Solve for x: e^x - e^-x = 1
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# Solve for x: e^x - e^-x = 1

[From: ] [author: ] [Date: 11-10-07] [Hit: ]
The only positive root is y = (sqrt(5)+1)/2.1/y = (sqrt(5)-1)/2.x = ln(y) = ln( (sqrt(5)+1)/2 ).-megaone-I think x is 0.......
That would be great if you could show the steps. Thanks! :D

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Note that since e^(-x) = 1/e^x, we have:
e^x - 1/e^x = 1.

Now, let u = e^x. Applying this substitution gives:
u - 1/u = 1.

Multiplying both sides by u yields:
u^2 - 1 = u ==> u^2 - u - 1 = 0.

u = (1 ± √5)/2.

However, note that since u = e^x and e^x > 0 for all x, we require that u > 0. Only the positive value of the plus-or-minus sign will make u > 0, so:
u = e^x = (1 + √5)/2.

Taking the natural logarithm of both sides yields the final answer to be:
x = ln[(1 + √5)/2].

I hope this helps!

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Let y = e^x. Note y>0. Then

y - 1/y = 1.

y^2 - y - 1 = 0

The only positive root is y = (sqrt(5)+1)/2. 1/y = (sqrt(5)-1)/2.

x = ln(y) = ln( (sqrt(5)+1)/2 ).

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megaone

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I think x is 0.
1
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