Question

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

• A. $\left(\frac{1}{2}\right)\left(a^x-a^{-x}\right)$
• B. not defined for all $x$
• C. defined for only positive $x$
• D. $\frac{a^{2x}-1}{2a^x}$

Answer 1

The correct option is D

$\frac{a^{2x}-1}{2a^x}$

Explanation for the correct option:

The inverse of a function is a function which outputs the input of the function given its result.

Given: $y={\log }_a\left(x+\sqrt{x^2+1}\right)\left(a\ne 1,a>0\right)$

$\begin{array}{cc}\Rightarrow & y={\log }_a\left(x+\sqrt{x^2+1}\right)\\ \Rightarrow & a^y=\left(x+\sqrt{x^2+1}\right)\\ \Rightarrow & a^y-x=\sqrt{x^2+1}\end{array}$

squaring both sides

$\begin{array}{cc}\Rightarrow & {\left(a^y-x\right)}^2=x^2+1\\ \Rightarrow & a^{2y}+x^2-2a^{y}x=x^2+1\\ \Rightarrow & a^{2y}-2a^{y}x=1\\ \Rightarrow & a^{2y}-1=2a^{y}x\\ \Rightarrow & \frac{a^{2y}-1}{2a^y}=x\end{array}$

Interchanging $y$ and $x$

$y=\frac{a^{2x}-1}{2a^x}$

Therefore, $y=\frac{a^{2x}-1}{2a^x}$ is the inverse of $y={\log }_a\left(x+\sqrt{x^2+1}\right)$.

Hence, option D is correct.