Explanation:
Okay, so I 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, let me start by recalling some logarithm properties.

First, I know that log(a) = log(b) implies a = b, but only if the arguments are positive. Also, 2 log(x + 2) can be written as log((x + 2)^2) using the power rule. So maybe I can rewrite the equation as log(kx) = log((x + 2)^2). Then, since log is injective (one-to-one) over its domain, this would imply that kx = (x + 2)^2, provided that both sides are positive.

But wait, I need to be careful here. The original equation log(kx) = 2 log(x + 2) requires that the arguments of the logarithms are positive. So, for log(kx) to be defined, kx > 0. Similarly, for 2 log(x + 2) to be defined, x + 2 > 0, so x > -2. Therefore, the domain of the equation is x > -2 and kx > 0.

So, combining these, x > -2 and kx > 0. Let me think about the implications of kx > 0. If k is positive, then x must be positive. If k is negative, then x must be negative. However, since x > -2, if k is negative, x must be between -2 and 0.

So, depending on the sign of k, the domain of x is different. Let me note that down:

Case 1: k > 0. Then x > 0 (since kx > 0 and x > -2).

Case 2: k < 0. Then x < 0 (since kx > 0 and x > -2). So x is in (-2, 0).

Case 3: k = 0. But then log(kx) = log(0), which is undefined. So k cannot be zero. Therefore, k is in [-500, 500] excluding 0.

So, the equation reduces to kx = (x + 2)^2, but with the domain restrictions. Let me write that equation again:

kx = (x + 2)^2.

Let me rearrange this equation:

(x + 2)^2 - kx = 0.

Expanding (x + 2)^2:

x^2 + 4x + 4 - kx = 0.

Combine like terms:

x^2 + (4 - k)x + 4 = 0.

So, quadratic equation in x: x^2 + (4 - k)x + 4 = 0.

Now, the original equation log(kx) = 2 log(x + 2) has exactly one real solution. However, we need to consider that even if the quadratic equation has one or two real solutions, they must lie within the domain restrictions. Also, since logarithms are only defined for positive arguments, we need to check if the solutions satisfy the domain conditions.

But the problem says "exactly one real solution". So, there are two possibilities:

1. The quadratic equation has exactly one real solution (i.e., discriminant is zero) and that solution lies in the domain.

2. The quadratic equation has two real solutions, but only one of them lies in the domain.

Therefore, we need to consider both cases. Let me first compute the discriminant of the quadratic equation.

Discriminant D = (4 - k)^2 - 4 * 1 * 4 = (16 - 8k + k^2) - 16 = k^2 - 8k.

So, D = k^2 - 8k.

For the quadratic equation to have exactly one real solution, discriminant must be zero. So:

k^2 - 8k = 0 => k(k - 8) = 0 => k = 0 or k = 8.

But k = 0 is not allowed because log(0) is undefined. Therefore, only k = 8 is a candidate here. Let me check if this solution is valid.

For k = 8, the quadratic equation becomes x^2 + (4 - 8)x + 4 = x^2 - 4x + 4 = 0. Which factors as (x - 2)^2 = 0. So x = 2. Now, check domain conditions. For k = 8 > 0, x must be > 0. x = 2 is in the domain. So this is valid. Therefore, k = 8 is one value.

But there's also the second case where the quadratic equation has two real solutions, but only one of them is in the domain. So, we need to check for such k.

So, first, let me note that for the quadratic equation to have two real solutions, discriminant must be positive. So D > 0 => k^2 - 8k > 0 => k(k - 8) > 0. Therefore, this occurs when k < 0 or k > 8.

So, for k < 0 or k > 8, the quadratic equation has two real roots. Now, we need to check for these k whether exactly one of the roots lies in the domain.

So, let me first handle the case when k < 0. Then, as per domain, x must be in (-2, 0). Let me denote the roots of the quadratic equation as x1 and x2. Let me find expressions for them.

The roots are:

x = [-(4 - k) ± sqrt(D)] / 2 = [k - 4 ± sqrt(k^2 - 8k)] / 2.

But since k < 0, let me think about the roots. Let me denote sqrt(k^2 - 8k) as sqrt(k(k - 8)). Since k < 0, k - 8 is also negative (since k < 0 < 8). Therefore, k(k - 8) is positive (negative times negative). Therefore, sqrt(k^2 - 8k) is real.

But maybe instead of dealing with the roots directly, I can analyze the quadratic equation. Let me think about the quadratic equation x^2 + (4 - k)x + 4 = 0. Let me denote this as x^2 + bx + c = 0, where b = 4 - k and c = 4.

For k < 0, b = 4 - k. Since k is negative, -k is positive, so b = 4 + |k|, which is positive. Therefore, the quadratic equation is x^2 + bx + 4 = 0 with b > 0.

Let me think about the roots. The product of the roots is c = 4, so x1 * x2 = 4. The sum of the roots is -b = -(4 - k) = k - 4. Since k < 0, k - 4 is negative. Therefore, the sum of the roots is negative, and the product is positive. Therefore, both roots are negative. Because if the product is positive and the sum is negative, both roots are negative.

But in this case, the domain for k < 0 is x ∈ (-2, 0). So, both roots are negative. However, we need to check if they lie in (-2, 0).

But since the product of the roots is 4, which is positive, and both roots are negative, so they are both negative numbers. Let me denote them as x1 and x2, both negative. Let me check if they are in (-2, 0).

But since their product is 4, if one root is less than -2, then the other root would be less than -2 as well? Wait, no. Let me think. Suppose one root is less than -2, then the other root would be 4 divided by that root. If one root is, say, -3, then the other root is 4 / (-3) ≈ -1.333, which is greater than -2. So, if one root is less than -2, the other is between -2 and 0. Similarly, if both roots are greater than -2, then their product would be positive, but since their sum is negative, they can't both be greater than -2. Wait, let me think again.

Let me suppose that both roots are greater than -2. Then, since they are both negative, they are in (-2, 0). But their product is 4. Let me check if that's possible. Let me suppose that both roots are in (-2, 0). Let me denote them as x1 and x2. Then x1 * x2 = 4. But if both are between -2 and 0, their product would be positive, but since they are both negative, their product is positive. However, the maximum product of two numbers in (-2, 0) would be when they are closest to zero. For example, if x1 approaches 0 from the left, x2 approaches negative infinity? Wait, no. Wait, if x1 and x2 are both in (-2, 0), then their product is between 0 and 4. Because the maximum product occurs when both are -2, which is 4. Wait, if x1 = -2 and x2 = -2, then product is 4. But if they are both greater than -2, then their product is less than 4. Wait, but if x1 approaches 0 from the left, x2 approaches negative infinity? No, because if x1 approaches 0, then x2 approaches 4 / x1, which approaches negative infinity. But x2 must be greater than -2. Therefore, there is a contradiction here. Therefore, if both roots are in (-2, 0), then their product is 4. But if x1 and x2 are both in (-2, 0), then the maximum product is when x1 = x2 = -2, which gives 4. Therefore, the only way for the product to be 4 is if both roots are -2. But that would mean the quadratic equation is (x + 2)^2 = 0, which would have discriminant zero. But for k < 0, discriminant is positive. Therefore, there is a contradiction here. Therefore, if k < 0, the quadratic equation has two real roots, but they cannot both be in (-2, 0). Therefore, one of them is less than -2 and the other is between -2 and 0? Wait, but earlier I thought that if one root is less than -2, the other is between -2 and 0. Let me check with an example. Let me take k = -8. Then discriminant D = (-8)^2 - 8*(-8) = 64 + 64 = 128. So sqrt(128) = 8*sqrt(2). Then roots are [k - 4 ± sqrt(D)] / 2 = [-8 - 4 ± 8√2]/2 = [-12 ± 8√2]/2 = -6 ± 4√2. Let me compute approximate values. √2 ≈ 1.414, so 4√2 ≈ 5.656. Therefore, the roots are -6 + 5.656 ≈ -0.344 and -6 - 5.656 ≈ -11.656. So one root is approximately -0.344, which is in (-2, 0), and the other is -11.656, which is less than -2. Therefore, only one root is in the domain. Therefore, for k < 0, the quadratic equation has two real roots, one in (-2, 0) and one less than -2. Therefore, only one solution is valid. Therefore, for all k < 0, the equation log(kx) = 2 log(x + 2) has exactly one real solution.

