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, the equation reduces to kx = (x + 2)^2, but only when x > -2 and kx > 0. Let me write that down:

Case 1: kx = (x + 2)^2, with x > -2 and kx > 0.

So, solving this equation will give me potential solutions, but I need to check if they satisfy the domain conditions. Also, the problem states that the equation should have exactly one real solution. Therefore, I need to find k such that the equation kx = (x + 2)^2 has exactly one solution in the domain x > -2 and kx > 0.

Let me first solve the equation kx = (x + 2)^2. Let's rearrange it:

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

Now, for this quadratic equation to have exactly one real solution, the discriminant must be zero. However, wait, but the problem says that the original equation has exactly one real solution. However, even if the quadratic equation has two solutions, maybe one of them is invalid due to domain restrictions. Therefore, maybe there are two cases: either the quadratic equation has exactly one solution (discriminant zero) which is valid, or the quadratic equation has two solutions, but one of them is invalid (due to domain restrictions), so that only one valid solution exists. Therefore, I need to consider both possibilities.

So, first, let me 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.

Case 1: D = 0. Then, k^2 - 8k = 0 => k(k - 8) = 0 => k = 0 or k = 8.

But wait, let me check if these values of k are valid. Let me check for k = 0 first. If k = 0, then the original equation becomes log(0 * x) = 2 log(x + 2). But log(0) is undefined, so k = 0 is invalid. Therefore, k = 0 is not allowed. So, only k = 8 is possible here. Let me check k = 8. Then, 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: x > -2 and kx > 0. For k = 8, x = 2. Then, 8*2 = 16 > 0, and x = 2 > -2. So, x = 2 is valid. Therefore, k = 8 gives exactly one solution. So, k = 8 is valid.

Case 2: D > 0. Then, quadratic equation has two real roots. However, maybe one of them is invalid. Therefore, we need to find k such that exactly one of the roots satisfies the domain conditions. So, for this case, we need to analyze when exactly one of the roots is valid.

Alternatively, maybe there are other cases where even though the quadratic equation has two roots, both are invalid, but that would mean no solutions. But since the problem requires exactly one real solution, we need to exclude those k where both roots are invalid or both are valid. Wait, but if D > 0, then there are two roots. If both roots are valid, then the equation would have two solutions. If one is valid and one is invalid, then exactly one solution. If both are invalid, then no solution. Therefore, we need to find k such that exactly one of the roots is valid. So, for D > 0, we need to find k such that exactly one of the roots satisfies x > -2 and kx > 0.

Alternatively, maybe there are other cases where even if D = 0, but the root is invalid? But we already considered D = 0 and found that k = 8 is valid. For k = 0, it's invalid. So, maybe only k = 8 from D = 0.

Therefore, the total number of k is the number of k from D = 0 (which is 1, k = 8) plus the number of k where D > 0 and exactly one root is valid. So, need to compute that.

So, first, let me handle the case when D > 0. So, k^2 - 8k > 0 => k(k - 8) > 0. Therefore, this inequality holds when k < 0 or k > 8. So, for k < 0 or k > 8, the quadratic equation has two real roots. Now, for each such k, we need to check how many roots satisfy the domain conditions.

So, let me first analyze the roots of the quadratic equation x^2 + (4 - k)x + 4 = 0. Let me denote the roots as x1 and x2. Let me recall that for a quadratic equation ax^2 + bx + c = 0, the sum of roots is -b/a and product is c/a. Here, a = 1, b = (4 - k), c = 4. Therefore, sum of roots x1 + x2 = -(4 - k) = k - 4, and product x1 * x2 = 4.

So, the product of the roots is 4. Therefore, both roots are either positive or both negative? Wait, product is positive, so they have the same sign. Therefore, if both roots are positive, then their product is positive. If both roots are negative, their product is positive. However, since the product is 4, which is positive, so both roots are positive or both negative. However, we have domain conditions: x > -2 and kx > 0. Let me think.

But first, let me note that x > -2. So, even if the roots are negative, they need to be greater than -2. Also, kx > 0. So, depending on the sign of k, the sign of x must be such that kx > 0.

So, let me consider two cases: k < 0 and k > 8.

Case 2a: k < 0. Then, since k is negative, kx > 0 implies x < 0. Therefore, x must be negative. Also, x > -2. Therefore, the domain for x is -2 < x < 0. Also, since the product of the roots is 4, which is positive, and the sum of the roots is k - 4. Since k < 0, then k - 4 < -4, so the sum of the roots is negative. Therefore, both roots are negative (since product is positive and sum is negative). Therefore, both roots are negative. However, we need them to be in (-2, 0). Therefore, we need to check if both roots are in (-2, 0), or only one of them is.

