So, for log(kx) to be defined, kx must be greater than 0. Similarly, for log(x + 2), the argument x + 2 must be greater than 0. Therefore, we have two conditions:

1. kx > 0
2. x + 2 > 0 ⇒ x > -2

So, the domain of x is x > -2, and also kx > 0. Depending on the sign of k, this will impose different constraints on x. Let me note that.

Now, moving on to the equation itself: log(kx) = 2 log(x + 2). Let me recall that 2 log(x + 2) can be rewritten as log((x + 2)^2). So, the equation becomes:

log(kx) = log((x + 2)^2)

Since logarithm is a one-to-one function, if log(a) = log(b), then a = b, provided that a and b are positive. Therefore, we can write:

kx = (x + 2)^2

But we also have to remember the domain restrictions. So, the solutions to this equation must satisfy kx > 0 and x > -2.

So, first, let me write the equation:

kx = (x + 2)^2 ⇒ kx = x² + 4x + 4 ⇒ x² + 4x + 4 - kx = 0 ⇒ x² + (4 - k)x + 4 = 0

So, this is a quadratic equation in terms of x: x² + (4 - k)x + 4 = 0.

Now, we need to find the number of real solutions to this equation, but considering the domain restrictions. However, the problem states that the original equation has exactly one real solution. So, even though the quadratic equation might have two solutions, we need to ensure that only one of them satisfies the domain conditions.

But first, let's analyze the quadratic equation. Let me recall that a quadratic equation ax² + bx + c = 0 has:

- Two real solutions if discriminant D = b² - 4ac > 0
- One real solution if D = 0
- No real solutions if D < 0

So, for our quadratic equation x² + (4 - k)x + 4 = 0, the discriminant D is:

D = (4 - k)² - 4 * 1 * 4 = (16 - 8k + k²) - 16 = k² - 8k

Therefore, D = k² - 8k.

So, depending on the value of k, the discriminant can be positive, zero, or negative.

But here's the thing: even if the quadratic equation has two real solutions, we need to check whether they satisfy the domain conditions. However, the problem says that the original equation has exactly one real solution, which could be due to:

1. The quadratic equation having exactly one real solution (i.e., discriminant zero), which would mean that the solution is valid (satisfies domain conditions)
2. The quadratic equation having two real solutions, but only one of them satisfies the domain conditions (i.e., one is in the domain and the other is not)

Therefore, we need to consider both cases.

So, let's break it down into two cases:

Case 1: The quadratic equation has exactly one real solution (D = 0). For this to happen, discriminant must be zero: k² - 8k = 0 ⇒ k(k - 8) = 0 ⇒ k = 0 or k = 8.

However, we need to check if these solutions satisfy the domain conditions.

First, let's check for k = 0.

If k = 0, the original equation becomes log(0 * x) = 2 log(x + 2). But log(0) is undefined, so k = 0 is not allowed. Therefore, k = 0 is invalid. Therefore, only k = 8 is a candidate here.

Wait, let's check for k = 8. If k = 8, then our quadratic equation becomes x² + (4 - 8)x + 4 = x² -4x + 4 = 0 ⇒ (x - 2)^2 = 0 ⇒ x = 2. So, x = 2. Let's check the domain conditions:

- kx = 8 * 2 = 16 > 0
- x + 2 = 4 > 0. So, both conditions are satisfied. Therefore, x = 2 is a valid solution. Therefore, for k = 8, the equation has exactly one real solution. So, k = 8 is valid.

So, Case 1 gives us k = 8.

Case 2: The quadratic equation has two real solutions, but only one of them satisfies the domain conditions.

So, in this case, we need D > 0, and exactly one of the roots satisfies the domain conditions.

So, first, D > 0 ⇒ k² - 8k > 0 ⇒ k(k - 8) > 0 ⇒ k < 0 or k > 8.

Therefore, for k < 0 or k > 8, discriminant D is positive, so quadratic has two real roots. Now, we need to check for each such k whether exactly one root is in the domain.

But before that, let me note that for k = 0, we saw that it's invalid. Similarly, for k = 8, we have discriminant zero, which we already considered.

So, now, for Case 2: k < 0 or k > 8.

Now, we need to analyze the roots of the quadratic equation and check which ones satisfy the domain conditions.

Let me recall that the quadratic equation is x² + (4 - k)x + 4 = 0.

Let me denote the roots as x₁ and x₂. Then, by Vieta's formula:

x₁ + x₂ = -(4 - k) = k - 4

x₁ * x₂ = 4

So, the product of the roots is 4.

Also, the domain conditions are:

For each root x, we need:

1. x > -2 (from x + 2 > 0)
2. kx > 0 (from log(kx) being defined)

So, for each root x, we need both x > -2 and kx > 0.

So, let's think about how these conditions interact.