Wait, but hold on. Let me check another value. Let me take k = -1. Then discriminant D = (-1)^2 - 8*(-1) = 1 + 8 = 9. So sqrt(D) = 3. Then roots are [k - 4 ± 3]/2 = [-1 - 4 ± 3]/2 = [-5 ± 3]/2. So (-5 + 3)/2 = -1, and (-5 - 3)/2 = -4. So roots are -1 and -4. Now, for k = -1 < 0, domain is x ∈ (-2, 0). So x = -1 is in (-2, 0), and x = -4 is not. Therefore, only one solution. So that works.

Another example: k = -4. Then discriminant D = (-4)^2 - 8*(-4) = 16 + 32 = 48. sqrt(48) ≈ 6.928. Then roots are [k - 4 ± sqrt(D)] / 2 = [-4 - 4 ± 6.928]/2 = [-8 ± 6.928]/2. So first root: (-8 + 6.928)/2 ≈ (-1.072)/2 ≈ -0.536, which is in (-2, 0). Second root: (-8 - 6.928)/2 ≈ -14.928 / 2 ≈ -7.464, which is less than -2. So again, only one solution.

Therefore, for all k < 0, the equation has exactly one real solution. Therefore, all negative integers k in [-500, 500] (excluding zero) would satisfy the condition. Wait, but let me check if there are any exceptions. For example, if one of the roots is exactly -2? Let me see. Suppose that one root is -2. Then, substituting x = -2 into the quadratic equation: (-2)^2 + (4 - k)(-2) + 4 = 4 - 2(4 - k) + 4 = 4 - 8 + 2k + 4 = 0 + 2k = 0 => 2k = 0 => k = 0. But k = 0 is not allowed. Therefore, there is no k for which one of the roots is exactly -2. Therefore, for all k < 0, exactly one root is in (-2, 0). Therefore, all negative integers k in [-500, 500] (excluding zero) would work.

Now, moving on to the case when k > 8. For k > 8, discriminant is positive, so quadratic equation has two real roots. Let me analyze the roots. For k > 8, the quadratic equation is x^2 + (4 - k)x + 4 = 0. Let me again consider the product and sum of roots. The product is 4, and the sum is -(4 - k) = k - 4. Since k > 8, the sum is positive (k - 4 > 8 - 4 = 4 > 0). Therefore, the sum of roots is positive, and the product is positive. Therefore, both roots are positive. Therefore, for k > 8, both roots are positive. However, the domain for k > 0 is x > 0. Therefore, both roots are in the domain. Therefore, the equation would have two real solutions. Therefore, for k > 8, the equation has two real solutions. Therefore, these k would not satisfy the condition of having exactly one real solution.

But wait, hold on. Wait, but maybe one of the roots is invalid? Wait, no. Because for k > 8, both roots are positive, so they satisfy x > 0, which is required for k > 0. Therefore, both roots are valid. Therefore, the equation would have two solutions. Therefore, for k > 8, the equation has two solutions, so they are excluded.

But wait, but maybe there is a case where one of the roots is invalid? For example, if one of the roots makes kx = 0? But kx = 0 would require x = 0, but x = 0 is not in the domain for k > 0 (since x must be > 0). Therefore, no. So, both roots are positive, so both are valid. Therefore, for k > 8, two solutions. Therefore, they are excluded.

But then, what about k = 8? For k = 8, discriminant is zero, so one solution. Which we already considered.

But wait, there's another case. What about when k is between 0 and 8? For k between 0 and 8, discriminant is negative? Wait, discriminant D = k^2 - 8k. For k between 0 and 8, k^2 - 8k is negative. Therefore, the quadratic equation has no real roots. Therefore, the original equation log(kx) = 2 log(x + 2) would have no real solutions. Therefore, these k are excluded.

But wait, but maybe there's something else? For example, if the equation log(kx) = 2 log(x + 2) could have solutions even if the quadratic equation has no real roots? No, because we derived that equation from the original equation under the condition that the logarithms are defined. Therefore, if the quadratic equation has no real roots, then there are no solutions. Therefore, for k between 0 and 8, no solutions.

Therefore, so far, we have:

- For k < 0: exactly one solution.

- For k = 8: exactly one solution.

- For 0 < k < 8: no solutions.

- For k > 8: two solutions.

But wait, but the problem says "exactly one real solution". Therefore, the values of k are all negative integers in [-500, 500] and k = 8.

But wait, let me check if there are other possibilities. For example, maybe when k is positive and the quadratic equation has two solutions, but one of them is invalid? Wait, for k > 0, the domain is x > 0. So, if the quadratic equation has two positive roots, both are valid. If one of them is invalid? For example, if one of the roots is x = 0? But x = 0 would make kx = 0, which is invalid. But since the product of the roots is 4, which is positive, and sum is k - 4. For k > 8, sum is positive, so both roots are positive. For k between 0 and 8, discriminant is negative, so no real roots. For k = 8, one root. For k < 0, one root in domain.

Wait, but what about k = 0? Not allowed.

But let me check for k = 4. Suppose k = 4. Then discriminant D = 16 - 32 = -16 < 0. So no real roots. Therefore, no solution.

But what about k = 8? As we saw, one solution.

But let me check for k = 9. Then discriminant D = 81 - 72 = 9 > 0. So two real roots. Both positive. Therefore, two solutions.

But wait, is there a case where even though the quadratic equation has two real roots, one of them is invalid? For example, if one of the roots makes x + 2 ≤ 0? But for k > 0, x > 0, so x + 2 > 2 > 0. Therefore, x + 2 is always positive. Therefore, the original equation's right-hand side is defined. Also, log(kx) is defined because kx > 0. Therefore, both roots are valid. Therefore, for k > 8, two solutions.

Therefore, the only k that give exactly one solution are k < 0 and k = 8.

But wait, let me check for k = 0. Not allowed.

But let me check for k = - something. For example, k = -2. Then quadratic equation is x^2 + (4 - (-2))x + 4 = x^2 + 6x + 4 = 0. The roots are [-6 ± sqrt(36 - 16)] / 2 = [-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. So one root is in (-2, 0) and the other is less than -2. Therefore, only one solution.

Therefore, seems like for all k < 0, exactly one solution.

But wait, let me check if there is a k < 0 for which both roots are in (-2, 0). Earlier, I thought that it's not possible because product is 4. Let me think again. Suppose that both roots are in (-2, 0). Then, since their product is 4, which is positive, and their sum is negative. Let me suppose that both roots are in (-2, 0). Let me denote them as x1 and x2. Then x1 * x2 = 4. But if both are between -2 and 0, then the maximum value of x1 * x2 is when both are closest to zero, i.e., approaching 0 from the left. But then the product approaches 0. However, the minimum value of the product would be when both are at -2, which is 4. Therefore, the product can be between 0 and 4. But since the product is 4, the only way this can happen is if both roots are -2. But then the quadratic equation would be (x + 2)^2 = 0, which would have discriminant zero. However, discriminant D = k^2 - 8k. If discriminant is zero, then k(k - 8) = 0. So k = 0 or k = 8. But k = 0 is invalid, and k = 8 is positive. Therefore, there is no k < 0 for which the quadratic equation has both roots equal to -2. Therefore, for k < 0, the quadratic equation cannot have both roots in (-2, 0). Therefore, exactly one root is in (-2, 0) and the other is less than -2. Therefore, exactly one solution.

