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Let $x\in\mathbb{R}$, how can $-e^{-x}+e^{x}=a$ be solved? I have already tried to use the sum of exponential formula.

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up vote 8 down vote accepted

The equation is equivalent to $$\sinh(x)=\frac a2\iff x=\operatorname{argsinh}\left(\frac a2\right)$$

Or if you want let $X=e^x$ and solve the quadratic equation


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Well, I think this will work: if $x \in \Bbb R$ satisfies

$-e^{-x} + e^x = a, \tag{1}$

then we can multiply through by $e^x$ to obtain

$-1 + e^{2x} = ae^x, \tag{2}$

and if we set $y = e^x$, then $y^2 = e^{2x}$, so (2) becomes

$-1 + y^2 = ay, \tag{3}$


$y^2 - ay -1 = 0. \tag{4}$.

We solve (4) via the quadratic formula:

$y = \dfrac{1}{2}(a \pm \sqrt{a^2 + 4}), \tag{5}$

and see that not only is the discriminant $a^2 + 4 > 0$, but in addition $\vert a^2 + 4 \vert > \vert a \vert^2$, which shows that exactly one root of (4) is positive, always, that root being

$y = \dfrac{1}{2}(a+ \sqrt{a^2 + 4}). \tag{6}$

Setting $x = \ln y$ with $y$ given by (6) does the trick, as may easily be checked by some simple algebraic maneuvers.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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It is always a pleasure to me to read your last words. They are so true ! Cheers. – Claude Leibovici Apr 6 '14 at 8:40
@Claude Leibovivi: indeed! And thanks! – Robert Lewis Apr 6 '14 at 8:42
@1950RobertLewis very nice. – kaka Apr 6 '14 at 8:49
@kaka: muchas gracias! – Robert Lewis Apr 6 '14 at 8:50
The simpler the better. Thanks. – Manolito Pérez Aug 13 '15 at 15:38

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