First, note that the product of the roots is 4. So, x₁ * x₂ = 4. Therefore, both roots are either positive or both negative? Wait, since the product is positive. Therefore, either both roots are positive or both are negative.

But let's check:

If the product is positive, then either both roots positive or both negative. However, since x > -2 is required, which is a lower bound, so if both roots are negative, but x > -2, then we need to check if they are greater than -2.

But let me think:

First, let's consider the case when k < 0. Then, for kx > 0, since k is negative, x must be negative. So, in this case, x must be negative. But also, x > -2, so x ∈ (-2, 0). However, if both roots are negative (since their product is positive) and they are in (-2, 0), then we need to check how many roots lie in (-2, 0).

Similarly, if k > 8, then k is positive. So, for kx > 0, since k is positive, x must be positive. Therefore, in this case, x > 0 (since kx > 0) and also x > -2 (which is automatically satisfied if x > 0). So, in this case, the domain is x > 0. Therefore, for k > 8, we need to check how many roots lie in x > 0.

So, let's handle k < 0 and k > 8 separately.

First, let's handle k < 0:

Case 2a: k < 0.

In this case, since product of roots is 4, both roots are negative. Also, from kx > 0, x must be negative. However, since x must also satisfy x > -2. So, the roots are negative numbers, but we need them to be greater than -2. Therefore, the roots must lie in (-2, 0). So, we need to check whether the quadratic equation has exactly one root in (-2, 0) or both roots in (-2, 0) or none.

But since the quadratic equation has two roots, and we need exactly one of them to be in (-2, 0). However, since both roots are negative (product positive) and their sum is k - 4. Since k < 0, then k - 4 is less than -4. Therefore, the sum of roots is negative. So, the sum of two negative numbers is negative, which is consistent.

But how do we know whether the roots lie in (-2, 0)?

Alternatively, perhaps we can analyze the quadratic function f(x) = x² + (4 - k)x + 4. Since the product of roots is 4, and both roots are negative.

Wait, but maybe we can use the fact that if a quadratic equation has two real roots, then to find how many roots lie in a certain interval, we can use the concept of sign changes or testing the function at the endpoints. But since we need to check whether the roots lie in (-2, 0), perhaps we can use the following approach:

For each root x, we need x ∈ (-2, 0). Also, since both roots are negative, and their product is 4. So, for example, if one root is in (-2, 0), then the other root is 4 divided by that root, which would be negative. Let's say one root is between -2 and 0, so the other root is 4 / x. Since x is between -2 and 0, 4 / x is less than -2 (since dividing a positive by a negative number gives negative, and dividing by a number between -2 and 0 gives a number less than -2). For example, if x is -1, then 4 / x is -4. If x approaches 0 from the left, 4 / x approaches -infty. If x approaches -2 from the right, 4 / x approaches -2.

Therefore, if one root is in (-2, 0), the other root is less than -2. Hence, only one root lies in (-2, 0). Therefore, for k < 0, the quadratic equation has two real roots: one in (-2, 0) and the other less than -2. Therefore, only the root in (-2, 0) satisfies the domain conditions (since the other one is less than -2, so x + 2 would be less than 0, which is invalid). Therefore, for k < 0, the equation log(kx) = 2 log(x + 2) would have exactly one real solution, as only one root is valid.

Wait, but hold on, is that always the case? Let me check with an example. Suppose k = -1. Then, quadratic equation becomes x² + (4 - (-1))x + 4 = x² + 5x + 4 = 0 ⇒ (x + 1)(x + 4) = 0 ⇒ x = -1 or x = -4. Now, check domain conditions:

For x = -1: kx = -1 * (-1) = 1 > 0, and x + 2 = -1 + 2 = 1 > 0. So valid. For x = -4: kx = -1 * (-4) = 4 > 0, but x + 2 = -4 + 2 = -2 < 0. So invalid. Hence, only one solution. So, that works.

Another example: k = -8. Then quadratic equation is x² + (4 - (-8))x + 4 = x² + 12x + 4 = 0. Let me compute the roots:

Discriminant D = 144 - 16 = 128. Roots are [-12 ± sqrt(128)] / 2 = [-12 ± 8*sqrt(2)] / 2 = -6 ± 4*sqrt(2). sqrt(2) ≈ 1.414, so 4*sqrt(2) ≈ 5.656. Therefore, the roots are approximately -6 + 5.656 ≈ -0.344 and -6 - 5.656 ≈ -11.656. So, the first root is approximately -0.344 which is in (-2, 0), and the second is less than -2. So again, only one valid solution. Hence, seems like for k < 0, regardless of the value, we have exactly one valid solution. Therefore, for all k < 0, the equation has exactly one real solution.

