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

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

Hey all,

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

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

Complete mind blank >.<

2. 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 ^-^

3. 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 $\displaystyle u=e^x\Rightarrow{\frac{1}{u}=e^{-x}}$

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

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

So we get

$\displaystyle u=1$

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

4. Originally Posted by Mathstud28
Let $\displaystyle u=e^x\Rightarrow{\frac{1}{u}=e^{-x}}$

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

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

So we get

$\displaystyle u=1$

Back subbing we get $\displaystyle e^x=1\Rightarrow{ln(1)=0}$
It's better seeing that you have the form $\displaystyle a^2-2ab+b^2$

5. Originally Posted by Moo
It's better seeing that you have the form $\displaystyle a^2-2ab+b^2$
I dont think so, but hey its all up to the individual, that is useful though, thanks

6. 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

7. 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

8. 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.

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 $\displaystyle e^x$:

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

10. Originally Posted by ebaines
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 $\displaystyle e^x$:

$\displaystyle e^{2x} + 1 =2 e^x$
$\displaystyle e^{2x} - 2 e^x + 1 = 0$
$\displaystyle (e^x -1)(e^x-1) = 0$
$\displaystyle e^x = 1$
$\displaystyle x = 0$
Yes, this is exactly what Moo was talking about in the earlier posts

11. 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 $\displaystyle u^2 + 1 -2u=0$ instead of the more familar $\displaystyle u^2 -2u+1 =0$.