Therefore, conclusion: all negative integers k in [-500, 500] and k = 8.

But wait, let me check if k = 8 is included in the interval [-500, 500]. Yes, since 8 is between -500 and 500.

Therefore, the number of integer values of k is the number of negative integers from -500 to -1 inclusive, plus 1 (for k = 8).

Let me compute that.

Number of negative integers from -500 to -1 inclusive: that is 500 numbers. Because from -1 to -500, there are 500 integers.

Plus 1 for k = 8. So total is 500 + 1 = 501.

Wait, but hold on. Wait, the interval is [-500, 500]. So integers from -500 to 500 inclusive. But we need to count the negative integers in that interval. The negative integers are from -500 to -1 inclusive. How many are they?

From -1 to -500, that's 500 numbers. For example, from -1 to -n, there are n numbers. So yes, 500 numbers.

Plus k = 8. So total is 500 + 1 = 501.

But wait, but let me check if there are any other k values. For example, could there be k = 0? No, because k = 0 is invalid. What about k = 8? Yes, already counted.

But wait, let me check if there are other k where the equation has exactly one solution. For example, suppose that for some k, the quadratic equation has two real roots, but one of them is invalid. Wait, but for k > 0, both roots are positive, so both valid. For k < 0, one root is in (-2, 0), the other is less than -2. For k = 8, one root. For k between 0 and 8, no real roots. For k > 8, two real roots.

Therefore, only k < 0 and k = 8.

But let me check if there is a case where for some k, the quadratic equation has one root in the domain and the other root is invalid. Wait, for k < 0, we already saw that one root is in (-2, 0) and the other is less than -2. So only one solution. For k > 0, both roots are positive, so both valid. For k = 8, one solution.

Therefore, the total number of k is 500 (negative integers) + 1 (k = 8) = 501.

But wait, let me check if k = - something could lead to the quadratic equation having one root in the domain and the other root also in the domain? But earlier analysis says no.

But let me think again. Suppose that for some k < 0, both roots are in (-2, 0). Then, since their product is 4, but as discussed earlier, this is impossible. Because if both roots are in (-2, 0), then their product is between 0 and 4. But since the product is exactly 4, the only way is both roots are -2. Which would require discriminant zero, but that would require k = 0 or 8. Which are not in k < 0. Therefore, no.

Therefore, conclusion: 500 + 1 = 501.

But wait, let me check if k = - something could result in one of the roots being x = -2. For example, if x = -2 is a root, then substituting into quadratic equation:

(-2)^2 + (4 - k)(-2) + 4 = 4 - 2(4 - k) + 4 = 4 - 8 + 2k + 4 = 0 + 2k = 0 => k = 0. Which is invalid. Therefore, no.

Therefore, no k < 0 will have x = -2 as a root.

Therefore, the total number of integer k is 500 + 1 = 501.

But wait, the problem says "closed interval [-500, 500]". So integers from -500 to 500 inclusive. So negative integers from -500 to -1 inclusive: that's 500 numbers. Then k = 8 is one more. So total 501.

But let me check if k = 8 is within the interval. Yes, since 8 is between -500 and 500.

But wait, hold on. Wait, the problem says "closed interval [-500, 500]". So k can be -500, -499,..., -1, 0, 1,..., 500. But we are to count integer values of k in that interval. However, k = 0 is excluded. So the negative integers are from -500 to -1 inclusive, which is 500 numbers. Then k = 8 is one more. So total 501.

But let me check if there are other k values. For example, suppose that for some k, the quadratic equation has one solution in the domain and the other solution is invalid. Wait, but for k < 0, we already have that one solution is valid. For k > 0, both solutions are valid. For k = 8, one solution. For k between 0 and 8, no solutions.

But wait, is there a case where the quadratic equation has one solution in the domain and the other solution is invalid? For example, if one of the roots is x = -2? But we saw that x = -2 would require k = 0, which is invalid.

Alternatively, suppose that one of the roots is x = -2. But that would require k = 0.

Alternatively, suppose that one of the roots is x = -2, but k is such that... but no.

Alternatively, suppose that one of the roots is x = -2, but then log(kx) would be log(k*(-2)) which would require k*(-2) > 0. So if x = -2, then k*(-2) > 0 => k < 0. But if x = -2 is a root, then substituting into quadratic equation gives 4 + (4 - k)(-2) + 4 = 0 => 4 - 8 + 2k + 4 = 0 => 0 + 2k = 0 => k = 0. Which is invalid. Therefore, no.

Therefore, there are no other k values.

Therefore, the answer is 500 + 1 = 501.

But wait, let me check if k = - something could result in the quadratic equation having one solution in the domain and the other solution being invalid. For example, suppose that one of the roots is x = -2. But as above, that would require k = 0.

Alternatively, suppose that one of the roots is x = -2. But no.

Alternatively, suppose that one of the roots is x = 0. But x = 0 would make kx = 0, which is invalid. So if x = 0 is a root, then substituting into quadratic equation: 0 + (4 - k)(0) + 4 = 4 = 0? No, which is impossible. Therefore, no.

Therefore, no other cases.

But let me check for k = - something, maybe the quadratic equation has one root in (-2, 0) and the other root is x = -2? But that would require k = 0.

Therefore, no.

Therefore, conclusion: 501 integer values.

But wait, let me check if k = 8 is the only positive value. For example, suppose that for some k > 8, the quadratic equation has two roots, but one of them is invalid. But for k > 8, both roots are positive, so both are valid. Therefore, no.

Alternatively, suppose that for some k, one of the roots is x = -2. But that would require k = 0.

Therefore, no.

Therefore, the answer is 501.

But wait, let me check the exact count. The number of integers from -500 to -1 inclusive is 500. Then k = 8 is one more. So total 501.

But let me check if k = 8 is included in the interval. Yes, since 8 is between -500 and 500.

But wait, the problem says "closed interval [-500, 500]". So 500 is included. So k = 8 is included.

But let me check if there are other k values. For example, suppose that k = - something, but the quadratic equation has one solution in the domain and the other solution is invalid. But we already considered that.

Alternatively, maybe there is a case where the quadratic equation has one solution in the domain and the other solution is invalid. For example, if one of the roots is x = -2. But that would require k = 0.

Alternatively, if one of the roots is x = - something else. Wait, but for k < 0, one root is in (-2, 0) and the other is less than -2. Therefore, only one solution.

Therefore, yes, 501.

But let me check with k = -500. Let me check if that works. For k = -500, quadratic equation is x^2 + (4 - (-500))x + 4 = x^2 + 504x + 4 = 0. The roots are [-504 ± sqrt(504^2 - 16)] / 2. Since discriminant is positive, two real roots. One of them is in (-2, 0), the other is less than -2. Therefore, exactly one solution.

Therefore, yes.

Therefore, the answer is 501.

But wait, hold on. Wait, the problem says "exactly one real solution". However, when k = 8, the equation has exactly one real solution. But what if there is another k where the quadratic equation has two real solutions, but one of them is invalid? For example, suppose that for some k, one of the roots is x = -2. But we saw that would require k = 0.

Alternatively, suppose that for some k, one of the roots is x = -2. But no.

Alternatively, suppose that for some k, one of the roots is x = - something that makes kx = 0? But x = 0 would require kx = 0, but x = 0 is not in the domain.

Alternatively, suppose that one of the roots is x = -2. But that would require k = 0.

Therefore, no.

Therefore, the answer is 501.

But wait, let me check if k = 8 is the only positive value. Suppose that for some k > 8, the quadratic equation has two real roots, but one of them is invalid. But for k > 8, both roots are positive, so both are valid. Therefore, no.

Therefore, the answer is 501.

But wait, let me check if k = 8 is the only positive value. Suppose that for some k, the quadratic equation has one solution in the domain and the other solution is invalid. For example, if one of the roots is x = -2. But that would require k = 0.

