Okay, let's see. I need to find the value of 'a' given that the cubic equation 8x³ + 4a x² + 2b x + a = 0 has three distinct positive roots, and the sum of the base-2 logarithms of the roots is 5. Hmm, logarithms... right, log base 2 of the roots summed up equals 5. 

First, I remember that for polynomials, there's something called Vieta's formulas which relates the coefficients of the polynomial to sums and products of its roots. Maybe that can help here. Let me recall Vieta's formulas for a cubic equation. For a general cubic equation like px³ + qx² + rx + s = 0, the sum of the roots is -q/p, the sum of the product of roots two at a time is r/p, and the product of the roots is -s/p. 

So in this case, the equation is 8x³ + 4a x² + 2b x + a = 0. Let me write down the coefficients:

Leading coefficient (p) = 8

Next coefficient (q) = 4a

Next (r) = 2b

Constant term (s) = a

Therefore, according to Vieta's formulas:

Sum of roots: -q/p = -4a/8 = -a/2

Sum of products two at a time: r/p = 2b/8 = b/4

Product of roots: -s/p = -a/8

But the problem states that all three roots are distinct and positive. So each root is positive. Therefore, the sum of the roots is positive, but according to Vieta, it's -a/2. Wait, that would mean that -a/2 is positive, so a must be negative. Because if -a/2 is positive, then a must be negative. Let me note that: a < 0.

Also, the product of the roots is -a/8. Since all roots are positive, their product is positive. Therefore, -a/8 must be positive. Which again implies that a is negative. So consistent with the previous conclusion. 

Now, the problem says the sum of the base-2 logarithms of the roots is 5. Let's denote the roots as r₁, r₂, r₃. Then log₂(r₁) + log₂(r₂) + log₂(r₃) = 5. 

Using logarithm properties, log₂(r₁) + log₂(r₂) + log₂(r₃) = log₂(r₁ r₂ r₃). Therefore, log₂(r₁ r₂ r₃) = 5. Which implies that r₁ r₂ r₃ = 2⁵ = 32.

But from Vieta's formula, the product of the roots is -a/8. Therefore, -a/8 = 32. Solving for a: -a = 32 * 8 = 256, so a = -256. 

Wait, is that all? Let me check. 

Given that the product of the roots is 32, which equals -a/8. Therefore, solving for a gives a = -8 * 32 = -256. 

But hold on, the problem states that the equation has three distinct positive roots. So we need to ensure that this value of 'a' doesn't lead to any multiple roots or negative roots. But since we derived the product of roots as 32, and the product is positive, and each root is positive, so their product is positive. Also, since all roots are positive, the coefficients should satisfy certain conditions. 

Wait, but let me think again. The sum of the roots is -a/2. If a is -256, then the sum is -(-256)/2 = 256/2 = 128. So the sum of the roots is 128. The sum of the products two at a time is b/4. But we don't know the value of b. 

But the problem doesn't mention anything about the sum of the roots or the sum of the products. So maybe we don't need to find 'b'. Since the problem only asks for 'a', and we can get 'a' directly from the product of the roots. 

But let me verify if this is correct. Let me go through the steps again.

Given the cubic equation: 8x³ + 4a x² + 2b x + a = 0

Let roots be r₁, r₂, r₃. All positive and distinct.

From Vieta:

r₁ + r₂ + r₃ = -4a/8 = -a/2

r₁ r₂ + r₁ r₃ + r₂ r₃ = 2b/8 = b/4

r₁ r₂ r₃ = -a/8

Sum of logs:

log₂ r₁ + log₂ r₂ + log₂ r₃ = log₂(r₁ r₂ r₃) = 5 ⇒ r₁ r₂ r₃ = 2⁵ = 32

Thus, -a/8 = 32 ⇒ a = -8 * 32 = -256

Therefore, a = -256. 

But wait, is there any possibility that this value of 'a' could lead to multiple roots or negative roots? Let's check the discriminant of the cubic. If the discriminant is positive, then all roots are real and distinct. However, calculating the discriminant for a cubic is a bit involved. Let me recall the discriminant formula for a cubic equation ax³ + bx² + cx + d = 0 is Δ = 18abcd - 4b³d + b²c² - 4ac³ - 27a²d². 

But in our case, the equation is 8x³ + 4a x² + 2b x + a = 0. So coefficients are:

A = 8, B = 4a, C = 2b, D = a

Therefore, discriminant Δ = 18*A*B*C*D - 4*B³*D + B²*C² - 4*A*C³ - 27*A²*D²

Plugging in:

Δ = 18*8*4a*2b*a - 4*(4a)³*a + (4a)²*(2b)² - 4*8*(2b)³ - 27*(8)²*a²

This seems complicated. Let's compute each term step by step.

