The value of left(left(log _2 9right)^2right) | vedclass

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Question

(B) Let the expression be $E = \left(\left(\log _2 9ight)^2ight)^{\frac{1}{\log _2\left(\log _2 9ight)}} 	imes(\sqrt{7})^{\frac{1}{\log _4 7}}$

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let the expression be $E = \left(\left(\log _2 9ight)^2ight)^{\frac{1}{\log _2\left(\log _2 9ight)}} 	imes(\sqrt{7})^{\frac{1}{\log _4 7}}$.Using the property $\frac{1}{\log_a b} = \log_b a$,the first term becomes $\left(\left(\log _2 9ight)^2ight)^{\log_{\log_2 9} 2} = (\log_2 9)^{2 \log_{\log_2 9} 2} = (\log_2 9)^{\log_{\log_2 9} 2^2} = 2^2 = 4$.For the second term,$\frac{1}{\log_4 7} = \log_7 4$. Thus,$(\sqrt{7})^{\log_7 4} = (7^{1/2})^{\log_7 4} = 7^{\frac{1}{2} \log_7 4} = 7^{\log_7 4^{1/2}} = 4^{1/2} = 2$.Therefore,$E = 4 	imes 2 = 8$.

The value of $\left(\left(\log _2 9\right)^2\right)^{\frac{1}{\log _2\left(\log _2 9\right)}} \times(\sqrt{7})^{\frac{1}{\log _4 7}}$ is . . . . . . .

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