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If 6^2k+1 = 18, then 6^k
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BrentGMATPrepNow
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If 6^2k+1 = 18, then 6^k [#permalink] New post  28 Jun 2022, 07:45
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If \(6^{2k+1} = 18\), then \(6^k =\)

(A) \(\sqrt{2}\)
(B) \(\sqrt{3}\)
(C) \(\sqrt{6}\)
(D) \(2\sqrt{3}\)
(E) \(6\)
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Re: If 6^2k+1 = 18, then 6^k [#permalink] New post  29 Jun 2022, 07:31
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Hint: Use the product rule in reverse.

Product Rule: \(n^a \times n^b = n^{a+b}\)
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Re: If 6^2k+1 = 18, then 6^k [#permalink] New post  29 Jun 2022, 09:11
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Set 6^k = X

6*X^2 = 18

X^2 = 3

X = (3)^.5 = 6^k

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Re: If 6^2k+1 = 18, then 6^k [#permalink] New post  29 Jun 2022, 17:17
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BrentGMATPrepNow wrote:
If \(6^{2k+1} = 18\), then \(6^k =\)

(A) \(\sqrt{2}\)
(B) \(\sqrt{3}\)
(C) \(\sqrt{6}\)
(D) \(2\sqrt{3}\)
(E) \(6\)

Given: \(6^{2k+1} = 18\)

Rewrite the left side as follows: \(6^{2k} \times 6^1 = 18\) [we're just applying the product law in reverse]

Divide both sides of the equation by \(6\) to get: \(6^{2k} = 3\)

Raise both sides of the equation to the power of \(\frac{1}{2}\) to get: \((6^{2k})^{\frac{1}{2}} = (3)^{\frac{1}{2}}\)

Simplify: \(6^k = \sqrt{3}\)

Answer: B
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Re: If 6^2k+1 = 18, then 6^k [#permalink]
29 Jun 2022, 17:17
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