Evaluate Logarithmic Power Expression with Double Logarithms

To find the value of the given expression:

$$E = \left((\log_2 9)^2\right)^{\frac{1}{\log_2 (\log_2 9)}} \times (\sqrt{7})^{\frac{1}{\log_4 7}}$$

We can evaluate the expression by breaking it into two separate parts:

Part 1: Evaluating the first term Let $x = \log_2 9$. The first term becomes:

$$T_1 = \left(x^2\right)^{\frac{1}{\log_2 x}} = x^{\frac{2}{\log_2 x}}$$

Using the base-change property of logarithms, $\frac{1}{\log_b a} = \log_a b$:

$$\frac{1}{\log_2 x} = \log_x 2$$

Substitute this back into $T_1$:

$$T_1 = x^{2\log_x 2} = \left(x^{\log_x 2}\right)^2$$

Using the fundamental logarithmic identity $a^{\log_a b} = b$:

$$T_1 = 2^2 = 4$$

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Part 2: Evaluating the second term Let the second term be $T_2$:

$$T_2 = (\sqrt{7})^{\frac{1}{\log_4 7}} = \left(7^{1/2}\right)^{\frac{1}{\log_4 7}}$$

Using the base-change property $\frac{1}{\log_4 7} = \log_7 4$:

$$T_2 = 7^{\frac{1}{2} \log_7 4} = 7^{\log_7 \left(4^{1/2}\right)} = 7^{\log_7 2}$$

Using the identity $a^{\log_a b} = b$:

$$T_2 = 2$$

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Combining both parts:

$$E = T_1 \times T_2 = 4 \times 2 = 8$$

Final Answer: The value of the expression is $8$ (or $8.00$).