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Bunuel
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We will get
3^(x-y+1) * 2^(x-y)

So power of 3 is 1 greater than power of 2.
If you convert the answer choices into power of 2 and 3.
Only 2/3 won’t follow this rule.
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Simplify this equation to get 3^(x-y+1) * 2^(1-y)
Match each answer choice to get the answer
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­Quick simplification:


­We're left with three times the ratio of two different powers of 6. The 6 part will give us some new power of 6: positive, negative, or 0. So our solution set looks like this: { .... 3/216, 3/36, 3/6, 3/1, 3*6, 3*36, 3*216, ....}

We can verify all the answers that work (e.g. 3/36 = 1/12), or just recognize that we'll never be left with a 2 in the numerator. We might cancel the 3 (as in 1/12 and 1/2), but none of the fractional results will bring new factors to the numerator.
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Eventually we got 3^(x-y+1) as numerator and 2^y as denominator, which means the denominator should be the multiples of 2, therefore answer is C
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didn't uunderstood the solution after first line andreagonzalez2k
andreagonzalez2k
\(\frac{3*2^x*18^x}{6^{(x+y)}}=\frac{3*(2*18)^x}{6^x*6^y}=\frac{3*6^{2x}}{6^x*6^y}=\frac{3*6^x}{6^y}=3*6^{(x-y)}\)

As x and y are integers, (x-y) can be for example:

-2 -> the expression ends up being \(\frac{3}{36}=\frac{1}{12}\)

-1 -> the expression ends up being \(\frac{3}{6}=\frac{1}{2}\)

0 -> the expression ends up being \(3*6^0=3\)

1 -> the expression ends up being \(3*6^1=18\)

As we see the expression rises as (x-y) rises and decreases as (x-y) decreases.

\(\frac{2}{3}\) is between \(\frac{1}{2}\) and \(3\) and the expression never has this value.

IMO C
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didn't uunderstood the solution after first line andreagonzalez2k

The expression simplifies to 3 * 6^(x - y). For example:

If x - y = -2, we get 3 * 1/36.

If x - y = -1, we get 3 * 1/6.

If x - y = 0, we get 3.

If x - y = 1, we get 3 * 6.

If x - y = 2, we get 3 * 6^2

and so on.

As you can see, 2/3 is impossible, as it cannot be written as 3 * 6^n for any integer n.
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