But since the product is 4, which is positive, and both roots are negative. Let me think. Let me denote the roots as x1 and x2, both negative. Let me suppose that x1 and x2 are both in (-2, 0). Then, their product is 4. But if both are in (-2, 0), then their product is positive, but let me check the maximum possible product. Let me think of two numbers between -2 and 0. Let me take x1 = -a, x2 = -b, where a, b > 0, and a < 2, b < 2. Then, product is ab = 4. But if a and b are both less than 2, then ab < 4. Therefore, ab = 4 is impossible. Therefore, there are no two numbers in (-2, 0) whose product is 4. Therefore, if both roots are negative, they cannot be in (-2, 0). Therefore, for k < 0, the roots are both negative, but since their product is 4, they cannot be in (-2, 0). Therefore, there are no solutions in the domain. Wait, but that seems contradictory. Wait, if the roots are negative, but not in (-2, 0), then they are either less than or equal to -2 or greater than or equal to 0. But since they are negative, they are either less than or equal to -2 or between -2 and 0. But we saw that if they are between -2 and 0, their product can't be 4. Therefore, the roots must be less than or equal to -2. Therefore, for k < 0, the roots are both less than or equal to -2. Therefore, x <= -2. However, the domain requires x > -2. Therefore, the roots are not in the domain. Therefore, for k < 0, there are no solutions? Wait, but that can't be. Let me check with an example. Let me take k = -1. Then, the quadratic equation becomes x^2 + (4 - (-1))x + 4 = x^2 + 5x + 4 = 0. Which factors as (x + 1)(x + 4) = 0. Therefore, roots at x = -1 and x = -4. Now, check domain conditions. For k = -1, x must satisfy x > -2 and kx > 0. Since k = -1 < 0, kx > 0 implies x < 0. Therefore, x must be in (-2, 0). The roots are x = -1 and x = -4. x = -1 is in (-2, 0), so valid. x = -4 is not in (-2, 0). Therefore, only x = -1 is valid. Therefore, for k = -1, there is exactly one solution. Wait, so my previous conclusion was wrong. So, even though the product of the roots is 4, which is positive, and the sum is k - 4 = -1 - 4 = -5. So, sum is -5. Therefore, the roots are -1 and -4. So, one of them is in (-2, 0), the other is not. Therefore, for k < 0, there is exactly one root in (-2, 0) and one root less than -2. Therefore, only one solution. Therefore, for k < 0, there is exactly one solution. Wait, but why did my previous reasoning say that the product can't be 4? Because if one root is in (-2, 0) and the other is less than -2, then their product is positive. For example, (-1) * (-4) = 4. So, that's possible. Therefore, my mistake was assuming that both roots have to be in (-2, 0), but actually, one can be in (-2, 0) and the other outside. Therefore, for k < 0, there is exactly one solution. Therefore, for all k < 0, the equation has exactly one solution? Wait, but let me check another value. Let me take k = -8. Then, quadratic equation is x^2 + (4 - (-8))x + 4 = x^2 + 12x + 4 = 0. The roots are [-12 ± sqrt(144 - 16)] / 2 = [-12 ± sqrt(128)] / 2 = [-12 ± 8*sqrt(2)] / 2 = -6 ± 4*sqrt(2). Let me approximate 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, 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 solution. Therefore, for k < 0, there is exactly one solution. Therefore, for all k < 0, the equation has exactly one solution. Therefore, all negative integers k in [-500, 500] would satisfy the condition? Wait, but need to check if the root in (-2, 0) is valid. Wait, for k < 0, the root in (-2, 0) is valid because x > -2 and kx > 0. So, yes. Therefore, for all k < 0, there is exactly one solution. Therefore, the number of integer k in [-500, 500] with k < 0 is from -500 to -1 inclusive. How many integers is that? From -500 to -1, that's 500 numbers. Wait, because from -1 to -500, inclusive, there are 500 numbers. So, 500 values. But wait, the interval is [-500, 500], so k can be -500, -499,..., -1, 0, 1,..., 500. So, the number of negative integers is 500 (from -500 to -1 inclusive). However, we need to check if k = 0 is included. But k = 0 is invalid, as we saw earlier. So, for k < 0, there are 500 values. Then, for k > 8, we need to check if the quadratic equation has exactly one solution. Wait, let me check.

