CUET Preparation Today
If $\left(\log _5 x\right)^2+\log _5 x<2$, then x belong to: |
$\left(\frac{1}{25}, 5\right)$ $\left(\frac{1}{5}, \frac{1}{\sqrt{5}}\right)$ $(1, \infty)$ none of these |
$\left(\frac{1}{25}, 5\right)$ |
We have $\left(\log _5 x\right)^2+\log _5 x<2$ Put $\log _5 x=a$ then $a^2+a<2$ $\Rightarrow a^2+a-2<0 \Rightarrow(a+2)(a-1)<0$ $\Rightarrow-2<a<1 \text { or }-2<\log _5 x<1$ ∴ $5^{-2}<x<5$ i.e. $1 / 25<x<5$ Hence (1) is the correct answer. |
