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

Question

If  a=x+\frac{1}{x}  and  b=x-\frac{1}{x} , then what is the value of  \frac{2^{a^2}}{2^{b^2}} ?

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

Kurtosis · Answer

2^a = (2^x)*(2^(1/x)) 
2^b = (2^x)/(2^(1/x))

(2^a)^2 = 2^2a = (2^2x)*(2^2/x)
(2^b)^2 = 2^2b = (2^2x)/(2^2/x)

((2^a)^2)/(2^b)^2) = (2^2/x) * (2^2/x) = 2^4/x

If x = 1 then Answer = 16
If x = 2 then Answer = 4
If x = 4 then Answer = 1
If x = 4/3 then Answer = 8
If x = 4/5 then Answer = 32

Thus we can have multiple answers for this question since value of x is not specified. Please correct the question or let me know if I have made a mistake.

sharmilaz87 · Answer

I think you are correct Vyshak. I have also got the same ans.

chetan2u · Answer

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..

Beixi88 · Answer

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.

Kurtosis · Answer

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.

MathRevolution · Answer

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.

chetan2u · Answer

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

MathRevolution · Answer

Sorry for the confusion.

I have modified the question and the solution.

Thank you.

bumpbot · Answer

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

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

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