Wait, but what if the quadratic equation has both roots in (-2, 0)? Is that possible? Let me see. Suppose that both roots are in (-2, 0). Then, since their product is 4, which is positive, but if both are between -2 and 0, then their product would be positive, but in that case, the product would be less than 4? Wait, for example, if both roots are -1, then their product is 1, which is less than 4. If both roots are approaching 0, then the product approaches 0. If both roots approach -2 from the right, then their product approaches (-2)*(-2)=4. So, the maximum product when both roots are in (-2, 0) is approaching 4. Therefore, if the product is exactly 4, which is when one root approaches -2 and the other approaches -2. But since the product is 4, if both roots are equal (i.e., discriminant zero), but discriminant zero would require k = 0 or k = 8. However, k < 0 here, so discriminant zero is not possible for k < 0. So, in the case of k < 0, the quadratic equation has two real roots, one in (-2, 0) and one less than -2. Therefore, only one solution is valid.

Wait, but let me check if there's any possible k < 0 where both roots are in (-2, 0). Suppose that both roots are in (-2, 0). Then, the sum of roots is k - 4. Let me denote the roots as α and β, both in (-2, 0). Then, α + β = k - 4, and αβ = 4.

But if α and β are both in (-2, 0), then α + β ∈ (-4, 0). Therefore, k - 4 ∈ (-4, 0) ⇒ k ∈ (0, 4). But we are considering k < 0 here. Therefore, there's a contradiction. Hence, α + β = k - 4, and for k < 0, k - 4 < -4. Therefore, the sum of roots is less than -4. But if both roots are in (-2, 0), their sum would be greater than -4 (since each root is greater than -2, so sum is greater than -4). Therefore, we have that k - 4 < -4, which is k < 0, but if both roots are in (-2, 0), then α + β > -4 ⇒ k - 4 > -4 ⇒ k > 0. Therefore, for k < 0, it's impossible for both roots to be in (-2, 0). Hence, only one root is in (-2, 0) and the other is less than -2, hence only one valid solution. Therefore, for all k < 0, the equation has exactly one real solution.

Therefore, Case 2a: k < 0 gives valid solutions (each k < 0 gives exactly one solution).

Case 2b: k > 8.

In this case, discriminant D = k² - 8k > 0. Also, k is positive, so for kx > 0, since k is positive, x must be positive. Therefore, the domain for x is x > 0. Also, we need to check whether the roots of the quadratic equation lie in x > 0.

So, the quadratic equation is x² + (4 - k)x + 4 = 0. Let's analyze its roots.

Since the product of the roots is 4 (positive) and the sum of the roots is k - 4. Since k > 8, the sum is k - 4 > 4.

So, the sum of roots is positive and greater than 4, and the product is positive (4). Therefore, both roots are positive. Because if both roots are positive, their sum and product are positive. Therefore, in this case, both roots are positive. Hence, both roots satisfy x > 0. However, we need to check if they satisfy x > -2, but since x > 0, that's automatically satisfied. Therefore, both roots are in the domain. Therefore, the original equation would have two real solutions. However, the problem requires the original equation to have exactly one real solution. Therefore, for k > 8, the equation would have two real solutions. Therefore, such k would not be acceptable.

Wait, but wait. Wait, unless one of the roots is invalid for some reason? But we have already considered that kx > 0 and x > -2. Since x > 0, both conditions are satisfied. Therefore, both roots are valid. Therefore, for k > 8, equation has two real solutions, which is not acceptable. Therefore, in Case 2b, we need to exclude k > 8. However, hold on, but maybe there is a case when one of the roots is invalid? But no, since both roots are positive and k is positive, so kx is positive. Therefore, both roots are valid. Hence, for k > 8, the equation has two solutions. Therefore, they are not acceptable.

But then, what about k = 8? For k = 8, discriminant is zero, so only one solution, which is valid as we saw. Therefore, in Case 2, only k < 0 gives valid solutions (each with exactly one solution), and k > 8 gives two solutions. Therefore, the only k's that satisfy the condition are k < 0 and k = 8.

Wait, but then wait, there might be another case where even though the quadratic equation has two solutions, one of them is invalid due to some other reason? Let me check.

Alternatively, maybe there are some k values where one of the roots is invalid even though k is positive. Wait, for example, if one of the roots is negative, but since for k > 8, we have both roots positive. Wait, but since the product is positive and sum is positive, both roots must be positive. Therefore, there's no way for one root to be negative. Therefore, for k > 8, both roots are positive, hence both valid. Hence, two solutions. Therefore, they are not acceptable. Therefore, only k < 0 and k = 8 are acceptable.

But hold on, let me check with an example. Suppose k = 9, which is greater than 8. Then quadratic equation is x² + (4 - 9)x + 4 = x² -5x + 4 = 0 ⇒ (x - 1)(x - 4) = 0. Therefore, roots at 1 and 4. Both positive, so both satisfy x > 0 and kx positive. Hence, two solutions. So, original equation would have two solutions, which is not acceptable. Therefore, k = 9 is excluded.