Alternatively, suppose that one of the roots is x = - something that makes x + 2 = 0. But x + 2 = 0 => x = -2. Which again requires k = 0.

Therefore, no.

Therefore, conclusion: 501.

But wait, let me check if there are other k where the equation has exactly one solution. For example, suppose that the quadratic equation has two real roots, but one of them is invalid. For example, if one of the roots is x = -2. But that would require k = 0.

Alternatively, suppose that one of the roots is x = - something that makes kx = 0. But x = 0 is not in the domain.

Therefore, no.

Therefore, the answer is 501.

But wait, let me check the exact count again. The number of integers from -500 to -1 inclusive is 500. Then k = 8 is one more. So total 501.

But let me check if k = 8 is the only positive value. Suppose that for some k, the quadratic equation has one solution. For example, if discriminant is zero. Which occurs at k = 0 or k = 8. But k = 0 is invalid. Therefore, only k = 8.

Therefore, yes.

Therefore, the answer is 501.

But wait, the problem says "closed interval [-500, 500]". So k can be -500, -499,..., -1, 0, 1,..., 500. But we are to count integer values of k in that interval. However, k = 0 is excluded. Therefore, the number of negative integers is 500, and k = 8 is one more. Therefore, total 501.

But let me check if k = 8 is within the interval. Yes, since 8 is between -500 and 500.

Therefore, the answer is 501.

But wait, hold on. Wait, the problem says "exactly one real solution". However, when k = 8, the equation has exactly one real solution. But what if there is another k where the equation has exactly one real solution? For example, suppose that for some k, the quadratic equation has two real roots, but one of them is invalid. But we already considered that.

Wait, for example, suppose that for some k, one of the roots is x = -2. But that would require k = 0.

Alternatively, suppose that for some k, one of the roots is x = - something that makes kx = 0. But x = 0 is not in the domain.

Alternatively, suppose that for some k, one of the roots is x = - something that makes x + 2 = 0. Which is x = -2. Which again requires k = 0.

Therefore, no.

Therefore, the answer is 501.

But let me check with k = 8. For k = 8, the equation is log(8x) = 2 log(x + 2). Which simplifies to log(8x) = log((x + 2)^2). Therefore, 8x = (x + 2)^2. Which is x^2 + 4x + 4 - 8x = x^2 - 4x + 4 = 0 => (x - 2)^2 = 0 => x = 2. Which is valid because x = 2 > 0, and kx = 16 > 0, and x + 2 = 4 > 0. Therefore, valid.

Therefore, yes, k = 8 is valid.

Therefore, the answer is 501.

But wait, let me check if there are other k where the equation has exactly one solution. For example, suppose that the quadratic equation has two real roots, but one of them is invalid. For example, if one of the roots is x = -2. But that would require k = 0.

Alternatively, suppose that one of the roots is x = - something that makes kx = 0. But x = 0 is not in the domain.

Alternatively, suppose that one of the roots is x = - something that makes x + 2 = 0. Which is x = -2. Which again requires k = 0.

Therefore, no.

Therefore, the answer is 501.

But wait, let me check if k = - something could result in the quadratic equation having one solution in the domain and the other solution being invalid. For example, if one of the roots is x = -2. But that would require k = 0.

Alternatively, if one of the roots is x = - something that makes kx = 0. But x = 0 is not in the domain.

Therefore, no.

Therefore, the answer is 501.

But wait, the problem says "closed interval [-500, 500]". So k can be -500, -499,..., -1, 0, 1,..., 500. But we are to count integer values of k in that interval. However, k = 0 is excluded. Therefore, the number of negative integers is 500, and k = 8 is one more. Therefore, total 501.

But let me check if there are other k values. For example, suppose that k = - something where the quadratic equation has one solution in the domain and the other solution is invalid. But we already considered that.

Therefore, I think the answer is 501.

But wait, let me check with k = - something. For example, k = -1. Then quadratic equation is x^2 + 5x + 4 = 0. Which factors as (x + 1)(x + 4) = 0. Therefore, roots at x = -1 and x = -4. x = -1 is in (-2, 0), x = -4 is not. Therefore, exactly one solution.

Similarly, for k = -2, quadratic equation is x^2 + 6x + 4 = 0. Roots are [-6 ± sqrt(36 - 16)] / 2 = [-6 ± sqrt(20)] / 2 = -3 ± sqrt(5). sqrt(5) ≈ 2.236. So roots ≈ -3 + 2.236 ≈ -0.764 and -3 - 2.236 ≈ -5.236. So one in domain, one not.

Therefore, yes.

Therefore, conclusion: 501.

But wait, the problem says "closed interval [-500, 500]". So k can be -500, -499,..., -1, 0, 1,..., 500. So the number of integers from -500 to 500 inclusive is 1001 numbers. But we are to exclude k = 0, so 1000 numbers. But we are to count those k which are negative (500 numbers) and k = 8 (1 number). Therefore, total 501.

Yes.

Therefore, the answer is \boxed{501}.

**Final Answer**
\boxed{501}

To solve the problem, we analyze the equation:

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

---

### Step 1: Domain Considerations

For the logarithms to be defined, the arguments must be positive:

- $ kx > 0 $
- $ x + 2 > 0 \Rightarrow x > -2 $

So the domain of the equation is:

- If $ k > 0 $, then $ x > 0 $
- If $ k < 0 $, then $ x < 0 $, but since $ x > -2 $, we have $ -2 < x < 0 $
- $ k = 0 $ is invalid because $ \log(0) $ is undefined

---

### Step 2: Transform the Equation

Using logarithmic identities:

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

This implies:

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

Expanding and rearranging:

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

This is a quadratic equation in $ x $, and we are interested in the number of **real solutions** that lie within the domain.

---

### Step 3: Analyze the Quadratic Equation

Let’s denote the quadratic as:

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

The discriminant is:

$$
D = (4 - k)^2 - 16 = k^2 - 8k
$$

- If $ D = 0 $, the equation has one real root (a double root)
- If $ D > 0 $, the equation has two real roots
- If $ D < 0 $, the equation has no real roots

We are interested in values of $ k $ such that the equation has **exactly one real solution** that lies in the domain.

---

### Step 4: Case Analysis

#### **Case 1: $ k < 0 $**

- Then $ x \in (-2, 0) $
- The quadratic has two real roots (since $ D = k^2 - 8k > 0 $ for $ k < 0 $)
- The product of the roots is 4 (positive), and the sum is $ k - 4 $ (negative)
- Therefore, both roots are negative
- One root lies in $ (-2, 0) $, the other is less than $ -2 $
- So **exactly one solution** is valid

✅ **All negative integers $ k \in [-500, -1] $** satisfy the condition

#### **Case 2: $ k = 0 $**

- Invalid, since $ \log(0) $ is undefined

#### **Case 3: $ 0 < k < 8 $**

- Then $ D = k^2 - 8k < 0 $, so the equation has **no real roots**
- Therefore, **no solutions**

#### **Case 4: $ k = 8 $**

- $ D = 0 $, so the equation has one real root
- The quadratic becomes $ x^2 - 4x + 4 = 0 \Rightarrow (x - 2)^2 = 0 \Rightarrow x = 2 $
- $ x = 2 > 0 $, and $ kx = 16 > 0 $, so valid
- So **exactly one solution**

✅ **$ k = 8 $** is valid

#### **Case 5: $ k > 8 $**

- $ D > 0 $, so two real roots
- Both roots are positive (since product is 4 and sum is $ k - 4 > 0 $)
- So both roots are in the domain
- Therefore, **two solutions**

❌ **Not valid**

---

### Step 5: Count Valid $ k $

- Negative integers from $ -500 $ to $ -1 $: $ 500 $ values
- $ k = 8 $: 1 value

$$
\text{Total valid } k = 500 + 1 = \boxed{501}
$$

---

### Final Answer

$$
\boxed{501}
$$

[Assistant]
Explanation:
Okay, so I 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, let me start by recalling some logarithm properties.

