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# If (1/2)^24*(1/81)^k = 1/18^24, then k =

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Joined: 02 Sep 2009
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If (1/2)^24*(1/81)^k = 1/18^24, then k = [#permalink]

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13 Dec 2017, 05:58
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If $$(\frac{1}{2})^{24}*(\frac{1}{81})^k=(\frac{1}{18})^{24}$$ then k =

A. 8
B. 12
C. 16
D. 24
E. 36
[Reveal] Spoiler: OA

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Posts: 1539
If (1/2)^24*(1/81)^k = 1/18^24, then k = [#permalink]

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13 Dec 2017, 09:36
Bunuel wrote:
If $$(\frac{1}{2})^{24}*(\frac{1}{81})^k=(\frac{1}{18})^{24}$$ then k =

A. 8
B. 12
C. 16
D. 24
E. 36

$$(\frac{1}{2})^{24}*(\frac{1}{81})^k=(\frac{1}{18})^{24}$$

$$(\frac{1}{2})^{24}*(\frac{1}{3^4})^k=(\frac{1}{2^13^2})^{24}$$

$$2^{-24} * (3^{-4})^{k} = (2^{-1})^{24}* (3^{-2})^{24}$$

$$2^{-24} * 3^{-4k} = 2^{-24} * 3^{-48}$$

$$2^{-24}$$ is factored out. Bases $$3$$ are identical. Set their exponents equal:

$$-4k = - 48$$

$$k = 12$$

[Reveal] Spoiler:
D

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If (1/2)^24*(1/81)^k = 1/18^24, then k = [#permalink]

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13 Dec 2017, 10:58
Bunuel wrote:
If $$(\frac{1}{2})^{24}*(\frac{1}{81})^k=(\frac{1}{18})^{24}$$ then k =

A. 8
B. 12
C. 16
D. 24
E. 36

Since $$(\frac{1}{2})^{24}*(\frac{1}{81})^k=(\frac{1}{18})^{24}$$, we can also say that $$(2)^{24}*(81)^k=(18)^{24}$$

When prime factorized, $$18 = 2 * 3^2$$
In the right hand side of the expression, 18 has been raised to the power of 24,
we need $$2^{24}$$ and $$3^{48}$$ in the left hand side to balance the equation out.

In order to balance the equation $$81^k = (3^4)^k = 3^{4*k}$$ must be equal to $$3^{48}$$

Therefore, 4k = 48 => k = 12(Option B)
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If (1/2)^24*(1/81)^k = 1/18^24, then k =   [#permalink] 13 Dec 2017, 10:58
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