vatiables

naaga wrote:. If y = 2x+1, what is the value of y – x?
(1) 2^(2x+2) = 64
(2) y = 2^(2x –1)


folks I got the answer A

but the OA is D

any one please explain

y = 2x + 1
y = x + x + 1
y - x = x + 1

Statement 1:
2^(2x+2) = 64
2^(2x+2) = 2^6

2x + 2 = 6
x + 1 = 3 = y - x Sufficient.

Statement 2:
y = 2^(2x-1) = 2x + 1

EDIT - There should be 2 possible values for x because 2^(2x-1) is nonlinear and will intersect 2x+1 twice : at x = 3/2, and at another point(I dont know how to work out this 2nd intersection point, someone please help!). So my guess is that statement 2 is not sufficient, since 2 possible values of y-x can be obtained.

-BM-

Statement 2:
y = 2^(2x-1) = 2x + 1

The only possible solution to this is x = 0. Any other value for x will not satisfy this equation (unless I missed something).

y - x = x + 1 = 1

Mr. Blue mentor how did you get y = 1 from bold statement, bit confused please clear my doubt. If we put x=0 we shall get 2^ -1

= y = 1/2 , then how did you get 1.

2x+1 = 2^(2x-1) has two solutions:

x = 3/2
x = -0.345

so statement 2 should be NOT SUFFICIENT

If you sketch both functions (this takes a bit of time) you can see they intersect at two points.... Is there a way to solve the equation analytically? how?

naaga wrote:Statement 2:
y = 2^(2x-1) = 2x + 1

The only possible solution to this is x = 0. Any other value for x will not satisfy this equation (unless I missed something).

y - x = x + 1 = 1

Mr. Blue mentor how did you get y = 1 from bold statement, bit confused please clear my doubt. If we put x=0 we shall get 2^ -1

= y = 1/2 , then how did you get 1.

I was wrong Naaga, thanks for pointing that out. I somewhat corrected my previous post, but I am not sure of my answer. Perhaps someone can help.

-BM-