Solve $2^x+2^{-x} = 2$ Need to solve:
$$2^x+2^{-x} = 2$$
I can't use substitution in this case. Which is the best approach?
Event in this form I do not have any clue:
$$2^x+\frac{1}{2^x} = 2$$
 A: Hint:
By AM-GM inequality , $2^x+2^{-x} \geq 2 \times \sqrt{2^x \times 2 ^{-x}} = 2 $
When $2 ^x = 2^{-x}$ the equality holds.
Alternative solution:
Easy to show that $x = 0$, is one possible solution.
$$\frac{d(2^x+2^{-x})}{dx} = (  2^x -2^{-x}) \log 2$$ 
which positive over $(0,+\infty)$, negative over $(-\infty,0)$
A: A contrived solution:
Write
$$\frac{e^{x\ln2}+e^{-x\ln2}}2=1=\cosh(x\ln2),$$
then
$$x\ln2=\cosh^{-1}1=0.$$

Another one:
By inspection, $x=0$ is a solution.
The derivative of the LHS is
$$2^x\ln2-2^{-x}\ln2,$$ which is positive for $x>0$ and negative for $x<0$, so the function is monotonic on both sides and there are no other roots.

Yet another:
Observe
$$2^x-2+2^{-x}=\sqrt2^{2x}-2\sqrt2^x\sqrt2^{-x}+\sqrt2^{-2x}=(\sqrt2^x-\sqrt2^{-x})^2.$$
The only root is when
$$\sqrt2^x=\sqrt2^{-x},$$ i.e. $$x=-x.$$

A last one:
The equation can be rewritten
$$2^x-1=1-2^{-x},$$ i.e. unless the RHS is zero
$$\frac{2^x-1}{1-2^{-x}}=2^x=1.$$
There is no other solution than $x=0$.
A: Elucidate the problem by using the substitution $u = 2^x$, then you have $$u + \frac{1}{u} = 2$$
Multiply throughout by $u \neq 0$ to get $$u^2 +1 = 2u \iff u^2 - 2u + 1 = 0$$
This is an easy quadratic to solve, you should get $u = 1$ and hence you need only solve $2^x = 1 \iff x = 0$. 
A: Substitute $y=2^x$. Hence, the equation is: $y+y^{-1}=2$ which is equivalent to: $y^2-2y+1=0$. Can you take it from here ? Once you find which $y$('s) satisfy the equation, try to find $x$ such that $2^x=y$.
A: $$2^x =y $$
$$y+ \frac{1}{y} =2$$ 
Using AM-GM INEQUALITY
$$\frac{({y+ \frac{1}{y}})}{2} \ge \sqrt {y.\frac{1}{y}}$$
$$y+ \frac{1}{y}\ge 2$$
A: Let,
$2^x=y$ then the equation becomes
$y+\frac{1}{y}=2\\
\implies y^2+1-2y=0\\
\implies (y-1)^2=0\\
\implies y=1\\
\implies 2^x=1\\
\implies 2^x=2^0\\
\implies x=0$
