# solving inequality.

• Nov 17th 2012, 03:42 AM
Vim
solving inequality.
hi everyone. first of all i have to apologize for my poor English.
I've got an inequality which I know is true but don't know how to solve,can anyone help?
$\displaystyle 2^{a}-2^{-a} \neq 2^{b}-2^{-b}$
• Nov 17th 2012, 05:52 AM
veileen
Re: solving inequality.
That isn't an inequality.

Anyway, $\displaystyle 2^a-2^{-a} = 2^b-2^{-b} \Leftrightarrow a=b$.

I.

Let's consider the functions $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}, f(x)=2^x-2^{-x}$, $\displaystyle f_1, f_2:\mathbb{R}\rightarrow \mathbb{R}, f_1(x)=2^x, f_2(x)=2^{-x}=\left ( \frac{1}{2} \right )^x$.
$\displaystyle f_1$ is strictly increasing;
$\displaystyle f_2$ is strictly decreasing, so $\displaystyle -f_2$ is strictly increasing;

$\displaystyle f=f_1+f_2$ is strictly increasing (being a sum of two strictly increasing functions) $\displaystyle \Rightarrow$ f is injective, and therefore $\displaystyle f(a)=f(b) \Leftrightarrow a=b$.

II.

$\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}, f(x)=2^x-2^{-x}$

$\displaystyle f'(x)=(2^x-2^{-x})'=2^x\ln 2+2^{-x}\ln 2=\ln2(2^x+2^{-x})> 0$, for any $\displaystyle x\in \mathbb{R} \Rightarrow$ f is injective, and therefore $\displaystyle f(a)=f(b) \Leftrightarrow a=b$.