Case 2b: k > 8. Then, discriminant D = k^2 - 8k > 0. So, quadratic equation has two real roots. Now, for k > 8, since k is positive, kx > 0 implies x > 0. Also, domain requires x > -2, which is automatically satisfied if x > 0. Therefore, the domain for x is x > 0. Also, the product of the roots is 4, and the sum of the roots is k - 4. Since k > 8, sum of roots is k - 4 > 8 - 4 = 4. Therefore, sum of roots is positive. Since product is positive, both roots are positive. Therefore, both roots are positive. Therefore, both roots satisfy x > 0, hence x > -2. Therefore, both roots are valid. Therefore, for k > 8, the equation has two solutions. Therefore, we need to exclude these k. However, the problem requires exactly one solution. Therefore, for k > 8, there are two solutions, so they are invalid. Therefore, only k = 8 from D = 0 and k < 0. Wait, but wait, let me check with an example. Let me take k = 9. Then, quadratic equation is x^2 + (4 - 9)x + 4 = x^2 -5x + 4 = 0. Which factors as (x - 1)(x - 4) = 0. Therefore, roots at x = 1 and x = 4. Both are positive, so both satisfy x > -2 and kx > 0 (since k = 9 > 0, x > 0). Therefore, two solutions. Therefore, k = 9 is invalid. Therefore, for k > 8, two solutions. Therefore, only k = 8 from D = 0 and k < 0. Wait, but earlier, for k < 0, there is exactly one solution. Therefore, the total number of k is the number of negative integers in [-500, 500] plus k = 8. Wait, but let me check if there are other k where the quadratic equation has two roots, but one is invalid. Wait, for k > 8, both roots are valid. For k < 0, one root is valid. For k = 0, invalid. For k between 0 and 8, discriminant D = k^2 - 8k. Let me check for k between 0 and 8, D is negative or zero. Wait, D = k^2 - 8k. For k between 0 and 8, D is negative except at k = 0 and k = 8. Therefore, for 0 < k < 8, D < 0, so no real roots. Therefore, equation has no solution. Therefore, for 0 < k < 8, no solution. For k = 0, invalid. For k = 8, one solution. For k > 8, two solutions. For k < 0, one solution. Therefore, the total number of k is the number of negative integers in [-500, 500] plus k = 8. Therefore, how many negative integers are there? From -500 to -1 inclusive. That is 500 numbers. Then, k = 8 is another one. Therefore, total number is 500 + 1 = 501. However, wait, need to check if k = 8 is within the interval [-500, 500]. Yes, 8 is between -500 and 500. Therefore, total number is 501. But wait, let me check if there are other k where the equation has exactly one solution. Wait, for example, when k = 0, invalid. For k between 0 and 8, no solution. For k = 8, one solution. For k < 0, one solution. For k > 8, two solutions. Therefore, total k is 500 (negative) + 1 (k=8) = 501. However, wait, but let me check if there are any other k where the equation has exactly one solution. For example, suppose that for some k, even though D > 0, but one of the roots is invalid. Wait, but for k < 0, we saw that one root is valid and one is invalid. For k > 8, both roots are valid. For k between 0 and 8, no real roots. For k = 0, invalid. For k = 8, one root. Therefore, only k < 0 and k = 8. Therefore, total number is 500 + 1 = 501. However, let me check if there are any other cases where the equation could have exactly one solution. For example, suppose that one of the roots is exactly at x = -2. But x = -2 is not allowed because log(x + 2) would be log(0), which is undefined. Therefore, x = -2 is excluded. Also, if one of the roots is x = 0, but then kx = 0, which is invalid because log(0) is undefined. Therefore, x = 0 is excluded. Therefore, no other cases. Therefore, the answer is 501. However, let me check if k = -500 is allowed. The interval is [-500, 500], so yes. So, from -500 to -1 inclusive, that's 500 numbers. Then, k = 8. So, total 501. However, wait, let me check if k = -500 is allowed. For k = -500, the equation log(-500x) = 2 log(x + 2). The domain requires that -500x > 0 and x + 2 > 0. So, -500x > 0 implies x < 0. And x + 2 > 0 implies x > -2. Therefore, x is in (-2, 0). Then, solving the quadratic equation 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 roots. But one of them is in (-2, 0) and the other is not. Therefore, exactly one solution. Therefore, k = -500 is valid. Similarly, all k from -500 to -1 inclusive are valid. Therefore, 500 values. Then, k = 8 is another. Therefore, total 501. However, 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 those k where the equation has exactly one solution. Which are k < 0 (500 numbers) and k = 8 (1 number). Therefore, total 501. However, let me check if k = 8 is the only one in the positive side. Wait, what about k = 0? No, k = 0 is invalid. What about k = 8? Yes, only k = 8. Wait, but let me check if there are other k where the quadratic equation has exactly one solution. For example, suppose that one of the roots is invalid, but the other is valid. Wait, but for k < 0, we already considered that. For k > 8, both roots are valid. For k between 0 and 8, no real roots. For k = 0, invalid. For k = 8, one root. Therefore, only k = 8. Therefore, total 501. However, let me check if there is a case where the quadratic equation has two roots, but one of them is invalid. Wait, for example, suppose that for some k, one root is x = -2, but that would make log(x + 2) undefined. But x = -2 is not in the domain. Similarly, if one root is x = 0, which would make log(kx) undefined. Therefore, no. Therefore, only k < 0 and k = 8. Therefore, answer is 501. However, let me check if k = 8 is the only one in positive numbers. Wait, suppose k = something else. For example, suppose that the quadratic equation has one root in the domain and one outside. Wait, but for k > 8, both roots are in the domain. For k between 0 and 8, no real roots. For k = 0, invalid. For k < 0, one root in domain. Therefore, only k = 8. Therefore, total 501. However, let me check if there is a mistake here. For example, suppose 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, exactly one solution. Therefore, k = -2 is valid. Similarly, for k = -1, as before. Therefore, all negative k are valid. Therefore, 500 negative integers. Then, k = 8. So, total 501. However, let me check if k = 8 is the only positive integer. Wait, suppose k = something else. For example, suppose k = 4. Then, discriminant D = 16 - 32 = -16 < 0. No solution. For k = 9, two solutions. For k = 8, one solution. For k = 7, D = 49 - 56 = -7 < 0. No solution. Therefore, only k = 8. Therefore, answer is 501. However, let me check if k = -500 is included. The interval is [-500, 500], so yes. Therefore, the number of integer values is 500 (negative) + 1 (k=8) = 501. However, wait, the problem says "closed interval [-500, 500]". So, k can be -500, -499,..., 0,..., 500. So, the number of negative integers is from -500 to -1 inclusive. How many numbers is that? From -1 to -500, that's 500 numbers. So, yes. Then, adding k = 8, total 501. Therefore, the answer is 501. However, let me check once more. Wait, but maybe there is a mistake here. For example, when k = -2, the equation log(-2x) = 2 log(x + 2). Let me check if x = -3 + sqrt(5) ≈ -0.764 is a solution. Let me compute left side: log(-2x) = log(-2*(-0.764)) = log(1.528). Right side: 2 log(x + 2) = 2 log(-0.764 + 2) = 2 log(1.236). Let me compute log(1.528) and 2 log(1.236). Since log(1.528) ≈ 0.184 and 2 log(1.236) ≈ 2 * 0.091 ≈ 0.182. Close enough, considering rounding errors. Therefore, it works. Therefore, k = -2 is valid. Therefore, seems correct. Therefore, answer is 501. However, let me check if k = 8 is the only positive integer. Suppose k = 8, which gives x = 2. Let me check: log(8*2) = log(16) and 2 log(2 + 2) = 2 log(4) = 2 * 1.386... = 2.772..., and log(16) is also 1.204... Wait, wait, no. Wait, log(16) is log(2^4) = 4 log(2) ≈ 4 * 0.3010 ≈ 1.204. 2 log(4) = 2 * log(2^2) = 2 * 2 log(2) = 4 log(2) ≈ 1.204. Therefore, they are equal. Therefore, correct. Therefore, k = 8 is valid. Therefore, answer is 501. However, let me check if there are other k where the equation has exactly one solution. For example, suppose that the quadratic equation has one root in the domain and one outside, but for some k > 8? Wait, for k > 8, both roots are positive. Therefore, both are in the domain. Therefore, two solutions. Therefore, no. For k between 0 and 8, no solution. For k = 0, invalid. Therefore, only k = 8. Therefore, answer is 501. However, let me check if k = -500 is allowed. The problem says "closed interval [-500, 500]". So, k = -500 is allowed. Therefore, yes. Therefore, total number is 500 + 1 = 501. Therefore, the answer is \boxed{501}. Wait, but wait, hold on. Wait, the problem says "integer values of k in the closed interval [-500, 500]". So, the interval includes -500 and 500. However, when I count the number of negative integers from -500 to -1 inclusive, that is 500 numbers. Then, k = 8 is another. However, what about k = 0? It's excluded. So, total is 500 + 1 = 501. However, let me check if there is a mistake in the count. For example, from -500 to -1 inclusive, how many integers? The number of integers from a to b inclusive is b - a + 1. So, from -500 to -1: (-1) - (-500) + 1 = 499 + 1 = 500. Yes, that's correct. Therefore, 500 numbers. Then, adding k = 8, total 501. Therefore, answer is 501. However, let me check if there is a mistake in considering k = 8. For example, suppose k = 8, then the equation has exactly one solution. But what if there is another k where the equation has exactly one solution? For example, suppose that the quadratic equation has two roots, but one of them is invalid. Wait, but for k > 8, both roots are valid. For k < 0, one root is valid. For k between 0 and 8, no solution. For k = 0, invalid. For k = 8, one solution. Therefore, only k = 8. Therefore, answer is 501. Therefore, I think that is correct. Therefore, the number of integer values is 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:

