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If \(x\) and \(y\) are integers and \(\frac{7^{x+1} - 7^{x-1}}{3}=(2^y)(49^6)\) what is the value of \(xy\)?

(A) 104 (B) 96 (C) 52 (D) 48 (E) 44

\([fraction]7^{x+1} - 7^{x-1}[/fraction]=(3)(2^y)(49^6)\)
in RHS 7^12
LHS x=13 we get power same as 7^12 (7^2-1) ; 7^12*2^4*3
we get

7^12*2^4*3= 3*2^y *7^12
y will be 4

so x*y ; 13*4 ; 52
option C

How you got x=13

how you got x=13?
how can we campare numerator with rhs of 7 ^12 only ...?
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hi
can someone please provide the explanation for this question? I didn't understand the given solution
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If xx and yy are integers and 7x+1−7x−13=(2y)(496)7x+1−7x−13=(2y)(496) what is the value of xyxy?
solution
(7^(x+1)-7^(x-1))/3=2^y(49)^6
=7^(x-1)(49-1)/3==2^y(49)^6
=7^(x-1)(48)/3=2^y(49)^6
=7^(x-1)(16)=2^y(49)^6
=here by comparing both sides
y=4,x-1=12
y=4,x=13
xy=52
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GMATNinja - I think you guys created this question for the Ukrain marathon but I cannot get to make headwind with this one, is there any solution? The posted solution are all over the place, I wonder if this question is correct, thanks!
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how did we get 7^x-1 . 7^2 -1 ?
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Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
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Zainali385
how did we get 7^x-1 . 7^2 -1 ?



Express \(7^{(x + 1)}\) as \(7^2*7^{(x - 1)}\):

\(7^{(x + 1)} - 7^{(x - 1)} =\\
\\
= 7^2*7^{(x - 1)} - 7^{(x - 1)} = \\
\\
= 7^{(x - 1)}(7^2 - 1) = \)

Hope it's clear.
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