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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)=?  [#permalink]

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New post Updated on: 08 Feb 2016, 21:27
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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


* A solution will be posted in two days.

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Originally posted by MathRevolution on 04 Feb 2016, 18:22.
Last edited by MathRevolution on 08 Feb 2016, 21:27, edited 1 time in total.
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Re: When a=x+(1/x) and b=x-(1/x), (2^a^2)/(2^b^2)=?  [#permalink]

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New post 04 Feb 2016, 20:03
2
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.
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Re: When a=x+(1/x) and b=x-(1/x), (2^a^2)/(2^b^2)=?  [#permalink]

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New post 04 Feb 2016, 20:30
I think you are correct Vyshak. I have also got the same ans.
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Re: When a=x+(1/x) and b=x-(1/x), (2^a^2)/(2^b^2)=?  [#permalink]

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New post 04 Feb 2016, 21:13
Vyshak wrote:
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.


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..
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Re: When a=x+(1/x) and b=x-(1/x), (2^a^2)/(2^b^2)=?  [#permalink]

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New post 04 Feb 2016, 22:09
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.
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Re: When a=x+(1/x) and b=x-(1/x), (2^a^2)/(2^b^2)=?  [#permalink]

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New post 04 Feb 2016, 22:31
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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.
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When a=x+(1/x) and b=x-(1/x), (2^a^2)/(2^b^2)=?  [#permalink]

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New post Updated on: 08 Feb 2016, 21:30
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.
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Originally posted by MathRevolution on 08 Feb 2016, 00:16.
Last edited by MathRevolution on 08 Feb 2016, 21:30, edited 1 time in total.
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Re: When a=x+(1/x) and b=x-(1/x), (2^a^2)/(2^b^2)=?  [#permalink]

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New post 08 Feb 2016, 00:32
MathRevolution wrote:
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
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When a=x+(1/x) and b=x-(1/x), (2^a^2)/(2^b^2)=?  [#permalink]

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New post 08 Feb 2016, 20:55
chetan2u wrote:
MathRevolution wrote:
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


Hi Bunuel,
since you are the math expert here..
May I request you to inform MathRevolution to be slightly careful in the wordings of his Q.
Inspite of observations raised by members, he just moves on to the next question without correcting the previous ones.

But this time he has gone beyond that.
He has changed the power rule of exponent... (a^m)^n= a^mn
..

I have nothing against the institute, but this attitude may harm the institute itself.
that's ofcourse is their outlook but NOW we are teaching wrong things to students..
As, There will be many who will take his answer as correct since he is the source..
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Re: When a=x+(1/x) and b=x-(1/x), (2^a^2)/(2^b^2)=?  [#permalink]

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New post 08 Feb 2016, 21:28
Sorry for the confusion.

I have modified the question and the solution.

Thank you.
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Re: When a=x+(1/x) and b=x-(1/x), (2^a^2)/(2^b^2)=?  [#permalink]

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New post 08 Feb 2016, 21:37
MathRevolution wrote:
Sorry for the confusion.

I have modified the question and the solution.

Thank you.


Hi,
thanks a lot...
pl dont mistake me. I did think there were some typo error..
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Re: When a=x+(1/x) and b=x-(1/x), (2^a^2)/(2^b^2)=?  [#permalink]

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