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

This implies that $ x $ must be in the interval $ (-2, \infty) $, and the sign of $ x $ must match the sign of $ k $ (since $ kx > 0 $).

---

### 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 satisfy the domain conditions.

---

### Step 3: Analyze the Quadratic Equation

Let’s denote the quadratic as:

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

The discriminant is:

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

We consider different cases based on the value of $ D $:

---

### Case 1: $ D = 0 $

This occurs when:

$$
k^2 - 8k = 0 \Rightarrow k(k - 8) = 0 \Rightarrow k = 0 \text{ or } k = 8
$$

- **For $ k = 0 $:** The equation becomes $ \log(0) = 2\log(x + 2) $, which is undefined. So, **invalid**.
- **For $ k = 8 $:** The quadratic becomes $ x^2 - 4x + 4 = 0 \Rightarrow (x - 2)^2 = 0 \Rightarrow x = 2 $

Check domain: $ x = 2 > -2 $, and $ kx = 8 \cdot 2 = 16 > 0 $. So, **valid**.

✅ **Valid solution: $ k = 8 $**

---

### Case 2: $ D > 0 $

This occurs when:

$$
k(k - 8) > 0 \Rightarrow k < 0 \text{ or } k > 8
$$

#### Subcase 2a: $ k < 0 $

- $ kx > 0 \Rightarrow x < 0 $
- Domain: $ x > -2 \Rightarrow x \in (-2, 0) $

Let’s analyze the roots of the quadratic:

- The product of the roots is $ 4 $, so both roots are negative.
- The sum of the roots is $ k - 4 $, which is negative (since $ k < 0 $).
- Therefore, one root lies in $ (-2, 0) $, and the other lies outside the domain.

✅ **Exactly one valid solution in this case**

#### Subcase 2b: $ k > 8 $

- $ kx > 0 \Rightarrow x > 0 $
- Domain: $ x > -2 $, so all $ x > 0 $ are valid.

The product of the roots is $ 4 $, and the sum is $ k - 4 > 4 $, so both roots are positive.

✅ **Two valid solutions in this case**

So, for $ k > 8 $, the equation has **two** real solutions, which is **not** what we want.

---

### Case 3: $ D < 0 $

This occurs when:

$$
k(k - 8) < 0 \Rightarrow 0 < k < 8
$$

In this case, the quadratic has **no real roots**, so the equation has **no solutions**.

---

### Case 4: $ k = 0 $

As discussed earlier, this leads to an undefined logarithm. So, **invalid**.

---

### Final Count

We now count the number of integer values of $ k $ in the interval $[-500, 500]$ that satisfy the condition of having **exactly one real solution**.

