0
$\begingroup$

I need to establish the inverse function of the hyperbolic sine: I am trying to do this by setting $y = \sinh(x)$ and solving for $x$, however I got stuck at this: $$ y=\frac{e^x -e^{-x}}{2} $$ $$ 2y=e^x - e^{-x} $$ I dont know how to solve for x at this point, though. Taking the logarithm seems nonsensical with a sum on the right side. The end goal is the arsinh given by $$ y = \log(x + \sqrt{x^2 +1}) $$

$\endgroup$
  • 1
    $\begingroup$ Note that $e^{-x} = \frac{1}{e^x}$ you have a quadratic equation in $e^x$. $\endgroup$ – user10354138 May 16 at 15:24
1
$\begingroup$

Hint: make replacement $e^x = z$ and use $e^{-x} = 1 / z$ to get quadratic equation on $z$.

$\endgroup$
  • 2
    $\begingroup$ You mean $e^{-x}=1/z$, not $e^{-x}=z$. $\endgroup$ – user10354138 May 16 at 15:25
0
$\begingroup$

Hint: Write $$2y=e^x-\frac{1}{e^x}$$, now substitute $$t=e^x$$ and solve the quadratic.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.