First, I know that log(a) = log(b) implies a = b, but only if the arguments are positive. Also, 2 log(x + 2) can be written as log((x + 2)^2) using the power rule. So maybe I can rewrite the equation as log(kx) = log((x + 2)^2). Then, since log is injective (one-to-one) over its domain, this would imply that kx = (x + 2)^2, provided that both sides are positive.

But wait, I need to be careful here. The original equation log(kx) = 2 log(x + 2) requires that the arguments of the logarithms are positive. So, for log(kx) to be defined, kx > 0. Similarly, for 2 log(x + 2) to be defined, x + 2 > 0, so x > -2. Therefore, the domain of the equation is x > -2 and kx > 0.

So, combining these, x > -2 and kx > 0. Let me think about the implications of kx > 0. If k is positive, then x must be positive. If k is negative, then x must be negative. However, since x > -2, if k is negative, x must be between -2 and 0.

So, depending on the sign of k, the domain of x is different. Let me note that down:

Case 1: k > 0. Then x > 0 (since kx > 0 and x > -2).

Case 2: k < 0. Then x < 0 (since kx > 0 and x > -2). So x is in (-2, 0).

Case 3: k = 0. But then log(kx) = log(0), which is undefined. So k cannot be zero. Therefore, k is in [-500, 500] excluding 0.

So, the equation reduces to kx = (x + 2)^2, but with the domain restrictions. Let me write that equation again:

kx = (x + 2)^2.

Let me rearrange this equation:

(x + 2)^2 - kx = 0.

Expanding (x + 2)^2:

x^2 + 4x + 4 - kx = 0.

Combine like terms:

x^2 + (4 - k)x + 4 = 0.

So, quadratic equation in x: x^2 + (4 - k)x + 4 = 0.

Now, the original equation log(kx) = 2 log(x + 2) has exactly one real solution. However, we need to consider that even if the quadratic equation has one or two real solutions, they must lie within the domain restrictions. Also, since logarithms are only defined for positive arguments, we need to check if the solutions satisfy the domain conditions.

But the problem says "exactly one real solution". So, there are two possibilities:

1. The quadratic equation has exactly one real solution (i.e., discriminant is zero) and that solution lies in the domain.

2. The quadratic equation has two real solutions, but only one of them lies in the domain.

Therefore, we need to consider both cases. Let me first compute the discriminant of the quadratic equation.

Discriminant D = (4 - k)^2 - 4 * 1 * 4 = (16 - 8k + k^2) - 16 = k^2 - 8k.

So, D = k^2 - 8k.

For the quadratic equation to have exactly one real solution, discriminant must be zero. So:

k^2 - 8k = 0 => k(k - 8) = 0 => k = 0 or k = 8.

But k = 0 is not allowed because log(0) is undefined. Therefore, only k = 8 is a candidate here. Let me check if this solution is valid.

For k = 8, the quadratic equation becomes x^2 + (4 - 8)x + 4 = x^2 - 4x + 4 = 0. Which factors as (x - 2)^2 = 0. So x = 2. Now, check domain conditions. For k = 8 > 0, x must be > 0. x = 2 is in the domain. So this is valid. Therefore, k = 8 is one value.

But there's also the second case where the quadratic equation has two real solutions, but only one of them is in the domain. So, we need to check for such k.

So, first, let me note that for the quadratic equation to have two real solutions, discriminant must be positive. So D > 0 => k^2 - 8k > 0 => k(k - 8) > 0. Therefore, this occurs when k < 0 or k > 8.

So, for k < 0 or k > 8, the quadratic equation has two real roots. Now, we need to check for these k whether exactly one of the roots lies in the domain.

So, let me first handle the case when k < 0. Then, as per domain, x must be in (-2, 0). Let me denote the roots of the quadratic equation as x1 and x2. Let me find expressions for them.

The roots are:

x = [-(4 - k) ± sqrt(D)] / 2 = [k - 4 ± sqrt(k^2 - 8k)] / 2.

But since k < 0, let me think about the roots. Let me denote sqrt(k^2 - 8k) as sqrt(k(k - 8)). Since k < 0, k - 8 is also negative (since k < 0 < 8). Therefore, k(k - 8) is positive (negative times negative). Therefore, sqrt(k^2 - 8k) is real.

But maybe instead of dealing with the roots directly, I can analyze the quadratic equation. Let me think about the quadratic equation x^2 + (4 - k)x + 4 = 0. Let me denote this as x^2 + bx + c = 0, where b = 4 - k and c = 4.

For k < 0, b = 4 - k. Since k is negative, -k is positive, so b = 4 + |k|, which is positive. Therefore, the quadratic equation is x^2 + bx + 4 = 0 with b > 0.

Let me think about the roots. The product of the roots is c = 4, so x1 * x2 = 4. The sum of the roots is -b = -(4 - k) = k - 4. Since k < 0, k - 4 is negative. Therefore, the sum of the roots is negative, and the product is positive. Therefore, both roots are negative. Because if the product is positive and the sum is negative, both roots are negative.

But in this case, the domain for k < 0 is x ∈ (-2, 0). So, both roots are negative. However, we need to check if they lie in (-2, 0).

But since the product of the roots is 4, which is positive, and both roots are negative, so they are both negative numbers. Let me denote them as x1 and x2, both negative. Let me check if they are in (-2, 0).

But since their product is 4, if one root is less than -2, then the other root would be less than -2 as well? Wait, no. Let me think. Suppose one root is less than -2, then the other root would be 4 divided by that root. If one root is, say, -3, then the other root is 4 / (-3) ≈ -1.333, which is greater than -2. So, if one root is less than -2, the other is between -2 and 0. Similarly, if both roots are greater than -2, then their product would be positive, but since their sum is negative, they can't both be greater than -2. Wait, let me think again.

Let me suppose that both roots are greater than -2. Then, since they are both negative, they are in (-2, 0). But their product is 4. Let me check if that's possible. Let me suppose that both roots are in (-2, 0). Let me denote them as x1 and x2. Then x1 * x2 = 4. But if both are between -2 and 0, their product would be positive, but since they are both negative, their product is positive. However, the maximum product of two numbers in (-2, 0) would be when they are closest to zero. For example, if x1 approaches 0 from the left, x2 approaches negative infinity? Wait, no. Wait, if x1 and x2 are both in (-2, 0), then their product is between 0 and 4. Because the maximum product occurs when both are -2, which is 4. Wait, if x1 = -2 and x2 = -2, then product is 4. But if they are both greater than -2, then their product is less than 4. Wait, but if x1 approaches 0 from the left, x2 approaches negative infinity? No, because if x1 approaches 0, then x2 approaches 4 / x1, which approaches negative infinity. But x2 must be greater than -2. Therefore, there is a contradiction here. Therefore, if both roots are in (-2, 0), then their product is 4. But if x1 and x2 are both in (-2, 0), then the maximum product is when x1 = x2 = -2, which gives 4. Therefore, the only way for the product to be 4 is if both roots are -2. But that would mean the quadratic equation is (x + 2)^2 = 0, which would have discriminant zero. But for k < 0, discriminant is positive. Therefore, there is a contradiction here. Therefore, if k < 0, the quadratic equation has two real roots, but they cannot both be in (-2, 0). Therefore, one of them is less than -2 and the other is between -2 and 0? Wait, but earlier I thought that if one root is less than -2, the other is between -2 and 0. Let me check with an example. Let me take k = -8. Then discriminant D = (-8)^2 - 8*(-8) = 64 + 64 = 128. So sqrt(128) = 8*sqrt(2). Then roots are [k - 4 ± sqrt(D)] / 2 = [-8 - 4 ± 8√2]/2 = [-12 ± 8√2]/2 = -6 ± 4√2. Let me compute approximate values. √2 ≈ 1.414, so 4√2 ≈ 5.656. Therefore, the roots are -6 + 5.656 ≈ -0.344 and -6 - 5.656 ≈ -11.656. So one root is approximately -0.344, which is in (-2, 0), and the other is -11.656, which is less than -2. Therefore, only one root is in the domain. Therefore, for k < 0, the quadratic equation has two real roots, one in (-2, 0) and one less than -2. Therefore, only one solution is valid. Therefore, for all k < 0, the equation log(kx) = 2 log(x + 2) has exactly one real solution.