First term: 18*8*4a*2b*a = 18*8*4*2*a²*b = 18*8*8*a²*b = 18*64*a²*b = 1152 a² b

Second term: -4*(4a)³*a = -4*64a³*a = -4*64a⁴ = -256a⁴

Third term: (4a)²*(2b)² = 16a²*4b² = 64a²b²

Fourth term: -4*8*(2b)³ = -32*(8b³) = -256b³

Fifth term: -27*(8)²*a² = -27*64*a² = -1728a²

Therefore, putting it all together:

Δ = 1152a²b - 256a⁴ + 64a²b² - 256b³ - 1728a²

This discriminant must be positive for the cubic to have three distinct real roots. However, since we don't know the value of 'b', it's hard to compute directly. But perhaps we can find a relationship between 'a' and 'b' from other given information. Wait, is there any other equation we can use?

We know the sum of the roots is 128 (from a = -256, so sum r₁ + r₂ + r₃ = 128). The sum of products two at a time is b/4. But without knowing 'b', we can't directly compute the discriminant. 

Alternatively, maybe we can use the fact that all roots are positive. If all roots are positive, then by Descartes' Rule of Signs, the number of positive roots is equal to the number of sign changes in the polynomial or less by a multiple of 2. Let's check the original polynomial: 8x³ + 4a x² + 2b x + a. Since a is negative, the constant term is negative. Let's substitute x with positive values. The coefficients are: 8, 4a, 2b, a. Since a is negative, 4a is negative, and a is negative. So the signs are: +, -, +, -. The sign changes: from + to -, then - to +, then + to -. So three sign changes. Therefore, there are three positive roots or one positive root. But the problem states there are three distinct positive roots, so the number of positive roots is three. So Descartes' Rule of Signs is satisfied.

But that doesn't necessarily help us with the discriminant. However, maybe the discriminant is positive? Let's see. If a = -256, then we can write the discriminant in terms of 'b' and check if it's positive.

But without knowing 'b', we can't compute the discriminant numerically. However, maybe there's another way. Since all roots are positive, we can use some inequalities related to symmetric sums. For example, using AM ≥ GM.

Given that the product of the roots is 32, and the sum is 128. The AM of the roots is 128/3 ≈ 42.67, and the GM is (32)^(1/3) ≈ 3.17. Since AM ≥ GM, which is true here. So that's consistent. Also, for three positive numbers, the discriminant of the cubic must be positive, so the cubic has three distinct real roots. Therefore, as long as the discriminant is positive, which we can assume here, because the problem states that there are three distinct positive roots. So maybe we don't need to check the discriminant.

Therefore, with a = -256, the product of the roots is 32, which satisfies the condition on the logarithms, and the sum of the roots is 128, which is positive. The other coefficients, like the sum of the products two at a time (b/4), we don't need to find because the problem doesn't mention anything about the sum of the products. So, unless there's a mistake in my reasoning, a should be -256.

Wait a minute, let me check the product of the roots again. The product is -a/8. So if a = -256, then product is -(-256)/8 = 256/8 = 32. Yes, that's correct. So log₂(32) is 5, which matches the given condition. So that seems right.

Therefore, the answer should be a = -256. 

But let me check once more for any possible mistakes. Maybe I missed a step or misapplied Vieta's formulas. Let's verify Vieta's formulas again.

Given the cubic equation 8x³ + 4a x² + 2b x + a = 0. Let's write the general form: px³ + qx² + rx + s = 0. Then:

Sum of roots: -q/p

Sum of products two at a time: r/p

Product of roots: -s/p

So here, q = 4a, r = 2b, s = a, p = 8.

Therefore:

Sum of roots: -4a/8 = -a/2

Sum of products two at a time: 2b/8 = b/4

Product of roots: -a/8

So that's correct.

Therefore, the product of roots is indeed -a/8. Given that the product is 32, so -a/8 = 32 ⇒ a = -256. 

So, unless there's a miscalculation here, the answer is -256.

But just to be thorough, let's think if there's another possible value of 'a'. Suppose someone thought that the sum of the logs is 5, so the product is 32, but maybe they thought the sum of the roots is 5? No, the problem says the sum of the base-2 logarithms, not the sum of the roots. So that's not an issue.

Alternatively, maybe someone could have confused the coefficients. For example, if they thought the coefficient of x² is 4a, but in Vieta's formula, the sum of roots is -q/p, which is -4a/8 = -a/2. So that's correct. 

Alternatively, if someone thought the product is a/8 instead of -a/8, but that would be incorrect. Vieta's formula for product of roots is -s/p, which in this case is -a/8. Since the roots are positive, their product is positive, so -a/8 must be positive. Therefore, a is negative, which we already considered.

Therefore, I think the answer is indeed -256.

**Final Answer**
The value of \( a \) is \boxed{-256}.
Therefore, the value of $a$ is $\boxed{-256}$.
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