Question

The values of $x$ satisfying $x^{{\log }_{5}x}>5$ lie in the interval

• A. $(0,\infty )$
• B. $(0,\frac{1}{5})\cup (5,\infty )$
• C. $(1,\infty )$
• D. $(1,2)$

Answer 1

The correct option is B $(0,\frac{1}{5})\cup (5,\infty )$

We have,
$x^{{\log }_{5}x}>5$
Clearly, it is meaningful for $x>0$.

Now, two cases arise:
Case 1: When $x>1$
In this case, we have
$x^{{\log }_{5}x>5}$
$\Rightarrow {\log }_{5}x>{\log }_{x}5[\because a^x>N\Rightarrow x>{\log }_{a}N\text{if}a>1]$
$\Rightarrow {\log }_{5}x>\frac{1}{{\log }_{5}x}$
$\Rightarrow ({\log }_{5}x-1)({\log }_{5}x+1)>0$
$\Rightarrow {\log }_{5}x-1>0\Rightarrow x>5\Rightarrow x\in (5,\infty )$

Case 2: When $0<x<1$
In this case, we have
$x^{{\log }_{5}x}>5$
$\Rightarrow {\log }_{5}x<{\log }_{x}5[\because a^x>N\Rightarrow x<{\log }_{a}N\text{if}0<a<1]$
$\Rightarrow {\log }_{5}x<\frac{1}{{\log }_{5}x}$
$\Rightarrow ({\log }_{5}x{)}^2-1>0(\because {\log }_{5}x<0)$
$\Rightarrow ({\log }_{5}x-1)({\log }_{5}x+1)>0$
$\Rightarrow ({\log }_{5}x+1)<0\Rightarrow x<5^{-1}\Rightarrow x\in (0,\frac{1}{5})$

Hence, $x\in (0,\frac{1}{5})\cup (5,\infty )$