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

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

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

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New post 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:

\(-4k = - 48\)

\(k = 12\)

Answer
[Reveal] Spoiler:
D

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

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New post 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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