Now, what about k = 8, which is the case where discriminant is zero? We saw that gives x = 2. Let me check if there are other k's where even though discriminant is positive, but one of the roots is invalid?

Wait, but for k > 8, both roots are valid. For k < 0, only one root is valid. For k = 0, invalid. For k between 0 and 8, discriminant is negative? Wait, D = k² - 8k. For k between 0 and 8, D = k(k - 8) < 0, so no real solutions. Therefore, for 0 < k < 8, equation has no real solutions. Therefore, the original equation would have no real solutions, so they are not acceptable.

Therefore, putting it all together:

The values of k for which the equation has exactly one real solution are:

- All k < 0 (since for each such k, exactly one solution)
- k = 8 (since discriminant zero and valid solution)

But wait, but hold on. Wait, let me check k = 0. If k = 0, original equation is log(0) = 2 log(x + 2). But log(0) is undefined, so k = 0 is invalid, as we saw. Similarly, for k = 8, we have one solution. So, so far, our candidates are k < 0 and k = 8.

But let me check if there are other possibilities. For example, could there be some k where the quadratic equation has one root valid and another invalid, even when discriminant is positive?

Wait, for k < 0: yes, we already considered that. For k > 8: no, both roots are valid. What about k between 0 and 8? The quadratic equation has no real roots. If k = 0, invalid. What about k = 8? discriminant zero. Therefore, seems like only k < 0 and k = 8.

But wait, hold on! Let me think again. Suppose there is some k where even though discriminant is positive (so two roots), but one of the roots is invalid. For example, suppose for some k, one of the roots is x = -2, which is on the boundary of the domain. However, x must be greater than -2, so x = -2 is not allowed. So, if one of the roots is x = -2, then that root is invalid. However, in that case, maybe there's another root that's valid?

But let's check when x = -2 is a root. Plugging x = -2 into the quadratic equation:

(-2)^2 + (4 - k)(-2) + 4 = 4 + (-8 + 2k) + 4 = 4 - 8 + 4 + 2k = 0 + 2k = 2k. Therefore, setting equal to zero: 2k = 0 ⇒ k = 0. But k = 0 is invalid because log(0) is undefined. So, only when k = 0, x = -2 is a root, but k = 0 is invalid. Therefore, no other k would have x = -2 as a root. Therefore, for other k, roots are either both valid (for k > 8), or one valid (for k < 0). Therefore, our previous conclusion holds.

But wait, there's another case. What if one of the roots is x = 0? Then, kx = 0, which is invalid. Let me check if x = 0 is a root. Plugging x = 0 into quadratic equation:

0 + (4 - k)*0 + 4 = 4 ≠ 0. Therefore, x = 0 is not a root. Therefore, no problem.

Alternatively, suppose that one of the roots is x = -2 + ε, very close to -2, but still greater than -2. However, even then, since for k < 0, we already have one root in (-2, 0) and one root less than -2, so as long as the root is in (-2, 0), it's valid. Therefore, no issues there.

Therefore, according to our analysis, the values of k are all negative integers in [-500, 500] and k = 8. But wait, the problem says "closed interval [-500, 500]". So, k is in [-500, 500], integers. Therefore, the number of integer values of k is:

Number of negative integers from -500 to -1 inclusive, plus 1 for k = 8.

But let me check if k = 8 is within the interval. Since 8 is in [-500, 500], yes.

So, first, count the number of integers k with k < 0 and in [-500, 500]. The integers from -500 up to -1 inclusive. The number of integers is 500. Because from -1 to -500, inclusive, that's 500 numbers. For example, from -1 to -n, there are n numbers. So, from -1 to -500, there are 500 numbers. Therefore, 500 values.

Then, plus k = 8, which is one more. Hence, total number of integer values is 500 + 1 = 501.

Wait, but hold on. Wait, the problem says "closed interval [-500, 500]". So, integers from -500 to 500 inclusive. So, the negative integers from -500 to -1 inclusive: that's 500 numbers. Then, k = 8 is one more. But wait, we need to check if k = 8 is the only value for Case 1 (discriminant zero). We had k = 8 and k = 0. But k = 0 is invalid, so only k = 8. So, total number of valid k is 500 + 1 = 501. But let me check again.

Wait, but wait. What about k = 8 is included in the interval [-500, 500], which it is. So, the answer would be 500 + 1 = 501. However, let me think again. Let me make sure that there are no other values of k where the equation has exactly one solution. For example, is there a possibility when the quadratic equation has two solutions, but one of them is invalid? But for k > 8, both are valid. For k < 0, one is valid. For 0 < k < 8, no solutions. For k = 0, invalid. For k = 8, one solution. So, yes, only those.

