The set {x in R: 16(2^x) 16^{-1/x}} = | vedclass

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(A) Given the inequality: $16(2^x) > 16^{-1/x}$ \ Since $16 = 2^4$,we can write: $2^4 \cdot 2^x > (2^4)^{-1/x}$ \ $2^{x+4} > 2^{-4/x}$ \ Since the

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$\{x \in R: x > 0\}$

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(A) Given the inequality: $16(2^x) > 16^{-1/x}$ \ Since $16 = 2^4$,we can write: $2^4 \cdot 2^x > (2^4)^{-1/x}$ \ $2^{x+4} > 2^{-4/x}$ \ Since the base $2 > 1$,the inequality holds for the exponents: $x + 4 > -4/x$ \ $x + 4 + 4/x > 0$ \ $\frac{x^2 + 4x + 4}{x} > 0$ \ $\frac{(x+2)^2}{x} > 0$ \ Since $(x+2)^2 > 0$ for all $x 
eq -2$,the inequality holds if $x > 0$ and $x 
eq -2$. \ Thus,the solution set is $\{x \in R: x > 0\}$.

The set $\{x \in R: 16(2^x) > 16^{-1/x}\} = $

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