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The solution is eluding me completely.

My attempt at a solution: Realize that$ y=e^{\ln(y)}$. Thus $e^x - e^{-x} = e^{\ln(y)}$ But this gets me nowhere.

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    $\begingroup$ With some practice MathJax doesn't take any longer than trying to make things readable without. $\endgroup$ Mar 25, 2016 at 23:07
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    $\begingroup$ Hint. Set $e^x=w$ and you get $w^2-2yw-1=0$, which is quadratic. $\endgroup$ Mar 25, 2016 at 23:07
  • $\begingroup$ do you mean $y=\frac{e^x-e^{-x}}{2}$? $\endgroup$ Mar 25, 2016 at 23:13
  • $\begingroup$ If you have a half-way decent scientific calculator, hit the keys for inverse hyperbolic sine... $\endgroup$ Mar 25, 2016 at 23:14
  • $\begingroup$ "$\ y=e^{\ln x}\ $" is false. $\endgroup$
    – user228113
    Mar 25, 2016 at 23:30

2 Answers 2

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multply by $e^x$ to give $$e^{2x} - 1 = 2ye^x \implies (e^x)^2 - 2y (e^x) - 1 = 0 \implies e^x = \frac{2y \pm \sqrt{4y^2 +4}}{2} = y \pm \sqrt{y^2+1}.$$ Since $e^x > 0$, we need to take the positive half of $\pm$. Then $$x = \log\left(y + \sqrt{y^2 + 1} \right).$$

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    $\begingroup$ a/k/a $ \ \sinh^{-1} y \ $ ... $\endgroup$ Mar 25, 2016 at 23:25
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    $\begingroup$ Yes of course. I assume the OP is asking for an alternative expression for $\text{arcsinh}$. That is, I think the answer $$\frac{e^{x} - e^{-x}}{2} = y \,\,\, \implies \,\,\, \sinh(x) = y \,\,\, \implies \,\,\, y = \text{arcsinh}(x)$$ is not the desired response. $\endgroup$
    – User8128
    Mar 25, 2016 at 23:27
  • $\begingroup$ i wasn't offering a criticism of any sort -- it isn't clear to me what the context of the post is. I just thought I would mention that this inversion is so widely used that it is a named function. $\endgroup$ Mar 25, 2016 at 23:29
  • $\begingroup$ Sure. I agree: it is worth it to point that out. $\endgroup$
    – User8128
    Mar 25, 2016 at 23:30
  • $\begingroup$ @User8128 Nice derivation of $\operatorname{arcsinh}$! $\endgroup$
    – flawr
    Mar 25, 2016 at 23:41
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Nikolaus provided an excellent hint.

Let w=$e^x$. Thus,

$w + 1/w = y$

Multiplying both sides by w, we get:

$w^2 + 1 = yw$

Thus,

$w^2 - yw + 1 = 0$ $w^2 - yw = -1$ $w^2 - yw + y^2 / 4 = -1 + y^2 / 4$

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  • $\begingroup$ There are two large mistakes here. When you multiply by $w$, you should get $w^2 + 1 = yw$ not $w^2 +2w = yw$. Next, even if the preceding work was correct, $w^2 = (2y-2)w$ does not imply $w = \pm (2y-2)$. Indeed, $w^2 = (2y-2)w$ implies that $w = 0$ or $w = 2y-2$. $\endgroup$
    – User8128
    Mar 26, 2016 at 1:33

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