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Does anyone have any thoughts on how to solve the following equation:

$$\exp(2x) - 2x\exp(x) - 1 = 0$$

If it helps, this equation is also equivalent to the following hyperbolic equation:

$$\sinh(x) = x$$

Thanks for any advice.

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You can use calculus to prove the last equation has zero as its only solution. –  Javier Badia Apr 11 '13 at 15:14

2 Answers 2

up vote 4 down vote accepted

It's clear that $x=0$ is a solution

On the other hand, $$\frac{d}{dx} \sinh x = \cosh x$$

We see that $$\cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots \geq 1$$

with equality if and only if $x = 0$.

If $\sinh x = x$ had a second non-zero solution, then mean value theorem would lead to a contradiction.

(Edit: As N.S. has said in the comments, we can just use the fact $\cosh^2 x = 1 + \sinh^2 x$ rather than Taylor series)

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Thanks very much for the solution. –  marshall Apr 11 '13 at 15:47
2  
You don't need Taylor series to prove $\cosh(x) \geq 1$. All you need is $\cosh^2(x)=1+\sinh^2(x)$. –  N. S. Apr 11 '13 at 17:55
    
Thanks @N.S., I'll add that detail to the solution –  muzzlator Apr 11 '13 at 18:02

I tried using simple algebra:

$$\exp(2x)-2x\exp(x)-1=0$$

Factor out $\exp(x)$: $$\exp(x)(\exp(x)-2x)=1$$

Take $\ln$ of both sides: $$x+\ln(e^x-2x)=0$$

So, the first obvious solution is $x=0$. Another possibility is that $\ln(e^x-2x)=0$: $$\ln(e^x-2x)=0 \\ e^x-2x=1 \\ e^x=1+2x$$

Again the most obvious solution for this expression is zero as well. However, my method does not rule out the possibility that there are other solutions. So, it serves as a supplement to muzzlator's answer.

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Hi dostre. A remark on one of the steps, $\log ab = \log a + \log b$ –  muzzlator Apr 11 '13 at 17:45
    
Thanks for the remark. –  Koba Apr 11 '13 at 17:48

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