If A = log _2 log _2 log _4 256 + 2 log _{sqr | vedclass

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(C) Given $A = \log _2 \log _2 \log _4 256 + 2 \log _{\sqrt{2}} 2$First,simplify $\log _4 256 = \log _4 (4^4) = 4 \log _4 4 = 4 	imes 1 = 4$.Next,sim

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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,simplify $\log _4 256 = \log _4 (4^4) = 4 \log _4 4 = 4 	imes 1 = 4$.Next,simplify $2 \log _{\sqrt{2}} 2 = 2 \log _{2^{1/2}} 2 = 2 	imes \frac{1}{1/2} \log _2 2 = 2 	imes 2 	imes 1 = 4$.Substitute these values back into the expression for $A$:$A = \log _2 \log _2 (4) + 4$Since $\log _2 4 = \log _2 (2^2) = 2$,we have:$A = \log _2 (2) + 4$Since $\log _2 2 = 1$,we get:$A = 1 + 4 = 5$.

If $A = \log _2 \log _2 \log _4 256 + 2 \log _{\sqrt{2}} 2$,then $A$ is equal to

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