If the function $f:[2,\infty ]\rightarrow [1,\infty ]$ is defined by $f(x)=3^{x(x-2)}$ then $f^{-1}(x)$ is-

$\begin{array}{l}f(x)=3^{x(x-2)} \\ \Rightarrow f(x)=3^{x^{2}-2 x} \\ \Rightarrow f(x)=3^{x^{2}-2 x+1-1} \\ \Rightarrow f(x)=3^{(x-1)^{2}} \cdot 3^{-1}\end{array}$

$\Rightarrow f(x)=\frac{1}{3} 3^{(x-1)^{2}}$

$\begin{aligned} \text { Let,}~ y & =f(x) \\ \Rightarrow x & =f^{-1}(y)\end{aligned}$

$\begin{aligned} & y=\frac{1}{3} 3^{(x-1)^{2}} \\ \Rightarrow & 3 y={3}^{(x-1)^{2}} \\ \Rightarrow & \log _{3}(3 y)=\log _{3} 3^{(x-1)^{2}} \\ \Rightarrow & \log _{3} 3+\log _{3} y=(x-1)^{2}\end{aligned}$

$\begin{array}{l}\Rightarrow 1+\log _{3} y=(x-1)^{2} \\ \Rightarrow(x-1)=\sqrt{1+\log _{3} y} \\ \Rightarrow x=1+\sqrt{1+\log _{3} y}\end{array}$

$\begin{array}{l}\Rightarrow f^{-1}(y)=1+\sqrt{1+\log _{3} y} \\ \Rightarrow f^{-1}(x)=1+\sqrt{1+\log _{3} x}\end{array}$