Question

If $f:R^{+}\to R$ such that $f\left(x\right)={\log }_{2}x,$ then $f^{-1}\left(x\right)=$

• A. ${\log }_{x}2$
• B. $2^x$
• C. $2^{-x}$
• D. doesn't exist

Answer 1

Answer: (B)

$2^x$

$f\left(x\right)={\log }_{2}x$

$y={\log }_{2}x$.

$2^y=x$.

$x=2^y$

$\therefore f^{-1}\left(x\right)=2^x$,

If Domain is $R$, then

Range is $R^{+}$

Hence $f^{-1}\left(x\right)=2^x$.