- **Negative integers $ k < 0 $:** From $ -500 $ to $ -1 $, inclusive → 500 values
- **$ k = 8 $:** 1 value

So, the total number of such $ k $ is:

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

---

### 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, the equation reduces to kx = (x + 2)^2, but only when x > -2 and kx > 0. Let me write that down:

Case 1: kx = (x + 2)^2, with x > -2 and kx > 0.

So, solving this equation will give me potential solutions, but I need to check if they satisfy the domain conditions. Also, the problem states that the equation should have exactly one real solution. Therefore, I need to find k such that the equation kx = (x + 2)^2 has exactly one solution in the domain x > -2 and kx > 0.

Let me first solve the equation kx = (x + 2)^2. Let's rearrange it:

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

Now, for this quadratic equation to have exactly one real solution, the discriminant must be zero. However, wait, but the problem says that the original equation has exactly one real solution. However, even if the quadratic equation has two solutions, maybe one of them is invalid due to domain restrictions. Therefore, maybe there are two cases: either the quadratic equation has exactly one solution (discriminant zero) which is valid, or the quadratic equation has two solutions, but one of them is invalid (due to domain restrictions), so that only one valid solution exists. Therefore, I need to consider both possibilities.

So, first, let me 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.

Case 1: D = 0. Then, k^2 - 8k = 0 => k(k - 8) = 0 => k = 0 or k = 8.

But wait, let me check if these values of k are valid. Let me check for k = 0 first. If k = 0, then the original equation becomes log(0 * x) = 2 log(x + 2). But log(0) is undefined, so k = 0 is invalid. Therefore, k = 0 is not allowed. So, only k = 8 is possible here. Let me check k = 8. Then, 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: x > -2 and kx > 0. For k = 8, x = 2. Then, 8*2 = 16 > 0, and x = 2 > -2. So, x = 2 is valid. Therefore, k = 8 gives exactly one solution. So, k = 8 is valid.

Case 2: D > 0. Then, quadratic equation has two real roots. However, maybe one of them is invalid. Therefore, we need to find k such that exactly one of the roots satisfies the domain conditions. So, for this case, we need to analyze when exactly one of the roots is valid.

Alternatively, maybe there are other cases where even though the quadratic equation has two roots, both are invalid, but that would mean no solutions. But since the problem requires exactly one real solution, we need to exclude those k where both roots are invalid or both are valid. Wait, but if D > 0, then there are two roots. If both roots are valid, then the equation would have two solutions. If one is valid and one is invalid, then exactly one solution. If both are invalid, then no solution. Therefore, we need to find k such that exactly one of the roots is valid. So, for D > 0, we need to find k such that exactly one of the roots satisfies x > -2 and kx > 0.

Alternatively, maybe there are other cases where even if D = 0, but the root is invalid? But we already considered D = 0 and found that k = 8 is valid. For k = 0, it's invalid. So, maybe only k = 8 from D = 0.

Therefore, the total number of k is the number of k from D = 0 (which is 1, k = 8) plus the number of k where D > 0 and exactly one root is valid. So, need to compute that.

So, first, let me handle the case when D > 0. So, k^2 - 8k > 0 => k(k - 8) > 0. Therefore, this inequality holds when k < 0 or k > 8. So, for k < 0 or k > 8, the quadratic equation has two real roots. Now, for each such k, we need to check how many roots satisfy the domain conditions.

So, let me first analyze the roots of the quadratic equation x^2 + (4 - k)x + 4 = 0. Let me denote the roots as x1 and x2. Let me recall that for a quadratic equation ax^2 + bx + c = 0, the sum of roots is -b/a and product is c/a. Here, a = 1, b = (4 - k), c = 4. Therefore, sum of roots x1 + x2 = -(4 - k) = k - 4, and product x1 * x2 = 4.

So, the product of the roots is 4. Therefore, both roots are either positive or both negative? Wait, product is positive, so they have the same sign. Therefore, if both roots are positive, then their product is positive. If both roots are negative, their product is positive. However, since the product is 4, which is positive, so both roots are positive or both negative. However, we have domain conditions: x > -2 and kx > 0. Let me think.

But first, let me note that x > -2. So, even if the roots are negative, they need to be greater than -2. Also, kx > 0. So, depending on the sign of k, the sign of x must be such that kx > 0.

So, let me consider two cases: k < 0 and k > 8.

Case 2a: k < 0. Then, since k is negative, kx > 0 implies x < 0. Therefore, x must be negative. Also, x > -2. Therefore, the domain for x is -2 < x < 0. Also, since the product of the roots is 4, which is positive, and the sum of the roots is k - 4. Since k < 0, then k - 4 < -4, so the sum of the roots is negative. Therefore, both roots are negative (since product is positive and sum is negative). Therefore, both roots are negative. However, we need them to be in (-2, 0). Therefore, we need to check if both roots are in (-2, 0), or only one of them is.

