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