e^x + e^(-x )= 2 : How?

**Kailiebe** · Jun 5th 2008, 10:47 AM

Hey all,

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

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

Complete mind blank >.<

**Moo** · Jun 5th 2008, 10:48 AM

Hello Kailiebe

Originally Posted by Kailiebe

Hey all,

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

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

Complete mind blank >.<

Multiply both sides by e^x.
Then substitute in an appropriate way to get a known form ^-^

**Mathstud28** · Jun 5th 2008, 10:51 AM

Originally Posted by Kailiebe

Hey all,

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

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

Complete mind blank >.<

Let $u=e^x\Rightarrow{\frac{1}{u}=e^{-x}}$

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

Quadratic formula gives

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

So we get

$u=1$

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

**Moo** · Jun 5th 2008, 10:53 AM

Originally Posted by Mathstud28

Let $u=e^x\Rightarrow{\frac{1}{u}=e^{-x}}$

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

Quadratic formula gives

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

So we get

$u=1$

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

It's better seeing that you have the form $a^2-2ab+b^2$

**Mathstud28** · Jun 5th 2008, 10:54 AM

Originally Posted by Moo

It's better seeing that you have the form $a^2-2ab+b^2$

I dont think so, but hey its all up to the individual, that is useful though, thanks

**Moo** · Jun 5th 2008, 10:56 AM

Originally Posted by Mathstud28

I dont think so, but hey its all up to the individual, that is useful though, thanks

Why you ?
I rather prefer one line to 3 lines

**Mathstud28** · Jun 5th 2008, 10:57 AM

Originally Posted by Moo

Why you ?
I rather prefer one line to 3 lines

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

**Moo** · Jun 5th 2008, 11:00 AM

Originally Posted by Mathstud28

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

Well, I'm just wondering eh ?
I think it's a good thing teaching people to get used to recognizing known forms

That's just my view too.

**ebaines** · Jun 5th 2008, 11:25 AM

You could solve it without needing to make the substitution of $u= e^x$ , as follows:

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

Multiply through by $e^x$ :

$ e^{2x} + 1 =2 e^x $
$ e^{2x} - 2 e^x + 1 = 0 $
$ (e^x -1)(e^x-1) = 0 $
$ e^x = 1 $
$ x = 0 $

**Mathstud28** · Jun 5th 2008, 11:38 AM

Originally Posted by ebaines

You could solve it without needing to make the substitution of $u= e^x$ , as follows:

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

Multiply through by $e^x$ :

$ e^{2x} + 1 =2 e^x $
$ e^{2x} - 2 e^x + 1 = 0 $
$ (e^x -1)(e^x-1) = 0 $
$ e^x = 1 $
$ x = 0 $

Yes, this is exactly what Moo was talking about in the earlier posts

**ebaines** · Jun 5th 2008, 11:45 AM

Originally Posted by Mathstud28

Yes, this is exactly what Moo was talking about in the earlier posts

I thought she was merely observing that you had an expression of the form $u^2 + 1 -2u=0$ instead of the more familar $u^2 -2u+1 =0$ .

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e^x + e^(-x )= 2 : How?