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

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Joined: 02 Sep 2009
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If (2^(-n)/3)(3^(-n)/2) = 1/36, what is the value of n ? [#permalink]

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31 Jul 2017, 22:16
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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
[Reveal] Spoiler: OA

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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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31 Jul 2017, 22:36
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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31 Jul 2017, 22:38
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 $$n = 1$$

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Joined: 03 May 2017
Posts: 115
If (2^(-n)/3)(3^(-n)/2) = 1/36, what is the value of n ? [#permalink]

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01 Aug 2017, 14:40
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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01 Aug 2017, 20:31
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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02 Aug 2017, 07:56
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 $$2^1$$ * $$3^1$$ (=$$6^1$$)

$$\frac{1}{6^{n}6^{1}} = \frac{1}{6^{2}}$$

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

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

$$-n - 1 = -2$$
$$n = 1$$

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Joined: 11 Sep 2015
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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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05 Oct 2017, 11:26
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, 11:26
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