Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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

$$-1+X^2=aX$$

share|improve this answer

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

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

or

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

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!!!

share|improve this answer
3  
It is always a pleasure to me to read your last words. They are so true ! Cheers. –  Claude Leibovici Apr 6 at 8:40
    
@Claude Leibovivi: indeed! And thanks! –  Robert Lewis Apr 6 at 8:42
    
@1950RobertLewis very nice. –  kaka Apr 6 at 8:49
1  
@kaka: muchas gracias! –  Robert Lewis Apr 6 at 8:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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