Hey all,

How would one solve e^x + e^(-x) = 2

(This is so basic, it's embarrassing :P)

Complete mind blank >.<

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- Jun 5th 2008, 10:47 AM #1

- Joined
- Mar 2008
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- Jun 5th 2008, 10:48 AM #2

- Jun 5th 2008, 10:51 AM #3
Let $\displaystyle u=e^x\Rightarrow{\frac{1}{u}=e^{-x}}$

So then you have $\displaystyle u+\frac{1}{u}-2=0\Rightarrow{u^2+1-2u=0}$

Quadratic formula gives

$\displaystyle u=\frac{2\pm\sqrt{4-4(1)(1)}}{2}=1$

So we get

$\displaystyle u=1$

Back subbing we get $\displaystyle e^x=1\Rightarrow{x=\ln(1)=0}$

- Jun 5th 2008, 10:53 AM #4

- Jun 5th 2008, 10:54 AM #5

- Jun 5th 2008, 10:56 AM #6

- Jun 5th 2008, 10:57 AM #7
Why do I? Why does it matter? But if you really want to know it is because I like showing the quadratic form when teaching/posting because the factoring form is a little less apparent, and you might get people attempting to factor everything like this, which sometiems is hard/impossible, But that is just my view

- Jun 5th 2008, 11:00 AM #8

- Jun 5th 2008, 11:25 AM #9
You could solve it without needing to make the substitution of $\displaystyle u= e^x$, as follows:

$\displaystyle

e^x +e^ {-x} = 2

$

Multiply through by $\displaystyle e^x$:

$\displaystyle

e^{2x} + 1 =2 e^x

$

$\displaystyle

e^{2x} - 2 e^x + 1 = 0

$

$\displaystyle

(e^x -1)(e^x-1) = 0

$

$\displaystyle

e^x = 1

$

$\displaystyle

x = 0

$

- Jun 5th 2008, 11:38 AM #10

- Jun 5th 2008, 11:45 AM #11