But since the product is 4, which is positive, and both roots are negative. Let me think. Let me denote the roots as x1 and x2, both negative. Let me suppose that x1 and x2 are both in (-2, 0). Then, their product is 4. But if both are in (-2, 0), then their product is positive, but let me check the maximum possible product. Let me think of two numbers between -2 and 0. Let me take x1 = -a, x2 = -b, where a, b > 0, and a < 2, b < 2. Then, product is ab = 4. But if a and b are both less than 2, then ab < 4. Therefore, ab = 4 is impossible. Therefore, there are no two numbers in (-2, 0) whose product is 4. Therefore, if both roots are negative, they cannot be in (-2, 0). Therefore, for k < 0, the roots are both negative, but since their product is 4, they cannot be in (-2, 0). Therefore, there are no solutions in the domain. Wait, but that seems contradictory. Wait, if the roots are negative, but not in (-2, 0), then they are either less than or equal to -2 or greater than or equal to 0. But since they are negative, they are either less than or equal to -2 or between -2 and 0. But we saw that if they are between -2 and 0, their product can't be 4. Therefore, the roots must be less than or equal to -2. Therefore, for k < 0, the roots are both less than or equal to -2. Therefore, x <= -2. However, the domain requires x > -2. Therefore, the roots are not in the domain. Therefore, for k < 0, there are no solutions? Wait, but that can't be. Let me check with an example. Let me take k = -1. Then, the quadratic equation becomes x^2 + (4 - (-1))x + 4 = x^2 + 5x + 4 = 0. Which factors as (x + 1)(x + 4) = 0. Therefore, roots at x = -1 and x = -4. Now, check domain conditions. For k = -1, x must satisfy x > -2 and kx > 0. Since k = -1 < 0, kx > 0 implies x < 0. Therefore, x must be in (-2, 0). The roots are x = -1 and x = -4. x = -1 is in (-2, 0), so valid. x = -4 is not in (-2, 0). Therefore, only x = -1 is valid. Therefore, for k = -1, there is exactly one solution. Wait, so my previous conclusion was wrong. So, even though the product of the roots is 4, which is positive, and the sum is k - 4 = -1 - 4 = -5. So, sum is -5. Therefore, the roots are -1 and -4. So, one of them is in (-2, 0), the other is not. Therefore, for k < 0, there is exactly one root in (-2, 0) and one root less than -2. Therefore, only one solution. Therefore, for k < 0, there is exactly one solution. Wait, but why did my previous reasoning say that the product can't be 4? Because if one root is in (-2, 0) and the other is less than -2, then their product is positive. For example, (-1) * (-4) = 4. So, that's possible. Therefore, my mistake was assuming that both roots have to be in (-2, 0), but actually, one can be in (-2, 0) and the other outside. Therefore, for k < 0, there is exactly one solution. Therefore, for all k < 0, the equation has exactly one solution? Wait, but let me check another value. Let me take k = -8. Then, quadratic equation is x^2 + (4 - (-8))x + 4 = x^2 + 12x + 4 = 0. The roots are [-12 ± sqrt(144 - 16)] / 2 = [-12 ± sqrt(128)] / 2 = [-12 ± 8*sqrt(2)] / 2 = -6 ± 4*sqrt(2). Let me approximate 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, 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 solution. Therefore, for k < 0, there is exactly one solution. Therefore, for all k < 0, the equation has exactly one solution. Therefore, all negative integers k in [-500, 500] would satisfy the condition? Wait, but need to check if the root in (-2, 0) is valid. Wait, for k < 0, the root in (-2, 0) is valid because x > -2 and kx > 0. So, yes. Therefore, for all k < 0, there is exactly one solution. Therefore, the number of integer k in [-500, 500] with k < 0 is from -500 to -1 inclusive. How many integers is that? From -500 to -1, that's 500 numbers. Wait, because from -1 to -500, inclusive, there are 500 numbers. So, 500 values. But wait, the interval is [-500, 500], so k can be -500, -499,..., -1, 0, 1,..., 500. So, the number of negative integers is 500 (from -500 to -1 inclusive). However, we need to check if k = 0 is included. But k = 0 is invalid, as we saw earlier. So, for k < 0, there are 500 values. Then, for k > 8, we need to check if the quadratic equation has exactly one solution. Wait, let me check.

