The set $\left\{x \in \mathbb{R}: 16\left(2^X\right)>16^{\frac{-1}{x}}\right\}=$
The set $\left\{x \in \mathbb{R}: 16\left(2^X\right)>16^{\frac{-1}{x}}\right\}=$
- $\{\mathrm{x} \in \mathbb{R}: \mathrm{x}>0\}$
- $\{\mathrm{x} \in \mathbb{R}: \mathrm{x} < 0\}$
- $\mathbb{R}$
- $\{x \in \mathbb{R}: x>2\}$
Solution
$\begin{aligned}
& \text {} 16\left(2^x\right)>16^{-\frac{1}{x}} \Rightarrow 2^{x+4}>2^{-4 / x} \\
& \Rightarrow x+4>-4 / x \Rightarrow x^2+4 x+4>0 \\
& \Rightarrow(x+2)^2>0
\end{aligned}$
which is true for all $x \in R$
Asked in: AP EAMCET 2023 (17 May Shift 1)
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