Question

If $f:R\to R^{+}$ then $f\left(x\right)={\left(\frac{1}{3}\right)}^x$, then $f^{-1}\left(x\right)=$

• A. ${\left(\frac{1}{3}\right)}^{-x}$
• B. $3^x$
• C. ${\log }_{1/3}$$x$
• D. ${\log }_x\left(\frac{1}{3}\right)$

Answer 1

Answer: (C)

${\log }_{1/3}$$x$

Let $f\left(x\right)={\left(\frac{1}{3}\right)}^x=y$
Taking logarithm on both sides,
$x\log \frac{1}{3}=\log y$
$\Rightarrow x=\frac{\log y}{\log \frac{1}{3}}$
$\Rightarrow x={\log }_{1/3}y[\because \frac{\log b}{\log a}={\log }_{a}b]$
$\therefore f^{-1}\left(x\right)={\log }_{1/3}x$