Find the value of $\left((\log_29)^2\right)^{1/\log_2(\log_29) }\times \left(\sqrt{7}\right)^{1/\log_47}$

Question

The value of$\left((\log_29)^2\right)^{1/\log_2(\log_29) }\times \left(\sqrt{7}\right)^{1/\log_47}$.

My approach :

I am able to solve this part : $(\sqrt{7})^{1/\log_47}$ by changing base :

$(\sqrt{7})^{1/\log_47} = (7)^{2\log_74} = 16$

But I am unable to figure how to solve this part :

$\Bigl((\log_29)^2\Bigr)^{1/\log_2(\log_29)} $

Please help on this … thanks

Answer 1

Your first part is incorrect. $$(\sqrt{7})^{1/\log_47} = (7^{1/2})^{\log_74} = 2$$ Use @00xxqhxx00's hint to solve for the 2nd part. Let $x = \log_29$ $$\Bigl((\log_29)^2\Bigr)^{1/\log_2(\log_29)} = (x^2)^{1/\log_2 x} = (x^2)^{\log_x 2} = 4$$ Hence the required product is $2 \times 4 = 8$.

Answer 2

There's a mistake.

$\begin{align}\sqrt7 ^ {\log_74} = 7^{\frac12\log_74} = 4^\frac12 = 2\end{align}$

Let $\log_29 = k$

So we have $(k^2)^{1/\log_2k} = 2^ 2 = 4$

Answer 3

An important relation is : $$\log_a b= \frac{1}{\log_b a}$$ And, $$a^{\log_a b}=b$$

Your question:$$ \Bigl((\log_2 9 * \log_2 9)^{\frac{1}{\log_2 (\log_2 9)}} \Bigr)* 7^{\frac{1}{2* \log_4 7}} $$ Can be written as: $$\Bigl((\log_2 9)^{\log_{\log_2 9}2}\Bigr)^{2} * \bigl(7^{\frac{\log_7 4}{2}}\bigr)$$ $$(2)^{2}*{2}=8$$

Hence, the answer is 8, using the above mentioned relations.