Case 2b: k > 8. Then, discriminant D = k^2 - 8k > 0. So, quadratic equation has two real roots. Now, for k > 8, since k is positive, kx > 0 implies x > 0. Also, domain requires x > -2, which is automatically satisfied if x > 0. Therefore, the domain for x is x > 0. Also, the product of the roots is 4, and the sum of the roots is k - 4. Since k > 8, sum of roots is k - 4 > 8 - 4 = 4. Therefore, sum of roots is positive. Since product is positive, both roots are positive. Therefore, both roots are positive. Therefore, both roots satisfy x > 0, hence x > -2. Therefore, both roots are valid. Therefore, for k > 8, the equation has two solutions. Therefore, we need to exclude these k. However, the problem requires exactly one solution. Therefore, for k > 8, there are two solutions, so they are invalid. Therefore, only k = 8 from D = 0 and k < 0. Wait, but wait, let me check with an example. Let me take k = 9. Then, quadratic equation is x^2 + (4 - 9)x + 4 = x^2 -5x + 4 = 0. Which factors as (x - 1)(x - 4) = 0. Therefore, roots at x = 1 and x = 4. Both are positive, so both satisfy x > -2 and kx > 0 (since k = 9 > 0, x > 0). Therefore, two solutions. Therefore, k = 9 is invalid. Therefore, for k > 8, two solutions. Therefore, only k = 8 from D = 0 and k < 0. Wait, but earlier, for k < 0, there is exactly one solution. Therefore, the total number of k is the number of negative integers in [-500, 500] plus k = 8. Wait, but let me check if there are other k where the quadratic equation has two roots, but one is invalid. Wait, for k > 8, both roots are valid. For k < 0, one root is valid. For k = 0, invalid. For k between 0 and 8, discriminant D = k^2 - 8k. Let me check for k between 0 and 8, D is negative or zero. Wait, D = k^2 - 8k. For k between 0 and 8, D is negative except at k = 0 and k = 8. Therefore, for 0 < k < 8, D < 0, so no real roots. Therefore, equation has no solution. Therefore, for 0 < k < 8, no solution. For k = 0, invalid. For k = 8, one solution. For k > 8, two solutions. For k < 0, one solution. Therefore, the total number of k is the number of negative integers in [-500, 500] plus k = 8. Therefore, how many negative integers are there? From -500 to -1 inclusive. That is 500 numbers. Then, k = 8 is another one. Therefore, total number is 500 + 1 = 501. However, wait, need to check if k = 8 is within the interval [-500, 500]. Yes, 8 is between -500 and 500. Therefore, total number is 501. But wait, let me check if there are other k where the equation has exactly one solution. Wait, for example, when k = 0, invalid. For k between 0 and 8, no solution. For k = 8, one solution. For k < 0, one solution. For k > 8, two solutions. Therefore, total k is 500 (negative) + 1 (k=8) = 501. However, wait, but let me check if there are any other k where the equation has exactly one solution. For example, suppose that for some k, even though D > 0, but one of the roots is invalid. Wait, but for k < 0, we saw that one root is valid and one is invalid. For k > 8, both roots are valid. For k between 0 and 8, no real roots. For k = 0, invalid. For k = 8, one root. Therefore, only k < 0 and k = 8. Therefore, total number is 500 + 1 = 501. However, let me check if there are any other cases where the equation could have exactly one solution. For example, suppose that one of the roots is exactly at x = -2. But x = -2 is not allowed because log(x + 2) would be log(0), which is undefined. Therefore, x = -2 is excluded. Also, if one of the roots is x = 0, but then kx = 0, which is invalid because log(0) is undefined. Therefore, x = 0 is excluded. Therefore, no other cases. Therefore, the answer is 501. However, let me check if k = -500 is allowed. The interval is [-500, 500], so yes. So, from -500 to -1 inclusive, that's 500 numbers. Then, k = 8. So, total 501. However, wait, let me check if k = -500 is allowed. For k = -500, the equation log(-500x) = 2 log(x + 2). The domain requires that -500x > 0 and x + 2 > 0. So, -500x > 0 implies x < 0. And x + 2 > 0 implies x > -2. Therefore, x is in (-2, 0). Then, solving the quadratic equation 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 roots. But one of them is in (-2, 0) and the other is not. Therefore, exactly one solution. Therefore, k = -500 is valid. Similarly, all k from -500 to -1 inclusive are valid. Therefore, 500 values. Then, k = 8 is another. Therefore, total 501. However, 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 those k where the equation has exactly one solution. Which are k < 0 (500 numbers) and k = 8 (1 number). Therefore, total 501. However, let me check if k = 8 is the only one in the positive side. Wait, what about k = 0? No, k = 0 is invalid. What about k = 8? Yes, only k = 8. Wait, but let me check if there are other k where the quadratic equation has exactly one solution. For example, suppose that one of the roots is invalid, but the other is valid. Wait, but for k < 0, we already considered that. For k > 8, both roots are valid. For k between 0 and 8, no real roots. For k = 0, invalid. For k = 8, one root. Therefore, only k = 8. Therefore, total 501. However, let me check if there is a case where the quadratic equation has two roots, but one of them is invalid. Wait, for example, suppose that for some k, one root is x = -2, but that would make log(x + 2) undefined. But x = -2 is not in the domain. Similarly, if one root is x = 0, which would make log(kx) undefined. Therefore, no. Therefore, only k < 0 and k = 8. Therefore, answer is 501. However, let me check if k = 8 is the only one in positive numbers. Wait, suppose k = something else. For example, suppose that the quadratic equation has one root in the domain and one outside. Wait, but for k > 8, both roots are in the domain. For k between 0 and 8, no real roots. For k = 0, invalid. For k < 0, one root in domain. Therefore, only k = 8. Therefore, total 501. However, let me check if there is a mistake here. For example, suppose 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, exactly one solution. Therefore, k = -2 is valid. Similarly, for k = -1, as before. Therefore, all negative k are valid. Therefore, 500 negative integers. Then, k = 8. So, total 501. However, let me check if k = 8 is the only positive integer. Wait, suppose k = something else. For example, suppose k = 4. Then, discriminant D = 16 - 32 = -16 < 0. No solution. For k = 9, two solutions. For k = 8, one solution. For k = 7, D = 49 - 56 = -7 < 0. No solution. Therefore, only k = 8. Therefore, answer is 501. However, let me check if k = -500 is included. The interval is [-500, 500], so yes. Therefore, the number of integer values is 500 (negative) + 1 (k=8) = 501. However, wait, the problem says "closed interval [-500, 500]". So, k can be -500, -499,..., 0,..., 500. So, the number of negative integers is from -500 to -1 inclusive. How many numbers is that? From -1 to -500, that's 500 numbers. So, yes. Then, adding k = 8, total 501. Therefore, the answer is 501. However, let me check once more. Wait, but maybe there is a mistake here. For example, when k = -2, the equation log(-2x) = 2 log(x + 2). Let me check if x = -3 + sqrt(5) ≈ -0.764 is a solution. Let me compute left side: log(-2x) = log(-2*(-0.764)) = log(1.528). Right side: 2 log(x + 2) = 2 log(-0.764 + 2) = 2 log(1.236). Let me compute log(1.528) and 2 log(1.236). Since log(1.528) ≈ 0.184 and 2 log(1.236) ≈ 2 * 0.091 ≈ 0.182. Close enough, considering rounding errors. Therefore, it works. Therefore, k = -2 is valid. Therefore, seems correct. Therefore, answer is 501. However, let me check if k = 8 is the only positive integer. Suppose k = 8, which gives x = 2. Let me check: log(8*2) = log(16) and 2 log(2 + 2) = 2 log(4) = 2 * 1.386... = 2.772..., and log(16) is also 1.204... Wait, wait, no. Wait, log(16) is log(2^4) = 4 log(2) ≈ 4 * 0.3010 ≈ 1.204. 2 log(4) = 2 * log(2^2) = 2 * 2 log(2) = 4 log(2) ≈ 1.204. Therefore, they are equal. Therefore, correct. Therefore, k = 8 is valid. Therefore, answer is 501. However, let me check if there are other k where the equation has exactly one solution. For example, suppose that the quadratic equation has one root in the domain and one outside, but for some k > 8? Wait, for k > 8, both roots are positive. Therefore, both are in the domain. Therefore, two solutions. Therefore, no. For k between 0 and 8, no solution. For k = 0, invalid. Therefore, only k = 8. Therefore, answer is 501. However, let me check if k = -500 is allowed. The problem says "closed interval [-500, 500]". So, k = -500 is allowed. Therefore, yes. Therefore, total number is 500 + 1 = 501. Therefore, the answer is \boxed{501}. Wait, but wait, hold on. Wait, the problem says "integer values of k in the closed interval [-500, 500]". So, the interval includes -500 and 500. However, when I count the number of negative integers from -500 to -1 inclusive, that is 500 numbers. Then, k = 8 is another. However, what about k = 0? It's excluded. So, total is 500 + 1 = 501. However, let me check if there is a mistake in the count. For example, from -500 to -1 inclusive, how many integers? The number of integers from a to b inclusive is b - a + 1. So, from -500 to -1: (-1) - (-500) + 1 = 499 + 1 = 500. Yes, that's correct. Therefore, 500 numbers. Then, adding k = 8, total 501. Therefore, answer is 501. However, let me check if there is a mistake in considering k = 8. For example, suppose k = 8, then the equation has exactly one solution. But what if there is another k where the equation has exactly one solution? For example, suppose that the quadratic equation has two roots, but one of them is invalid. Wait, but for k > 8, both roots are valid. For k < 0, one root is valid. For k between 0 and 8, no solution. For k = 0, invalid. For k = 8, one solution. Therefore, only k = 8. Therefore, answer is 501. Therefore, I think that is correct. Therefore, the number of integer values is 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:

