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If 3^k = 16, and 2^j = 27, then kj =
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11 Jul 2022, 11:07
11 Jul 2022, 11:07
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Question Stats:
80% (01:34) correct
20% (02:40) wrong based on 10 sessions
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If \(3^k\) = 16, and \(2^j\) = 27, then kj =
A) 8
B) 9
C) 10
D) 12
E) 15
A) 8
B) 9
C) 10
D) 12
E) 15
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If 3^k = 16, and 2^j = 27, then kj =
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11 Jul 2022, 11:26
11 Jul 2022, 11:26
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Carcass wrote:
If \(3^k\) = 16, and \(2^j\) = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
A) 8
B) 9
C) 10
D) 12
D) 15
One approach is to isolate the 3 in both equations. Here’s what I mean:
Given: \(3^k = 16\)
Rewrite \(16\) as \(2^4\) to get: \(3^k = 2^4\)
Raise both sides to the power of \(\frac{1}{k}\) to get: \((3^k)^{(\frac{1}{k})} = (2^4)^{(\frac{1}{k})}\)
Use power of power law to simplify: \(3 = 2^{(\frac{4}{k})}\)
Given: \(2^j = 27\)
Rewrite \(27\) as \(3^3\) to get: \(2^j = 3^3\)
Raise both sides to the power of \(\frac{1}{3}\) to get: \( (2^j)^{(\frac{1}{3})} = (3^3)^{(\frac{1}{3})}\)
Use power of power law to simplify: \(2^(\frac{j}{3}) = 3\)
We now have two equations:
\(3 = 2^{(\frac{4}{k})}\)
\(2^{(\frac{j}{3})} = 3\)
Since both equations are set equal to 3, we can write: \(2^{(\frac{4}{k})}\) \(=\) \(2^{(\frac{j}{3})}\)
Since the bases both equal 2, we can conclude that \(\frac{4}{k} = \frac{j}{3}\)
Cross multiply to get: \(jk = (4)(3) \)
So, \(jk = 12\)
Answer: D
Cheers,
Brent
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Status:GRE Quant Tutor
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Given Kudos: 9
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Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
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WE:Engineering (Computer Software)
Re: If 3^k = 16, and 2^j = 27, then kj =
[#permalink]
15 Jan 2023, 01:42
15 Jan 2023, 01:42
Given that \(3^k\) = 16, and \(2^j\) = 27 and we need to find the value of kj
\(3^k\) = 16 = \(2^4\)
Raising both the sides to the power of \(\frac{1}{k}\) we get
\(3^{k/k}\) = \(2^{\frac{4}{k}}\)
=> 3 = \(2^{\frac{4}{k}}\) ...(1)
\(2^j\) = 27 = \(3^3\)
Raising both the sides to the power of \(\frac{1}{3}\) we get
\(2^{\frac{j}{3}}\) = \(3^{\frac{3}{3}}\)
=> \(2^{\frac{j}{3}}\) = 3 ...(2)
From (1) and (2) we get
3 = \(2^{\frac{4}{k}}\) = \(2^{\frac{j}{3}}\)
=> \(\frac{4}{k}\) = \(\frac{j}{3}\)
=> 4 * 3 = k*j
=> kj = 12
So, Answer will be D
Hope it helps!
Watch the following video to learn the Basics of Exponents
\(3^k\) = 16 = \(2^4\)
Raising both the sides to the power of \(\frac{1}{k}\) we get
\(3^{k/k}\) = \(2^{\frac{4}{k}}\)
=> 3 = \(2^{\frac{4}{k}}\) ...(1)
\(2^j\) = 27 = \(3^3\)
Raising both the sides to the power of \(\frac{1}{3}\) we get
\(2^{\frac{j}{3}}\) = \(3^{\frac{3}{3}}\)
=> \(2^{\frac{j}{3}}\) = 3 ...(2)
From (1) and (2) we get
3 = \(2^{\frac{4}{k}}\) = \(2^{\frac{j}{3}}\)
=> \(\frac{4}{k}\) = \(\frac{j}{3}\)
=> 4 * 3 = k*j
=> kj = 12
So, Answer will be D
Hope it helps!
Watch the following video to learn the Basics of Exponents
