Search any question & find its solution
Question:
Answered & Verified by Expert
The set $\left\{x \in \mathbb{R}: 16\left(2^X\right)>16^{\frac{-1}{x}}\right\}=$
Options:
Solution:
2252 Upvotes
Verified Answer
The correct answer is:
$\{\mathrm{x} \in \mathbb{R}: \mathrm{x}>0\}$
$\begin{aligned}
& \text {} 16\left(2^x\right)>16^{-\frac{1}{x}} \Rightarrow 2^{x+4}>2^{-4 / x} \\
& \Rightarrow x+4>-4 / x \Rightarrow x^2+4 x+4>0 \\
& \Rightarrow(x+2)^2>0
\end{aligned}$
which is true for all $x \in R$
& \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 $x \in R$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.