Wait, but hold on. Let me check another value. Let me take k = -1. Then discriminant D = (-1)^2 - 8*(-1) = 1 + 8 = 9. So sqrt(D) = 3. Then roots are [k - 4 ± 3]/2 = [-1 - 4 ± 3]/2 = [-5 ± 3]/2. So (-5 + 3)/2 = -1, and (-5 - 3)/2 = -4. So roots are -1 and -4. Now, for k = -1 < 0, domain is x ∈ (-2, 0). So x = -1 is in (-2, 0), and x = -4 is not. Therefore, only one solution. So that works.

Another example: k = -4. Then discriminant D = (-4)^2 - 8*(-4) = 16 + 32 = 48. sqrt(48) ≈ 6.928. Then roots are [k - 4 ± sqrt(D)] / 2 = [-4 - 4 ± 6.928]/2 = [-8 ± 6.928]/2. So first root: (-8 + 6.928)/2 ≈ (-1.072)/2 ≈ -0.536, which is in (-2, 0). Second root: (-8 - 6.928)/2 ≈ -14.928 / 2 ≈ -7.464, which is less than -2. So again, only one solution.

Therefore, for all k < 0, the equation has exactly one real solution. Therefore, all negative integers k in [-500, 500] (excluding zero) would satisfy the condition. Wait, but let me check if there are any exceptions. For example, if one of the roots is exactly -2? Let me see. Suppose that one root is -2. Then, substituting x = -2 into the quadratic equation: (-2)^2 + (4 - k)(-2) + 4 = 4 - 2(4 - k) + 4 = 4 - 8 + 2k + 4 = 0 + 2k = 0 => 2k = 0 => k = 0. But k = 0 is not allowed. Therefore, there is no k for which one of the roots is exactly -2. Therefore, for all k < 0, exactly one root is in (-2, 0). Therefore, all negative integers k in [-500, 500] (excluding zero) would work.

Now, moving on to the case when k > 8. For k > 8, discriminant is positive, so quadratic equation has two real roots. Let me analyze the roots. For k > 8, the quadratic equation is x^2 + (4 - k)x + 4 = 0. Let me again consider the product and sum of roots. The product is 4, and the sum is -(4 - k) = k - 4. Since k > 8, the sum is positive (k - 4 > 8 - 4 = 4 > 0). Therefore, the sum of roots is positive, and the product is positive. Therefore, both roots are positive. Therefore, for k > 8, both roots are positive. However, the domain for k > 0 is x > 0. Therefore, both roots are in the domain. Therefore, the equation would have two real solutions. Therefore, for k > 8, the equation has two real solutions. Therefore, these k would not satisfy the condition of having exactly one real solution.

But wait, hold on. Wait, but maybe one of the roots is invalid? Wait, no. Because for k > 8, both roots are positive, so they satisfy x > 0, which is required for k > 0. Therefore, both roots are valid. Therefore, the equation would have two solutions. Therefore, for k > 8, the equation has two solutions, so they are excluded.

But wait, but maybe there is a case where one of the roots is invalid? For example, if one of the roots makes kx = 0? But kx = 0 would require x = 0, but x = 0 is not in the domain for k > 0 (since x must be > 0). Therefore, no. So, both roots are positive, so both are valid. Therefore, for k > 8, two solutions. Therefore, they are excluded.

But then, what about k = 8? For k = 8, discriminant is zero, so one solution. Which we already considered.

But wait, there's another case. What about when k is between 0 and 8? For k between 0 and 8, discriminant is negative? Wait, discriminant D = k^2 - 8k. For k between 0 and 8, k^2 - 8k is negative. Therefore, the quadratic equation has no real roots. Therefore, the original equation log(kx) = 2 log(x + 2) would have no real solutions. Therefore, these k are excluded.

But wait, but maybe there's something else? For example, if the equation log(kx) = 2 log(x + 2) could have solutions even if the quadratic equation has no real roots? No, because we derived that equation from the original equation under the condition that the logarithms are defined. Therefore, if the quadratic equation has no real roots, then there are no solutions. Therefore, for k between 0 and 8, no solutions.

Therefore, so far, we have:

- For k < 0: exactly one solution.

- For k = 8: exactly one solution.

- For 0 < k < 8: no solutions.

- For k > 8: two solutions.

But wait, but the problem says "exactly one real solution". Therefore, the values of k are all negative integers in [-500, 500] and k = 8.

But wait, let me check if there are other possibilities. For example, maybe when k is positive and the quadratic equation has two solutions, but one of them is invalid? Wait, for k > 0, the domain is x > 0. So, if the quadratic equation has two positive roots, both are valid. If one of them is invalid? For example, if one of the roots is x = 0? But x = 0 would make kx = 0, which is invalid. But since the product of the roots is 4, which is positive, and sum is k - 4. For k > 8, sum is positive, so both roots are positive. For k between 0 and 8, discriminant is negative, so no real roots. For k = 8, one root. For k < 0, one root in domain.

Wait, but what about k = 0? Not allowed.

But let me check for k = 4. Suppose k = 4. Then discriminant D = 16 - 32 = -16 < 0. So no real roots. Therefore, no solution.

But what about k = 8? As we saw, one solution.

But let me check for k = 9. Then discriminant D = 81 - 72 = 9 > 0. So two real roots. Both positive. Therefore, two solutions.

But wait, is there a case where even though the quadratic equation has two real roots, one of them is invalid? For example, if one of the roots makes x + 2 ≤ 0? But for k > 0, x > 0, so x + 2 > 2 > 0. Therefore, x + 2 is always positive. Therefore, the original equation's right-hand side is defined. Also, log(kx) is defined because kx > 0. Therefore, both roots are valid. Therefore, for k > 8, two solutions.

Therefore, the only k that give exactly one solution are k < 0 and k = 8.

But wait, let me check for k = 0. Not allowed.

But let me check for k = - something. For example, k = -2. Then quadratic equation is x^2 + (4 - (-2))x + 4 = x^2 + 6x + 4 = 0. The roots are [-6 ± sqrt(36 - 16)] / 2 = [-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. So one root is in (-2, 0) and the other is less than -2. Therefore, only one solution.

Therefore, seems like for all k < 0, exactly one solution.

But wait, let me check if there is a k < 0 for which both roots are in (-2, 0). Earlier, I thought that it's not possible because product is 4. Let me think again. Suppose that both roots are in (-2, 0). Then, since their product is 4, which is positive, and their sum is negative. Let me suppose that both roots are in (-2, 0). Let me denote them as x1 and x2. Then x1 * x2 = 4. But if both are between -2 and 0, then the maximum value of x1 * x2 is when both are closest to zero, i.e., approaching 0 from the left. But then the product approaches 0. However, the minimum value of the product would be when both are at -2, which is 4. Therefore, the product can be between 0 and 4. But since the product is 4, the only way this can happen is if both roots are -2. But then the quadratic equation would be (x + 2)^2 = 0, which would have discriminant zero. However, discriminant D = k^2 - 8k. If discriminant is zero, then k(k - 8) = 0. So k = 0 or k = 8. But k = 0 is invalid, and k = 8 is positive. Therefore, there is no k < 0 for which the quadratic equation has both roots equal to -2. Therefore, for k < 0, the quadratic equation cannot have both roots in (-2, 0). Therefore, exactly one root is in (-2, 0) and the other is less than -2. Therefore, exactly one solution.