Wait, but hold on! There is another case. What if the quadratic equation has two solutions, but one of them is invalid due to kx > 0? Wait, but for k > 8, we established both roots are positive, so kx positive, so valid. For k < 0, one root is negative (but x > -2) and the other is less than -2 (so invalid). So, only one valid. But is there a case where for some k < 0, the quadratic equation's roots are such that both are in (-2, 0), hence both valid? Earlier, we showed that's impossible. Because for k < 0, the sum of roots is k - 4, which is less than -4. However, if both roots are in (-2, 0), the sum would be greater than -4, so conflicting. Therefore, not possible. Therefore, our conclusion holds.

Therefore, the answer is 500 (for k < 0) + 1 (for k = 8) = 501. However, wait, but hold on. Let me check again the count of negative integers. The interval is from -500 to 500 inclusive. How many integers are there from -500 to -1 inclusive? Well, from -1 to -500, that's 500 numbers. Because if you take 500 - (-1) = 501? Wait, no. Wait, the number of integers from a to b inclusive is b - a + 1. So, if we have from -500 to -1 inclusive, that is, a = -500, b = -1. Then number of integers is (-1) - (-500) + 1 = 499 + 1 = 500. Yes. So that's correct. So 500 negative integers. So total is 500 + 1 = 501. So is 501 the answer?

But wait, wait. Wait, hold on. Wait, k = 8. But what if k = 8 is not in the interval? Wait, no, the interval is [-500, 500], so 8 is within it. Therefore, answer is 501.

But let me check again with another example. Let me check if k = -500 is included. Yes, since the interval is closed. So k = -500 is included. So, number of negative integers from -500 to -1 inclusive is 500 numbers. Then, adding k = 8. Therefore, total 501.

But wait, but maybe there is a mistake here. Because for k < 0, we need to check that the equation actually has exactly one solution. But we already established that for each k < 0, the equation has exactly one solution. Therefore, 500 values. Plus k = 8. Therefore, 501.

But wait, but in the problem statement, it says "closed interval [-500, 500]". So, k can be from -500 to 500 inclusive. But let me check if k = -500 is allowed. For k = -500, the equation is log(-500x) = 2 log(x + 2). Then, domain requires -500x > 0, so x < 0, and x > -2. So x ∈ (-2, 0). Then, solving the quadratic equation:

x² + (4 - (-500))x + 4 = x² + 504x + 4 = 0. The roots are [-504 ± sqrt(504² - 16)] / 2. Which are real numbers. As we saw, one root is in (-2, 0) and another is less than -2, so only one valid solution. So, k = -500 is acceptable.

Similarly, k = -1 is acceptable. So, yes, all negative integers from -500 to -1 inclusive are acceptable, which is 500 values. Then k = 8 is another value, so total of 501.

But wait, wait. Wait, hold on. Wait, when k = 8, we have exactly one solution. But what about k = 8? But what if there is another k where the equation has exactly one solution? For example, suppose that for some k > 8, one of the roots is invalid? Wait, we had considered that for k > 8, both roots are positive. Suppose that one of the roots is x = -2? Wait, but we saw that only when k = 0, x = -2 is a root, but k = 0 is invalid. Therefore, no. Therefore, for k > 8, both roots are positive. Hence, both valid. Therefore, they give two solutions. Hence, they are not counted.

But here's a thought. What if the quadratic equation has one root which is x = -2? But we saw that only happens when k = 0. But k = 0 is invalid. Therefore, no.

Alternatively, maybe there's some k where one of the roots is invalid even for k < 0? For example, suppose for some k < 0, the root in (-2, 0) actually makes log(kx) undefined? Wait, no. Because for k < 0, and x in (-2, 0), then kx is positive? Let me check. If k is negative and x is negative, then kx is positive. Yes. So, log(kx) is defined. Also, x + 2 > 0 because x is in (-2, 0). Therefore, both sides are defined. Therefore, yes, valid. Therefore, the solution is valid. Therefore, all k < 0 are okay.

Therefore, the answer should be 501.

But wait, hold on. But the problem says "closed interval [-500, 500]". Therefore, the integers k from -500 to 500 inclusive. But k = 8 is included, so total of 500 negative integers, plus 1 for k = 8. So, total 501. But wait, maybe there is a mistake here. Let me check the count again.

Number of integers from -500 to -1 inclusive: Let me think of it as the number of integers from 1 to 500, which is 500. Similarly, from -500 to -1 is also 500 numbers. Therefore, yes. Then, adding k = 8. So 501.

However, let me check once again if there are other values of k where the equation has exactly one solution. For example, if discriminant is zero. We had discriminant zero when k = 0 or k = 8. But k = 0 is invalid, so only k = 8. Therefore, only one such value.

