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# If a and n are positive, does 2a^(2x) = n?

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If a and n are positive, does 2a^(2x) = n? [#permalink]  14 Apr 2007, 14:56
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How would you approach solving the below beauty?

If a and n are positive, does 2a^(2x) = n?
1) a^x +1/a^x = sqrt(n+2)
2)x>0

After a few manipulations, I have done it using trial and error method.

Is there any neater way of approaching similar probs?
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Is it A?

1) a^x +1/a^x = sqrt(n+2)
Using algebra,
(a^(2x)+1)/a^x=sqrt(n+2)
[(a^(2x)+1)/a^x]^2=n+2
Some nasty algebra gives me,
(a^(4x)+1)/a^(2x)=n
So the answer is no and 1) is SUFF.

2) x>0
We don't know a or n. Not enough info. INSUFF.

So A.
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kevin, unfor, asnwer is not A
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ricokevin wrote:
Is it A?

1) a^x +1/a^x = sqrt(n+2)
Using algebra,
(a^(2x)+1)/a^x=sqrt(n+2)
[(a^(2x)+1)/a^x]^2=n+2
Some nasty algebra gives me,
(a^(4x)+1)/a^(2x)=n
So the answer is no and 1) is SUFF.

2) x>0
We don't know a or n. Not enough info. INSUFF.

So A.

(C)

The problem with 1 alone is that if x=0, then
o (a^(4x)+1)/a^(2x)= (a^(4*0)+1)/a^(2*0)= (1+1)/1 = 2 = n
o 2a^(2x) = 2*a^(2*0) = 2*1 = 2 = (a^(4x)+1)/a^(2x) = n

We have to be sure to remove the case of x=0. Thus, x > 0 concludes on it.

To be sure, is there another possible value for x to make equal a:
2a^(2x) = (a^(4x)+1)/a^(2x) (1) ?

We can try to solve it:
(1) <=> 2a^(2x) = (a^(4x)+1)/a^(2x)
<=> 2*a^(4*x) = a^(4x) + 1
<=> a^(4*x) = 1 >>> only possible if x = 0.
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