Question

If $f:(\frac{1}{2},\infty )\to R$ defined as $f\left(x\right)={\log }_5(2x-1),$ then $f^{-1}\left(x\right)=$

• A. $\frac{1}{2}(5^x+1)$
• B. $2^x+1$
• C. $5^x+1$
• D. $\frac{1}{5}(2^x+1)$

Answer 1

The correct option is A $\frac{1}{2}(5^x+1)$

It can be inferred from the options that $f$ is bijective function.
$y={\log }_5(2x-1)$
$\Rightarrow 5^y=2x-1$
$\Rightarrow 5^y+1=2x$
$\Rightarrow \frac{5^y+1}{2}=x$
Swapping $x$ with $y$,
$\Rightarrow y=\frac{1}{2}(5^x+1)$
$\therefore f^{-1}\left(x\right)=\frac{1}{2}(5^x+1)$