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Retired Moderator
Joined: 10 Apr 2015
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If 6^2k+1 = 18, then 6^k
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28 Jun 2022, 07:45
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) \(6\)
(A) \(\sqrt{2}\)
(B) \(\sqrt{3}\)
(C) \(\sqrt{6}\)
(D) \(2\sqrt{3}\)
(E) \(6\)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
If 6^2k+1 = 18, then 6^k
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29 Jun 2022, 07:28
29 Jun 2022, 07:28
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GreenlightTestPrep wrote:
If \(6^{2k+1} = 18\), then \(6^k =\)
(A) \(\sqrt{2}\)
(B) \(\sqrt{3}\)
(C) \(\sqrt{6}\)
(D) \(2\sqrt{3}\)
(E) \(6\)
(A) \(\sqrt{2}\)
(B) \(\sqrt{3}\)
(C) \(\sqrt{6}\)
(D) \(2\sqrt{3}\)
(E) \(6\)
No takers???? Here's my solution....
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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Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
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Re: If 6^2k+1 = 18, then 6^k
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14 Jan 2023, 23:31
14 Jan 2023, 23:31
Given that \(6^{2k+1}\) = 18 and we need to find the value of \(6^k\)
\(6^{2k+1}\) = 18
=> \(6^{2k} *6^1\) = 18
=> \(6^{2k}\) = \(\frac{18}{6}\) = 3
=> \(6^{2k/2}\) = \(3^{1/2}\)
=> \(6^k\) = \(\sqrt{3}\)
So, Answer will be B
Hope it helps!
Watch the following video to learn the Basics of Exponents
\(6^{2k+1}\) = 18
=> \(6^{2k} *6^1\) = 18
=> \(6^{2k}\) = \(\frac{18}{6}\) = 3
=> \(6^{2k/2}\) = \(3^{1/2}\)
=> \(6^k\) = \(\sqrt{3}\)
So, Answer will be B
Hope it helps!
Watch the following video to learn the Basics of Exponents
