May or may not be Binary

Easy. As ${ x }^{ m+n }={ x }^{ m }·{ x }^{ n }$ ,

${ 2 }^{ 2009 }-{ 2 }^{ 2007 }={ 2 }^{ 2007+2 }-{ 2 }^{ 2007 }={ 2 }^{ 2007 }·{ 2 }^{ 2 }-{ 2 }^{ 2007 }$ .

Then, the result is:

${ 2 }^{ 2007 }·{ 2 }^{ 2 }-{ 2 }^{ 2007 }={ 2 }^{ 2007 }\left( { 2 }^{ 2 }-1 \right) ={ 2 }^{ 2007 }\left( 4-1 \right) =\boxed { 3·{ 2 }^{ 2007 } }$ .

2^2007(2^2-1)=2^2007(4-1)=2^2007(3)

This is my stupid solution..... Consider this.. If 2^4 - 2^2 = 12 , which is 3 x 2^2 , therefore it has the same scenarion with the original problem.. So , 2^2009 - 2^2007 = 3 x 2^2007 .. It is an easier solution for me.

As we have to find,

2^2009 - 2^2007

=2^2007 (2^2 - 1) (taking 2^2007 common)

=2^2007 (4 - 1)

=2^2007 ( 3)

Thus giving us the answer,

=3×2^2007

(2^2009) - (2^2007) = (2^2007x2x2) - (2^2007)

(2^2007x4) - (2^2007) = 2^2007x3

By taking common 2^2007 ,

2^2007 (2^2 - 1 ) = 3 * 2^2007

2^2009 - 2^2007 = 4 X 2^2007 - 2^2007 = 2^2007 (4 - 1) = 3 X 2^2007

2^2009-2^2007=2^2007.2^2-2^2007----->2^2007(2^2-1)=3X2^2007

2^2009 -2^2007 2^2007(2^2-2^0) 2^2009 -2^2007 3*2^2007

2^{2009}-2^{2007}=2^{2007+2}-2^{2007}=2^{2007}\times(2^2-1)=3\times2^{2007}.

=2^2009 - 2^2007 =2^2007(2^2 - 1 ) = 2^2007 × (4-1) = 2^2007 × (3) = 3 × 2^2007

2^2009-2^2007=2^2007(2^1-1)=2^2007(4-1)=3 . 2^2007

I think it would be rather easy if we take an example like this. 2 ^ 4 (-) 2 ^ 2 = 12 So 3 * 2 ^ 2 gives us the answer 12. So the final answer should be 3 * (2 ^ smaleer number) which is 3 * (2 ^ 2007).

this is a simple trick....if 2^x-2^(x-2)=3*2^(x-2)

We can write this as 2raise to power 20007( 2 square-1) that is 3 X 2 raise topower 2007 answer. K.K.GARG,India

2^2009 - 2^2007 = 2^2007(2^2-1) = 3*2^2007

Let 2009 be a, then 2^a-2^a-2, then 2^a-1/4(2^a), after it is simplified an expression of 3(2^a-2) is formed. Consequently, the answer C suits the expression when it is substituited.