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If 2^{-n}/3*3^{-n}/2=1/36, what is the value of n

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If 2^{-n}/3*3^{-n}/2=1/36, what is the value of n [#permalink] New post 01 Oct 2017, 04:35
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If \(\frac{2^{(-n)}}{3}*\frac{3^{(-n)}}{2}=\frac{1}{36}\), what is the value of n?

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Re: If 2^{-n}/3*3^{-n}/2=1/36, what is the value of n [#permalink] New post 01 Oct 2017, 06:59
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Bunuel wrote:
If \(\frac{2^{(-n)}}{3}*\frac{3^{(-n)}}{2}=\frac{1}{36}\), what is the value of n?

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from the given equ

we have \(2^{-n} * 3^{-n} = 6^{-1}\)

or \(2^{-n} * 3^{-n} =2^{-1} * 3^{-1}\)

or n=1
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Re: If 2^{-n}/3*3^{-n}/2=1/36, what is the value of n [#permalink] New post 05 Oct 2017, 11:14
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Bunuel wrote:
If \(\frac{2^{(-n)}}{3}*\frac{3^{(-n)}}{2}=\frac{1}{36}\), what is the value of n?


Given: \(\frac{2^{(-n)}}{3}*\frac{3^{(-n)}}{2}=\frac{1}{36}\)

On the left side of this equation, let's multiply the numerators together, and then we'll multiply the denominators.

NUMERATORS:
\(2^{-n}\) x \(3^{-n}\) = \(6^{-n}\)

DENOMINATORS:
3 x 2 = 6

So, when we simplify the left side of the equation, we get: \(\frac{6^{(-n)}}{6} = \frac{1}{36}\)

From here, we can multiply both sides by 6 to get: \(6^{(-n)} = \frac{1}{6}\)


Next, recognize that \(\frac{1}{6} = 6^{(-1)}\)


So, we can write: \(6^{(-n)} = 6^{(-1)}\)

From this, we can conclude that -n = -1, which means n = 1

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Re: If 2^{-n}/3*3^{-n}/2=1/36, what is the value of n   [#permalink] 05 Oct 2017, 11:14
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