Question

If ${\left({\log }_{5}x\right)}^2+{\log }_{5}x<2$, then $x$ belong to the interval

• A. $(\frac{1}{25},5)$
• B. $(\frac{1}{5},\frac{1}{\sqrt{5}})$
• C. $(1,\infty )$
• D. None of these

Answer 1

The correct option is A $(\frac{1}{25},5)$

We have, ${\left({\log }_{5}x\right)}^2+{\log }_{5}x<2$
Substitute ${\log }_{5}x=a$ then $a^2+a<2$
$\Rightarrow a^2+a-2<0\Rightarrow (a+2)(a-1)<0$
$\Rightarrow -2<a<1$ or $-2<{\log }_{5}x<1$
$\therefore 5^{-2}<x<5\Rightarrow \frac{1}{25}<x<5.$

Hence option (A) is correct.