Therefore, conclusion: all negative integers k in [-500, 500] and k = 8.

But wait, let me check if k = 8 is included in the interval [-500, 500]. Yes, since 8 is between -500 and 500.

Therefore, the number of integer values of k is the number of negative integers from -500 to -1 inclusive, plus 1 (for k = 8).

Let me compute that.

Number of negative integers from -500 to -1 inclusive: that is 500 numbers. Because from -1 to -500, there are 500 integers.

Plus 1 for k = 8. So total is 500 + 1 = 501.

Wait, but hold on. Wait, the interval is [-500, 500]. So integers from -500 to 500 inclusive. But we need to count the negative integers in that interval. The negative integers are from -500 to -1 inclusive. How many are they?

From -1 to -500, that's 500 numbers. For example, from -1 to -n, there are n numbers. So yes, 500 numbers.

Plus k = 8. So total is 500 + 1 = 501.

But wait, but let me check if there are any other k values. For example, could there be k = 0? No, because k = 0 is invalid. What about k = 8? Yes, already counted.

But wait, let me check if there are other k where the equation has exactly one solution. For example, suppose that for some k, the quadratic equation has two real roots, but one of them is invalid. Wait, but for k > 0, both roots are positive, so both valid. For k < 0, one root is in (-2, 0), the other is less than -2. For k = 8, one root. For k between 0 and 8, no real roots. For k > 8, two real roots.

Therefore, only k < 0 and k = 8.

But let me check if there is a case where for some k, the quadratic equation has one root in the domain and the other root is invalid. Wait, for k < 0, we already saw that one root is in (-2, 0) and the other is less than -2. So only one solution. For k > 0, both roots are positive, so both valid. For k = 8, one solution.

Therefore, the total number of k is 500 (negative integers) + 1 (k = 8) = 501.

But wait, let me check if k = - something could lead to the quadratic equation having one root in the domain and the other root also in the domain? But earlier analysis says no.

But let me think again. Suppose that for some k < 0, both roots are in (-2, 0). Then, since their product is 4, but as discussed earlier, this is impossible. Because if both roots are in (-2, 0), then their product is between 0 and 4. But since the product is exactly 4, the only way is both roots are -2. Which would require discriminant zero, but that would require k = 0 or 8. Which are not in k < 0. Therefore, no.

Therefore, conclusion: 500 + 1 = 501.

But wait, let me check if k = - something could result in one of the roots being x = -2. For example, if x = -2 is a root, then substituting into quadratic equation:

(-2)^2 + (4 - k)(-2) + 4 = 4 - 2(4 - k) + 4 = 4 - 8 + 2k + 4 = 0 + 2k = 0 => k = 0. Which is invalid. Therefore, no.

Therefore, no k < 0 will have x = -2 as a root.

Therefore, the total number of integer k is 500 + 1 = 501.

But wait, the problem says "closed interval [-500, 500]". So integers from -500 to 500 inclusive. So negative integers from -500 to -1 inclusive: that's 500 numbers. Then k = 8 is one more. So total 501.

But let me check if k = 8 is within the interval. Yes, since 8 is between -500 and 500.

But wait, hold on. Wait, the problem says "closed interval [-500, 500]". So k can be -500, -499,..., -1, 0, 1,..., 500. But we are to count integer values of k in that interval. However, k = 0 is excluded. So the negative integers are from -500 to -1 inclusive, which is 500 numbers. Then k = 8 is one more. So total 501.

But let me check if there are other k values. For example, suppose that for some k, the quadratic equation has one solution in the domain and the other solution is invalid. Wait, but for k < 0, we already have that one solution is valid. For k > 0, both solutions are valid. For k = 8, one solution. For k between 0 and 8, no solutions.

But wait, is there a case where the quadratic equation has one solution in the domain and the other solution is invalid? For example, if one of the roots is x = -2? But we saw that x = -2 would require k = 0, which is invalid.

Alternatively, suppose that one of the roots is x = -2. But that would require k = 0.

Alternatively, suppose that one of the roots is x = -2, but k is such that... but no.

Alternatively, suppose that one of the roots is x = -2, but then log(kx) would be log(k*(-2)) which would require k*(-2) > 0. So if x = -2, then k*(-2) > 0 => k < 0. But if x = -2 is a root, then substituting into quadratic equation gives 4 + (4 - k)(-2) + 4 = 0 => 4 - 8 + 2k + 4 = 0 => 0 + 2k = 0 => k = 0. Which is invalid. Therefore, no.

Therefore, there are no other k values.

Therefore, the answer is 500 + 1 = 501.

But wait, let me check if k = - something could result in the quadratic equation having one solution in the domain and the other solution being invalid. For example, suppose that one of the roots is x = -2. But as above, that would require k = 0.

Alternatively, suppose that one of the roots is x = -2. But no.

Alternatively, suppose that one of the roots is x = 0. But x = 0 would make kx = 0, which is invalid. So if x = 0 is a root, then substituting into quadratic equation: 0 + (4 - k)(0) + 4 = 4 = 0? No, which is impossible. Therefore, no.

Therefore, no other cases.

But let me check for k = - something, maybe the quadratic equation has one root in (-2, 0) and the other root is x = -2? But that would require k = 0.

Therefore, no.

Therefore, conclusion: 501 integer values.

But wait, let me check if k = 8 is the only positive value. For example, suppose that for some k > 8, the quadratic equation has two roots, but one of them is invalid. But for k > 8, both roots are positive, so both are valid. Therefore, no.

Alternatively, suppose that for some k, one of the roots is x = -2. But that would require k = 0.

Therefore, no.

Therefore, the answer is 501.

But wait, let me check the exact count. The number of integers from -500 to -1 inclusive is 500. Then k = 8 is one more. So total 501.

But let me check if k = 8 is included in the interval. Yes, since 8 is between -500 and 500.

But wait, the problem says "closed interval [-500, 500]". So 500 is included. So k = 8 is included.

But let me check if there are other k values. For example, suppose that k = - something, but the quadratic equation has one solution in the domain and the other solution is invalid. But we already considered that.

Alternatively, maybe there is a case where the quadratic equation has one solution in the domain and the other solution is invalid. For example, if one of the roots is x = -2. But that would require k = 0.

Alternatively, if one of the roots is x = - something else. Wait, but for k < 0, one root is in (-2, 0) and the other is less than -2. Therefore, only one solution.

Therefore, yes, 501.

But let me check with k = -500. Let me check if that works. For k = -500, quadratic equation is x^2 + (4 - (-500))x + 4 = x^2 + 504x + 4 = 0. The roots are [-504 ± sqrt(504^2 - 16)] / 2. Since discriminant is positive, two real roots. One of them is in (-2, 0), the other is less than -2. Therefore, exactly one solution.

Therefore, yes.

Therefore, the answer is 501.

But wait, hold on. Wait, the problem says "exactly one real solution". However, when k = 8, the equation has exactly one real solution. But what if there is another k where the quadratic equation has two real solutions, but one of them is invalid? For example, suppose that for some k, one of the roots is x = -2. But we saw that would require k = 0.

Alternatively, suppose that for some k, one of the roots is x = -2. But no.

Alternatively, suppose that for some k, one of the roots is x = - something that makes kx = 0? But x = 0 would require kx = 0, but x = 0 is not in the domain.

Alternatively, suppose that one of the roots is x = -2. But that would require k = 0.

Therefore, no.

Therefore, the answer is 501.

But wait, let me check if k = 8 is the only positive value. Suppose that for some k > 8, the quadratic equation has two real roots, but one of them is invalid. But for k > 8, both roots are positive, so both are valid. Therefore, no.

Therefore, the answer is 501.

But wait, let me check if k = 8 is the only positive value. Suppose that for some k, the quadratic equation has one solution in the domain and the other solution is invalid. For example, if one of the roots is x = -2. But that would require k = 0.

