If $\left(\log _5 x\right)^2+\log _5 x<2$, then x belong to:
Answer & explanation
Correct answer: option 1
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.