Consider the function y = log_{a}(x + sqrt{x^ | vedclass

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(C) Given,$y = \log_{a}(x + \sqrt{x^{2} + 1})$,$a > 0, a 
eq 1$. Taking exponential on both sides,we get $a^{y} = x + \sqrt{x^{2} + 1}$. Now,consider

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is $x = \sinh(y \log a)$

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(C) Given,$y = \log_{a}(x + \sqrt{x^{2} + 1})$,$a > 0, a 
eq 1$. Taking exponential on both sides,we get $a^{y} = x + \sqrt{x^{2} + 1}$. Now,consider $a^{-y} = \frac{1}{x + \sqrt{x^{2} + 1}}$. Rationalizing the denominator,we get $a^{-y} = \sqrt{x^{2} + 1} - x$. Subtracting the two equations: $a^{y} - a^{-y} = (x + \sqrt{x^{2} + 1}) - (\sqrt{x^{2} + 1} - x) = 2x$. Thus,$x = \frac{a^{y} - a^{-y}}{2}$. Since $a^{y} = e^{y \ln a}$,we have $x = \frac{e^{y \ln a} - e^{-y \ln a}}{2}$. Using the definition $\sinh(u) = \frac{e^{u} - e^{-u}}{2}$,we get $x = \sinh(y \ln a)$. Therefore,the inverse function is $f^{-1}(y) = \sinh(y \log a)$.

Consider the function $y = \log_{a}(x + \sqrt{x^{2} + 1})$ where $a > 0$ and $a \neq 1$. The inverse of the function:

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