The value of ((log₂9)²)/(log₂(log₂9)) × (√7)^(1/log₄7) is ________?

((log₂9)²)/(log₂(log₂9)) × (√7)^(1/log₄7) = (log₂9)²/log₂(log₂9) × 7^(1/(2log₄7))
= (log₂9)²/log(log₂9)₂ × 7^(1/(2(log₂7/log₂4)))
= (log₂9)²/log₂(log₂9) × 7^(log₂4/(2log₂7))
= (log₂9)²/log₂(log₂9) × 7^(2log₂2/(2log₂7))
= (log₂9)²/log₂(log₂9) × 7^(log₂2/log₂7)
= (log₂9)²/log₂(log₂9) × 7^(log₂7⁻¹)
= (log₂9)²/log₂(log₂9) × 7^(log₂(7⁻¹))
Let x = log₂9. Then we have:
= x²/log₂x × 7^(log₂(1/7))
Using the change of base rule: log₂(1/7) = log(1/7)/log2
= x²/log₂x × (7^log₂(1/7))
Since a^(logₐb) = b, we have 7^(log₇(1/7)) = 1/7. So this calculation is incorrect.
Let's try a different approach.
((log₂9)²)/(log₂(log₂9)) × (√7)^(1/log₄7) = (log₂9)²/log₂(log₂9) × (7^1/2)^(1/(log₂7/log₂4))
= (log₂9)²/log₂(log₂9) × 7^((1/2)/(log₂7/2log₂2))
= (log₂9)²/log₂(log₂9) × 7^(log₂2/log₂7)
= (log₂9)²/log₂(log₂9) × 7^log₇2
This also doesn't seem correct. Let's re-examine the original expression:
((log₂9)²)/(log₂(log₂9)) × (√7)^(1/log₄7)
Let's use the change of base formula: logₐb = logₓb/logₓa
Then 1/log₄7 = log₇4 = 2log₇2
(√7)^(2log₇2) = (7^(1/2))^(2log₇2) = 7^(log₇2) = 2
Now let's consider (log₂9)²/log₂(log₂9)
Let x = log₂9
x²/log₂x = x²/log₂x
This expression is difficult to simplify without numerical approximation. The initial solution is also incorrect.
Let's revisit the calculation:
((log₂9)²)/(log₂(log₂9)) × (√7)^(1/log₄7)
The problem is likely in the provided solution. There seems to be a mistake in the simplification steps. A more rigorous approach, possibly using numerical methods or properties of logarithms beyond what's been attempted here, is required to solve this problem accurately.