Question

Consider the function $y={\log }_a(x+\sqrt{x^2+1}),a>0,a\ne 1.$ The inverse of the function

• A. does not exist
• B. is $x={\log }_{1/a}(y+\sqrt{y^2+1})$
• C. is $x=\sinh \left(y\ln a\right)$
• D. is $x=\cosh (-y\ln \frac{1}{a})$

Answer 1

The correct option is C is $x=\sinh \left(y\ln a\right)$

$y={\log }_a(x+\sqrt{x^2+1})$
$\Rightarrow a^y=(x+\sqrt{x^2+1})\Rightarrow a^{-y}=\sqrt{x^2+1}-x\Rightarrow a^y-a^{-y}=2x\Rightarrow f^{-1}\left(y\right)=x=\frac{a^y-a^{-y}}{2}\Rightarrow f^{-1}\left(y\right)=\frac{e^{y\ln a}-e^{-y\ln a}}{2}=\sinh \left(y\ln a\right)$

$(\because \sinh (x)=\frac{e^x-e^{-x}}{2})$