Question

The solution set of the inequality ${ 5^{x + 2} } > \left( \frac{1}{25} \right)^{ {1 /x}}$ is

• A. (-2, 0)
• B. (-2, 2)
• C. (-5, 5)
• D. (0, $\infty$)

Answer 1

Answer: (D)

(0, $\infty$)

Answer 2

We have ${ 5^{x + 2} } > \bigg( \frac{1}{25} \bigg)^{ {1 /x}} \Rightarrow { 5^{x + 2}} > 5^{- \frac{2}{x}} \Rightarrow \, {x + 2 > \- \frac{2}{x}}$ ${ [ \because \, base \, 5 > 1]}$ ${\Rightarrow x + 2 + \frac{2}{x} > 0 \Rightarrow \frac{x^{2} + 2x + 2}{x}> 0 \Rightarrow \frac{1}{x} > 0 }$ $[\because \: { x^2 + 2x + 2 > 0 \, \forall \, x \, \in R }]$ ${ \Rightarrow \, x > 0 \: or \: x \, \in ( 0, \infty)}$