Question

The inverse of the function $y=\frac{10^x-10^{-x}}{10^x+10^{-x}}$ is

• A. $lo{g}_{10}(2-x)$
• B. $\frac{1}{2}lo{g}_{10}\left(\frac{1+x}{1-x}\right)$
• C. $\frac{1}{2}lo{g}_{10}\left(2x-1\right)$
• D. $\frac{1}{4}log\left(\frac{2x}{2-x}\right)$

Answer 1

Answer: (B)

Since, $y=\frac{10^{2x}-1}{10^{2x}+1}$
$\Rightarrow y10^{2x}+y=10^{2x}-1$
$\Rightarrow 10^{2x}=\frac{1+y}{1-y}$
$\therefore 2xlo{g}_{10}10=lo{g}_{10}\frac{1+y}{1-y}$
$\Rightarrow x=\frac{1}{2}lo{g}_{10}\left(\frac{1+y}{1-y}\right)\left(\because lo{g}_{a}a=1\right)$
Hence, the inverse of $y$ is $\frac{1}{2}lo{g}_{10}\left(\frac{1+x}{1-x}\right)$.