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Bunuel
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If ab=1
\(a = \frac{1}{b }= b^{-1}\)

This relation holds true whether a or b is an integer.

happiness10
Kinshook what if we have fraction values of A or B .
let say a=1/2 b = 2 then ans would be E because a and b can be non integer also.
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Bunuel
What is the value of x ?

(1) \(a^{(2x)}=b^{(x+6)}\)

(2) \(ab = 1\)

1) \(a^{(2x)}=b^{(x+6)}\)
Not sufficient to find the value of x.

2) \(ab = 1\)
No info about x.

Combining 1 & 2:
\(ab = 1\)
\(b = \frac{1}{a}\) -- (1)

\(a^{(2x)}=b^{(x+6)}\) -- (2)
Putting the value of b in eq(2):
\(a^{(2x)}=\frac{1}{a}^{(x+6)}\)
\(a^{(2x)}=a^{-(x+6)}\)
\(2x = -x - 6\)
\(x = -2\) (C)
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Bunuel

I think the OA should be C. Please check my solution and take action.

Bunuel
What is the value of x ?

(1) \(a^{(2x)}=b^{(x+6)}\)

(2) \(ab = 1\)
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The OA is E.
Taking a=b=1, then 2x=x+6, giving X=6, but
Taking a=2 and b=1/2 satisfies the second statement and putting in first statement gives,
2x= -(X+6)
2x= -x-6
3x=-6
X=-2
Both statements together are insufficient.
Kinshook I guess assuming that a and b are integers is the trap you're falling for.

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I have not considered the case when a = b = 1.
Otherwise the logic was correct. There is no difference whether a and/or b are integers.

Rahulprabhu
The OA is E.
Taking a=b=1, then 2x=x+6, giving X=6, but
Taking a=2 and b=1/2 satisfies the second statement and putting in first statement gives,
2x= -(X+6)
2x= -x-6
3x=-6
X=-2
Both statements together are insufficient.
Kinshook I guess assuming that a and b are integers is the trap you're falling for.

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If a=1, b=1
ab=1
and x can take any value as 1^2x =1^(x+6) for any value of x.
Hence,(1)+(2) insufficient. => Option (E).
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