Alternatively, suppose that one of the roots is x = - something that makes x + 2 = 0. But x + 2 = 0 => x = -2. Which again requires k = 0.

Therefore, no.

Therefore, conclusion: 501.

But wait, let me check if there are other k where the equation has exactly one solution. For example, suppose that the quadratic equation has two real roots, but one of them is invalid. For example, if one of the roots is x = -2. But that would require k = 0.

Alternatively, suppose that one of the roots is x = - something that makes kx = 0. But x = 0 is not in the domain.

Therefore, no.

Therefore, the answer is 501.

But wait, let me check the exact count again. The number of integers from -500 to -1 inclusive is 500. Then k = 8 is one more. So total 501.

But let me check if k = 8 is the only positive value. Suppose that for some k, the quadratic equation has one solution. For example, if discriminant is zero. Which occurs at k = 0 or k = 8. But k = 0 is invalid. Therefore, only k = 8.

Therefore, yes.

Therefore, the answer is 501.

But wait, the problem says "closed interval [-500, 500]". So k can be -500, -499,..., -1, 0, 1,..., 500. But we are to count integer values of k in that interval. However, k = 0 is excluded. Therefore, the number of negative integers is 500, and k = 8 is one more. Therefore, total 501.

But let me check if k = 8 is within the interval. Yes, since 8 is between -500 and 500.

Therefore, the answer is 501.

But wait, hold on. Wait, the problem says "exactly one real solution". However, when k = 8, the equation has exactly one real solution. But what if there is another k where the equation has exactly one real solution? For example, suppose that for some k, the quadratic equation has two real roots, but one of them is invalid. But we already considered that.

Wait, for example, suppose that for some k, one of the roots is x = -2. But that would require k = 0.

Alternatively, suppose that for some k, one of the roots is x = - something that makes kx = 0. But x = 0 is not in the domain.

Alternatively, suppose that for some k, one of the roots is x = - something that makes x + 2 = 0. Which is x = -2. Which again requires k = 0.

Therefore, no.

Therefore, the answer is 501.

But let me check with k = 8. For k = 8, the equation is log(8x) = 2 log(x + 2). Which simplifies to log(8x) = log((x + 2)^2). Therefore, 8x = (x + 2)^2. Which is x^2 + 4x + 4 - 8x = x^2 - 4x + 4 = 0 => (x - 2)^2 = 0 => x = 2. Which is valid because x = 2 > 0, and kx = 16 > 0, and x + 2 = 4 > 0. Therefore, valid.

Therefore, yes, k = 8 is valid.

Therefore, the answer is 501.

But wait, let me check if there are other k where the equation has exactly one solution. For example, suppose that the quadratic equation has two real roots, but one of them is invalid. For example, if one of the roots is x = -2. But that would require k = 0.

Alternatively, suppose that one of the roots is x = - something that makes kx = 0. But x = 0 is not in the domain.

Alternatively, suppose that one of the roots is x = - something that makes x + 2 = 0. Which is x = -2. Which again requires k = 0.

Therefore, no.

Therefore, the answer is 501.

But wait, let me check if k = - something could result in the quadratic equation having one solution in the domain and the other solution being invalid. For example, if one of the roots is x = -2. But that would require k = 0.

Alternatively, if one of the roots is x = - something that makes kx = 0. But x = 0 is not in the domain.

Therefore, no.

Therefore, the answer is 501.

But wait, the problem says "closed interval [-500, 500]". So k can be -500, -499,..., -1, 0, 1,..., 500. But we are to count integer values of k in that interval. However, k = 0 is excluded. Therefore, the number of negative integers is 500, and k = 8 is one more. Therefore, total 501.

But let me check if there are other k values. For example, suppose that k = - something where the quadratic equation has one solution in the domain and the other solution is invalid. But we already considered that.

Therefore, I think the answer is 501.

But wait, let me check with k = - something. For example, k = -1. Then quadratic equation is x^2 + 5x + 4 = 0. Which factors as (x + 1)(x + 4) = 0. Therefore, roots at x = -1 and x = -4. x = -1 is in (-2, 0), x = -4 is not. Therefore, exactly one solution.

Similarly, for k = -2, quadratic equation is x^2 + 6x + 4 = 0. Roots are [-6 ± sqrt(36 - 16)] / 2 = [-6 ± sqrt(20)] / 2 = -3 ± sqrt(5). sqrt(5) ≈ 2.236. So roots ≈ -3 + 2.236 ≈ -0.764 and -3 - 2.236 ≈ -5.236. So one in domain, one not.

Therefore, yes.

Therefore, conclusion: 501.

But wait, the problem says "closed interval [-500, 500]". So k can be -500, -499,..., -1, 0, 1,..., 500. So the number of integers from -500 to 500 inclusive is 1001 numbers. But we are to exclude k = 0, so 1000 numbers. But we are to count those k which are negative (500 numbers) and k = 8 (1 number). Therefore, total 501.

Yes.

Therefore, the answer is \boxed{501}.

**Final Answer**
\boxed{501}

To solve the problem, we analyze the equation:

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

---

### Step 1: Domain Considerations

For the logarithms to be defined, the arguments must be positive:

- $ kx > 0 $
- $ x + 2 > 0 \Rightarrow x > -2 $

So the domain of the equation is:

- If $ k > 0 $, then $ x > 0 $
- If $ k < 0 $, then $ x < 0 $, but since $ x > -2 $, we have $ -2 < x < 0 $
- $ k = 0 $ is invalid because $ \log(0) $ is undefined

---

### Step 2: Transform the Equation

Using logarithmic identities:

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

This implies:

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

Expanding and rearranging:

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

This is a quadratic equation in $ x $, and we are interested in the number of **real solutions** that lie within the domain.

---

### Step 3: Analyze the Quadratic Equation

Let’s denote the quadratic as:

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

The discriminant is:

$$
D = (4 - k)^2 - 16 = k^2 - 8k
$$

- If $ D = 0 $, the equation has one real root (a double root)
- If $ D > 0 $, the equation has two real roots
- If $ D < 0 $, the equation has no real roots

We are interested in values of $ k $ such that the equation has **exactly one real solution** that lies in the domain.

---

### Step 4: Case Analysis

#### **Case 1: $ k < 0 $**

- Then $ x \in (-2, 0) $
- The quadratic has two real roots (since $ D = k^2 - 8k > 0 $ for $ k < 0 $)
- The product of the roots is 4 (positive), and the sum is $ k - 4 $ (negative)
- Therefore, both roots are negative
- One root lies in $ (-2, 0) $, the other is less than $ -2 $
- So **exactly one solution** is valid

✅ **All negative integers $ k \in [-500, -1] $** satisfy the condition

#### **Case 2: $ k = 0 $**

- Invalid, since $ \log(0) $ is undefined

#### **Case 3: $ 0 < k < 8 $**

- Then $ D = k^2 - 8k < 0 $, so the equation has **no real roots**
- Therefore, **no solutions**

#### **Case 4: $ k = 8 $**

- $ D = 0 $, so the equation has one real root
- The quadratic becomes $ x^2 - 4x + 4 = 0 \Rightarrow (x - 2)^2 = 0 \Rightarrow x = 2 $
- $ x = 2 > 0 $, and $ kx = 16 > 0 $, so valid
- So **exactly one solution**

✅ **$ k = 8 $** is valid

#### **Case 5: $ k > 8 $**

- $ D > 0 $, so two real roots
- Both roots are positive (since product is 4 and sum is $ k - 4 > 0 $)
- So both roots are in the domain
- Therefore, **two solutions**

❌ **Not valid**

---

### Step 5: Count Valid $ k $

- Negative integers from $ -500 $ to $ -1 $: $ 500 $ values
- $ k = 8 $: 1 value

$$
\text{Total valid } k = 500 + 1 = \boxed{501}
$$

---

### 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.