The solution set of $\large{(2\sqrt{2} - 2)^{x} + (6-4\sqrt{2})^{x} \ge 2^{1+x}}$ is ??

Question

vishweshshrimali5 · Answer

@iambatman @ganeshie8 @dan815

mathslover · Answer

You always forget one user to tag ... @mathslover

mathslover · Answer

This is easy.

mathslover · Answer

First step is clear, cancel 2^x from both sides.

mathslover · Answer

$(\sqrt{2} -1)^x + (3-2\sqrt{2})^x \ge 2$

vishweshshrimali5 · Answer

You got the first step right !!

mathslover · Answer

Now... use some logic. 

Clearly : $\sqrt{2} - 1 < 1$ | and $(3-2\sqrt{2}) < 1$

vishweshshrimali5 · Answer

Soo ?

mathslover · Answer

So, : 

for x not to be negative : 

$(\sqrt{2} - 1)^x < 1 $ and  $(3-2\sqrt{2})^x < 1$ 

Add these two equations, you get what you are not given. So, x is not positive. 

Now,

it should be clear that x is negative.

vishweshshrimali5 · Answer

You are wrong ;)

mathslover · Answer

Hmm , wait, let me recheck what I did.

mathslover · Answer

Are you pointing towards the condition of x = 0 ?

vishweshshrimali5 · Answer

nope

mathslover · Answer

x is negative or x = 0 ...

mathslover · Answer

And I know, that I'm not wrong here.

vishweshshrimali5 · Answer

Nope

mathslover · Answer

Wolfram verifies it here : http://www.wolframalpha.com/input/?i=(sqrt(2)+-+1)%5Ex+%2B+(3-2sqrt2)%5Ex+greater+than+or+equal+to+2

vishweshshrimali5 · Answer

Yeah sorry you are right.

mathslover · Answer

Okay, so, we can say that x is negative or x = 0

mathslover · Answer

Now, to find the solution set, we will have to do something creative.