Alternatively, is there another case when the equation has exactly one solution? For example, when one root is valid and another is invalid, but discriminant is positive. But for k < 0, as we saw, one is valid and one is invalid. For k > 8, both are valid. For other k, either no solutions or one solution (k=8). Therefore, yes, only the cases considered.

Thus, answer is 501. However, wait, wait! Wait, the problem says "exactly one real solution". However, for k < 0, we have exactly one real solution. For k = 8, exactly one real solution. So total 501.

But hold on, let me check if there are any overlaps or something. For example, is k = 8 included in the k < 0? No, since 8 is positive. Therefore, no overlap. Therefore, the total number is 500 + 1 = 501. Therefore, the answer is 501.

But now, wait. Wait, the problem says "closed interval [-500, 500]". So, k can be -500, -499,..., -1, 0, 1,..., 500. But we are to count the number of integer values of k in this interval for which the equation has exactly one real solution. So, as per our analysis, all negative integers (from -500 to -1, inclusive) and k = 8. So, the total number is 500 + 1 = 501.

But wait, let me check the problem statement again. It says "exactly one real solution". Wait, but for k < 0, even though the quadratic equation has two real roots, only one of them is valid. Therefore, the equation has exactly one real solution. For k = 8, the quadratic equation has one real solution. Therefore, both cases satisfy the condition. Hence, 501.

However, I'm a bit unsure because sometimes there can be edge cases. Let me check for k = -2. Let me take k = -2. Then quadratic equation is x² + (4 - (-2))x + 4 = x² + 6x + 4 = 0. Discriminant D = 36 - 16 = 20 > 0. So roots are [-6 ± sqrt(20)] / 2 = [-6 ± 2*sqrt(5)] / 2 = -3 ± sqrt(5). sqrt(5) ≈ 2.236. Therefore, roots are approximately -3 + 2.236 ≈ -0.764 and -3 - 2.236 ≈ -5.236. Therefore, x ≈ -0.764 is in (-2, 0), valid. The other root is less than -2, invalid. So, exactly one solution. Correct. Similarly for k = -1. So, seems okay.

Wait, but what about k = -4? Let me try k = -4. Then quadratic equation is x² + (4 - (-4))x + 4 = x² + 8x + 4 = 0. Discriminant D = 64 - 16 = 48 > 0. Roots are [-8 ± sqrt(48)] / 2 = [-8 ± 4*sqrt(3)] / 2 = -4 ± 2*sqrt(3). sqrt(3) ≈ 1.732. So, roots ≈ -4 + 3.464 ≈ -0.536 and -4 - 3.464 ≈ -7.464. So, again, one valid solution. Correct.

Another check for k = 8: equation is log(8x) = 2 log(x + 2). Which reduces to 8x = (x + 2)^2. Which is x² + 4x + 4 - 8x = x² -4x +4 = (x - 2)^2 = 0. So x = 2. Check domain: 8*2 = 16 > 0, and x + 2 = 4 > 0. Valid. So, one solution. Correct.

So, seems like our reasoning is correct.

Therefore, the answer is 501. Therefore, the number of integer values of k is 501.

Wait, but wait! Wait, the problem says "closed interval [-500, 500]". So, k can be -500, -499,..., 0,..., 500. So, the number of integers from -500 to 500 inclusive is 1001 numbers. But we are to count how many of them satisfy the condition. As per our analysis, it's 500 negative numbers plus 1 (for k = 8). So 501. But let me check once more if k = 8 is the only case where discriminant is zero.

Yes, discriminant D = k² - 8k. Setting D = 0: k² - 8k = 0 ⇒ k(k - 8) = 0 ⇒ k = 0 or k = 8. As before, k = 0 invalid, so only k = 8. Therefore, 501 total.

But the problem is in the original equation log(kx) = 2 log(x + 2). Wait, does the equation have any restrictions on k? For example, k can't be zero, as log(0) is undefined. But k is allowed to be negative, as long as kx is positive. Which is considered in our analysis.

But wait, in our analysis, k can be any negative integer from -500 to -1 inclusive. And k = 8. So, total of 500 + 1 = 501 integers.

But I'm a bit uncertain because I recall sometimes problems like this might have more cases. Wait, let me think again. Maybe there's a case when the quadratic equation has two solutions, both valid, but one of them is extraneous? For example, when solving log(kx) = 2 log(x + 2), we assumed that kx > 0 and x + 2 > 0, but perhaps there's another extraneous solution when we did the logarithm manipulation?

Wait, when we started with log(kx) = 2 log(x + 2), we converted it to kx = (x + 2)^2. However, this step is valid only if both sides are defined, i.e., log(kx) and log(x + 2) are defined. So, we have already considered that kx > 0 and x + 2 > 0. Therefore, all solutions to kx = (x + 2)^2 that are in the domain are valid. Therefore, there are no extraneous solutions. Therefore, our analysis holds.

