Question

Let $f:[4,\infty )\to [4,\infty )$ be a function defined by $f\left(x\right)=5^{x(x-4)}$, then $f^{-1}\left(x\right)$ is

• A. $2-\sqrt{4+{\log }_{5}x}$
• B. $2+\sqrt{4+{\log }_{5}x}$
• C. ${\left(\frac{1}{5}\right)}^{x(x-4)}$
• D. None of these

Answer 1

Answer: (B)

$2+\sqrt{4+{\log }_{5}x}$

Let $y=5^{x(x-4)}\Rightarrow x(x-4)={\log }_{5}y\Rightarrow x^2-4x-{\log }_{5}y=0$

$\Rightarrow x=\frac{4\pm \sqrt{16+4{\log }_{5}y}}{2}=(2\pm \sqrt{4+{\log }_{5}y})$

But $x\ge 4$, so $x=(2+\sqrt{4+{\log }_{5}y})$

$\therefore f^{-1}\left(y\right)=2+\sqrt{4+{\log }_{5}y}$

$\therefore f^{-1}\left(x\right)=2+\sqrt{4+{\log }_{5}x}$