If A = log_2 log_2 log_4 256 + 2 log_{sqrt{2} | vedclass

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(C) Given $A = \log_2 \log_2 \log_4 256 + 2 \log_{\sqrt{2}} 2$. First,evaluate $\log_4 256$: Since $256 = 4^4$,$\log_4 256 = 4$. Next,evaluate $\log_2

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$5$

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(C) Given $A = \log_2 \log_2 \log_4 256 + 2 \log_{\sqrt{2}} 2$. First,evaluate $\log_4 256$: Since $256 = 4^4$,$\log_4 256 = 4$. Next,evaluate $\log_2 \log_2 4$: Since $\log_2 4 = 2$,we have $\log_2 2 = 1$. Now,evaluate $2 \log_{\sqrt{2}} 2$: Since $\sqrt{2} = 2^{1/2}$,$\log_{2^{1/2}} 2 = \frac{1}{1/2} \log_2 2 = 2 	imes 1 = 2$. Thus,$2 \log_{\sqrt{2}} 2 = 2 	imes 2 = 4$. Finally,$A = 1 + 4 = 5$.

If $A = \log_2 \log_2 \log_4 256 + 2 \log_{\sqrt{2}} 2$,then the value of $A$ is:

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