Therefore, I think the answer is 501. However, let me check an example where k is positive but less than 8, but maybe some edge case?

Wait, for example, take k = 4. Then quadratic equation is x² + (4 - 4)x + 4 = x² + 0x + 4 = x² + 4 = 0 ⇒ x² = -4, which has no real solutions. So, no solution. Correct. For k = 5, quadratic equation x² + (4 - 5)x + 4 = x² - x + 4 = 0. Discriminant D = 1 - 16 = -15 < 0. No solution. So, correct.

Another example, k = 8: one solution. Correct.

So, yes, seems okay.

Therefore, I think the answer is 501. However, to make sure, let me check if k = -500 is allowed. We said that yes, since x must be in (-2, 0) for k = -500. For k = -500, x ≈ -0.764 is valid. Therefore, yes.

But let me check if there's a mistake in the count. The interval from -500 to 500 inclusive has 1001 integers. The number of integers k with k < 0 is 500 (from -500 to -1 inclusive). Then, k = 8 is another value, so total 501. Therefore, answer is 501. Therefore, I think that is correct.

But wait, just to check if k = -500 is included. Yes, since -500 is in [-500, 500]. Similarly, -1 is included. So, numbers from -500 to -1 inclusive is 500 numbers. Then k = 8. So, 501.

But let me check if there are any other values of k where the equation has exactly one solution. For example, suppose k = something where the quadratic equation has two solutions, but one of them is invalid? But for k > 8, both solutions are valid. For k < 0, one is valid, one invalid. For k between 0 and 8, no solution. For k = 0, invalid. For k = 8, one solution. Therefore, yes, only the above cases. Therefore, 501.

But wait, just another thought. What if k = -2? Wait, we checked that already. So, yes, seems okay.

Therefore, I think the answer is 501.

Wait, but let me check again the count of negative integers. Let me think of the number of integers between -500 and -1 inclusive. The number is equal to 500. Because from -1 to -500, that's 500 numbers. For example, from -1 to -n, there are n numbers. Therefore, from -1 to -500, there are 500 numbers. Therefore, correct.

Therefore, the answer is 501. But to cross-verify, let me check how many integers are there in total. From -500 to 500 inclusive, it's 500 - (-500) + 1 = 1001 numbers. If we have 500 negative, 1 zero, and 500 positive numbers. But we are only counting the negative and k = 8. Therefore, 500 + 1 = 501. Yes. Therefore, answer is 501.

But hold on! Wait, in the interval [-500, 500], the number of negative integers is 500 (from -1 to -500), and then k = 8 is one specific positive integer. Therefore, total of 501. Therefore, I think that is correct.

But, wait, the problem says "closed interval [-500, 500]". So, 500 is included. Therefore, k = 8 is included. Therefore, answer is 501. So, I think the answer is \boxed{501}.

But wait, I remember sometimes in these problems, there's a possibility that when k = 8, maybe there's another solution or something. But no, we checked that k = 8 gives exactly one solution. Therefore, I think 501 is correct.

Alternatively, maybe I made a mistake in assuming that for all k < 0, exactly one solution exists. Let me think if there is any k < 0 where the quadratic equation has two solutions in (-2, 0). But as per earlier reasoning, since the sum of roots is k - 4, and k < 0, so sum is less than -4. But if two roots are in (-2, 0), then their sum would be greater than -4. Contradiction. Hence, no. Therefore, only one solution. Hence, correct.

Therefore, I think the answer is 501.

**Final Answer**
\boxed{501}

To solve the given problem, we are asked to find how many integer values of $ k $ in the interval $[-500, 500]$ satisfy the condition that the equation

$$
\log(kx) = 2\log(x + 2)
$$

has **exactly one real solution**.

---

### Step 1: Domain Analysis

The logarithmic equation is defined only when:

- $ kx > 0 $, because $ \log(kx) $ is defined only if its argument is positive.
- $ x + 2 > 0 $, because $ \log(x + 2) $ is defined only if the argument is positive.

So the domain of $ x $ is:

$$
x > -2 \quad \text{and} \quad kx > 0
$$

This gives us a restriction on how $ x $ behaves relative to $ k $:

- If $ k > 0 $, then $ x > 0 $
- If $ k < 0 $, then $ x < 0 $, but we also need $ x > -2 $, so $ x \in (-2, 0) $

If $ k = 0 $, $ kx = 0 $, which makes $ \log(0) $ undefined, so $ k = 0 $ is **not** allowed.

