When a=x+(1/x) and b=x-(1/x), (2^a^2)/(2^b^2)=?

hi,
you are absolutely correct in your observation..
looking at the equation and the OA,
the equation could have been:-
(2^a)^(a+b)/(2^b)^(a+b)..
try out this Q ..

just a point, in these Q to save time, take the power to the numerator and then substitute for a and b..
example
2^a/2^b= 2^(a-b)= 2^(x + 1/x - x - (1/x))
= 2^(2/x)...
you should do this when you see a and b have common terms with change of sign..
Ofcourse, the way you have done is also absolutely correct..

I have an itching feeling that my method is grasping at too many straws, but it gave me the correct answer. Can anyone give me some feedback on my method for solving the question?

Solve for a to get a = 2:

1. Given: \(a = x + \frac{1}{x}\)
2. \(ax = x^2 + \frac{x}{x}\) -> \(ax = x^2 + 1\)
3. \(0 = x^2 - ax + 1\)
4. \(x^2 - ax + 1 = (x-1)^2\). To complete the expression, a = 2

Solve for b to get b = 0:

1. Given: \(b = x - \frac{1}{x}\)
2. \(bx = x^2 - \frac{x}{x}\) -> \(bx = x^2 - 1\)
3. \(0 = x^2 - bx - 1\)
4. \(x^2 - bx - 1 = (x - 1)(x + 1)\). To complete the expression, b = 0

We plug in a = 2 and b = 0 into \(\frac{(2^a)^2}{(2^b)^2}=?\):
1. \(\frac{(2^a)^2}{(2^b)^2}=?\)
2. \(\frac{{2^{2a}}}{{2^{2b}}}=?\)
3. \(2^{2a} * 2^{-2b}\)
4. \(2^{2(a - b)}\)
5. \(2^{2(2 - 0)}\)
6. \(2^4 = 16\)

My answer is D, 16.

Please let me know if the method I used is incorrect or if you have any improvements on the method used.

Hi @Beixi88,

Your solution focuses only on one of the solutions for 'a'. But a can also vary with the value of x.

According to your solution, we have one of the equations as x^2 - ax + 1 = 0. Here we have two unknowns x and a.. X and a can be plugged in many ways to solve the equation.
If x = 2, 5 - 2a = 0 --> a = 5/2. Thus we can have many values of a depending on the value of x.

When a=x+(1/x) and b=x-(1/x), (2^a)^2/(2^b)^2=?

A. 2
B. 4
C. 8
D. 16
E. 32


--> (2^a^2)/(2^b^2) ={(2)^(a^2-b^2)}=2^(a-b)(a+b). Since a-b=2/x and a+b=2x, 2^(a-b)(a+b)=2^(2/x)(2x)=2^4=16.
Therefore, the answer is D.

HI,

I am sorry, your solution is wrong,...
(2^a)^2/(2^b)^2 = 2^(2a-2b) and not {(2)^(a^2-b^2)}

(2^a)^2=(2^a)*(2^a)=2^(a+a)= 2^2a..
REQUEST rectify the Question or the answer accordingly

Sorry for the confusion.

I have modified the question and the solution.

Thank you.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.