If x^ 7^x > 7 where x > 0 , then which one of the following is...

x^{\log } 7^x > 7

where

x > 0

, then which one of the following is correct?

Options

C. $x \in\left(0, \frac{1}{7}\right) \cup(7, \infty)$

x^{\log _7^x} > 7 ; x > 0

taking log both sides

\Rightarrow \log _7^x \cdot \log x > \log 7

\begin{aligned} & \Rightarrow \frac{\log x}{\log 7} \cdot \log x > \log 7 \\ & \Rightarrow(\log x)^2 > (\log 7)^2 \\ & \Rightarrow(\log x)^2-(\log 7)^2 > 0 \\ & \Rightarrow(\log x+\log 7)(\log x-\log 7) > 0 \\ & \Rightarrow x=\frac{1}{7}, 7\end{aligned}

<br/>

\therefore \quad x \in\left(0, \frac{1}{7}\right) \cup(7, \infty)$