Question

The solution of $3^{x+2}>{\left(\frac{1}{9}\right)}^{\frac{1}{x}}$ is

• A. $x\in (0,\infty )$
• B. $x\in (-\infty ,\infty )$
• C. $x\in (1,\infty )$
• D. None of these

Answer 1

The correct option is B

$x\in (-\infty ,\infty )$

Find the range for the given expression

Given, $3^{x+2}>{\left(\frac{1}{9}\right)}^{\frac{1}{x}}$

We can also write as,

$$\begin{gathered} 3^{x+2}>{\left(3^{-2}\right)}^{\frac{1}{x}} \\ \Rightarrow 3^{x+2}>3^{\frac{-2}{x}} \end{gathered}$$

Here, base$3>1$

$$\begin{gathered} \Rightarrow x+2>\frac{-2}{x} \\ \Rightarrow x^2+2x>-2 \\ \Rightarrow x^2+2x+2>0 \end{gathered}$$

Now,

$$\begin{gathered} D=b^2-4ac \\ =2^2-4\left(1\right)\left(2\right) \\ =-4 \end{gathered}$$

As, $D<0$, that means $x\in \mathrm{ℝ}$ or $x\in \left(-\infty ,\infty \right)$

Hence, option (B), $x\in \left(-\infty ,\infty \right)$ is the correct answer.