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

This implies that $ x $ must be in the interval $ (-2, \infty) $, and the sign of $ x $ must match the sign of $ k $ (since $ kx > 0 $).

---

### 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 satisfy the domain conditions.

---

### Step 3: Analyze the Quadratic Equation

Let’s denote the quadratic as:

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

The discriminant is:

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

We consider different cases based on the value of $ D $:

---

### Case 1: $ D = 0 $

This occurs when:

$$
k^2 - 8k = 0 \Rightarrow k(k - 8) = 0 \Rightarrow k = 0 \text{ or } k = 8
$$

- **For $ k = 0 $:** The equation becomes $ \log(0) = 2\log(x + 2) $, which is undefined. So, **invalid**.
- **For $ k = 8 $:** The quadratic becomes $ x^2 - 4x + 4 = 0 \Rightarrow (x - 2)^2 = 0 \Rightarrow x = 2 $

Check domain: $ x = 2 > -2 $, and $ kx = 8 \cdot 2 = 16 > 0 $. So, **valid**.

✅ **Valid solution: $ k = 8 $**

---

### Case 2: $ D > 0 $

This occurs when:

$$
k(k - 8) > 0 \Rightarrow k < 0 \text{ or } k > 8
$$

#### Subcase 2a: $ k < 0 $

- $ kx > 0 \Rightarrow x < 0 $
- Domain: $ x > -2 \Rightarrow x \in (-2, 0) $

Let’s analyze the roots of the quadratic:

- The product of the roots is $ 4 $, so both roots are negative.
- The sum of the roots is $ k - 4 $, which is negative (since $ k < 0 $).
- Therefore, one root lies in $ (-2, 0) $, and the other lies outside the domain.

✅ **Exactly one valid solution in this case**

#### Subcase 2b: $ k > 8 $

- $ kx > 0 \Rightarrow x > 0 $
- Domain: $ x > -2 $, so all $ x > 0 $ are valid.

The product of the roots is $ 4 $, and the sum is $ k - 4 > 4 $, so both roots are positive.

✅ **Two valid solutions in this case**

So, for $ k > 8 $, the equation has **two** real solutions, which is **not** what we want.

---

### Case 3: $ D < 0 $

This occurs when:

$$
k(k - 8) < 0 \Rightarrow 0 < k < 8
$$

In this case, the quadratic has **no real roots**, so the equation has **no solutions**.

---

### Case 4: $ k = 0 $

As discussed earlier, this leads to an undefined logarithm. So, **invalid**.

---

### Final Count

We now count the number of integer values of $ k $ in the interval $[-500, 500]$ that satisfy the condition of having **exactly one real solution**.

- **Negative integers $ k < 0 $:** From $ -500 $ to $ -1 $, inclusive → 500 values
- **$ k = 8 $:** 1 value

So, the total number of such $ k $ 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.