---

### Step 2: Solve the Equation

Given:
$$
\log(kx) = 2\log(x + 2)
$$

Using logarithmic identities:

$$
\log(kx) = \log((x + 2)^2)
\Rightarrow kx = (x + 2)^2
$$

Expanding the right-hand side:

$$
(x + 2)^2 = x^2 + 4x + 4
\Rightarrow x^2 + (4 - k)x + 4 = 0
$$

So the quadratic equation is:
$$
x^2 + (4 - k)x + 4 = 0
$$

Let’s denote this as $ f(x) = x^2 + (4 - k)x + 4 $

---

### Step 3: Nature of the Roots

We analyze the number of real roots of the quadratic equation.

**Discriminant** $ D $ is:
$$
D = (4 - k)^2 - 16 = k^2 - 8k
$$

So:

- $ D > 0 $ when $ k^2 - 8k > 0 $ ⟹ $ k < 0 $ or $ k > 8 $
- $ D = 0 $ when $ k = 0 $ or $ k = 8 $
- $ D < 0 $ when $ 0 < k < 8 $

---

### Step 4: Determine Validity of Roots for Each Case

Let’s consider the three cases:

#### Case 1: $ k = 0 $ → Invalid
As noted earlier, $ k = 0 $ leads to an undefined equation, so this case is **excluded**.

#### Case 2: $ k = 8 $
Then $ D = 0 $, so the quadratic has one real solution (double root):

$$
x = \frac{-(4 - 8)}{2} = \frac{4}{2} = 2
$$

Check domain:
- $ kx = 8 \cdot 2 = 16 > 0 $
- $ x + 2 = 4 > 0 $

✅ Valid. So **this value of $ k = 8 $ is acceptable**.

#### Case 3: $ k < 0 $ (i.e., $ k \in [-500, -1] $)
- The discriminant is $ D = k^2 - 8k > 0 $
- The quadratic has two real roots
- Since $ k < 0 $, for $ kx > 0 $, $ x < 0 $

This implies $ x $ must be negative, and from domain: $ x > -2 $. So $ x \in (-2, 0) $

However, the product of the roots is 4 (from the constant term of the quadratic), so both roots must have the same sign. Since $ k < 0 $, both roots are negative.

Let’s denote the roots as $ x_1 $ and $ x_2 $. Since their product is 4 and both are negative, and their sum is $ k - 4 < -4 $, the sum of two negative numbers must be less than -4.

This forces one root to be in $ (-2, 0) $ and the other to be **less than -2**.

Only the root in $ (-2, 0) $ satisfies the domain, so for all $ k < 0 $, the equation has exactly **one valid solution**.

#### Case 4: $ 0 < k < 8 $
Then $ D < 0 $, so the quadratic has **no real roots**.
So, the equation has **no real solutions** → not acceptable.

#### Case 5: $ k > 8 $
Then $ D > 0 $, and the quadratic has two real roots.

- Product of roots = 4 ⇒ both roots are positive
- Sum of roots = $ k - 4 $, which is greater than 4

Hence, **both roots are positive**, and since $ x > -2 $ is already satisfied for them, **both are valid**.

Thus, this leads to **two real solutions**, which is invalid.

---

### Step 5: Count the Valid Integer Values of $ k $

From the above, we have two sets of valid values:

1. All $ k < 0 $ in the interval $[-500, 500]$ → these are the integers from **-500 to -1**, inclusive.
   - Total: $ 500 $ values

2. $ k = 8 $ → **1 additional** value

So, the total number of valid values is:

$$
500 + 1 = \boxed{501}
$$

---

### ✅ Final Answer

$$
\boxed{501}
$$

Solution:
First, note that if $k < 0,$ then $\log(kx)$ is defined for $x \in (-\infty, 0),$ and is strictly decreasing on that interval. Since $2\log(x+2)$ is defined for $x \in (-2, \infty)$ and is strictly increasing on that interval, it follows that $\log(kx) = 2\log(x+2)$ has exactly one real solution, which must lie in the interval $(-2, 0).$ Therefore, all the values $k = -500, -499, \ldots, -2, -1$ satisfy the condition.

If $k = 0,$ then the left-hand side is never defined, so we may assume now that $k > 0.$ In this case, converting to exponential form, we have \[ kx = (x+2)^2\]or \[x^2 + (4-k)x + 4 = 0.\]Any solution of this equation satisfies $\log(kx) = 2\log(x+2)$ as well, as long as the two logarithms are defined; since $k > 0,$ the logarithms are defined exactly when $x > 0.$ Therefore, this quadratic must have exactly one positive root.

But by Vieta's formulas, the product of the roots of this quadratic is $4,$ which is positive, so the only way for it to have exactly one positive root is if it has $\sqrt{4} = 2$ as a double root. That is, \[x^2 + (4-k)x + 4 = (x-2)^2 = x^2 - 4x + 4\]for all $x,$ so $4-k=-4,$ and $k=8,$ which is the only positive value of $k$ satisfying the condition.

In total, there are $500 + 1 = \boxed{501}$ values of $k$ satisfying the condition.