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If x > y > 0, does 3^(x + 1) + 3(2^y) = 12v?

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If x > y > 0, does 3^(x + 1) + 3(2^y) = 12v?  [#permalink]

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If \(x > y > 0\), does \(3^{x + 1}\) + \(3(2^y)\) = \(12v\)?


(1) \(\frac{(3^{2x} - 2^{2y})}{(3^{x} - 2^{y})}\) = \(4v\)

(2) \(2(3^{x+2}) + 9(2^{y+1})\) = \(72v\)

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If x > y > 0, does 3^(x + 1) + 3(2^y) = 12v?  [#permalink]

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New post 26 Jun 2018, 10:53
CAMANISHPARMAR wrote:
If \(x > y > 0\), does \(3^{x + 1}\) + \(3(2^y)\) = \(12v\)?


(1) \(\frac{(3^{2x} - 2^{2y})}{(3^{x} - 2^{y})}\) = \(4v\)

(2) \(2(3^{x+2}) + 9(2^{y+1})\) = \(72v\)


This is a little more complex than most algebra DS problems I've seen.

That said, since it has algebra in it, my first thought is to simplify both the question and the statements as much as possible.

The \(3^{x + 1}\) in the question, along with the other 3 in the question, makes me think about factoring out a 3 term. That would get rid of the 'x + 1' and make things look simpler, so let's do it.

does \(3^{x + 1}\) + \(3(2^y)\) = \(12v\)?
does \(3(3^{x} + 2^{y}) = 12v\)?
does \(3^{x} + 2^{y} = 4v\)?

Much nicer.

Now, go to the statements.

Statement 1: \(\frac{(3^{2x} - 2^{2y})}{(3^{x} - 2^{y})}\) = \(4v\)

What I noticed is that both of the values on the top of the fraction are perfect squares. I know that because both of the exponents are even (they both have 2s in the exponents.) I know a rule about the difference of perfect squares - here's the rule:

\(a^2 - b^2 = (a + b)(a - b)\)

Let's rewrite the top of the fraction as a difference of squares, then simplify it like that.

\((3^{2x} - 2^{2y})\)
\((3^{x})^2 - (2^{y})^2\)
\((3^{x} - 2^{y})(3^{x}+2^{y})\)

Now, there's a common term in the top and bottom of the fraction. Divide it out of both the top and the bottom:

\(\frac{(3^{x} - 2^{y})(3^{x}+2^{y})}{(3^{x} - 2^{y})}\)
\(3^{x}+2^{y}\)

The whole statement simplifies like this:

\(3^{x}+2^{y} = 4v\)

That's our question exactly, so it's sufficient.

Statement 2: This statement looks similar to the question, but it's not exactly the same. I want to get rid of the 'x + 2' and the 'y + 1', since I don't have those in the original question. To turn the x + 2 exponent into an x, like in the question, I need to factor out \(3^2\) To turn the y + 1 into a y, I need to factor out a 2. Here's how it looks:

\(2(3^{x+2}) + 9(2^{y+1})\) = \(72v\)
\(2*3^2*(3^{x}) + 9*2*(2^{y})\) = \(72v\)
\(18(3^{x}) + 18(2^{y})\) = \(72v\)

Divide both sides by 18...

\(3^{x} + 2^{y} = 4v\)

Again, that's our question, so this statement is sufficient as well.
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If x > y > 0, does 3^(x + 1) + 3(2^y) = 12v? &nbs [#permalink] 26 Jun 2018, 10:53
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