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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]

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New post 31 Jul 2017, 23:16
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A
B
C
D
E

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  35% (medium)

Question Stats:

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Re: If (2^(-n)/3)(3^(-n)/2) = 1/36, what is the value of n ? [#permalink]

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New post 31 Jul 2017, 23:36
Bunuel wrote:
If \(\frac{2^{(-n)}}{3}*\frac{3^{(-n)}}{2} = \frac{1}{36}\) ?

A. -1
B. 0
C. 1
D. 2
E. 3


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

\(-n-1 =-2\)
\(-n=-1\)
\(n=1\)

C
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If (2^(-n)/3)(3^(-n)/2) = 1/36, what is the value of n ? [#permalink]

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New post 31 Jul 2017, 23:38
Bunuel wrote:
If \(\frac{2^{(-n)}}{3}*\frac{3^{(-n)}}{2} = \frac{1}{36}\), what is the value of n ?

A. -1
B. 0
C. 1
D. 2
E. 3


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

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

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

\(\frac{1}{2^{(n+1)}*3^{(n+1)}} = \frac{1}{2^2*3^2}\)

\(\frac{1}{2^{(n+1)}*3^{(n+1)}} = \frac{1}{2^2*3^2}\)

\(2^{(n+1)} = 2^2\)

\(n + 1 = 2\) \(=> n = 2-1 = 1\)

\(3^{(n+1)} = 3^2\)

\(n + 1 = 2\) \(=> n = 2-1 = 1\)

Therefore \(n = 1\)

Answer (C)...

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If (2^(-n)/3)(3^(-n)/2) = 1/36, what is the value of n ? [#permalink]

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New post 01 Aug 2017, 15:40
By just dropping the numerators down. We see that we get the 1/ (36)^(n)^2 = 1/36, n=1. C

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Re: If (2^(-n)/3)(3^(-n)/2) = 1/36, what is the value of n ? [#permalink]

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New post 01 Aug 2017, 21:31
We need 36 in denominator of LHS such that it equals the RHS. It can be only gotten when the value of n = 1. Hence C.
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If (2^(-n)/3)(3^(-n)/2) = 1/36, what is the value of n ? [#permalink]

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New post 02 Aug 2017, 08:56
Bunuel wrote:
If \(\frac{2^{(-n)}}{3}*\frac{3^{(-n)}}{2} = \frac{1}{36}\), what is the value of n ?

A. -1
B. 0
C. 1
D. 2
E. 3

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

\(\frac{1}{2^{n}3^{1}}*\frac{1}{3^{n}2^{1}} = \frac{1}{6^{2}}\)

Different base, same exponent rule: \(a^{n}*b^{n} = (ab)^{n}\). Combine \(2^{n}\) * \(3^{n}\) (=\(6^{n}\)), as well as \(2^1\) * \(3^1\) (=\(6^1\))

\(\frac{1}{6^{n}6^{1}} = \frac{1}{6^{2}}\)

\(\frac{6^{(-n)}}{6^{1}} = 6^{(-2)}\)

\(6^{(-n-1)} = 6^{(-2)}\)

\(-n - 1 = -2\)
\(n = 1\)

Answer C

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Re: If (2^(-n)/3)(3^(-n)/2) = 1/36, what is the value of n ? [#permalink]

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New post 05 Oct 2017, 12:26
Expert's post
Top Contributor
Bunuel wrote:
If \(\frac{2^{(-n)}}{3}*\frac{3^{(-n)}}{2} = \frac{1}{36}\), what is the value of n ?

A. -1
B. 0
C. 1
D. 2
E. 3


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, 12:26
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